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Tag: teaching functions
An algebraic function is a function created by applying the operation of addition, subtraction, multiplication, division, and extracting the nth root. Let me give an example. Suppose you have the function f and g where f is a linear function and g is a constant function. Let f(x)=x and g(x) = -3. We can create…
In What is an inverse function? I proposed a way of teaching this concept starting with its graphical representations using GeoGebra applets. Al-Zboun Lilliana in our Linkedin group shares her idea for introducing the inverse function. She says that the most difficult part in teaching this concept is to make it make sense to students and not…
In many history texts, algebra is considered to have three stages in its historical development: The rhetorical stage – the stage where are all statements and arguments are made in words and sentences The syncopated stage – the stage where some abbreviations are used when dealing with algebraic expressions. The symbolic stage – the stage where there… | 677.169 | 1 |
Math
Students are required to take math every year
All high school math courses are taught one on one. The instruction can be very direct and targeted for an individual student's way of learning math concepts. We also have a variety of math teachers and we match the students with a math teacher based on teaching and instructional style. Students meet multiple times during the week and have access to a math teacher between classes and during study periods.
Our students follow a traditional math sequence of Algebra 1, Geometry, Algebra 2, Math analysis or Pre Calculus, Calculus and Advanced Calculus or statistics. Mastery of the material and understanding of the concepts is considered more important than just covering the material. Students must be ready to complete assignments regularly, outside of class, to take a math class in this sequence beyond Geometry.
Integrated Math courses contain the basic concepts of algebra and geometry but allows for a personal, modified design and pace.
Algebra I:
The student will meet individually with the instructor for the study of concepts in algebra including: equations, inequalities, exponents, polynomials, linear equations, functions, factoring, and rational expressions.
Algebra IIAlgebra II TrigCalculus:
The student will meet with the teacher individually to study concepts in calculus including: limits, differentiation, related rates, maxima and minima, techniques of integration, volume, area, and natural logs.
Calculus, Advanced:
This second-year individual course continues the study of calculus. After a brief review of topics covered in first-year calculus, advanced topics including differential equations, sequence and series, parametric and polar equations, three dimensional coordinate systems, and vectors will be explored.
Consumer Math:
The student will work individually with the teacher to apply the basic concepts of arithmetic to everyday life. Topics may include: statistics, measurement, income, banking, credit, buying and maintaining a car and a house, taxes, insurance, and investments.
Geometry:
The student and teacher will focus on the major concepts of geometry including lines, angles, triangles, proofs, constructions, area, volume, circles, coordinates, and conic sections.
Integrated Math I:
The student will meet individually with the teacher for a study of the basic concepts of Algebra 1 and geometry which may include integers, equations, inequalities, exponents, factoring, linear equations, points, lines, planes & angles, deductive reasoning, parallel lines & planes, and congruent triangles.
Integrated Math II:
The student will meet individually with the teacher for a continuation of Integrated Math I. Topics may include: rational expressions and equations, radical expressions and equations, quadratic functions, trigonometry, probability, quadrilaterals, similar polygons, right triangles, circles, area, surface area, and volume.
Integrated Math III:
The student will meet individually with the instructor for the study of concepts in advanced algebra with exposure to trigonometry including: equations, functions, polynomials, rational expressions, exponents, quadratic equations, conic sections, and logarithmic functions.
Math Analysis:
The student will continue the study of advanced algebra and trigonometry concepts, but at a slower pace than Pre-calculus. The teacher may use a variety of materials to review previous material and introduce the pre-calculus elements.
Pre-Algebra:
The student will meet with a teacher individually for a review of the concepts of arithmetic and an introduction to the basic concepts of algebra including variables, expressions, and simple equations.
Pre-Calculus:
The student and teacher will meet individually for a continuation of concepts of advanced algebra and trigonometry. New concepts include: graphs of higher-order polynomials, probability, vectors, complicated trigonometric identities and equations.
SAT Math Prep:
This course focuses on the review of math concepts from arithmetic, algebra, geometry, and miscellaneous topics including ratios and percents, equation solving, word problems, graphs, counting and probability, and special triangles. Test taking skills will be reviewed and the student will take practice tests.
Statistics:
This course covers data collection and sampling, and the organization and summarization of data using graphs and measures of central tendency and variation. The study of probability is an integral part of the course and probability is applied to statistics concepts. Topics include normal distributions, discrete distributions, confidence intervals, hypothesis testing, correlation and regression. | 677.169 | 1 |
Autograph Newsletter 8 – Calculus: Integration
I hope you enjoy the eighth Autograph Newsletter! Each jam-packed edition looks at a specific topic in mathematics and how Autograph can help engage students and enable them to understand the key concepts better.
The topic of integration remains a mystery for lots of students. Many can understand, if introduced to it from first principles, that differentiation can be used to find the gradient of a curve. But why on earth should doing the opposite of differentiation magically find us the area underneath a curve? Once again, Autograph can help. Not only can you help students' visualise their answers to standard textbook integration questions to check their validity, but more importantly you can dynamically introduce the process of numerical approximations to integration, which can be used to unlock the mystery of why integration works. And once you have mastered that, why not venture into the world of 3D for some volume of revolution?The Trapezium Rule
What happens to the value of the area when you increase the number of strips? Is the area given an under or over estimate of the actual value? How can we change that this topic require the full version of Autograph.
Challenge your students to work out the area between the line and the curve.
•
How many different ways can they think to do this?
– Using integration and the area of a triangle
– Calculating two integrations and then subtracting the answer
– Can they do it using a single integration?
•
What happens if you do the subtraction the other way around?
•
Once students have reached an answer, click on the shaded area and select Text Box to display the size of the area.
•
You can double click on either of the lines to change the equation and test your students on other situations.
The way Autograph brings the concept of volume of revolution to life in 3D is quite something!
•
Challenge your students to picture the solid that will form if we rotate the shaded area 360° around the x-axis.
•
When you are ready, select Slow Plot, left-click on the shaded area, right-click and choose Find Volume
•
Hold down left-click and drag the cursor around the screen to see the solid emerge.
•
Now you can challenge your students to work out the volume.
•
A good way to start is to select the point at x = 3 and drag it closer to the other point (see Handy Autograph Tip below).
•
Students should hopefully see that the solid begins to resemble a cylinder the closer it gets, and following on from this the entire volume itself can be thought of as the sum of lots of these thin cylinders.
Video Tutorials
The following video takes you through, step-by-step, how you can use Autograph to investigate the area underneath a curve.
Handy Autograph Tip
The Animation Controller is very handy for moving the position of points in 2D. But how about in 3D? Fear not!
Open the Volume of Revolution file.
Click on the point at x = 3.
Use the left arrow key on the keyboard to move the point.
Now, imagine you wanted things moving a bit quicker – do the same again, but this time hold down theShift key.
And if you want more precision – simply hold down the Ctrl key whilst hitting the arrow key. | 677.169 | 1 |
Matlab is a package for scientific and engineering computation
that is heavily used in teaching (both formal and informal) and research.
We run the newest versions of the above software, bringing in new programmes
wherever appropriate. We also maintain a wide range of freely available
software and utilities.
As well as providing a platform for course-related software,
the Teaching System is used heavily by the
research community outside term time, so many research-orientated
programs are installed too. The system has been designed so that research
systems are easily accessible (in particular for 4th year students).
This all means that the undergraduates have
the option of using the same tools as researchers. | 677.169 | 1 |
Understanding the Problem:
This Understanding the Problem: lesson plan also includes:
Learners use iteration, recursion and algebra to model and analyze a changing fish population. They also use graphs, equations, tables, and technology to investigate the effect of varying parameters on the long-term population. | 677.169 | 1 |
Mathematics
Often the solution to a problem will require you to think outside its original framing. This is true here, and you will see the second problem solved in your course; the first is far too deep and was famously solved by Andrew Wiles.
In applied mathematics we use mathematics to explain phenomena that occur in the real world. You can learn how a leopard gets its spots, explore quantum theory and relativity, or study the mathematics of stock markets.
We will encourage you to ask questions and find solutions for yourself. You will need to think mathematically and we begin by teaching you careful definitions so that you can construct theorems and proofs. Above all, mathematics is a logical subject, so you will need to argue clearly and concisely as you solve problems. For some of you, this way of thinking or solving problems will be your goal. Others will want to see what further can be discovered. Either way, it is a subject we want you to enjoy. For more information on this course please visit ox.ac.uk/ugmaths. | 677.169 | 1 |
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Mathematical Sciences
As one of the oldest and most fundamental sciences...
Mathematics is still the basis for understanding and solving the planet's economic, scientific, engineering, physics, and business problems. It is also one of the most dynamic fields in which to work. Mathematicians use computational techniques, algorithms, and the latest technology to meet the ever-increasing demand for data analysis and statistical information. At NIU, you'll not only learn the fundamentals, but will also choose electives that prepare you for graduate or professional school or one of dozens of exciting careers in this challenging and vital field.
Reflecting the dual teaching/research role of a major university, our professors are not only nationally recognized researchers, but award-winning teachers, as well. Most mathematics majors will never have a math class with more than 25 students, and they will be taught by regular faculty, not graduate assistants.
In addition to NIU's Honors Program, we have our own Honors program, which recognizes students who excel in their studies. We also offer master's and doctoral degrees, programs for which our graduates are very well prepared, whether they pursue these degrees with us or at another institution.
There are plenty of opportunities to learn beyond the classroom, too. You might work with faculty on research projects or complete an internship with a leading corporation. You could compete in local and national mathematics competitions or join our Math Club for stimulating discussions, tours of local labs, movie nights and socializing with other majors. Or, come to one of our Friday afternoon seminars to hear distinguished professors and graduate students present on a variety of topics.
Recognizing that mathematics is one of the most challenging subjects, we offer one-on-one and group tutoring in our Mathematics Tutoring Center. Topic sessions are offered throughout the week, summarizing main points and answering questions posed in the previous week's material. Exam review sessions and practice exams are also offered.
"I wouldn't be the student I am if it wasn't for the Mathematics Tutoring Center. It has significantly improved my grade and helped me understand the concepts of calculus."
What can I do with this degree?
Actuarial and Insurance
Actuary
Benefits Specialist
Financial or Investment Analyst
Applications Programmer
Computational Scientist
Computer Applications Software Engineer
Computer Programmer
Computer Systems Analyst
Database Administrator
Operations Research
Operations Research Analyst
Systems Engineer
Statistics
Biometrician
Demographer
Econometrician
Psychometrician
Quality Control Analyst
Statistician
Business/Industry
Applied Mathematician
Business Analyst
Communications Engineer
Economic Analyst
Investment Manager
Market Researcher
Product Developer
Health
Biomathematican
Biomedical Engineer
Ecologist
Emerging Fields
Computational Biology and Genomics
Data Mining
Neuroscience
Materials Science
Computer Animation and Digital Imaging
Degree Information
The Department of Mathematical Sciences offers the B.S. degree with a major in mathematical sciences with emphases in general mathematical sciences, applied mathematics, computational mathematics, probability and statistics, mathematics education, and actuarial science. Successful completion of the emphasis in mathematics education leads to certification to teach at the 6-12 grade levels.
The department also offers minors in mathematical sciences, elementary mathematics education, applied probability and statistics, and actuarial science. These minors should be of interest to students majoring in the physical or social sciences or in business. In addition, the department offers an honors program in mathematical sciences and participates in the University Honors Program. | 677.169 | 1 |
fifth edition continues to improve on the features that have made it the market leader. The text offers a flexible organization, enabling instructors to adapt the book to their particular courses. The book is both complete and careful, and it continues to maintain its emphasis on algorithms and applications. Excellent exercise sets allow students to perfect skills as they practice. This new edition continues to feature numerous computer science applications-making this the ideal text for preparing students for advanced study as a supplement to a summer course in Discrete Math, and since this was my first ever exposure to mathematical proof and dialog, I first thought this book mostly alien, with occaisional sections of brevity; it did help me fill in some gaps left behind in Rosen's book, especially on some basic proofs dealing with integers and with combinatorial reasoning--something this book is REALLY good at... I'm in my first course of Combinatorics with a teacher that assumes we know alot more calculus than we do. We use Tucker's Applied combinatorics 5th, and I was cruising along just fine until we hit Generating Functions. Brick wall. Rosen's book didn't cover it (well; there's a great page of known identities, but not an intro-level version), neither did Epp, so I dusted this tome off my shelf and cracked it open... section 9.1 presents Generating functions on such an easy to use language and analytic explanation that I went from getting every problem wrong in Tucker's book to getting them all right; all due to the clarity of exposition.
I've also found that as my 'mathematical maturity' has grown in the last year, so has the comprehensibility of this text. It may be too deep for a beginner--I would agree that it would be too much for all but your brightest minus an excellent teacher--but this book teaches 'real math' and does so *very* well.
In conclusion, if you have the available student loan $$ and want a very good supplementary book that you really can take with you to higher classes, put this at the top of your list.
I also own Epp and Rosen's discrete math texts, and have to say that for me ultimately I needed all three as a beginner; plus a few extra books from the library for special topics. But what I learned stayed with me and all three have their positives and negatives, but if I were to choose only one to stay on my shelf, THIS would be the one.
Gives a good overview of abstract mathematics in a very well presented fashion. Gives an introduction and some practice applying first order logic, set theory, algebras, solutions to recurrence equations, and more. It really has nothing to do with computer science and makes no attempt to show anything about it, as any language is just a language.
Out of the three main discrete math texts, Rosen, Epp, and this one--Grimaldi--this text unites the best parts of both; Epp has some really great explanations, but suffers from not having enough solutions and lacks depth. Rosen's book manages to write hundreds of words per concept while completely confusing new students in dense mathematical jargon.
I used this book as a supplement to my discrete math class in summer and as a supplement for a combinatorics class this past fall.
My mathematical 'maturity' when approaching discrete math was business calculus. (Yeah, I know that sucks, and all you mathematicians and engineers can laugh your hind off about it. Don't remind me.) So basically, I was behind the class in both this and in the combinatorics class this fall.
This book is best approached if you take the explanations it uses *while trying to solve the problems.* It seemed pitched high to me because Epp is focused on giving you concepts and Rosen is concerned with making sure you learn theory.
Grimaldi is interested in teaching you to solve problems.
This book also has the one of the *best* sections on recurrence relations. I thought Chen's book was king here, but this book, when working through gobs of problems, helps you learn them inside and out. It has two charts detailing what happens in a non-homegeneous recurrence relation, one that states general solutions, another that gives you a relation, its homogeneous counterpart, and changes the NH part and shows you how the general form changes.
Brilliant, and blows Tucker's "Applied Combinatorics" out of the water in clarity when solving recurrence relations.
To begin, I have absolutely no experience with discrete math and generally learn through doing a lot of practice problems. This book is the book our professor is using for our discrete math class and it is atrocious. While it has a a lot of examples, they cover the bare basics and don't go into enough detail to then understand what you need to do for later problems or even the early problems. The author uses the cop out of "you need to use creativity to solve some of these problems," to cover his tracks and not just accept that he is lazy and is not going into enough detail to really understand the problems. This means you'll just get a whole bunch of problems wrong in the process of learning and will eventually run out of problems to practice because after looking at the answer, the problems lose their ability to test your understanding. Or even worse, somethings are not highlighted in a very clear manner (for the sake of being "concise") leading to severe frustration when you in essence have to basically learn discrete math by trying every option under the sun. If you're a discrete math prodigy and can learn through one reading of the material, then you may find this book good, but for the rest of us who have to work for our knowledge, this book is about as pleasurable as eating one ton of bear poop, while getting your eyes gouged and taking roundhouse kicks from Chuck Norris. | 677.169 | 1 |
Calculus
Calculus I is where mathematics really starts to get interesting because its focus is on being able to describe the real, dynamic systems we see in real life that change over space and time. As such, this class will have a significant project-based component that will allow students to see the actual application of the techniques of differentiation and integration.
It starts with an introduction to limits. Finding limits helps in finding values that might not exist themselves but can be inferred by taking closer and closer approximations of reality.
We next use our knowledge of limits to derive the basic rules of differentiation: the power rule, basic trigonometric rules, product rule, quotient rule, chain rule and the rules implicit differentiation.
These rules are then used to in problem solving using differentiation. Mathematical problems include finding maximum, minimum and inflection points on curves, while computing the motion of projectiles (position, speed, acceleration) is particularly tractable when using differential calculus.
After differentiation we must necessarily study its inverse, integration. Using our knowledge of limits, we determine the basic rules of integration, then go into more complex rules. This will include an introduction to numerical techniques.
Our knowledge of integration is then applied to solving initial-value problems and other mathematical and real-life problems. | 677.169 | 1 |
Visual Algebra
This Visual Algebra lesson plan also includes:
Learners solve basic simple single-variable algebraic equations using a Windows Application, "Visual Algebra." The instructional activity is introduced using a set of familiar objects to balance boxes. The transition of program skills to autonomous abilities is the foc | 677.169 | 1 |
L.J., Utah
The Algebrator could replace teachers, sometime in the future. It is more detailed and more patient than my current math teacher. I, personally, understand algebra better. Thank you for creating it! S.D., Oregon
Algebrator is a wonderful tool for algebra teacher who wants to easily create math lessons. Students will love its step-by-step solution of their algebra homework. Explanations given by the math tutor are excellent.07-06:
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The problems in this book are suggested for evaluating the concepts taught in the intermediate geometry class. The problems are of a highly visual nature and meant to be challenging. The problems are designed to lead to a merging of geometry and art at the middle school level. The problems presented in this book include: · Visual problems to determine area of various iterative polygon based shapes · Visual representation of solid objects to determine their volume · Visual medley of circles, squares, and triangles to determine their relationships · Determining properties of angles, triangles, square, and rhombus · Visual problems for determining equivalence of geometric properties of polygonal shapes · Determination of area of objects using reference objects as basic elements · Visual representations of lines and triangles to solve problems based on equations · Identifying intersection points for an underlying visual diagram · Application of Pythagorean Theorem to problems represented visually · Applications of factorization and LCM to problems on area and volume · Changes to area of triangles based on various construction techniques · Inferences for area or angle measures of unknown elements in constructed diagrams Also available at CreateSpace eStore:
"synopsis" may belong to another edition of this title.
About the Author:
Kiran R. Desai received a Ph.D. in Computer Science from Binghamton University, New York, in 1996, specializing in Parallel Processing Interconnection Networks. Even after working in the computer industry for more than a decade, he has an inclination to contribute to mathematics education and problem solving skills. As an elementary student, he discovered a simple way to find the next squared number if he started from squares of increasing numbers. (1, (1+3) = 4, (4+5)= 9, (9+7) = 16, (16+9) = 25, ...). As a high school student he developed the ability to solve the Rubik'sTM cube on his own when it became popular. As a doctoral candidate, he continued to solve various graph and mathematics problems by analysis and using computers. During his Ph.D., he also taught a course on Computer Algorithms. During his graduate studies, he authored and co-authored 10 refereed papers. He believes that developing a love for mathematics and problem solving at an early age helps build a strong foundation for the later years in life. | 677.169 | 1 |
" Pain is temporary, MATH is 4ever.
Paano ba gumaling sa math?"
Search Your Lesson
Basic Calculus for Grade 11
On this course, the students must know how to determine the limit of a function, differentiate, and integrate algebraic, exponential, logarithmic, and trigonometric functions in one variable, and to formulate and solve problems involving continuity, extreme values, related rates, population models, and areas of plane regions.
This course is patterend under the Teaching Guide for Teaching Senior High School "SHS" Framework, which stands for "Saysay-Husay-Sarili for Senior High School." | 677.169 | 1 |
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Just watching my students, one after another, easily grasp these higher mathematical concepts and really, truly understand what they are doing, makes Algebra Helper worth the price of admission. Furthermore, for the gains, the price is tremendously inexpensive!12:
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Algebra is about the unknown and putting real life problems into equations and solving them. Unfortunately, Math textbooks go straight to the rules, proceduress, and formulas forgetting that these are real life problems being solved. We will discuss many Algebraic topics that will engage students and develop students' thinking, logic, pattern, and problem-solving by using hands-on activities.
Class Notes
Start an Application for the Class
If you wish to submit an application for this class, please enter your email address in the space below and click the submit button. If your email address is found in the database, you will be asked to enter your password. If you have forgotten your password, you can have it emailed to the email address you entered. If your email address is not found, you will need to create a profile.
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Asian students are famous for their mathematical prowess, and these are the books that they use to gain those skills. If you want your child to learn to think mathematically, not just be able to solve math problems, you will be pleased and surprised with how well your student progresses. The textbooks teach the concepts and the workbooks are designed to be done independently by your student, and are paperback, with happy, color illustrations-very kid friendly. We offer the U.S. edition (which has been minimally modified to teach American money and U.S. spellings). Test scores conclusively and repeatedly put Singapore Primary Mathematics in top rank in international tests! Singapore's genius is in a unique pattern of moving from hands-on demonstrations, to picture drawings (concrete examples with pictures) to the abstract (numbers and symbols) in a natural, easy-to-understand progression. Students go right from Singapore level 6 into Saxon Algebra 1 with ease
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Specifications
Country
USA
Binding
Unknown Binding
Edition
US Edition
Label
Marshall Cavendish International
Manufacturer
Marshall Cavendish International
PublicationDate
2008
Publisher
Marshall Cavendish International
Studio
Marshall Cavendish | 677.169 | 1 |
MTH317 Mathematical Modeling
Course Description
An introductory to mathematical modeling, utilizing a variety of diverse applications from physical, biological, business, social, and computer sciences. Discuss the limitations, as well as the capabilities, of mathematics as applied to understanding of our world. Teaches problem identification, models of solutions and model implementation. Graphing calculator is required.
Learning Outcomes
Carefully define problems and detail the parameters.
Identify variables and design suitable experiments to secure data.
Learn to apply mathematical skills and tools to find structure in the data.
Develop skills in the display of data and mathematical conclusions | 677.169 | 1 |
Jeff Ply, CO
I am a 9th grade student and always wondered how some students always got good marks in mathematics but could never imagine that Ill be one of them. Hats off to Algebrator! Now I have a firm grasp over algebra and my approach to problem solving is more methodical. Laura Keller, MD14:
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Rating:543214.4/5 from 3119 votes.
Book Descriptions:
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SSC CGL Maths syllabus 2018
December 15, 2017
SSC CGL Maths syllabus 2018 with Maths exam pattern can be checked now. SSC CGL Maths syllabus 2018 is available on this page and you can download them here. Are you ready to start their preparation through SSC CGL math's syllabus, then this is the right place where you can get each & every information related SSC CGL Exam syllabus.
SSC CGL Maths syllabus 2018
In this article, we are sharing SSC CGL maths syllabus section wise. Our website will help you and guide you to give your SSC CGL preparation a right direction. As we all know that, achieving SSC CGL exam is not an easy task, it required great dedication passion and hard work.
Below we have provided a detailed syllabus for SSC CGL math's section which will cover all topics related Quantitative aptitude. Without the syllabus, you can't able to qualify this section. Those candidates who are having outstanding knowledge of Quantitative aptitude section then it is easy to crack this section. After knowing the complete syllabus the second step should be of choosing the best SSC CGL math's books that cover each and every topic and these books should be written in an easy to understand manner.
SSC CGL Maths exam pattern 2018
There are several books available in marked and online mode. If you purchase waste material it does not benefit you, it is totally wasted on you. The candidates have a look at SSC CGL Books store in which candidates are offered by the various coaching institute faculty. If you also want to download SSC CGL math's syllabus in pdf format then you can also download it to the pdf file from the below-given link.
Subject name
Questions
Total marks
Quantitative Aptitude
25
50
Candidates who want to appear for the SSC CGL 2018 exam will be searching for the SSC CGL 2018 syllabus and exam pattern to check in pdf format. We are providing the SSC CGL syllabus for the guidance of candidates.
SSC CGL 2018 Maths Syllabus
Ratio and ProportionSimple
Topic name
Important Topics
Simplification
Questions on BODMAS Rule
Fractions
Surds and Indices
Interest
Simple Interest
Compound Interest
Installments
Percentage
Calculation oriented basic Percentage Problems
Ratio
CompProportion-Simple
Direct Proportion
Indirect Proportion
Average
Average Age
Average Height
Average marks
Problems on Ages
Basic Problems on Ages
Speed, time and distance
Speed of boats, trains
Average Speed
Relative speed
Profit & loss
Problems on Profit/Loss
Dishonest/Successive dealings
Partnership
Number Series
Completing the series
Finding Missing/Wrong Term
Number System
LCM & HCF
Irrational or Rational Numbers
Mensuration
Problem on geometric structures
Data Interpretation
Problems on Bars
Line Graphs
Pie Charts
Tables
Time & Work
Work Efficiency
Wages
Pipes
Mixture Problems
Mixture of two or more entities
Algebra
Equation on one or two variables
Quadratic & Algebraic Identities
Geometry
Triangles congruency
Similarity theorems
Circle chords & tangent theorems
Co-ordinate geometry
Trigonometry
Trigonometric ratios and Identities
Height and Distance
Staff selection commission releases various notifications every year for recruitment of efficient candidates into various posts. Various positions include Tax inspector, Audit Officer, Accountants etc. All eligible candidates who are interested in working in SSC organization can apply for this recruitment through online mode after the release the notification. Before going to examination candidates need to prepare through SSC CGL Maths syllabus. If you need any help then you can share with me | 677.169 | 1 |
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Maths Resources
The School of Mathematical Sciences has been involved in and has supported many different projects which have produced resources for teaching mathematics and highlighting career opportunities for mathematics graduates. You can access many resources from this page.
Parents' Guide to Studying Mathematics
Aimed at parents whose children are thinking about studying mathematics, articles cover what the subject is like at university, the finance arrangements from 2012 onwards, and what careers graduates can expect to go into.
This booklet looks at the impact of mathematics beyond the subject itself and provides a clear link from the syllabus to applications in a range of different industries. It is hoped therefore that provides extra impetus for students in their studies and starts to provide some thoughts to how a student will want to one day use his/her mathematics degree. This was developed for undergraduate students, but the content is accessible for A-level students. View the booklet here:
This series of videos came into the same project as the "Where the Maths You Learn is Used" booklet. A range of employers talk about their experiences and offer advice to mathematics undergraduates. They can be viewed online here:
We are grateful to the Institute of Maths and its Applications and the HE STEM programme for their support.
Mathematical Magic
Mathematics and magic may seem a strange combination, but many of the most powerful magical effects performed today have a mathematical basis. Maths is also the secret behind the technologies we use, the products we buy and the jobs we do. Written jointly by our Mathematics Fellow Matt Parker and Professor of Computer Science Peter McOwan, the Manual gives young mathematicians the chance to be creative, finding new ways to solve problems and discovering the key to the perfect magic trick. Along the journey they will also uncover the skills of a good mathematician, one with the useful employment skills you get from being good at mathematics.
The following website contains step by step guides on how to do some magic tricks. This is another project from our Outreach Officer Matt Parker and Professor of Computer Science Peter McOwan but this time science, engineering, maths and technology are tackled. It also includes the history of science and engineering in magic and numerous films showing a variety of mind-bending tricks.
This is a guide for sixth form students to aid them in their university choices. It covers mathematics in addition to other science and engineering subjects. You can download all the issues below as PDFs. | 677.169 | 1 |
modern mathematical physics : groups, Hilbert space, and differential geometry
"This book is suitable for advanced undergraduate and beginning graduate students in mathematical and theoretical physics. It includes numerous exercises and worked examples to test the reader's understanding of the various concepts, as well as extending the themes covered in the main text. The only prerequisites are elementary calculus and linear algebra. No prior knowledge of group theory, abstract vector spaces or topology is required."--Jacket.Read more...
Abstract:
This textbook, first published in 2004, provides an introduction to the major mathematical structures used in physics today.Read more...
Reviews
Editorial reviews
Publisher Synopsis
'This is a beautifully crafted book. ... Peter Szekeres presents in the most elegant and compelling manner a magnificent overview of how classic areas such as algebra, topology, vector spaces and differential geometry form a consistent and unified language that has enabled us to develop a description of the physical world reaching a truly profound level of comprehension. ... Szekeres's style is clear, thorough and immensely readable. His selection of topics concentrates on areas where a fully developed rigorous mathematical exposition is possible. ... One cannot help but be slightly awed by the beauty and the capability with which seemingly abstract concepts, often developed in the realms of pure mathematics, turn out to be applicable ... I recommend that you get hold of this book for yourself or for your library.' The Times Higher Education Supplement 'The superb layout and an index contribute to the excellent overall impression of this book ...'. Zentralblatt MATH ' ... the book may serve as an easily accessible introductory text on a wide range of the standard and more basic topics in mathematics and mathematical physics for the beginner, with an emphasis on differential geometry. a nice feature is that a considerable number of examples and exercises is provided, together with numerous suggestions for further reading: there is also an extensive index which will be particularly helpful for beginners in the subject.' General Relativity and Gravitation JournalRead more... | 677.169 | 1 |
...fellowships by national and international bodies. The course has well-structured syllabi and students are aded by well equipped libraries. Objective... Learn about: Institute Mathematics, Mathematics Series...
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...finance, the civil service, scientific research, after completing the course. One can opt for building career in these different fields, because mathematics... Learn about: Mathematical Economics, Mathematics Algebra, Mathematics Series...
More
...prestigious fellowships by national and international bodies. The course has well-structured syllabi and students are aded by well equipped libraries... Learn about: Mathematical Statistics, Mathematics Series...
More
...fellowships by national and international bodies. The course has well-structured syllabi and students are aided by well equipped libraries. Objective... Learn about: Mathematics Series, Mathematical Economics...
More
...or in health and fitness, which will help learners of all levels feel more confident about Maths. Also, learn how to use Maths when undertaking home improvement... Learn about: Mathematical Physics, General Mathematics, General Maths...
More
..., measurement, and the systematic study of the shapes and motions of physical objects. Practical mathematics has been a human activity for as far back as written records exist. The research required to solve mathematical problems can take years or even centuries of sustained inquiry... Learn about: Mathematics Teacher, Mathematics Series, Mathematics Algebra...
More | 677.169 | 1 |
In this breakthrough text, authors Holtfrerich and Haughn speak as teachers, offering passionate yet patient emphasis on concepts, but not at the expense of teaching algebraic skills. The authors build on and connect topics by discussing the concepts with students in a unique yet logical way. Speaking to students as if they were having a discussion with them in the classroom, the authors use common, everyday applications that make sense to the average student to demonstrate concepts. The authors' approach about where and when topics should be covered is unique, as is the capstone chapter on Modeling/Data Analysis.
"synopsis" may belong to another edition of this title.
About the Author:
Michael Holtfrerich received his BS and MS from Northern Arizona University. He has taught at Glendale Community College in Glendale Arizona, a member of the Maricopa Community College District, for the past 16 years. Besides teaching mathematics he is currently the chair of the Mathematics/Computer Science Department which employs 35 full time and 61 part time instructors. He has spoken at a variety of math conferences and is a member of the American Mathematics Association of Two-Year Colleges and the Mathematics Association of America. He has received several awards among those, the Paul Pair Endowed Chair award for integrating technology into education.
Review:
"Â...the author has written the text in easy to understand detail. It is not only student friendly, but mathematically correct."
"I am definitely excited about this bookÂ...Thanks for the opportunity to review. Please keep me attuned to how this project is going. IÂ'm really excited about the potential here."
"I really like the conversational style of the text. many places the explanations are done the way I do them in class."
"I am often concerned about the formality of math texts that are used by a wide segment of the student population such as college algebra. When I find it difficult to follow the language of a text on a topic I am quite familiar with, I worry about a student even standing a chance of understanding it. These authors have found a middle ground between informal and formal language." | 677.169 | 1 |
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What a great step-by-step explanations. As a father, sometimes it helps me explaining things to my children more clearly, and sometimes it shows me a better way to solve problems | 677.169 | 1 |
Review info & popularity
Brief course description
Derivatives and Integrals are at the HEART of calculus and this course enables you to Differentiate and Integrate in 45 minutes. It is a short dense course designed to get the student mastery over the rules and shortcuts of differentiation and Integration.
This course is for those who are studying Calculus 1, or have basic knowledge of functions, limits, and slopes. No ambiguous and complicated terminology is used, and the concepts are taught in simple language to make the learning process easy.
The course is divided into two sections, the first one is: Differentiation, and the second is: Integration. In each lecture, one rule of differentiation or integration is discussed, with examples, followed by a small session of five question quiz that aims at testing your understanding of the concepts.
(Read more about this course on the official course page.)
Miran Fattah bio
Fattah has B.S. in Mathematics and Geophysics from theUniversity of Oklahoma in Oklahoma, USA. He has taught and tutored many college students both in the United States and Iraq. His love for teaching made him one of four students in Iraq to receive a full scholarship to pursue a B.S. degree in the States so to return back to his home country and teach. | 677.169 | 1 |
Functions (Zambak)
This is an introductory book covering functions and graphs together with some applications. Functions are used in ordinary life applications, including economics and engineering. Although we do not face functions directly, problems that we face are most frequently solved by modeling them as a function. This book is divided into four sections. The first section, relations, prepares the student for the concept of functions and introduces the cartesian coordinate system uhich uill be consistent throughout the remaining part of the book. The heart of the book is the second section in uhich we study the main concepts of functions. The third section goes more in depth with further operations of functions in a more arithmetic and theoretical way. The last section concentrates on visualization of functional terminology throughout the graphs and introduces a very important technique called transformation uhich ? ill be necessary in future concepts of algebra and mathematical analysis. Teachers may choose four different plans of teaching (sections 1+2, sections 2+2+3, sections 1+2+4, or sections 1+2+3+4) enabling you to use the book depending on the level of the students and the total hours allowed. DOWNLOADREAD ONLINE NOW | 677.169 | 1 |
Students make conversions between different units of measurement. Undoubtedly, Precalculus course prepares for students entering into Calculus-the mathematics of changes. Not being ready, students can't study calculus successfully. | 677.169 | 1 |
Book 9th ncert maths
Ncert books — ncert books free download for class 1,2,3,4,5,6,7,8,9,10,11,12 and for the upsc and cbse candidates here are the list of ncert books covered in learncbse.in app. ncert 9th maths book students who are studying in class 9th & want to download chapter wise answers pdf can get it through this page. please visit our new website here learn more at search for videos in friends, rbi has released notification for grade "b" officers exam batch 2016.
Book ncert 9th maths
Site editor, [email protected] this site is best viewed in 1024 x 768 resolution. to download the entire book in one nice package goto we cover subjects like maths, science. for ncert solutions of class 8 science click here. ncert 9th maths book. | 677.169 | 1 |
Algebra
Designed to provide students with practice in the necessary skill areas involved in mastering algebra concepts such as polynomials, linear equations, inequalities, and radicals. A review of basic skills is presented in the first part of the book and more specific algebra topics are introduced on a gradual basis throughout. 128 pages. Grades 5-8. | 677.169 | 1 |
This ebook is available for the following devices:
iPad
Windows
Mac
Sony Reader
Cool-er Reader
Nook
Kobo Reader
iRiver Story
more
Advanced Mathematical Thinking has played a central role in the development of human civilization for over two millennia. Yet in all that time the serious study of the nature of advanced mathematical thinking – what it is, how it functions in the minds of expert mathematicians, how it can be encouraged and improved in the developing minds of students – has been limited to the reflections of a few significant individuals scattered throughout the history of mathematics. In the twentieth century the theory of mathematical education during the compulsory years of schooling to age 16 has developed its own body of empirical research, theory and practice. But the extensions of such theories to more advanced levels have only occurred in the last few years. In 1976 The International Group for the Psychology of Mathematics (known as PME) was formed and has met annually at different venues round the world to share research ideas. In 1985 a Working Group of PME was formed to focus on Advanced Mathematical Thinking with a major aim of producing this volume. The text begins with an introductory chapter on the psychology of advanced mathema- cal thinking, with the remaining chapters grouped under three headings: • the nature of advanced mathematical thinking, • cognitive theory, and • reviews of the progress of cognitive research into different areas of advanced mathematics. less
In the press
This book can help mathematics educators place their research within a broader perspective of the development of mathematical reasoning, and it can help mathematicians see the deep connections between mathematics education research and their own teaching. (Patrick W. Thompson inJournal for Research in Mathematics Education) It promises to be the standard reference work for many years to come, and a source of both information and inspiration ... The book is a most welcome distillation and resource for further development in our collective sense of what constitutes advanced mathematical thinking, and how we can make it accessible to more people. (John Mason inZentralblatt für Didaktik der Mathematik) | 677.169 | 1 |
Math 13 Study Suggestions:
Many incoming freshman have not developed good study habits, and find the
transition to college work to be quite difficult. You may find some
aspects of this course to be frustrating to you; this is natural, but you
should try not to let that block you from succeeding at the course. Try
to remember that part of what you are learning is how to learn
effectively; find out what works for you and what doesn't.
Here are some hints that you should consider:
Come to my office hours. If you are having trouble doing the
homework problems, please come see me during my office hours or ask questions in class
(everyone will be thankful that you did).
If you have trouble with a homework problem, do more problems like it
from the book. There are usually others nearby that are similar. If not,
there are lots of calculus books in the library and in the lounge in the
Math Department; these will have problems similar to the ones we do, so you
can use these as additional practice.
Make sure you continue to work on a problem until you get it
right and understand why it's right. It does not do you much good
to work through many problems but get them all wrong. It is better to
get one right in the end than to do ten incorrectly.
Don't get behind in the homework. You have many demands on your time,
and you will not be able to catch up once you start to fall behind.
Don't put off the homework until the last minute. Some of the
problems may require considerable thinking, and you will need to come back
to them several times. This is to be expected, and does not mean
you are not understanding the material. It is simply part of the process
of doing hard problems.
Don't turn in work that you know is wrong but pretend that you think it
is right. It is far better not to finish a problem than to continue on and
produce a wrong answer, or make some other error to compensate so as to get
a reasonable-looking answer; that just makes you look like you
really don't know what you are doing. No answer at all is better
than an answer you know to be wrong. You should say something like: "I
know that something has gone wrong at this point, but I can't figure out
what." You might also say why you think something is not right, and what
you would have done if you could have continued on. There is no shame in
admitting you don't know how to do something.
Review your notes between classes. This is a critical part of the
learning process, and should not be overlooked. This will help you
organize the material yourself for yourself, and to locate areas
where you may have questions.
Come to class prepared. Do the homework before the next class,
if you can, and be prepared with questions if not. Look over your
notes briefly in the few minutes before class starts, so that you
remember where we left off and what we were trying to accomplish.
When determining how much time you should spend on your course work,
consider the following analogy: if you were working a full-time job, you
would be putting in 40 hours a week (at least). Think of your class work
here as your job. Since you are probably taking three courses, that means
you should spend about 13 hours a week on each course. Of that, 3 will be
in class, so you should expect to put in about 10 hours of work outside of
class each week. I expect the weekly homework to take you about 6 hourts a
week, so the remainder should be spent reviewing your notes, coming to
office hours with questions, and trying to internalize the material from
class. On the course evaluation at the end of the term you will be asked
to estimate how much time you spent per week on the course. Many students
end up saying 2 to 4 hours, which is not enough; however, if you find you
are spending more than 12 hours a week, then something is probably wrong,
and you should come see me.
Finally, I want to remind you that (although it may not always seem like
it), I do want you to succeed at this course, and I will do whatever
I can to make that happen for you. Your role in that is to do
whatever you can to make that happen, and to come to me for
help when it's not working. It is far better to admit that there is
a problem than to brazen it out and hope it gets better; it usually
doesn't on its own. | 677.169 | 1 |
Numerical Methods are systems, or algorithms, for solving equations that cannot be solved using normal techniques. There are a number of different types of numerical methods available. The ones you need for your exams are listed below and are shown in more detail in the next Learn-It: | 677.169 | 1 |
Computational Geometry (Winterterm 13/14)
Computational geometry is a branch of computer science that deals with algorithmic questions on (discrete) geometric objects, for instance points, line segments, polygons, circles and their higher-dimensional counterparts. Algorithms in computational geometry are a backbone in Computer graphics,
Computer-Aided Geometric Design, Computer Vision, Robotics, and others. Computational geometry also includes theoretic aspects of discrete geometry that are connected to computational problems.
Our course gives a broad introduction to the topic: we will start with a selection of "classical topics" in the first half of the course. Then, we will look at several topics that are crucial for the applicability of geometric algorithms (Geometric computing and geometric approximation algorithms). Finally, we touch on the modern field of computational topology which emerged from computational geometry and various applications in high-dimensional data analysis.
Target Audience:
The course is for all bachelor and master and PhD students enrolled to a "computer science" program. It counts as advanced course ("Vertiefungsvorlesung").
The course requires basic knowledge in algorithms and data structures, combinatorics and complexity theory (in particular O-notation). All required background should be covered by "Mathematik fuer Informatiker" and "Programmierung 2" at Saarland University. | 677.169 | 1 |
10th Grade Math School Books – learningthings.com 10th grade math students typically study geometry, although that is not always the case because students have progressed at different rates.. Regardless of the subject . | 677.169 | 1 |
Mathematics Textbook For Class IX has been designed in a way so as to enable students to tackle difficult mathematical problems with ease. This book comprehensively covers the curriculum for the CBSE board and provides convenient ways of solutions to various numerical problems. The textbook has copious diagrams and solved examples to facilitate easier understanding and better adaptability among students. Unnecessary jargon is kept at bay, making mathematics more interesting for the students. Every topic is supported with abundant solved examples and practice sums to constitute a complete learning experience.
Mathematics maybe one of the most dreaded subjects for students all over, but this book is aimed to curb that fear and build the love for the subject in the students. While a book does not qualify by the mere virtue of being appealing and student-friendly, it also needs to be relevant, useful and bear appropriate content. This book has been thoroughly revised by reputed teachers and educationists from the field of Mathematics.
In keeping with the prescribed syllabus for CBSE, the book covers topics broadly ranging from Pure Arithmetic, Commercial Mathematics, Algebra, Geometry and Statistics to Mensuration, Trigonometry and Coordinate Geometry. The edition has been upgraded in accordance to the latest syllabus. Each chapter records attest series with problems of varying degree of difficulty to enable students to recapitulate what they have learnt and internalise the methods for the purpose of solving similar sums in the future. The step-by-step illustrations have proven extremely helpful to the students.
Topics such as Compound Interest, Surds, Indices, Formulae, Logarithms, Mean and Median, Simultaneous Linear Equations, Volume, Cube and Cuboid, Histogram and Frequency Polygon, Rational and Irrational Numbers, Trigonometrical Ratios are focused on with an eye for recognising the potential problem areas and help students to deal with them better. Mathematics Textbook For Class IX was published by NCERT in the year 2006 and is available in paperback.
Key Features:
Won impressive reviews from students and teachers nationwide.
Contains question papers from previous years' as well as sample question papers.
Enter your mobile number or email address below and we'll send you a link to download the free Kindle App. Then you can start reading Kindle books on your smartphone, tablet, or computer - no Kindle device required. | 677.169 | 1 |
Suitable for a one-semester course in general relativity for senior undergraduates or beginning graduate students, this text clarifies the mathematical aspects of Einstein's theory of relativity without sacrificing physical understanding. | 677.169 | 1 |
Essential Engineering Mathematics
Description: The aim is to highlight and explain some areas commonly found difficult, such as calculus, and to ease the transition from school level to university level mathematics, where sometimes the subject matter is similar, but the emphasis is usually different.61 views)4 views) | 677.169 | 1 |
Precalculus: With Unit Circle Trigonometry
Precalculus - University Blogs
Student Solutions Manual For Cohens Precalculus With Unit Circle Trigonometry 4th By David Cohen Published By Cengage Learning 4th Fourth Edition 2006 Paperback.Trigonometry studies triangles, circles, and the oscillation (rise and fall) of natural phenomena (examples include the tides or populations of predators and prey).
Buy Precalculus With Unit Circle Trigonometry with CD-ROM and iLrnTM Tutorial by David Cohen ISBN 9780534402303 0534402305.Buy or Rent Precalculus: With Unit Circle Trigonometry as an eTextbook and get instant access.David Cohens PRECALCULUS: A PROBLEMS-ORIENTED APPROACH, Sixth Edition, focuses on teaching.Train your worship teams and musicians online at WorshipTraining.Unit Circle: Sine and Cosine Functions. and practice with sine and cosine functions.Download Ebook: precalculus with unit circle trigonometry in PDF Format. also available for mobile reader.
Precalculus With Unit Circle Trigonometry Solutions Manual
The unit circle -- Topics in trigonometry - TheMathPage
Learn vocabulary, terms, and more with flashcards, games, and other study tools.Buy a cheap copy of Precalculus: With Unit Circle. book by David Sklar.Free download precalculus with unit circle trigonometry book which is Mathematics book that wrote by David Cohen.Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry, Third Edition focuses on the fundamentals: preparation for class.College Algebra and Trigonometry: A Unit Circle Approach (5th Edition) Precalculus.
In this video I will introduce and explain the unit circle and the relationship of the.Cohen, Kundu Fluid Mechanics 4th Edition Instructors Solutions Manual.Oct 23, 2013.Buy Precalculus With Unit-Circle Trigonometry - With CD 4th edition (9780534402303) by David Cohen for up to 90% off at Textbooks.com.
Precalculus With Unit Circle Trigonometry - mybookdir.com
Precalculus With Unit-Circle Trigonometry | Rent
Precalculus Trigonometry Angles THE UNIT CIRCLE
Pre-Calculus Unit Circle - dummies
Precalculus Unit 03 – Introduction to Unit Circle Trigonometry
Learn trigonometry for free—right triangles, the unit circle, graphs, identities, and more.Precalculus david cohen pdf. in stock more on the way.Precalculus: With Unit Circle Trigonometry with Interactive. for Cal Poly.Precalculus David Cohen. | 677.169 | 1 |
Math IR - Final Exam
This review worksheet touches on a number of different Algebra 1 topics. Learners solve and complete 38 different types of problems. First, they find the inequality that is best represented in the graph illustrated. Then, they graph the expression and label the axis of symmetry and the vertex. And, finally they create a stem-and-leaf plot and find the median and mode. | 677.169 | 1 |
Lessons Learned from Years with Resources […] and up or down. The weight of a person keeps on changing from low to high or from high to low making weight an active variable. The movement of the Sun is a result of the Earths motion around the sun. Every time there is a change in the stock market from high to low. The salaries of employees keep changing according to the number of hours they have worked. 6 Facts About Lessons Everyone Thinks Are True
It can also be said that Algebra Is the study of events that keep on varying with time. Therefore, Algebra is nearly everywhere around us. Questions About Resources You Must Know the Answers To
Before beginning Algebra, these are the basic concept to be learned. Elementary multiplication, addition subtraction and divisions. Times tables at least from 1 to 10. Know how to write all the factors of a number, finding the least common multiple (LCM), finding the greatest common factor (GFC). Fractions. Numerals. Command of operations. Multiples are added to many students in the fourth grade. After beginning, the learning of multiplication students should learn about multiples. The next thing after the students start learning multiples is knowing where and how to use multiples in mathematics. Students can gain the capacity to predict times of numbers accurately without any delay with the knowledge of multiples which enhances the core mathematics skills. Main topics earned in Algebra. Familiarity with variables. Know the constants and the coefficients. Script expressions in Algebra. Simple linear equations in one variable. Balanced expression and Factorization. Patterns like series and sequences in overall. Why is Algebra hard? It is not tough. You can make it a difficult course. There are extremely general terminologies used in Algebra. A generic term means a proper name that is used to distinguish a person from a crowd. Algebra has some procedures to be followed. If you follow these procedures, Algebra is not that tough. In the elementary theory of numbers, finding the greatest common factor that divides two or more numbers without leaving a remainder is an important thing. Students in some schools are introduced to factor finding while in fifth grade, and some are introduced to it during the late fourth grade. Having the knowledge about composite and prime numbers is the key to learning factoring of numbers. | 677.169 | 1 |
Students will review the basic skills learned in Algebra 1 while at the same time going into greater depth. Quadratic functions are studied in depth. Other polynomial functions, radicals, exponential growth and decay, and basic trigonometric functions are also studied. A graphing calculator is used throughout the course. A TI-83+ or higher is required. Algebra 2 is a graduation requirement. Prerequisite: Algebra I & Geometry
This is a one semester course that shows the link between statistics and our world. Examine averages, sampling of data, normal distributions, regression and probability. Overcome your fear of statistics, learn how to analyze statistics you see or hear and learn how businesses use statistics to increase their success. A graphing calculator is required.Prerequisite: Algebra I & II, Geometry
Students will learn an appreciation of various shapes in our world. They will also learn to reason deductively throughout the course and apply this reasoning process to real life situations. See how Geometry fits into other courses of mathematics. The integration of Algebra along with the use of coordinates and transformations is presented in 2 dimensions. The idea of proof is built up slowly throughout the course. Geometry is a graduation requirement. Prerequisite: Algebra I
The GHEC School District does not discriminate on the basis of race, color, national origin, sex, disability, or age in its programs and
activities and provides equal access to the Boy Scouts and other designated youth groups.
The following person has been designated to handle inquiries regarding the non-discrimination policies:
Mandy Fletcher/Superintendent/Principal, 300 Reynolds Street, Granada, MN; 507-447-2211. | 677.169 | 1 |
This 17-slide PowerPoint presentation has been made using quality graphics and lesson-enhancing animations. I strongly recommend downloading the preview file to see examples of the way my animations help bring the Geometry concepts to life! The preview file also shows screenshots of every slide included in this purchase, so you can see exactly what you are buying beforehand. Your students will have fun learning while you save valuable time with this ready-to-go lesson!
Topics within this Lesson PowerPoint include:
- Overview and real-world applications of slope
- Common sense approach to the sloe formula
- Slope-Intercept Form & Point-Slope Form of a linear equation (Identifying slope, y-intercept or a point on the line; Graphing lines in different forms; Writing equations given characteristics or given a graph)
- Equations of Vertical and Horizontal Lines
(Due to its size, this lesson is best taught over two days or in a block-period).
This download includes one teacher PPT file with complete notes and step-by-step solutions. Also included is an accompanying student skeleton note PPT file, which can either be printed for students on which to write notes, or the PPT can be given directly to your students for use with student tablet laptops.
Note: This lesson is also available as part of a Chapter 3 Bundle (click here), or all lessons are available as part of a progressive 13-Chapter Geometry Course bundle. (A progressive bundle is compiled and updated over a period of time. My promise to you is to have the full 13-chapter Geometry bundle completed by no later than Jan 1, 2016. Currently it includes 3 chapters.)
Tip: To edit the files you will need to first copy the files from this compressed (zipped) folder onto your computer, then you should be able to edit and print the files as you wish. Please message me with any questions. Thanks!
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Here are some other resources you might like!
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Today, we share our final blog from our series highlighting courses available on edX that you can take as your first steps towards exploring and learning STEM – science, technology, engineering and math – skills. Read the previous blogs featuring science, technology and engineering courses.
Math is the science of numbers, quantity, structure, change, and space and we use it in our everyday lives. From using technology to cooking and paying bills, having basic math skills helps you effectively solve problems and work through daily tasks. Not only does it help you in your personal life, but it is integrated into nearly all fields of study, jobs, and industries. Biologists, photographers, analysts, engineers, marketing and finance professional all use mathematics.
There are two main divisions of mathematics, pure and applied. Pure mathematics is the study of the basic concepts and structures that underlie mathematics, such as algebra, geometry and calculus. Applied mathematics focuses on the study of mathematical methods that can be applied to real-world problems in the areas of, for example, science, engineering, business and computer science. Learning about these branches not only helps you gain foundational, practical skills but gives you a competitive edge in the workplace.
EdX offers an extensive variety of math courses, ranging from basic introductory to more advanced topics. Here are six courses you can take to advance your math skills: | 677.169 | 1 |
Understanding Multivariable Calculus: Problems, Solutions, and Tips
Gain a profound understanding of multivariable calculus with this excellent and clear guide that is useful for students, professionals, and lovers of mathematics.
Watch First Lecture
Review key concepts from basic calculus, then immediately jump into three dimensions with a brief overview of what you'll be learning. Apply distance and midpoint formulas to three-dimensional objects in your very first of many extrapolations from two-dimensional to multidimensional calculus, and observe some of the new curiosities unique to functions of more than one variable.
Reviews
Awesome course. Very concise, upbeat, and the visual aids helped me a lot.
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k********m
February 22, 2017
Great teacher! I just wish they would use bracket notation instead of ijk fo vectors.
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e********m
August 6, 2016
Just finished a Multivariable Calculus course and I must say that this series was a large part of the success I had in it. Prof. Edwards passion for the material is evident in his presentation style. He makes interesting things which could otherwise just be drudgery and tedious. Well done!! | 677.169 | 1 |
Quaternions and Clifford Geometric Algebras
Quaternions and Clifford Geometric Algebras
Book Details:
pos
Global
pos
Category
Year:
2015
Publisher:
Autoedición
Pages:
187 pages
Language:
english
Since:
01/07/2015
Size:
783 KB
License:
Pending review
Content:
In mathematics, the quaternions are a number system that extends the complex numbers. They were first described by Irish mathematician William Rowan Hamilton in 1843 and applied to mechanics in three-dimensional space. A feature of quaternions is that multiplication of two quaternions is noncommutative. Hamilton defined a quaternion as the quotient of two directed lines in a three-dimensional space or equivalently as the quotient of two vectors.
Quaternions find uses in both theoretical and applied mathematics, in particular for calculations involving three-dimensional rotations such as in three-dimensional computer graphics, computer vision and crystallographic texture analysis. In practical applications, they can be used alongside other methods, such as Euler angles and rotation matrices, or as an alternative to them, depending on the application.
As a first rough draft that has been put together very quickly, this book is likely to contain errata and disorganization. The references list and inline citations are very incompete, so the reader should search around for more references.
I do not claim to be the inventor of any of the mathematics found here. However, some parts of this book may be considered new in some sense and were in small parts my own original research.
Much of the contents was originally written by me as contributions to a web encyclopedia project just for fun, but for various reasons was inappropriate in an encyclopedic volume. I did not originally intend to write this book. This is not a dissertation, nor did its development receive any funding or proper peer review. | 677.169 | 1 |
Pre-calculus concepts. Can identify student shortfalls and address them through from | 677.169 | 1 |
incompetence prompts maths review
Concern over the mathematical competence of students starting degree courses in mathematics, science and engineering has spurred the London Mathematical Society to launch an inquiry into the National Curriculum, A levels and their impact on the needs of universities.
The working group is to be chaired by Geoffrey Howson, emeritus professor of mathematics at Southampton University. Professor Howson stressed that the aim is not to consider the whole of the National Curriculum and that the emphasis "will not be on standards as such but on whether high attainers who go on to do A levels are being set suitable goals".
He said that the group will also examine whether schools and colleges are properly promoting mathematics and its intellectual demands.
"We want to see what can be done to help ease the path of students into science, engineering and mathematics."
Professor Howson said that the problem of students not being adequately prepared mathematically is felt most acutely by engineering and physics departments. Creating time for improving technique on mathematics courses was a little easier than on engineering and science courses.
Another member of the working group, David Crighton, head of the department of applied mathematics and theoretical physics at Cambridge University, has already voiced concern about mathematics in schools and colleges. He believes that the combination of the National Curriculum and the A-level system has allowed standards to "drop very considerably" and that many students are not well prepared for entry on to science, mathematics and engineering courses.
He said his concerns were widely shared in higher education: "A common concern is that there is far too much emphasis on self-discovery rather than the presentation of material as a body of knowledge. Such knowledge is the culmination of the work of very smart people over a very long period of time. It is laughable that pupils can achieve mastery of such work through self-discovery."
Students are also lacking mathematical "fluency". Professor Crighton said that this stems from an overemphasis on numerical problems in applied mathematics, at the expense of technique for manipulation of equations rather than numbers.
"That emphasis falsifies the whole nature of applied mathematics." Much greater attention should be paid to arming pupils with a conceptual framework and understanding of mathematics, he argued.
Professor Crighton will be highlighting his concerns next month as part of a campaign to promote mathematics by the Institute of Mathematics and its Applications, of which he is to be the president.
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Learning a new language is a fun and challenging feat for students at every level. Perfect for those just starting out or returning to Spanish after time away, Spanish Essentials For Dummies focuses on core …
Passing grades in two years of algebra courses are required for high school graduation. Algebra II Essentials For Dummies covers key ideas from typical second-year Algebra coursework to help students get up to speed.
Many students worry about starting algebra. Pre-Algebra Essentials For Dummies provides an overview of critical pre-algebra concepts to help new algebra students (and their parents) take the next step without fear.
Just the essential concepts you need to get ahead in statistics This practical guide sticks to the point with discrete explanations of essential concepts taught in a typical first semester statistics course. | 677.169 | 1 |
What Is the Input & Output in Math?
By Amy Harris; Updated April 24, 2017
Students learn about input and output in math as part of a pre-algebra course, or in preparation for one. Simply put, inputs are numeric values to which a procedure is applied, producing an output, which is also a numeric value. Students typically learn about inputs and outputs during a wider study of the topic of functions.
Relationship to Functions
Understanding the concept of a function is crucial to understanding inputs and outputs, because functions are the bridges that connect inputs to outputs. Functions are rules or processes in which every input value is assigned to an output value. The input is the independent variable, and the output is the dependent variable -- that is, the output's value depends upon the input's value. There can be multiple inputs, but each input results in exactly one output.
Notation and Presentation
Inputs and outputs may be presented either in table or equation form. Tables have two columns or rows, which may be oriented either vertically or horizontally. In the case of the first, input values appear on the left side and output values appear on the right; in the case of the latter, inputs appear on top and outputs on the bottom. In equation form, an x usually represents the input quantity, while a y or f(x) -- where the "f" stands for function -- usually represents the output. For instance, in the function y = x – 3, y is the output value, while x is the input value.
References
About the Author
Based in western New York, Amy Harris began writing for Demand Media and Great Lakes Brewing News in 2010. Harris holds a Bachelor of Science in Mathematics from Penn State University; she taught high school math for several years and has also worked in the field of instructional design. | 677.169 | 1 |
The purpose of this study is to review previous studies regarding the relationship between graphing calculators usage and teaching and learning school mathematics and to suggest practical implications for further research in mathematics education. Through reviewing the total of 21 studies five subsections are divided in order to gain the answers to the research questions. The results of this study are as follows: students typically used graphing calculators in drawing graphs of functions, and they used graphing calculators as a tool for calculations, transformations, data collection and analysis, visualization, and checking. The implications for further research are suggested corresponding to these results. | 677.169 | 1 |
An introduction to the Calculus, with an excellent balance between theory and technique. Integration is treated before differentiation—this is a departure from most modern texts, but it is historically correct, and it is the best way to establish the...
Professor Pearson's book starts with an introduction to the area and an explanation of the most commonly used functions. It then moves on through differentiation, special functions, derivatives, integrals and onto full differential equations. As with...
A new version of this classic math primer explains the timeless theories of calculus in a contemporary and comprehensible way, with updated method and terminology, and twenty new recreational problems for practice and enjoyment. 20,000 firstFor freshman-level, two-semester or three-semester courses in Calculus for Life Sciences. Shows students how calculus is used to analyze phenomena in nature -- while providing flexibility for instructors to teach at their desired level...
This introduction to calculus is designed for beginning college undergraduates majoring in mathematics as well as undergraduates pursuing other areas of science and engineering for whom calculus will be a vital tool.The three-part treatment begins by...
Your light-hearted, practical approach to conquering calculus Does the thought of calculus give you a coronary? You aren't alone. Thankfully, this new edition of Calculus Workbook For Dummies makes it infinitely easier. Focusing 'beyond the classroom...
Calculus For Dummies, 2nd Edition (9781119293491) was previously published as Calculus For Dummies, 2nd Edition (9781118791295). While this version features a new Dummies cover and design, the content is the same as the prior release and should notThis comprehensive but concise introductory workbook is less rigorous than most calculus texts. Its informal and accessible treatment explains functions, derivatives, differentiation of algebraic functions and transcendental functions, partial...
Master calculus from the comfort of home!Want to 'know it ALL' when it comes to calculus? This book gives you the expert, one-on-one instruction you need, whether you're new to calculus or you're looking to ramp up your skills. Providing...
Many colleges and universities require students to take at least one math course, and Calculus I is often the chosen option. Calculus Essentials For Dummies provides explanations of key concepts for students who may have taken calculus in high school...
This concise text offers an introduction to the fundamentals and standard methods of the calculus of variations. In addition to surveys of problems with fixed and movable boundaries, its subjects include practical direct methods for solution of...
This introductory text offers a far-reaching, rigorous, application-oriented approach to variational theory that will increase students' understanding of more specialized books and research papers in the field. The treatment acquaints readers with...
Authored by two distinguished researchers/teachers and an experiences, successful textbook author, Calculus for Life Sciences is a valuable resource for Life Science courses. As life-science departments increase the math requirements for their majorsBased on undergraduate courses in advanced calculus, the treatment covers a wide range of topics, from soft functional analysis and finite-dimensional linear algebra to differential equations on submanifolds of Euclidean space.
Building on previous texts in the Modular Mathematics series, in particular 'Vectors in Two or Three Dimensions' and 'Calculus and ODEs', this book introduces the student to the concept of vector calculus. It provides an overview of some of the key... | 677.169 | 1 |
AQA Business for A Level 2 by Andrew Gillespie, Malcolm Surridge textbook has been fully revised to reflect the 2015 AQA Business specification, giving you up-to-date material that supports your teaching and student's learning. - Builds up quantitative skills with 'Maths moment' features and assesses them in the end of chapter activities - Ensures students have the kwledge of real life businesses so they can apply their theoretical understanding with the 'Business in focus' feature - Helps students get to grips with the content and tests key skills with activities at the end of every chapter
Author Biography
Andrew Gillespie and Malcolm Surridge are both experienced authors and this is the 5th edition of their successful textbook. | 677.169 | 1 |
great introduction to the tricky concept of differentiation. Fully animated graphical display of how we use tangents to curves to differentiate a function, with regular examples for the teacher to explain key concepts. Solomon Worksheet Questions and Answers provided and embedded throughout the Powerpoint. If you like these resource, please review so I can see if there would be interest for my other A-Level / IB resources.
Subject: Calculus
Grade Levels: 10th, 11th, 12th, Higher Education, Adult Education, Homeschool
Resource Type: Examinations/Quizzes
This is the second quiz in the Integration by Algebraic Substitution series. These problems are dealing with integrals that have only Examinations/Quizzes (3 pages)
This is the first quiz in the Integration by Algebraic Substitution series. These problems are dealing with integrals that have Test Preparation
This is a complete table of the derivatives of all algebraic, trigonometric, exponential, logarithmic and hyperbolic functions that students taking AP Calculus through sophomore Linear Algebra and Ordinary Differential Equations are likely to see. The table contains 49 items, including basic properties and derivative formulas with some variations. This Table of Derivatives was designed as a companion to the Short and longer-form Table of Integrals.
.pdf will not open on phone, use a computer or tablet.
This carefully selected compilation of exam questions has fully-worked solutions designed for students to go through at home. This can save a lot of valuable time in class.
Click --> Question Practice to download question compilations on 40 other topics.
--
This series of 'booklets' cover the full range of skills expected at GCSE/IGCSE for each topic. The questions in this particular compilation gives examples of the types of differentiation questions that have come up in IGCSE exams in the past. This includes calculations of gradients and turning points, but also includes working out velocities and accelerations by differentiating the distance-time equation. There are also a couple of examples of using differentiation within a different context, such as finding a maximum/minimum area or perimeter.
I usually print these questions as an A5 booklet and issue them in class or give them out as a homework. I also make them available for a student who wants to do independent study
A revision sheet (with answers) containing IGCSE exam-type questions, which require the students to differentiate to work out equations for velocity and acceleration.
This sheet is designed for International GCSE revision (IGCSE), but could also be used as a homework for first-year A-level students | 677.169 | 1 |
Algebra can be exciting!
Basic Algebra Skills 1
This video shows the basic structure of how to perform algebra at the beginner level for highschoolers | 677.169 | 1 |
Content: The modern notion of measure, developed in the late 19th century, is an extension of the notions of length, area or volume. A measurem is a law which assigns a number to certain subsets A of a given space and is a natural generalization of the following notions: 1) length of an interval, 2) area of a plane figure, 3) volume of a solid, 4) amount of mass contained in a region, 5) probability that an event from A occurs, etc.
It originated in the real analysis and is used now in many areas of mathematics like, for instance, geometry, probability theory, dynamical systems, functional analysis, etc.
Given a measure m, one can define the integral of suitable real valued functions with respect to m. Riemann integral is applied to continuous functions or functions with ``few`` points of discontinuity. For measurable functions that can be discontinuous ``almost everywhere'' Riemann integral does not make sense. However it is possible to define more flexible and powerful Lebesgue's integral (integral with respect to Lebesgue's measure) which is one of the key notions of modern analysis.
Aims: To introduce the concepts of measure and integral with respect to a measure, to show their basic properties, and to provide a basis for further studies in Analysis, Probability, and Dynamical Systems.
Objectives: To gain understanding of the abstract measure theory and definition and main properties of the integral. To construct Lebesgue's measure on the real line and in n-dimensional Euclidean space. To explain the basic advanced directions of the theory.
Books: There is no official textbook for the course. The list below contains some of many books that may be used to complement the lectures. | 677.169 | 1 |
Introduction to mathematical logic by Elliott Mendelson(
Book
) 168
editions published
between
1963
and
2015
in
5
languages
and held by
2,183 WorldCat member
libraries
worldwide
"Retaining all the key features of the previous editions, Introduction to Mathematical Logic, Fifth Edition explores the principal
topics of mathematical logic. It covers propositional logic, first-order logic, first-order number theory, axiomatic set theory,
and the theory of computability. The text also discusses the major results of Godel, Church, Kleene, Rosser, and Turing."--Jacket
Schaum's outline of theory and problems of differential and integral calculus by Frank Ayres(
Book
) 75
editions published
between
1964
and
2013
in
3
languages
and held by
1,672 WorldCat member
libraries
worldwide
More than 40 million students have trusted Schaum's to help them succeed in the classroom and on exams. Schaum's is the key
to faster learning and higher grades in every subject. Each Outline presents all the essential course information in an easy-to-follow,
topic-by-topic format. You also get hundreds of examples, solved problems, and practice exercises to test your skills
Schaum's outline of beginning calculus by Elliott Mendelson(
Book
) 50
editions published
between
1985
and
2011
in
English
and held by
1,270 WorldCat member
libraries
worldwide
A concise study guide for beginning calculus students that provides step-by-step instructions in solving problems; covers
fundamentals; and contains 226 fully solved problems as well as over 1500 supplementary problems with answers
Schaum's outline of calculus by Frank Ayres(
Book
) 16
editions published
between
1999
and
2002
in
3
languages
and held by
582 WorldCat member
libraries
worldwide
1103 fully solved problems; expert tips on the graphing calculaor; many math review sections; covers all course fundamentals,
supplements any class text; the perfect aid for higher grades
3000 solved problems in calculus by Elliott Mendelson(
Book
) 28
editions published
between
1988
and
2009
in
English
and held by
480 WorldCat member
libraries
worldwide
Introducing game theory and its applications by Elliott Mendelson(
Book
) 12
editions published
between
2004
and
2016
in
English
and held by
448 WorldCat member
libraries
worldwide
"Introducing Game Theory and its Applications presents an introduction to the basic ideas and techniques of game theory. This
book prepares its readers for more advanced study of game theory's applications in economics, business, and the physical,
biological, and social sciences."--Jacket
Schaum's outline of 3000 solved problems in calculus by Elliott Mendelson(
Book
) 10
editions published
between
1988
and
2009
in
English
and held by
189 WorldCat member
libraries
worldwide
A collection of problems in various fields of calculus and their solutions
McGraw-Hill's 500 calculus questions : ace your college exams by Elliott Mendelson(
Book
) 3
editions published
in
2013
in
English
and held by
82 WorldCat member
libraries
worldwide
Contains questions and answers designed to prepare the reader for a college calculus exam, including such topics as inequalities,
trigonometric functions, and improper integrals
Foundations--logic, language, and mathematics by Hugues Leblanc(
Book
) 10
editions published
in
1984
in
English
and held by
82 WorldCat member
libraries
worldwide
The more traditional approaches to the history and philosophy of science and technology continue as well, and probably will
continue as long as there are skillful practitioners such as Carl Hempel, Ernest Nagel, and th~ir students. Finally, there
are still other approaches that address some of the technical problems arising when we try to provide an account of belief
and of rational choice. - These include efforts to provide logical frameworks within which we can make sense of these notions.
This series will attempt to bring together work from all of these approaches to the history and philosophy of science and
technology in the belief that each has something to add to our understanding. The volumes of this series have emerged either
from lectures given by authors while they served as honorary visiting professors at the City College of New York or from conferences
sponsored by that institution. The City College Program in the History and Philosophy of Science and Technology oversees and
directs these lectures and conferences with the financial aid of the Association for Philosophy of Science, Psychotheraphy,
and Ethics. MARTIN TAMNY RAPHAEL STERN PREFACE The papers in this collection stem largely from the conference 'Foun dations:
Logic, Language, and Mathematics' held at the Graduate Center of the City University of New York on 14-15 November 1980 | 677.169 | 1 |
Martindale Calculators
Martindale Calculators is a Web-based tool collection that contains over 19,000 online calculators created by over "3,450" very "creative" individuals, businesses and "tax supported entities world wide." The collection is organized by the following topics: mathematics; statistics; science A-Z; chemistry; physics, astrophysics and astronomy; engineering A-Z; and electrical engineering, computer engineering, & computer science. Each section includes a wealth of websites to explore, all related to mathematical calculations, mostly course materials and articles. Another section lists online calculators relevant for various industries, such as aviation, cosmetics, insurance, and library science. The list is organized alphabetically and creatively stretches the meaning of "calculator" to include such things as name translators and databases on animal breeds. | 677.169 | 1 |
01308610College Mathematics for Technology (5th Edition)
For Technical Math courses that do not include calculus.This text addresses curriculum and pedagogy standards that are initiatives of the American Mathematical Association of two-year Colleges (AMATYC), the National Council of Teachers of Mathematics (NCTM), and the Mathematics Association of America (MAA). It uses simplified language that appeals to a variety of student learning styles, promotes active and independent learning, and strengthens critical thinking and writing skills. A "six-step approach" to problem solving, numerous tips, and clear, concise explanations throughout the text enable students to understand the concepts underlying mathematical processes | 677.169 | 1 |
Riphah FM 102.2 App
MAPLE works just like a high-level language and also provides an interactive environmental for numerical computing. The MAPLE tools and built-in math functions enable the users to explore multiple approaches and reach a solution faster than with spread sheets or traditional programming languages. | 677.169 | 1 |
For a one-quarter/semester, sophomore-level transitional
("bridge") course that supplies background for students going
from calculus to the more abstract, upper-division mathematics courses.
Also appropriate as a supplement for junior-level courses such as
abstract algebra or real analysis.
Focused on "What Every Mathematician Needs to Know,"
this text provides material necessary for students to succeed in upper-division
mathematics courses, and more importantly, the analytical tools necessary
for thinking like a mathematician. It begins with a natural progression
from elementary logic, methods of proof, and set theory, to relations
and functions; then provides application examples, theorems, and student
projects.
NEW—"Find the Flaw" problems—At
the beginning of exercise sets in Chs. 1-5.
Help students learn to read proofs critically.
NEW—A collection of True/False questions—Begins
each set of review exercises in Chs. 2-5.
NEW—Many new exercises of all kinds—More
than in any other textbook of its kind.
Five core chapters (Chs. 1-5)—In a natural progression:
elementary logic, methods of proof, set theory, functions, and relations.
Each chapter contains a full exposition of topics with many examples
and practice problems to reinforce the concepts as they are introduced.
Anticipates many of the questions students might have
and develops the subject slowly and carefully. Students are then able
to work more independently—and with much greater understanding
of the material.
Four chapters of examples, theorems, and projects (Chs.
6-9)—Many theorems have no proof or only a hint or outline
for the proof. Likewise, the examples may have no solutions or just
a hint for the solution.
The intent is that the material be used as a basis for
students to construct their own proofs or solutions and perhaps present
them to the class.
Clearly written examples and practice problems—
Provides solutions to practice problems, odd-numbered exercises, and
review problems.
Supplementary exercises—Extend or relate to some
of the concepts discussed in the text but are not necessary for the
continuity of the subject matter. | 677.169 | 1 |
Activity Overview
In this activity, students use a motion detector to collect "linear" motion data and examine the relationship between a physical action and a mathematical and/or graphic model of that action. The students will use the "eyeball" method to find the mathematical model.
Before the Activity
Install the Transformation Graphing application on the students' graphing calculators using one of these two methods: | 677.169 | 1 |
Elements of
Abstract and Linear Algebra
E. H. Connell
ii
E.H. Connell
Department of Mathematics
University of Miami
P.O. Box 249085
Coral Gables, Florida 33124 USA
ec@math.miami.edu
Mathematical Subject Classifications (1991): 12-01, 13-01, 15-01, 16-01, 20-01
c (1999 E.H. Connell
March 20, 2004
iii
Introduction
In 1965 I first taught an undergraduate course in abstract algebra. It was fun to
teach because the material was interesting and the class was outstanding. Five of
those students later earned a Ph.D. in mathematics. Since then I have taught the
course about a dozen times from various texts. Over the years I developed a set of
lecture notes and in 1985 I had them typed so they could be used as a text. They
now appear (in modified form) as the first five chapters of this book. Here were some
of my motives at the time.
1) To have something as short and inexpensive as possible. In my experience,
students like short books.
2) To avoid all innovation. To organize the material in the most simple-minded
straightforward manner.
3) To order the material linearly. To the extent possible, each section should use
the previous sections and be used in the following sections.
4) To omit as many topics as possible. This is a foundational course, not a topics
course. If a topic is not used later, it should not be included. There are three
good reasons for this. First, linear algebra has top priority. It is better to go
forward and do more linear algebra than to stop and do more group and ring
theory. Second, it is more important that students learn to organize and write
proofs themselves than to cover more subject matter. Algebra is a perfect place
to get started because there are many "easy" theorems to prove. There are
many routine theorems stated here without proofs, and they may be considered
as exercises for the students. Third, the material should be so fundamental
that it be appropriate for students in the physical sciences and in computer
science. Zillions of students take calculus and cookbook linear algebra, but few
take abstract algebra courses. Something is wrong here, and one thing wrong
is that the courses try to do too much group and ring theory and not enough
matrix theory and linear algebra.
5) To offer an alternative for computer science majors to the standard discrete
mathematics courses. Most of the material in the first four chapters of this text
is covered in various discrete mathematics courses. Computer science majors
might benefit by seeing this material organized from a purely mathematical
viewpoint.
iv
Over the years I used the five chapters that were typed as a base for my algebra
courses, supplementing them as I saw fit. In 1996 I wrote a sixth chapter, giving
enough material for a full first year graduate course. This chapter was written in the
same "style" as the previous chapters, i.e., everything was right down to the nub. It
hung together pretty well except for the last two sections on determinants and dual
spaces. These were independent topics stuck on at the end. In the academic year
1997-98 I revised all six chapters and had them typed in LaTeX. This is the personal
background of how this book came about.
It is difficult to do anything in life without help from friends, and many of my
friends have contributed to this text. My sincere gratitude goes especially to Marilyn
Gonzalez, Lourdes Robles, Marta Alpar, John Zweibel, Dmitry Gokhman, Brian
Coomes, Huseyin Kocak, and Shulim Kaliman. To these and all who contributed,
this book is fondly dedicated.
This book is a survey of abstract algebra with emphasis on linear algebra. It is
intended for students in mathematics, computer science, and the physical sciences.
The first three or four chapters can stand alone as a one semester course in abstract
algebra. However they are structured to provide the background for the chapter on
linear algebra. Chapter 2 is the most difficult part of the book because groups are
written in additive and multiplicative notation, and the concept of coset is confusing
at first. After Chapter 2 the book gets easier as you go along. Indeed, after the
first four chapters, the linear algebra follows easily. Finishing the chapter on linear
algebra gives a basic one year undergraduate course in abstract algebra. Chapter 6
continues the material to complete a first year graduate course. Classes with little
background can do the first three chapters in the first semester, and chapters 4 and 5
in the second semester. More advanced classes can do four chapters the first semester
and chapters 5 and 6 the second semester. As bare as the first four chapters are, you
still have to truck right along to finish them in one semester.
The presentation is compact and tightly organized, but still somewhat informal.
The proofs of many of the elementary theorems are omitted. These proofs are to
be provided by the professor in class or assigned as homework exercises. There is a
non-trivial theorem stated without proof in Chapter 4, namely the determinant of the
product is the product of the determinants. For the proper flow of the course, this
theorem should be assumed there without proof. The proof is contained in Chapter 6.
The Jordan form should not be considered part of Chapter 5. It is stated there only
as a reference for undergraduate courses. Finally, Chapter 6 is not written primarily
for reference, but as an additional chapter for more advanced courses.
v
This text is written with the conviction that it is more effective to teach abstract
and linear algebra as one coherent discipline rather than as two separate ones. Teach-
ing abstract algebra and linear algebra as distinct courses results in a loss of synergy
and a loss of momentum. Also with this text the professor does not extract the course
from the text, but rather builds the course upon it. I am convinced it is easier to
build a course from a base than to extract it from a big book. Because after you
extract it, you still have to build it. The bare bones nature of this book adds to its
flexibility, because you can build whatever course you want around it. Basic algebra
is a subject of incredible elegance and utility, but it requires a lot of organization.
This book is my attempt at that organization. Every effort has been extended to
make the subject move rapidly and to make the flow from one topic to the next as
seamless as possible. The student has limited time during the semester for serious
study, and this time should be allocated with care. The professor picks which topics
to assign for serious study and which ones to "wave arms at". The goal is to stay
focused and go forward, because mathematics is learned in hindsight. I would have
made the book shorter, but I did not have any more time.
When using this text, the student already has the outline of the next lecture, and
each assignment should include the study of the next few pages. Study forward, not
just back. A few minutes of preparation does wonders to leverage classroom learning,
and this book is intended to be used in that manner. The purpose of class is to
learn, not to do transcription work. When students come to class cold and spend
the period taking notes, they participate little and learn little. This leads to a dead
class and also to the bad psychology of "OK, I am here, so teach me the subject."
Mathematics is not taught, it is learned, and many students never learn how to learn.
Professors should give more direction in that regard.
Unfortunately mathematics is a difficult and heavy subject. The style and
approach of this book is to make it a little lighter. This book works best when
viewed lightly and read as a story. I hope the students and professors who try it,
enjoy it.
E. H. Connell
Department of Mathematics
University of Miami
Coral Gables, FL 33124
ec@math.miami.edu
vi
Outline
Chapter 1 Background and Fundamentals of Mathematics
Sets, Cartesian products 1
Relations, partial orderings, Hausdorff maximality principle, 3
equivalence relations
Functions, bijections, strips, solutions of equations, 5
right and left inverses, projections
Notation for the logic of mathematics 13
Integers, subgroups, unique factorization 14
Chapter 2 Groups
Groups, scalar multiplication for additive groups 19
Subgroups, order, cosets 21
Normal subgroups, quotient groups, the integers mod n 25
Homomorphisms 27
Permutations, the symmetric groups 31
Product of groups 34
Chapter 3 Rings
Rings 37
Units, domains, fields 38
The integers mod n 40
Ideals and quotient rings 41
Homomorphisms 42
Polynomial rings 45
Product of rings 49
The Chinese remainder theorem 50
Characteristic 50
Boolean rings 51
Chapter 4 Matrices and Matrix Rings
Addition and multiplication of matrices, invertible matrices 53
Transpose 56
Triangular, diagonal, and scalar matrices 56
Elementary operations and elementary matrices 57
Systems of equations 59
vii
Determinants, the classical adjoint 60
Similarity, trace, and characteristic polynomial 64
Chapter 5 Linear Algebra
Modules, submodules 68
Homomorphisms 69
Homomorphisms on R
n
71
Cosets and quotient modules 74
Products and coproducts 75
Summands 77
Independence, generating sets, and free basis 78
Characterization of free modules 79
Uniqueness of dimension 82
Change of basis 83
Vector spaces, square matrices over fields, rank of a matrix 85
Geometric interpretation of determinant 90
Linear functions approximate differentiable functions locally 91
The transpose principle 92
Nilpotent homomorphisms 93
Eigenvalues, characteristic roots 95
Jordan canonical form 96
Inner product spaces, Gram-Schmidt orthonormalization 98
Orthogonal matrices, the orthogonal group 102
Diagonalization of symmetric matrices 103
Chapter 6 Appendix
The Chinese remainder theorem 108
Prime and maximal ideals and UFD
s
109
Splitting short exact sequences 114
Euclidean domains 116
Jordan blocks 122
Jordan canonical form 123
Determinants 128
Dual spaces 130
viii
1 2 3 4
5 6 7 8
9 11 10
Abstract algebra is not only a major subject of science, but it is also
magic and fun. Abstract algebra is not all work and no play, and it is
certainly not a dull boy. See, for example, the neat card trick on page
18. This trick is based, not on sleight of hand, but rather on a theorem
in abstract algebra. Anyone can do it, but to understand it you need
some group theory. And before beginning the course, you might first try
your skills on the famous (some would say infamous) tile puzzle. In this
puzzle, a frame has 12 spaces, the first 11 with numbered tiles and the
last vacant. The last two tiles are out of order. Is it possible to slide the
tiles around to get them all in order, and end again with the last space
vacant? After giving up on this, you can study permutation groups and
learn the answer!
Chapter 1
Background and Fundamentals of
Mathematics
This chapter is fundamental, not just for algebra, but for all fields related to mathe-
matics. The basic concepts are products of sets, partial orderings, equivalence rela-
tions, functions, and the integers. An equivalence relation on a set A is shown to be
simply a partition of A into disjoint subsets. There is an emphasis on the concept
of function, and the properties of surjective, injective, and bijective. The notion of a
solution of an equation is central in mathematics, and most properties of functions
can be stated in terms of solutions of equations. In elementary courses the section
on the Hausdorff Maximality Principle should be ignored. The final section gives a
proof of the unique factorization theorem for the integers.
Notation Mathematics has its own universally accepted shorthand. The symbol
∃ means "there exists" and ∃! means "there exists a unique". The symbol ∀ means
"for each" and ⇒ means "implies". Some sets (or collections) are so basic they have
their own proprietary symbols. Five of these are listed below.
N = Z
+
= the set of positive integers = ¦1, 2, 3, ...¦
Z = the ring of integers = ¦..., −2, −1, 0, 1, 2, ...¦
Q = the field of rational numbers = ¦a/b : a, b ∈ Z, b = 0¦
R = the field of real numbers
C = the field of complex numbers = ¦a +bi : a, b ∈ R¦ (i
2
= −1)
Sets Suppose A, B, C,... are sets. We use the standard notation for intersection
and union.
A ∩ B = ¦x : x ∈ A and x ∈ B¦ = the set of all x which are elements
1
2 Background Chapter 1
of A and B.
A ∪ B = ¦x : x ∈ A or x ∈ B¦ = the set of all x which are elements of
A or B.
Any set called an index set is assumed to be non-void. Suppose T is an index set and
for each t ∈ T, A
t
is a set.
¸
t∈T
A
t
= ¦x : ∃ t ∈ T with x ∈ A
t
¦
¸
t∈T
A
t
= ¦x : if t ∈ T, x ∈ A
t
¦ = ¦x : ∀t ∈ T, x ∈ A
t
¦
Let ∅ be the null set. If A∩ B = ∅, then A and B are said to be disjoint.
Definition Suppose each of A and B is a set. The statement that A is a subset
of B (A ⊂ B) means that if a is an element of A, then a is an element of B. That
is, a ∈ A ⇒a ∈ B. If A ⊂ B we may say A is contained in B, or B contains A.
Exercise Suppose each of A and B is a set. The statement that A is not a subset
of B means .
Theorem (De Morgan's laws) Suppose S is a set. If C ⊂ S (i.e., if C is a subset
of S), let C
, the complement of C in S, be defined by C
= S−C = ¦x ∈ S : x ∈ C¦.
Then for any A, B ⊂ S,
(A ∩ B)
= A
∪ B
and
(A ∪ B)
= A
∩ B
Cartesian Products If X and Y are sets, X Y = ¦(x, y) : x ∈ X and y ∈ Y ¦.
In other words, the Cartesian product of X and Y is defined to be the set of all
ordered pairs whose first term is in X and whose second term is in Y .
Example RR = R
2
= the plane.
Chapter 1 Background 3
Definition If each of X
1
, ..., X
n
is a set, X
1
X
n
= ¦(x
1
, ..., x
n
) : x
i
∈ X
i
for 1 ≤ i ≤ n¦ = the set of all ordered n-tuples whose i-th term is in X
i
.
Example R R = R
n
= real n-space.
Question Is (RR
2
) = (R
2
R) = R
3
?
Relations
If A is a non-void set, a non-void subset R ⊂ A A is called a relation on A. If
(a, b) ∈ R we say that a is related to b, and we write this fact by the expression a ∼ b.
Here are several properties which a relation may possess.
1) If a ∈ A, then a ∼ a. (reflexive)
2) If a ∼ b, then b ∼ a. (symmetric)
2
) If a ∼ b and b ∼ a, then a = b. (anti-symmetric)
3) If a ∼ b and b ∼ c, then a ∼ c. (transitive)
Definition A relation which satisfies 1), 2
), and 3) is called a partial ordering.
In this case we write a ∼ b as a ≤ b. Then
1) If a ∈ A, then a ≤ a.
2
) If a ≤ b and b ≤ a, then a = b.
3) If a ≤ b and b ≤ c, then a ≤ c.
Definition A linear ordering is a partial ordering with the additional property
that, if a, b ∈ A, then a ≤ b or b ≤ a.
Example A = R with the ordinary ordering, is a linear ordering.
Example A = all subsets of R
2
, with a ≤ b defined by a ⊂ b, is a partial ordering.
Hausdorff Maximality Principle (HMP) Suppose S is a non-void subset of A
and ∼ is a relation on A. This defines a relation on S. If the relation satisfies any
of the properties 1), 2), 2
), or 3) on A, the relation also satisfies these properties
when restricted to S. In particular, a partial ordering on A defines a partial ordering
4 Background Chapter 1
on S. However the ordering may be linear on S but not linear on A. The HMP is
that any linearly ordered subset of a partially ordered set is contained in a maximal
linearly ordered subset.
Exercise Define a relation on A = R
2
by (a, b) ∼ (c, d) provided a ≤ c and
b ≤ d. Show this is a partial ordering which is linear on S = ¦(a, a) : a < 0¦. Find
at least two maximal linearly ordered subsets of R
2
which contain S.
One of the most useful applications of the HMP is to obtain maximal monotonic
collections of subsets.
Definition A collection of sets is said to be monotonic if, given any two sets of
the collection, one is contained in the other.
Corollary to HMP Suppose X is a non-void set and A is some non-void
collection of subsets of X, and S is a subcollection of A which is monotonic. Then ∃
a maximal monotonic subcollection of A which contains S.
Proof Define a partial ordering on A by V ≤ W iff V ⊂ W, and apply HMP.
The HMP is used twice in this book. First, to show that infinitely generated
vector spaces have free bases, and second, in the Appendix, to show that rings have
maximal ideals (see pages 87 and 109). In each of these applications, the maximal
monotonic subcollection will have a maximal element. In elementary courses, these
results may be assumed, and thus the HMP may be ignored.
Equivalence Relations A relation satisfying properties 1), 2), and 3) is called
an equivalence relation.
Exercise Define a relation on A = Z by n ∼ m iff n − m is a multiple of 3.
Show this is an equivalence relation.
Definition If ∼ is an equivalence relation on A and a ∈ A, we define the equiva-
lence class containing a by cl(a) = ¦x ∈ A : a ∼ x¦.
Chapter 1 Background 5
Theorem
1) If b ∈ cl(a) then cl(b) = cl(a). Thus we may speak of a subset of A
being an equivalence class with no mention of any element contained
in it.
2) If each of U, V ⊂ A is an equivalence class and U ∩ V = ∅, then
U = V .
3) Each element of A is an element of one and only one equivalence class.
Definition A partition of A is a collection of disjoint non-void subsets whose union
is A. In other words, a collection of non-void subsets of A is a partition of A provided
any a ∈ A is an element of one and only one subset of the collection. Note that if A
has an equivalence relation, the equivalence classes form a partition of A.
Theorem Suppose A is a non-void set with a partition. Define a relation on A by
a ∼ b iff a and b belong to the same subset of the partition. Then ∼ is an equivalence
relation, and the equivalence classes are just the subsets of the partition.
Summary There are two ways of viewing an equivalence relation — one is as a
relation on A satisfying 1), 2), and 3), and the other is as a partition of A into
disjoint subsets.
Exercise Define an equivalence relation on Z by n ∼ m iff n −m is a multiple
of 3. What are the equivalence classes?
Exercise Is there a relation on R satisfying 1), 2), 2
) and 3) ? That is, is there
an equivalence relation on R which is also a partial ordering?
Exercise Let H ⊂ R
2
be the line H = ¦(a, 2a) : a ∈ R¦. Consider the collection
of all translates of H, i.e., all lines in the plane with slope 2. Find the equivalence
relation on R
2
defined by this partition of R
2
.
Functions
Just as there are two ways of viewing an equivalence relation, there are two ways
of defining a function. One is the "intuitive" definition, and the other is the "graph"
or "ordered pairs" definition. In either case, domain and range are inherent parts of
the definition. We use the "intuitive" definition because everyone thinks that way.
6 Background Chapter 1
Definition If X and Y are (non-void) sets, a function or mapping or map with
domain X and range Y , is an ordered triple (X, Y, f) where f assigns to each x ∈ X
a well defined element f(x) ∈ Y . The statement that (X, Y, f) is a function is written
as f : X →Y or X
f
→Y .
Definition The graph of a function (X, Y, f) is the subset Γ ⊂ X Y defined
by Γ = ¦(x, f(x)) : x ∈ X¦. The connection between the "intuitive" and "graph"
viewpoints is given in the next theorem.
Theorem If f : X → Y , then the graph Γ ⊂ X Y has the property that each
x ∈ X is the first term of one and only one ordered pair in Γ. Conversely, if Γ is a
subset of X Y with the property that each x ∈ X is the first term of one and only
ordered pair in Γ, then ∃! f : X → Y whose graph is Γ. The function is defined by
"f(x) is the second term of the ordered pair in Γ whose first term is x."
Example Identity functions Here X = Y and f : X → X is defined by
f(x) = x for all x ∈ X. The identity on X is denoted by I
X
or just I : X →X.
Example Constant functions Suppose y
0
∈ Y . Define f : X → Y by f(x) =
y
0
for all x ∈ X.
Restriction Given f : X →Y and a non-void subset S of X, define f [ S : S →Y
by (f [ S)(s) = f(s) for all s ∈ S.
Inclusion If S is a non-void subset of X, define the inclusion i : S → X by
i(s) = s for all s ∈ S. Note that inclusion is a restriction of the identity.
Composition Given W
f
→X
g
→Y define g ◦ f : W →Y by
(g ◦ f)(x) = g(f(x)).
Theorem (The associative law of composition) If V
f
→ W
g
→ X
h
→ Y , then
h ◦ (g ◦ f) = (h ◦ g) ◦ f. This may be written as h ◦ g ◦ f.
Chapter 1 Background 7
Definitions Suppose f : X →Y .
1) If T ⊂ Y , the inverse image of T is a subset of X, f
−1
(T) = ¦x ∈ X :
f(x) ∈ T¦.
2) If S ⊂ X, the image of S is a subset of Y , f(S) = ¦f(s) : s ∈ S¦ =
¦y ∈ Y : ∃s ∈ S with f(s) = y¦.
3) The image of f is the image of X , i.e., image (f) = f(X) =
¦f(x) : x ∈ X¦ = ¦y ∈ Y : ∃x ∈ X with f(x) = y¦.
4) f : X →Y is surjective or onto provided image (f) = Y i.e., the image
is the range, i.e., if y ∈ Y , f
−1
(y) is a non-void subset of X.
5) f : X →Y is injective or 1-1 provided (x
1
= x
2
) ⇒f(x
1
) = f(x
2
), i.e.,
if x
1
and x
2
are distinct elements of X, then f(x
1
) and f(x
2
) are
distinct elements of Y .
6) f : X →Y is bijective or is a 1-1 correspondence provided f is surjective
and injective. In this case, there is function f
−1
: Y →X with f
−1
◦ f =
I
X
: X →X and f ◦ f
−1
= I
Y
: Y →Y . Note that f
−1
: Y →X is
also bijective and (f
−1
)
−1
= f.
Examples
1) f : R →R defined by f(x) = sin(x) is neither surjective nor injective.
2) f : R →[−1, 1] defined by f(x) = sin(x) is surjective but not injective.
3) f : [0, π/2] →R defined by f(x) = sin(x) is injective but not surjective.
4) f : [0, π/2] →[0, 1] defined by f(x) = sin(x) is bijective. (f
−1
(x) is
written as arcsin(x) or sin
−1
(x).)
5) f : R →(0, ∞) defined by f(x) = e
x
is bijective. (f
−1
(x) is written as
ln(x).)
Note There is no such thing as "the function sin(x)." A function is not defined
unless the domain and range are specified.
8 Background Chapter 1
Exercise Show there are natural bijections from (R R
2
) to (R
2
R) and
from (R
2
R) to R R R. These three sets are disjoint, but the bijections
between them are so natural that we sometimes identify them.
Exercise Suppose X is a set with 6 elements and Y is a finite set with n elements.
1) There exists an injective f : X →Y iff n .
2) There exists a surjective f : X →Y iff n .
3) There exists a bijective f : X →Y iff n .
Pigeonhole Principle Suppose X is a finite set with m elements, Y is a finite
set with n elements, and f : X →Y is a function.
1) If m = n, then f is injective iff f is surjective iff f is bijective.
2) If m > n, then f is not injective.
3) If m < n, then f is not surjective.
If you are placing 6 pigeons in 6 holes, and you run out of pigeons before you fill
the holes, then you have placed 2 pigeons in one hole. In other words, in part 1) for
m = n = 6, if f is not surjective then f is not injective. Of course, the pigeonhole
principle does not hold for infinite sets, as can be seen by the following exercise.
Exercise Show there is a function f : Z
+
→ Z
+
which is injective but not
surjective. Also show there is one which is surjective but not injective.
Exercise Suppose f : [−2, 2] → R is defined by f(x) = x
2
. Find f
−1
(f([1, 2])).
Also find f(f
−1
([3, 5])).
Exercise Suppose f : X → Y is a function, S ⊂ X and T ⊂ Y . Find the
relationship between S and f
−1
(f(S)). Show that if f is injective, S = f
−1
(f(S)).
Also find the relationship between T and f(f
−1
(T)). Show that if f is surjective,
T = f(f
−1
(T)).
Strips We now define the vertical and horizontal strips of X Y .
If x
0
∈ X, ¦(x
0
, y) : y ∈ Y ¦ = (x
0
Y ) is called a vertical strip.
If y
0
∈ Y, ¦(x, y
0
) : x ∈ X¦ = (X y
0
) is called a horizontal strip.
Chapter 1 Background 9
Theorem Suppose S ⊂ X Y . The subset S is the graph of a function with
domain X and range Y iff each vertical strip intersects S in exactly one point.
This is just a restatement of the property of a graph of a function. The purpose
of the next theorem is to restate properties of functions in terms of horizontal strips.
Theorem Suppose f : X →Y has graph Γ. Then
1) Each horizontal strip intersects Γ in at least one point iff f is .
2) Each horizontal strip intersects Γ in at most one point iff f is .
3) Each horizontal strip intersects Γ in exactly one point iff f is .
Solutions of Equations Now we restate these properties in terms of solutions of
equations. Suppose f : X → Y and y
0
∈ Y . Consider the equation f(x) = y
0
. Here
y
0
is given and x is considered to be a "variable". A solution to this equation is any
x
0
∈ X with f(x
0
) = y
0
. Note that the set of all solutions to f(x) = y
0
is f
−1
(y
0
).
Also f(x) = y
0
has a solution iff y
0
∈ image(f) iff f
−1
(y
0
) is non-void.
Theorem Suppose f : X →Y .
1) The equation f(x) = y
0
has at least one solution for each y
0
∈ Y iff
f is .
2) The equation f(x) = y
0
has at most one solution for each y
0
∈ Y iff
f is .
3) The equation f(x) = y
0
has a unique solution for each y
0
∈ Y iff
f is .
Right and Left Inverses One way to understand functions is to study right and
left inverses, which are defined after the next theorem.
Theorem Suppose X
f
→Y
g
→W are functions.
1) If g ◦ f is injective, then f is injective.
10 Background Chapter 1
2) If g ◦ f is surjective, then g is surjective.
3) If g ◦ f is bijective, then f is injective and g is surjective.
Example X = W = ¦p¦, Y = ¦p, q¦, f(p) = p, and g(p) = g(q) = p. Here
g ◦ f is the identity, but f is not surjective and g is not injective.
Definition Suppose f : X → Y is a function. A left inverse of f is a function
g : Y → X such that g ◦ f = I
X
: X → X. A right inverse of f is a function
h : Y →X such that f ◦ h = I
Y
: Y →Y .
Theorem Suppose f : X →Y is a function.
1) f has a right inverse iff f is surjective. Any such right inverse must be
injective.
2) f has a left inverse iff f is injective. Any such left inverse must be
surjective.
Corollary Suppose each of X and Y is a non-void set. Then ∃ an injective
f : X → Y iff ∃ a surjective g : Y → X. Also a function from X to Y is bijective
iff it has a left inverse and a right inverse iff it has a left and right inverse.
Note The Axiom of Choice is not discussed in this book. However, if you worked
1) of the theorem above, you unknowingly used one version of it. For completeness,
we state this part of 1) again.
The Axiom of Choice If f : X → Y is surjective, then f has a right inverse
h. That is, for each y ∈ Y , it is possible to choose an x ∈ f
−1
(y) and thus to define
h(y) = x.
Note It is a classical theorem in set theory that the Axiom of Choice and the
Hausdorff Maximality Principle are equivalent. However in this text we do not go
that deeply into set theory. For our purposes it is assumed that the Axiom of Choice
and the HMP are true.
Exercise Suppose f : X → Y is a function. Define a relation on X by a ∼ b if
f(a) = f(b). Show this is an equivalence relation. If y belongs to the image of f,
then f
−1
(y) is an equivalence class and every equivalence class is of this form. In the
next chapter where f is a group homomorphism, these equivalence classes will be
called cosets.
Chapter 1 Background 11
Projections If X
1
and X
2
are non-void sets, we define the projection maps
π
1
: X
1
X
2
→X
1
and π
2
: X
1
X
2
→X
2
by π
i
(x
1
, x
2
) = x
i
.
Theorem If Y, X
1
, and X
2
are non-void sets, there is a 1-1 correspondence
between ¦functions f: Y →X
1
X
2
¦ and ¦ordered pairs of functions (f
1
, f
2
) where
f
1
: Y →X
1
and f
2
: Y →X
2
¦.
Proof Given f, define f
1
= π
1
◦ f and f
2
= π
2
◦ f. Given f
1
and f
2
define
f : Y → X
1
X
2
by f(y) = (f
1
(y), f
2
(y)). Thus a function from Y to X
1
X
2
is
merely a pair of functions from Y to X
1
and Y to X
2
. This concept is displayed in
the diagram below. It is summarized by the equation f = (f
1
, f
2
).
X
1
X
2
X
1
X
2
Y
·
,
»
f
1
f
2
f
π
1
π
2
One nice thing about this concept is that it works fine for infinite Cartesian
products.
Definition Suppose T is an index set and for each t ∈ T, X
t
is a non-void set.
Then the product
¸
t∈T
X
t
=
¸
X
t
is the collection of all sequences ¦x
t
¦
t∈T
= ¦x
t
¦
where x
t
∈ X
t
. Formally these sequences are functions α from T to
¸
X
t
with each
α(t) in X
t
and written as α(t) = x
t
. If T = ¦1, 2, . . . , n¦ then ¦x
t
¦ is the ordered
n-tuple (x
1
, x
2
, . . . , x
n
). If T = Z
+
then ¦x
t
¦ is the sequence (x
1
, x
2
, . . .). For any T
and any s in T, the projection map π
s
:
¸
X
t
→X
s
is defined by π
s
(¦x
t
¦) = x
s
.
Theorem If Y is any non-void set, there is a 1-1 correspondence between
¦functions f : Y →
¸
X
t
¦ and ¦sequences of functions ¦f
t
¦
t∈T
where f
t
: Y → X
t
¦.
Given f, the sequence ¦f
t
¦ is defined by f
t
= π
t
◦ f. Given ¦f
t
¦, f is defined by
f(y) = ¦f
t
(y)¦.
12 Background Chapter 1
A Calculus Exercise Let A be the collection of all functions f : [0, 1] → R
which have an infinite number of derivatives. Let A
0
⊂ A be the subcollection of
those functions f with f(0) = 0. Define D : A
0
→A by D(f) = df/dx. Use the mean
value theorem to show that D is injective. Use the fundamental theorem of calculus
to show that D is surjective.
Exercise This exercise is not used elsewhere in this text and may be omitted. It
is included here for students who wish to do a little more set theory. Suppose T is a
non-void set.
1) If Y is a non-void set, define Y
T
to be the collection of all functions with domain
T and range Y . Show that if T and Y are finite sets with m and n elements, then
Y
T
has n
m
elements. In particular, when T = ¦1, 2, 3¦, Y
T
= Y Y Y has
n
3
elements. Show that if n ≥ 3, the subset of Y
{1,2,3}
of all injective functions has
n(n − 1)(n − 2) elements. These injective functions are called permutations on Y
taken 3 at a time. If T = N, then Y
T
is the infinite product Y Y . That is,
Y
N
is the set of all infinite sequences (y
1
, y
2
, . . .) where each y
i
∈ Y . For any Y and
T, let Y
t
be a copy of Y for each t ∈ T. Then Y
T
=
¸
t∈T
Y
t
.
2) Suppose each of Y
1
and Y
2
is a non-void set. Show there is a natural bijection
from (Y
1
Y
2
)
T
to Y
T
1
Y
T
2
. (This is the fundamental property of Cartesian products
presented in the two previous theorems.)
3) Define {(T), the power set of T, to be the collection of all subsets of T (including
the null set). Show that if T is a finite set with m elements, {(T) has 2
m
elements.
4) If S is any subset of T, define its characteristic function χ
S
: T → ¦0, 1¦ by
letting χ
S
(t) be 1 when t ∈ S, and be 0 when t ∈[ S. Define α : {(T) → ¦0, 1¦
T
by
α(S) = χ
S
. Define β : ¦0, 1¦
T
→ {(T) by β(f) = f
−1
(1). Show that if S ⊂ T then
β ◦ α(S) = S, and if f : T → ¦0, 1¦ then α ◦ β(f) = f. Thus α is a bijection and
β = α
−1
.
{(T) ←→¦0, 1¦
T
5) Suppose γ : T →¦0, 1¦
T
is a function and show that it cannot be surjective. If
t ∈ T, denote γ(t) by γ(t) = f
t
: T → ¦0, 1¦. Define f : T → ¦0, 1¦ by f(t) = 0 if
f
t
(t) = 1, and f(t) = 1 if f
t
(t) = 0. Show that f is not in the image of γ and thus
γ cannot be surjective. This shows that if T is an infinite set, then the set ¦0, 1¦
T
represents a "higher order of infinity than T".
6) An infinite set Y is said to be countable if there is a bijection from the positive
Chapter 1 Background 13
integers N to Y. Show Q is countable but the following three collections are not.
i) {(N), the collection of all subsets of N.
ii) ¦0, 1¦
N
, the collection of all functions f : N →¦0, 1¦.
iii) The collection of all sequences (y
1
, y
2
, . . .) where each y
i
is 0 or 1.
We know that ii) and iii) are equal and there is a natural bijection between i)
and ii). We also know there is no surjective map from N to ¦0, 1¦
N
, i.e., ¦0, 1¦
N
is
uncountable. Finally, show there is a bijection from ¦0, 1¦
N
to the real numbers R.
(This is not so easy. To start with, you have to decide what the real numbers are.)
Notation for the Logic of Mathematics
Each of the words "Lemma", "Theorem", and "Corollary" means "true state-
ment". Suppose A and B are statements. A theorem may be stated in any of the
following ways:
Theorem Hypothesis Statement A.
Conclusion Statement B.
Theorem Suppose A is true. Then B is true.
Theorem If A is true, then B is true.
Theorem A ⇒B (A implies B ).
There are two ways to prove the theorem — to suppose A is true and show B is
true, or to suppose B is false and show A is false. The expressions "A ⇔ B", "A is
equivalent to B", and "A is true iff B is true " have the same meaning (namely, that
A ⇒B and B ⇒A).
The important thing to remember is that thoughts and expressions flow through
the language. Mathematical symbols are shorthand for phrases and sentences in the
English language. For example, "x ∈ B " means "x is an element of the set B." If A
is the statement "x ∈ Z
+
" and B is the statement "x
2
∈ Z
+
", then "A ⇒ B"means
"If x is a positive integer, then x
2
is a positive integer".
Mathematical Induction is based upon the fact that if S ⊂ Z
+
is a non-void
subset, then S contains a smallest element.
14 Background Chapter 1
Theorem Suppose P(n) is a statement for each n = 1, 2, ... . Suppose P(1) is
true and for each n ≥ 1, P(n) ⇒P(n + 1). Then for each n ≥ 1, P(n) is true.
Proof If the theorem is false, then ∃ a smallest positive integer m such that
P(m) is false. Since P(m−1) is true, this is impossible.
Exercise Use induction to show that, for each n ≥ 1, 1+2+ +n = n(n+1)/2.
The Integers
In this section, lower case letters a, b, c, ... will represent integers, i.e., elements
of Z. Here we will establish the following three basic properties of the integers.
1) If G is a subgroup of Z, then ∃ n ≥ 0 such that G = nZ.
2) If a and b are integers, not both zero, and G is the collection of all linear
combinations of a and b, then G is a subgroup of Z, and its
positive generator is the greatest common divisor of a and b.
3) If n ≥ 2, then n factors uniquely as the product of primes.
All of this will follow from long division, which we now state formally.
Euclidean Algorithm Given a, b with b = 0, ∃! m and r with 0 ≤ r <[b[ and
a = bm + r. In other words, b divides a "m times with a remainder of r". For
example, if a = −17 and b = 5, then m = −4 and r = 3, −17 = 5(−4) + 3.
Definition If r = 0, we say that b divides a or a is a multiple of b. This fact is
written as b [ a. Note that b [ a ⇔ the rational number a/b is an integer ⇔ ∃! m
such that a = bm ⇔ a ∈ bZ.
Note Anything (except 0) divides 0. 0 does not divide anything.
± 1 divides anything . If n = 0, the set of integers which n divides
is nZ = ¦nm : m ∈ Z¦ = ¦..., −2n, −n, 0, n, 2n, ...¦. Also n divides
a and b with the same remainder iff n divides (a −b).
Definition A non-void subset G ⊂ Z is a subgroup provided (g ∈ G ⇒ −g ∈ G)
and (g
1
, g
2
∈ G ⇒(g
1
+g
2
) ∈ G). We say that G is closed under negation and closed
under addition.
Chapter 1 Background 15
Theorem If n ∈ Z then nZ is a subgroup. Thus if n = 0, the set of integers
which n divides is a subgroup of Z.
The next theorem states that every subgroup of Z is of this form.
Theorem Suppose G ⊂ Z is a subgroup. Then
1) 0 ∈ G.
2) If g
1
and g
2
∈ G, then (m
1
g
1
+m
2
g
2
) ∈ G for all integers m
1
, m
2
.
3) ∃! non-negative integer n such that G = nZ. In fact, if G = ¦0¦
and n is the smallest positive integer in G, then G = nZ.
Proof Since G is non-void, ∃ g ∈ G. Now (−g) ∈ G and thus 0 = g + (−g)
belongs to G, and so 1) is true. Part 2) is straightforward, so consider 3). If G = 0,
it must contain a positive element. Let n be the smallest positive integer in G. If
g ∈ G, g = nm +r where 0 ≤ r < n. Since r ∈ G, it must be 0, and g ∈ nZ.
Now suppose a, b ∈ Z and at least one of a and b is non-zero.
Theorem Let G be the set of all linear combinations of a and b, i.e., G =
¦ma +nb : m, n ∈ Z¦. Then
1) G contains a and b.
2) G is a subgroup. In fact, it is the smallest subgroup containing a and b.
It is called the subgroup generated by a and b.
3) Denote by (a, b) the smallest positive integer in G. By the previous
theorem, G = (a, b)Z, and thus (a, b) [ a and (a, b) [ b. Also note that
∃ m, n such that ma +nb = (a, b). The integer (a, b) is called
the greatest common divisor of a and b.
4) If n is an integer which divides a and b, then n also divides (a, b).
Proof of 4) Suppose n [ a and n [ b i.e., suppose a, b ∈ nZ. Since G is the
smallest subgroup containing a and b, nZ ⊃ (a, b)Z, and thus n [ (a, b).
Corollary The following are equivalent.
1) a and b have no common divisors, i.e., (n [ a and n [ b) ⇒n = ±1.
16 Background Chapter 1
2) (a, b) = 1, i.e., the subgroup generated by a and b is all of Z.
3) ∃ m, n ∈Z with ma +nb = 1.
Definition If any one of these three conditions is satisfied, we say that a and b
are relatively prime.
This next theorem is the basis for unique factorization.
Theorem If a and b are relatively prime with a not zero, then a[bc ⇒a[c.
Proof Suppose a and b are relatively prime, c ∈ Z and a [ bc. Then there exist
m, n with ma + nb = 1, and thus mac + nbc = c. Now a [ mac and a [ nbc. Thus
a [ (mac +nbc) and so a [ c.
Definition A prime is an integer p > 1 which does not factor, i.e., if p = ab then
a = ±1 or a = ±p. The first few primes are 2, 3, 5, 7, 11, 13, 17,... .
Theorem Suppose p is a prime.
1) If a is an integer which is not a multiple of p, then (p, a) = 1. In other
words, if a is any integer, (p, a) = p or (p, a) = 1.
2) If p [ ab then p [ a or p [ b.
3) If p [ a
1
a
2
a
n
then p divides some a
i
. Thus if each a
i
is a prime,
then p is equal to some a
i
.
Proof Part 1) follows immediately from the definition of prime. Now suppose
p [ ab. If p does not divide a, then by 1), (p, a) = 1 and by the previous theorem, p
must divide b. Thus 2) is true. Part 3) follows from 2) and induction on n.
The Unique Factorization Theorem Suppose a is an integer which is not 0,1,
or -1. Then a may be factored into the product of primes and, except for order, this
factorization is unique. That is, ∃ a unique collection of distinct primes p
1
, p
2
, ..., p
k
and positive integers s
1
, s
2
, ..., s
k
such that a = ±p
s
1
1
p
s
2
2
p
s
k
k
.
Proof Factorization into primes is obvious, and uniqueness follows from 3) in the
theorem above. The power of this theorem is uniqueness, not existence.
Chapter 1 Background 17
Now that we have unique factorization and part 3) above, the picture becomes
transparent. Here are some of the basic properties of the integers in this light.
Theorem (Summary)
1) Suppose [ a[> 1 has prime factorization a = ±p
s
1
1
p
s
k
k
. Then the only
divisors of a are of the form ±p
t
1
1
p
t
k
k
where 0 ≤ t
i
≤ s
i
for i = 1, ..., k.
2) If [ a [> 1 and [ b [> 1, then (a, b) = 1 iff there is no common prime in
their factorizations. Thus if there is no common prime in their
factorizations, ∃ m, n with ma +nb = 1, and also (a
2
, b
2
) = 1.
3) Suppose [ a[> 1 and [ b[> 1. Let ¦p
1
, . . . , p
k
¦ be the union of the distinct
primes of their factorizations. Thus a = ±p
s
1
1
p
s
k
k
where 0 ≤ s
i
and
b = ±p
t
1
1
p
t
k
k
where 0 ≤ t
i
. Let u
i
be the minimum of s
i
and t
i
. Then
(a, b) = p
u
1
1
p
u
k
k
. For example (2
3
5 11, 2
2
5
4
7) = 2
2
5.
3
) Let v
i
be the maximum of s
i
and t
i
. Then c = p
v
1
1
p
v
k
k
is the least
(positive) common multiple of a and b. Note that c is a multiple of
a and b, and if n is a multiple of a and b, then n is a multiple of c.
Finally, if a and b are positive, their least common multiple is
c = ab/(a, b), and if in addition a and b are relatively prime,
then their least common multiple is just their product.
4) There is an infinite number of primes. (Proof: Suppose there were only
a finite number of primes p
1
, p
2
, ..., p
k
. Then no prime would divide
(p
1
p
2
p
k
+ 1).)
5) Suppose c is an integer greater than 1. Then
√
c is rational iff
√
c is an
integer. In particular,
√
2 and
√
3 are irrational. (Proof: If
√
c is
rational, ∃ positive integers a and b with
√
c = a/b and (a, b) = 1.
If b > 1, then it is divisible by some prime, and since cb
2
= a
2
, this
prime will also appear in the prime factorization of a. This is a
contradiction and thus b = 1 and
√
c is an integer.) (See the fifth
exercise below.)
Exercise Find (180,28), i.e., find the greatest common divisor of 180 and 28,
i.e., find the positive generator of the subgroup generated by ¦180,28¦. Find integers
m and n such that 180m + 28n = (180, 28). Find the least common multiple of 180
and 28, and show that it is equal to (180 28)/(180, 28).
18 Background Chapter 1
Exercise We have defined the greatest common divisor (gcd) and the least com-
mon multiple (lcm) of a pair of integers. Now suppose n ≥ 2 and S = ¦a
1
, a
2
, .., a
n
¦
is a finite collection of integers with [a
i
[ > 1 for 1 ≤ i ≤ n. Define the gcd and the
lcm of the elements of S and develop their properties. Express the gcd and the lcm
in terms of the prime factorizations of the a
i
. When is the lcm of S equal to the
product a
1
a
2
a
n
? Show that the set of all linear combinations of the elements of
S is a subgroup of Z, and its positive generator is the gcd of the elements of S.
Exercise Show that the gcd of S = ¦90, 70, 42¦ is 2, and find integers n
1
, n
2
, n
3
such that 90n
1
+ 70n
2
+ 42n
3
= 2. Also find the lcm of the elements of S.
Exercise Show that if each of G
1
, G
2
, ..., G
m
is a subgroup of Z, then
G
1
∩ G
2
∩ ∩ G
m
is also a subgroup of Z. Now let G = (90Z) ∩ (70Z) ∩ (42Z)
and find the positive integer n with G = nZ.
Exercise Show that if the nth root of an integer is a rational number, then it
itself is an integer. That is, suppose c and n are integers greater than 1. There is a
unique positive real number x with x
n
= c. Show that if x is rational, then it is an
integer. Thus if p is a prime, its nth root is an irrational number.
Exercise Show that a positive integer is divisible by 3 iff the sum of its digits is
divisible by 3. More generally, let a = a
n
a
n−1
. . . a
0
= a
n
10
n
+ a
n−1
10
n−1
+ + a
0
where 0 ≤ a
i
≤ 9. Now let b = a
n
+a
n−1
+ +a
0
, and show that 3 divides a and b
with the same remainder. Although this is a straightforward exercise in long division,
it will be more transparent later on. In the language of the next chapter, it says that
[a] = [b] in Z
3
.
Card Trick Ask friends to pick out seven cards from a deck and then to select one
to look at without showing it to you. Take the six cards face down in your left hand
and the selected card in your right hand, and announce you will place the selected
card in with the other six, but they are not to know where. Put your hands behind
your back and place the selected card on top, and bring the seven cards in front in
your left hand. Ask your friends to give you a number between one and seven (not
allowing one). Suppose they say three. You move the top card to the bottom, then
the second card to the bottom, and then you turn over the third card, leaving it face
up on top. Then repeat the process, moving the top two cards to the bottom and
turning the third card face up on top. Continue until there is only one card face
down, and this will be the selected card. Magic? Stay tuned for Chapter 2, where it
is shown that any non-zero element of Z
7
has order 7.
Chapter 2
Groups
Groups are the central objects of algebra. In later chapters we will define rings and
modules and see that they are special cases of groups. Also ring homomorphisms and
module homomorphisms are special cases of group homomorphisms. Even though
the definition of group is simple, it leads to a rich and amazing theory. Everything
presented here is standard, except that the product of groups is given in the additive
notation. This is the notation used in later chapters for the products of rings and
modules. This chapter and the next two chapters are restricted to the most basic
topics. The approach is to do quickly the fundamentals of groups, rings, and matrices,
and to push forward to the chapter on linear algebra. This chapter is, by far and
above, the most difficult chapter in the book, because group operations may be written
as addition or multiplication, and also the concept of coset is confusing at first.
Definition Suppose G is a non-void set and φ : G G → G is a function. φ is
called a binary operation, and we will write φ(a, b) = a b or φ(a, b) = a+b. Consider
the following properties.
1) If a, b, c ∈ G then a (b c) = (a b) c. If a, b, c ∈ G then a + (b +c) = (a +b) +c.
2) ∃ e = e
G
∈ G such that if a ∈ G ∃ 0
¯
=0
¯
G
∈ G such that if a ∈ G
e a = a e = a. 0
¯
+a = a+0
¯
= a.
3) If a ∈ G, ∃b ∈ G with a b = b a = e If a ∈ G, ∃b ∈ G with a +b = b +a = 0
¯
(b is written as b = a
−1
). (b is written as b = −a).
4) If a, b ∈ G, then a b = b a. If a, b ∈ G, then a +b = b +a.
Definition If properties 1), 2), and 3) hold, (G, φ) is said to be a group. If we
write φ(a, b) = a b, we say it is a multiplicative group. If we write φ(a, b) = a + b,
19
20 Groups Chapter 2
we say it is an additive group. If in addition, property 4) holds, we say the group is
abelian or commutative.
Theorem Let (G, φ) be a multiplicative group.
(i) Suppose a, c, ¯ c ∈ G. Then a c = a ¯ c ⇒ c = ¯ c.
Also c a = ¯ c a ⇒ c = ¯ c.
In other words, if f : G →G is defined by f(c) = a c, then f is injective.
Also f is bijective with f
−1
given by f
−1
(c) = a
−1
c.
(ii) e is unique, i.e., if ¯ e ∈ G satisfies 2), then e = ¯ e. In fact,
if a, b ∈ G then (a b = a) ⇒(b = e) and (a b = b) ⇒ (a = e).
Recall that b is an identity in G provided it is a right and left
identity for any a in G. However, group structure is so rigid that if
∃ a ∈ G such that b is a right identity for a, then b = e.
Of course, this is just a special case of the cancellation law in (i).
(iii) Every right inverse is an inverse, i.e., if a b = e then b = a
−1
.
Also if b a = e then b = a
−1
. Thus inverses are unique.
(iv) If a ∈ G, then (a
−1
)
−1
= a.
(v) The multiplication a
1
a
2
a
3
= a
1
(a
2
a
3
) = (a
1
a
2
) a
3
is well-defined.
In general, a
1
a
2
a
n
is well defined.
(vi) If a, b ∈ G, (a b)
−1
= b
−1
a
−1
. Also (a
1
a
2
a
n
)
−1
=
a
−1
n
a
−1
n−1
a
−1
1
.
(vii) Suppose a ∈ G. Let a
0
= e and if n > 0, a
n
= a a (n times)
and a
−n
= a
−1
a
−1
(n times). If n
1
, n
2
, ..., n
t
∈ Z then
a
n
1
a
n
2
a
nt
= a
n
1
+···+nt
. Also (a
n
)
m
= a
nm
.
Finally, if G is abelian and a, b ∈ G, then (a b)
n
= a
n
b
n
.
Exercise. Write out the above theorem where G is an additive group. Note that
part (vii) states that G has a scalar multiplication over Z. This means that if a is in
G and n is an integer, there is defined an element an in G. This is so basic, that we
state it explicitly.
Theorem. Suppose G is an additive group. If a ∈ G, let a0 =0
¯
and if n > 0,
let an = (a + +a) where the sum is n times, and a(−n) = (−a) + (−a) +(−a),
Chapter 2 Groups 21
which we write as (−a − a − a). Then the following properties hold in general,
except the first requires that G be abelian.
(a +b)n = an +bn
a(n +m) = an +am
a(nm) = (an)m
a1 = a
Note that the plus sign is used ambiguously — sometimes for addition in G
and sometimes for addition in Z. In the language used in Chapter 5, this theorem
states that any additive abelian group is a Z-module. (See page 71.)
Exercise Suppose G is a non-void set with a binary operation φ(a, b) = a b which
satisfies 1), 2) and [ 3
) If a ∈ G, ∃b ∈ G with a b = e]. Show (G, φ) is a group,
i.e., show b a = e. In other words, the group axioms are stronger than necessary.
If every element has a right inverse, then every element has a two sided inverse.
Exercise Suppose G is the set of all functions from Z to Z with multiplication
defined by composition, i.e., f g = f ◦ g. Note that G satisfies 1) and 2) but not 3),
and thus G is not a group. Show that f has a right inverse in G iff f is surjective,
and f has a left inverse in G iff f is injective (see page 10). Also show that the set
of all bijections from Z to Z is a group under composition.
Examples G = R, G = Q, or G = Z with φ(a, b) = a +b is an additive
abelian group.
Examples G = R−0 or G = Q−0 with φ(a, b) = ab is a multiplicative
abelian group.
G = Z −0 with φ(a, b) = ab is not a group.
G = R
+
= ¦r ∈ R : r > 0¦ with φ(a, b) = ab is a multiplicative
abelian group.
Subgroups
Theorem Suppose G is a multiplicative group and H ⊂ G is a non-void subset
satisfying
1) if a, b ∈ H then a b ∈ H
and 2) if a ∈ H then a
−1
∈ H.
22 Groups Chapter 2
Then e ∈ H and H is a group under multiplication. H is called a subgroup of G.
Proof Since H is non-void, ∃a ∈ H. By 2), a
−1
∈ H and so by 1), e ∈ H. The
associative law is immediate and so H is a group.
Example G is a subgroup of G and e is a subgroup of G. These are called the
improper subgroups of G.
Example If G = Z under addition, and n ∈ Z, then H = nZ is a subgroup of
Z. By a theorem in the section on the integers in Chapter 1, every subgroup of Z
is of this form (see page 15). This is a key property of the integers.
Exercises Suppose G is a multiplicative group.
1) Let H be the center of G, i.e., H = ¦h ∈ G : g h = h g for all g ∈ G¦. Show
H is a subgroup of G.
2) Suppose H
1
and H
2
are subgroups of G. Show H
1
∩ H
2
is a subgroup of G.
3) Suppose H
1
and H
2
are subgroups of G, with neither H
1
nor H
2
contained in
the other. Show H
1
∪ H
2
is not a subgroup of G.
4) Suppose T is an index set and for each t ∈ T, H
t
is a subgroup of G.
Show
¸
t∈T
H
t
is a subgroup of G.
5) Furthermore, if ¦H
t
¦ is a monotonic collection, then
¸
t∈T
H
t
is a subgroup of G.
6) Suppose G= ¦all functions f : [0, 1] →R¦. Define an addition on G by
(f +g)(t) = f(t) +g(t) for all t ∈ [0, 1]. This makes G into an abelian group.
Let K be the subset of G composed of all differentiable functions. Let H
be the subset of G composed of all continuous functions. What theorems
in calculus show that H and K are subgroups of G? What theorem shows
that K is a subset (and thus subgroup) of H?
Order Suppose G is a multiplicative group. If G has an infinite number of
Chapter 2 Groups 23
elements, we say that o(G), the order of G, is infinite. If G has n elements, then
o(G) = n. Suppose a ∈ G and H = ¦a
i
: i ∈ Z¦. H is an abelian subgroup of G
called the subgroup generated by a. We define the order of the element a to be the
order of H, i.e., the order of the subgroup generated by a. Let f : Z → H be the
surjective function defined by f(m) = a
m
. Note that f(k + l) = f(k) f(l) where
the addition is in Z and the multiplication is in the group H. We come now to the
first real theorem in group theory. It says that the element a has finite order iff f
is not injective, and in this case, the order of a is the smallest positive integer n
with a
n
= e.
Theorem Suppose a is an element of a multiplicative group G, and
H = ¦a
i
: i ∈ Z¦. If ∃ distinct integers i and j with a
i
= a
j
, then a has some finite
order n. In this case H has n distinct elements, H = ¦a
0
, a
1
, . . . , a
n−1
¦, and a
m
= e
iff n[m. In particular, the order of a is the smallest positive integer n with a
n
= e,
and f
−1
(e) = nZ.
Proof Suppose j < i and a
i
= a
j
. Then a
i−j
= e and thus ∃ a smallest positive
integer n with a
n
= e. This implies that the elements of ¦a
0
, a
1
, ..., a
n−1
¦ are distinct,
and we must show they are all of H. If m ∈ Z, the Euclidean algorithm states that
∃ integers q and r with 0 ≤ r < n and m = nq + r. Thus a
m
= a
nq
a
r
= a
r
, and
so H = ¦a
0
, a
1
, ..., a
n−1
¦, and a
m
= e iff n[m. Later in this chapter we will see that
f is a homomorphism from an additive group to a multiplicative group and that,
in additive notation, H is isomorphic to Z or Z
n
.
Exercise Write out this theorem for G an additive group. To begin, suppose a is
an element of an additive group G, and H = ¦ai : i ∈ Z¦.
Exercise Show that if G is a finite group of even order, then G has an odd number
of elements of order 2. Note that e is the only element of order 1.
Definition A group G is cyclic if ∃ an element of G which generates G.
Theorem If G is cyclic and H is a subgroup of G, then H is cyclic.
Proof Suppose G = ¦a
i
: i ∈ Z¦ is a cyclic group and H is a subgroup
of G. If H = e, then H is cyclic, so suppose H = e. Now there is a small-
est positive integer m with a
m
∈ H. If t is an integer with a
t
∈ H, then by
the Euclidean algorithm, m divides t, and thus a
m
generates H. Note that in
the case G has finite order n, i.e., G = ¦a
0
, a
1
, . . . , a
n−1
¦, then a
n
= e ∈ H,
and thus the positive integer m divides n. In either case, we have a clear picture
of the subgroups of G. Also note that this theorem was proved on page 15 for the
additive group Z.
24 Groups Chapter 2
Cosets Suppose H is a subgroup of a group G. It will be shown below that H
partitions G into right cosets. It also partitions G into left cosets, and in general
these partitions are distinct.
Theorem If H is a subgroup of a multiplicative group G, then a ∼ b defined by
a ∼ b iff a b
−1
∈ H is an equivalence relation. If a ∈ G, cl(a) = ¦b ∈ G : a ∼ b¦ =
¦h a : h ∈ H¦ = Ha. Note that a b
−1
∈ H iff b a
−1
∈ H.
If H is a subgroup of an additive group G, then a ∼ b defined by a ∼ b iff
(a − b) ∈ H is an equivalence relation. If a ∈ G, cl(a) = ¦b ∈ G : a ∼ b¦ = ¦h + a :
h ∈ H¦ = H +a. Note that (a −b) ∈ H iff (b −a) ∈ H.
Definition These equivalence classes are called right cosets. If the relation is
defined by a ∼ b iff b
−1
a ∈ H, then the equivalence classes are cl(a) = aH and
they are called left cosets. H is a left and right coset. If G is abelian, there is no
distinction between right and left cosets. Note that b
−1
a ∈ H iff a
−1
b ∈ H.
In the theorem above, H is used to define an equivalence relation on G, and thus
a partition of G. We now do the same thing a different way. We define the right
cosets directly and show they form a partition of G. You might find this easier.
Theorem Suppose H is a subgroup of a multiplicative group G. If a ∈ G, define
the right coset containing a to be Ha = ¦h a : h ∈ H¦. Then the following hold.
1) Ha = H iff a ∈ H.
2) If b ∈ Ha, then Hb = Ha, i.e., if h ∈ H, then H(h a) = (Hh)a = Ha.
3) If Hc ∩ Ha = ∅, then Hc = Ha.
4) The right cosets form a partition of G, i.e., each a in G belongs to one and
only one right coset.
5) Elements a and b belong to the same right coset iff a b
−1
∈ H iff b a
−1
∈ H.
Proof There is no better way to develop facility with cosets than to prove this
theorem. Also write this theorem for G an additive group.
Theorem Suppose H is a subgroup of a multiplicative group G.
Chapter 2 Groups 25
1) Any two right cosets have the same number of elements. That is, if a, b ∈ G,
f : Ha →Hb defined by f(h a) = h b is a bijection. Also any two left cosets
have the same number of elements. Since H is a right and left coset, any
two cosets have the same number of elements.
2) G has the same number of right cosets as left cosets. The function F defined
by F(Ha) = a
−1
H is a bijection from the collection of right cosets to the left
cosets. The number of right (or left) cosets is called the index of H in G.
3) If G is finite, o(H) (index of H) = o(G) and so o(H) [ o(G). In other words,
o(G)/o(H) = the number of right cosets = the number of left cosets.
4) If G is finite, and a ∈ G, then o(a) [ o(G). (Proof: The order of a is the order
of the subgroup generated by a, and by 3) this divides the order of G.)
5) If G has prime order, then G is cyclic, and any element (except e) is a generator.
(Proof: Suppose o(G) = p and a ∈ G, a = e. Then o(a) [ p and thus o(a) = p.)
6) If o(G) = n and a ∈ G, then a
n
= e. (Proof: a
o(a)
= e and n = o(a) (o(G)/o(a)) .)
Exercises
i) Suppose G is a cyclic group of order 4, G = ¦e, a, a
2
, a
3
¦ with a
4
= e. Find the
order of each element of G. Find all the subgroups of G.
ii) Suppose G is the additive group Z and H = 3Z. Find the cosets of H.
iii) Think of a circle as the interval [0, 1] with end points identified. Suppose G = R
under addition and H = Z. Show that the collection of all the cosets of H
can be thought of as a circle.
iv) Let G = R
2
under addition, and H be the subgroup defined by
H = ¦(a, 2a) : a ∈ R¦. Find the cosets of H. (See the last exercise on p 5.)
Normal Subgroups
We would like to make a group out of the collection of cosets of a subgroup H. In
26 Groups Chapter 2
general, there is no natural way to do that. However, it is easy to do in case H is a
normal subgroup, which is described below.
Theorem If H is a subgroup of a group G, then the following are equivalent.
1) If a ∈ G, then aHa
−1
= H
2) If a ∈ G, then aHa
−1
⊂ H
3) If a ∈ G, then aH = Ha
4) Every right coset is a left coset, i.e., if a ∈ G, ∃ b ∈ G with Ha = bH.
Proof 1) ⇒ 2) is obvious. Suppose 2) is true and show 3). We have (aHa
−1
)a ⊂
Ha so aH ⊂ Ha. Also a(a
−1
Ha) ⊂ aH so Ha ⊂ aH. Thus aH = Ha.
3) ⇒ 4) is obvious. Suppose 4) is true and show 3). Ha = bH contains a, so
bH = aH because a coset is an equivalence class. Thus aH = Ha.
Finally, suppose 3) is true and show 1). Multiply aH = Ha on the right by a
−1
.
Definition If H satisfies any of the four conditions above, then H is said to be a
normal subgroup of G. (This concept goes back to Evariste Galois in 1831.)
Note For any group G, G and e are normal subgroups. If G is an abelian group,
then every subgroup of G is normal.
Exercise Show that if H is a subgroup of G with index 2, then H is normal.
Exercise Show the intersection of a collection of normal subgroups of G is a
normal subgroup of G. Show the union of a monotonic collection of normal subgroups
of G is a normal subgroup of G.
Exercise Let A ⊂ R
2
be the square with vertices (−1, 1), (1, 1), (1, −1), and
(−1, −1), and G be the collection of all "isometries" of A onto itself. These are
bijections of A onto itself which preserve distance and angles, i.e., which preserve dot
product. Show that with multiplication defined as composition, G is a multiplicative
group. Show that G has four rotations, two reflections about the axes, and two
reflections about the diagonals, for a total of eight elements. Show the collection of
rotations is a cyclic subgroup of order four which is a normal subgroup of G. Show
that the reflection about the x-axis together with the identity form a cyclic subgroup
of order two which is not a normal subgroup of G. Find the four right cosets of this
subgroup. Finally, find the four left cosets of this subgroup.
Chapter 2 Groups 27
Quotient Groups Suppose N is a normal subgroup of G, and C and D are
cosets. We wish to define a coset E which is the product of C and D. If c ∈ C and
d ∈ D, define E to be the coset containing c d, i.e., E = N(c d). The coset E does
not depend upon the choice of c and d. This is made precise in the next theorem,
which is quite easy.
Theorem Suppose G is a multiplicative group, N is a normal subgroup, and
G/N is the collection of all cosets. Then (Na) (Nb) = N(a b) is a well de-
fined multiplication (binary operation) on G/N, and with this multiplication, G/N
is a group. Its identity is N and (Na)
−1
= (Na
−1
). Furthermore, if G is finite,
o(G/N) = o(G)/o(N).
Proof Multiplication of elements in G/N is multiplication of subsets in G.
(Na) (Nb) = N(aN)b = N(Na)b = N(a b). Once multiplication is well defined,
the group axioms are immediate.
Exercise Write out the above theorem for G an additive group. In the additive
abelian group R/Z, determine those elements of finite order.
Example Suppose G = Z under +, n > 1, and N = nZ. Z
n
, the group of
integers mod n is defined by Z
n
= Z/nZ. If a is an integer, the coset a + nZ is
denoted by [a]. Note that [a] + [b] = [a + b], −[a] = [−a], and [a] = [a + nl] for any
integer l. Any additive abelian group has a scalar multiplication over Z, and in this
case it is just [a]m = [am]. Note that [a] = [r] where r is the remainder of a divided
by n, and thus the distinct elements of Z
n
are [0], [1], ..., [n − 1]. Also Z
n
is cyclic
because each of [1] and [−1] = [n −1] is a generator. We already know that if p is a
prime, any non-zero element of Z
p
is a generator, because Z
p
has p elements.
Theorem If n > 1 and a is any integer, then [a] is a generator of Z
n
iff (a, n) = 1.
Proof The element [a] is a generator iff the subgroup generated by [a] contains
[1] iff ∃ an integer k such that [a]k = [1] iff ∃ integers k and l such that ak +nl = 1.
Exercise Show that a positive integer is divisible by 3 iff the sum of its digits is
divisible by 3. Note that [10] = [1] in Z
3
. (See the fifth exercise on page 18.)
Homomorphisms
Homomorphisms are functions between groups that commute with the group op-
erations. It follows that they honor identities and inverses. In this section we list
28 Groups Chapter 2
the basic properties. Properties 11), 12), and 13) show the connections between coset
groups and homomorphisms, and should be considered as the cornerstones of abstract
algebra. As always, the student should rewrite the material in additive notation.
Definition If G and
¯
G are multiplicative groups, a function f : G →
¯
G is a
homomorphism if, for all a, b ∈ G, f(a b) = f(a) f(b). On the left side, the group
operation is in G, while on the right side it is in
¯
G. The kernel of f is defined by
ker(f) = f
−1
(¯ e) = ¦a ∈ G : f(a) = ¯ e¦. In other words, the kernel is the set of
solutions to the equation f(x) = ¯ e. (If
¯
G is an additive group, ker(f) = f
−1
(0
¯
).)
Examples The constant map f : G →
¯
G defined by f(a) = ¯ e is a homomorphism.
If H is a subgroup of G, the inclusion i : H → G is a homomorphism. The function
f : Z → Z defined by f(t) = 2t is a homomorphism of additive groups, while the
function defined by f(t) = t +2 is not a homomorphism. The function h : Z →R−0
defined by h(t) = 2
t
is a homomorphism from an additive group to a multiplicative
group.
We now catalog the basic properties of homomorphisms. These will be helpful
later on in the study of ring homomorphisms and module homomorphisms.
Theorem Suppose G and
¯
G are groups and f : G →
¯
G is a homomorphism.
1) f(e) = ¯ e.
2) f(a
−1
) = f(a)
−1
. The first inverse is in G, and the second is in
¯
G.
3) f is injective ⇔ ker(f) = e.
4) If H is a subgroup of G, f(H) is a subgroup of
¯
G. In particular, image(f) is
a subgroup of
¯
G.
5) If
¯
H is a subgroup of
¯
G, f
−1
(
¯
H) is a subgroup of G. Furthermore, if
¯
H is
normal in
¯
G, then f
−1
(
¯
H) is normal in G.
6) The kernel of f is a normal subgroup of G.
7) If ¯ g ∈
¯
G, f
−1
(¯ g) is void or is a coset of ker(f), i.e., if f(g) = ¯ g then
f
−1
(¯ g) = Ng where N= ker(f). In other words, if the equation f(x) = ¯ g has a
Chapter 2 Groups 29
solution, then the set of all solutions is a coset of N= ker(f). This is a key fact
which is used routinely in topics such as systems of equations and linear
differential equations.
8) The composition of homomorphisms is a homomorphism, i.e., if h :
¯
G →
=
G
is
a homomorphism, then h ◦ f : G →
=
G
is a homomorphism.
9) If f : G →
¯
G is a bijection, then the function f
−1
:
¯
G →G is a homomorphism.
In this case, f is called an isomorphism, and we write G ≈
¯
G. In the case
G =
¯
G, f is also called an automorphism.
10) Isomorphisms preserve all algebraic properties. For example, if f is an
isomorphism and H ⊂ G is a subset, then H is a subgroup of G
iff f(H) is a subgroup of
¯
G, H is normal in G iff f(H) is normal in
¯
G, G is
cyclic iff
¯
G is cyclic, etc. Of course, this is somewhat of a cop-out, because an
algebraic property is one that, by definition, is preserved under isomorphisms.
11) Suppose H is a normal subgroup of G. Then π : G →G/H defined by
π(a) = Ha is a surjective homomorphism with kernel H. Furthermore, if
f : G →
¯
G is a surjective homomorphism with kernel H, then G/H ≈
¯
G
(see below).
12) Suppose H is a normal subgroup of G. If H ⊂ ker(f), then
¯
f : G/H →
¯
G
defined by
¯
f(Ha) = f(a) is a well-defined homomorphism making
the following diagram commute.
G
¯
G
G/H
f
·
π
¯
f
Thus defining a homomorphism on a quotient group is the same as defining a
homomorphism on the numerator which sends the denominator to ¯ e. The
image of
¯
f is the image of f and the kernel of
¯
f is ker(f)/H. Thus if H = ker(f),
¯
f is injective, and thus G/H ≈ image(f).
13) Given any group homomorphism f, domain(f)/ker(f) ≈ image(f). This is
the fundamental connection between quotient groups and homomorphisms.
30 Groups Chapter 2
14) Suppose K is a group. Then K is an infinite cycle group iff K is isomorphic to
the integers under addition, i.e., K ≈ Z. K is a cyclic group of order n iff
K ≈ Z
n
.
Proof of 14) Suppose
¯
G = K is generated by some element a. Then f : Z →K
defined by f(m) = a
m
is a homomorphism from an additive group to a multiplicative
group. If o(a) is infinite, f is an isomorphism. If o(a) = n, ker(f) = nZ and
¯
f : Z
n
→K is an isomorphism.
Exercise If a is an element of a group G, there is always a homomorphism from Z
to G which sends 1 to a. When is there a homomorphism from Z
n
to G which sends [1]
to a? What are the homomorphisms from Z
2
to Z
6
? What are the homomorphisms
from Z
4
to Z
8
?
Exercise Suppose G is a group and g is an element of G, g = e.
1) Under what conditions on g is there a homomorphism f : Z
7
→G with
f([1]) = g ?
2) Under what conditions on g is there a homomorphism f : Z
15
→G with
f([1]) = g ?
3) Under what conditions on G is there an injective homomorphism f : Z
15
→G ?
4) Under what conditions on G is there a surjective homomorphism f : Z
15
→G ?
Exercise We know every finite group of prime order is cyclic and thus abelian.
Show that every group of order four is abelian.
Exercise Let G = ¦h : [0, 1] → R : h has an infinite number of derivatives¦.
Then G is a group under addition. Define f : G → G by f(h) =
dh
dt
= h
. Show f
is a homomorphism and find its kernel and image. Let g : [0, 1] → R be defined by
g(t) = t
3
−3t + 4. Find f
−1
(g) and show it is a coset of ker(f).
Exercise Let G be as above and g ∈ G. Define f : G →G by f(h) = h
+ 5h
+
6t
2
h. Then f is a group homomorphism and the differential equation h
+5h
+6t
2
h =
g has a solution iff g lies in the image of f. Now suppose this equation has a solution
and S ⊂ G is the set of all solutions. For which subgroup H of G is S an H-coset?
Chapter 2 Groups 31
Exercise Suppose G is a multiplicative group and a ∈ G. Define f : G → G to
be conjugation by a, i.e., f(g) = a
−1
g a. Show that f is a homomorphism. Also
show f is an automorphism and find its inverse.
Permutations
Suppose X is a (non-void) set. A bijection f : X → X is called a permutation
on X, and the collection of all these permutations is denoted by S = S(X). In this
setting, variables are written on the left, i.e., f = (x)f. Therefore the composition
f ◦g means "f followed by g". S(X) forms a multiplicative group under composition.
Exercise Show that if there is a bijection between X and Y , there is an iso-
morphism between S(X) and S(Y ). Thus if each of X and Y has n elements,
S(X) ≈ S(Y ), and these groups are called the symmetric groups on n elements.
They are all denoted by the one symbol S
n
.
Exercise Show that o(S
n
) = n!. Let X = ¦1, 2, ..., n¦, S
n
= S(X), and H =
¦f ∈ S
n
: (n)f = n¦. Show H is a subgroup of S
n
which is isomorphic to S
n−1
. Let
g be any permutation on X with (n)g = 1. Find g
−1
Hg.
The next theorem shows that the symmetric groups are incredibly rich and com-
plex.
Theorem (Cayley's Theorem) Suppose G is a multiplicative group with n
elements and S
n
is the group of all permutations on the set G. Then G is isomorphic
to a subgroup of S
n
.
Proof Let h : G →S
n
be the function which sends a to the bijection h
a
: G →G
defined by (g)h
a
= g a. The proof follows from the following observations.
1) For each given a, h
a
is a bijection from G to G.
2) h is a homomorphism, i.e., h
a·b
= h
a
◦ h
b
.
3) h is injective and thus G is isomorphic to image(h) ⊂ S
n
.
The Symmetric Groups Now let n ≥ 2 and let S
n
be the group of all permu-
tations on ¦1, 2, ..., n¦. The following definition shows that each element of S
n
may
32 Groups Chapter 2
be represented by a matrix.
Definition Suppose 1 < k ≤ n, ¦a
1
, a
2
, ..., a
k
¦ is a collection of distinct inte-
gers with 1 ≤ a
i
≤ n, and ¦b
1
, b
2
, ..., b
k
¦ is the same collection in some different order.
Then the matrix
a
1
a
2
... a
k
b
1
b
2
... b
k
represents f ∈ S
n
defined by (a
i
)f = b
i
for 1 ≤ i ≤ k,
and (a)f = a for all other a. The composition of two permutations is computed by
applying the matrix on the left first and the matrix on the right second.
There is a special type of permutation called a cycle. For these we have a special
notation.
Definition
a
1
a
2
...a
k−1
a
k
a
2
a
3
...a
k
a
1
is called a k-cycle, and is denoted by (a
1
, a
2
, ..., a
k
).
A 2-cycle is called a transposition. The cycles (a
1
, ..., a
k
) and (c
1
, ..., c
) are disjoint
provided a
i
= c
j
for all 1 ≤ i ≤ k and 1 ≤ j ≤ .
Listed here are eight basic properties of permutations. They are all easy except
4), which takes a little work. Properties 9) and 10) are listed solely for reference.
Theorem
1) Disjoint cycles commute. (This is obvious.)
2) Every nonidentity permutation can be written uniquely (except for order) as
the product of disjoint cycles. (This is easy.)
3) Every permutation can be written (non-uniquely) as the product of transposi-
tions. (Proof: I = (1, 2)(1, 2) and (a
1
, ..., a
k
) = (a
1
, a
2
)(a
1
, a
3
) (a
1
, a
k
). )
4) The parity of the number of these transpositions is unique. This means that if
f is the product of p transpositions and also of q transpositions, then p is
even iff q is even. In this case, f is said to be an even permutation. In the other
case, f is an odd permutation.
5) A k-cycle is even (odd) iff k is odd (even). For example (1, 2, 3) = (1, 2)(1, 3) is
an even permutation.
6) Suppose f, g ∈ S
n
. If one of f and g is even and the other is odd, then g ◦ f is
Chapter 2 Groups 33
odd. If f and g are both even or both odd, then g ◦ f is even. (Obvious.)
7) The map h : S
n
→Z
2
defined by h(even)= [0] and h(odd)= [1] is a
homomorphism from a multiplicative group to an additive group. Its kernel (the
subgroup of even permutations) is denoted by A
n
and is called the alternating
group. Thus A
n
is a normal subgroup of index 2, and S
n
/A
n
≈ Z
2
.
8) If a, b, c and d are distinct integers in ¦1, 2, . . . , n¦, then (a, b)(b, c) = (a, c, b)
and (a, b)(c, d) = (a, c, d)(a, c, b). Since I = (1, 2, 3)
3
, it follows that for
n ≥ 3, every even permutation is the product of 3-cycles.
The following parts are not included in this course. They are presented here merely
for reference.
9) For any n = 4, A
n
is simple, i.e., has no proper normal subgroups.
10) S
n
can be generated by two elements. In fact, ¦(1, 2), (1, 2, ..., n)¦ generates S
n
.
(Of course there are subgroups of S
n
which cannot be generated by two
elements).
Proof of 4) It suffices to prove if the product of t transpositions is the identity I
on ¦1, 2, . . . , n¦, then t is even. Suppose this is false and I is written as t transposi-
tions, where t is the smallest odd integer this is possible. Since t is odd, it is at least 3.
Suppose for convenience the first transposition is (a, n). We will rewrite I as a prod-
uct of transpositions σ
1
σ
2
σ
t
where (n)σ
i
= (n) for 1 ≤ i < t and (n)σ
t
= n, which
will be a contradiction. This can be done by inductively "pushing n to the right"
using the equations below. If a, b, and c are distinct integers in ¦1, 2, . . . , n − 1¦,
then (a, n)(a, n) = I, (a, n)(b, n) = (a, b)(a, n), (a, n)(a, c) = (a, c)(c, n), and
(a, n)(b, c) = (b, c)(a, n). Note that (a, n)(a, n) cannot occur here because it would
result in a shorter odd product. (Now you may solve the tile puzzle on page viii.)
Exercise
1) Write
1 2 3 4 5 6 7
6 5 4 3 1 7 2
as the product of disjoint cycles.
Write (1,5,6,7)(2,3,4)(3,7,1) as the product of disjoint cycles.
Write (3,7,1)(1,5,6,7)(2,3,4) as the product of disjoint cycles.
Which of these permutations are odd and which are even?
34 Groups Chapter 2
2) Suppose (a
1
, . . . , a
k
) and (c
1
, . . . , c
) are disjoint cycles. What is the order of
their product?
3) Suppose σ ∈ S
n
. Show that σ
−1
(1, 2, 3)σ = ((1)σ, (2)σ, (3)σ). This shows
that conjugation by σ is just a type of relabeling. Also let τ = (4, 5, 6) and
find τ
−1
(1, 2, 3, 4, 5)τ.
4) Show that H = ¦σ ∈ S
6
: (6)σ = 6¦ is a subgroup of S
6
and find its right
cosets and its left cosets.
5) Let A ⊂ R
2
be the square with vertices (−1, 1), (1, 1), (1, −1), and (−1, −1),
and G be the collection of all isometries of A onto itself. We know from a
previous exercise that G is a group with eight elements. It follows from Cayley's
theorem that G is isomorphic to a subgroup of S
8
. Show that G is isomorphic
to a subgroup of S
4
.
6) If G is a multiplicative group, define a new multiplication on the set G by
a ◦ b = b a. In other words, the new multiplication is the old multiplication
in the opposite order. This defines a new group denoted by G
op
, the opposite
group. Show that it has the same identity and the same inverses as G, and
that f : G →G
op
defined by f(a) = a
−1
is a group isomorphism. Now consider
the special case G = S
n
. The convention used in this section is that an element
of S
n
is a permutation on ¦1, 2, . . . , n¦ with the variable written on the left.
Show that an element of S
op
n
is a permutation on ¦1, 2, . . . , n¦ with the variable
written on the right. (Of course, either S
n
or S
op
n
may be called the symmetric
group, depending on personal preference or context.)
Product of Groups
The product of groups is usually presented for multiplicative groups. It is pre-
sented here for additive groups because this is the form that occurs in later chapters.
As an exercise, this section should be rewritten using multiplicative notation. The
two theorems below are transparent and easy, but quite useful. For simplicity we
first consider the product of two groups, although the case of infinite products is only
slightly more difficult. For background, read first the two theorems on page 11.
Theorem Suppose G
1
and G
2
are additive groups. Define an addition on G
1
G
2
by (a
1
, a
2
) +(b
1
, b
2
) = (a
1
+b
1
, a
2
+b
2
). This operation makes G
1
G
2
into a group.
Its "zero" is (0
¯
1
, 0
¯
2
) and −(a
1
, a
2
) = (−a
1
, −a
2
). The projections π
1
: G
1
G
2
→G
1
Chapter 2 Groups 35
and π
2
: G
1
G
2
→G
2
are group homomorphisms. Suppose G is an additive group.
We know there is a bijection from ¦functions f : G →G
1
G
2
¦ to ¦ordered pairs of
functions (f
1
, f
2
) where f
1
: G → G
1
and f
2
: G → G
2
¦. Under this bijection, f is a
group homomorphism iff each of f
1
and f
2
is a group homomorphism.
Proof It is transparent that the product of groups is a group, so let's prove
the last part. Suppose G, G
1
, and G
2
are groups and f = (f
1
, f
2
) is a function
from G to G
1
G
2
. Now f(a + b) = (f
1
(a + b), f
2
(a + b)) and f(a) + f(b) =
(f
1
(a), f
2
(a)) +(f
1
(b), f
2
(b)) = (f
1
(a) +f
1
(b), f
2
(a) +f
2
(b)). An examination of these
two equations shows that f is a group homomorphism iff each of f
1
and f
2
is a group
homomorphism.
Exercise Suppose G
1
and G
2
are groups. Show that G
1
G
2
and G
2
G
1
are
isomorphic.
Exercise If o(a
1
) = m and o(a
2
) = n, find the order of (a
1
, a
2
) in G
1
G
2
.
Exercise Show that if G is any group of order 4, G is isomorphic to Z
4
or Z
2
Z
2
.
Show Z
4
is not isomorphic to Z
2
Z
2
. Show Z
12
is isomorphic to Z
4
Z
3
. Finally,
show that Z
mn
is isomorphic to Z
m
Z
n
iff (m, n) = 1.
Exercise Suppose G
1
and G
2
are groups and i
1
: G
1
→ G
1
G
2
is defined by
i
1
(g
1
) = (g
1
, 0
¯
2
). Show i
1
is an injective group homomorphism and its image is a
normal subgroup of G
1
G
2
. Usually G
1
is identified with its image under i
1
, so G
1
may be considered to be a normal subgroup of G
1
G
2
. Let π
2
: G
1
G
2
→ G
2
be the projection map defined in the Background chapter. Show π
2
is a surjective
homomorphism with kernel G
1
. Therefore (G
1
G
2
)/G
1
≈ G
2
as you would expect.
Exercise Let R be the reals under addition. Show that the addition in the
product RR is just the usual addition in analytic geometry.
Exercise Suppose n > 2. Is S
n
isomorphic to A
n
G where G is a multiplicative
group of order 2 ?
One nice thing about the product of groups is that it works fine for any finite
number, or even any infinite number. The next theorem is stated in full generality.
36 Groups Chapter 2
Theorem Suppose T is an index set, and for any t ∈ T, G
t
is an additive
group. Define an addition on
¸
t∈T
G
t
=
¸
G
t
by ¦a
t
¦ + ¦b
t
¦ = ¦a
t
+ b
t
¦. This op-
eration makes the product into a group. Its "zero" is ¦0
¯
t
¦ and −¦a
t
¦ = ¦−a
t
¦.
Each projection π
s
:
¸
G
t
→ G
s
is a group homomorphism. Suppose G is an ad-
ditive group. Under the natural bijection from ¦functions f : G →
¸
G
t
¦ to
¦sequences of functions ¦f
t
¦
t∈T
where f
t
: G → G
t
¦, f is a group homomorphism
iff each f
t
is a group homomorphism. Finally, the scalar multiplication on
¸
G
t
by integers is given coordinatewise, i.e., ¦a
t
¦n = ¦a
t
n¦.
Proof The addition on
¸
G
t
is coordinatewise.
Exercise Suppose s is an element of T and π
s
:
¸
G
t
→G
s
is the projection map
defined in the Background chapter. Show π
s
is a surjective homomorphism and find
its kernel.
Exercise Suppose s is an element of T and i
s
: G
s
→
¸
G
t
is defined by i
s
(a) =
¦a
t
¦ where a
t
= 0
¯
if t = s and a
s
= a. Show i
s
is an injective homomorphism
and its image is a normal subgroup of
¸
G
t
. Thus each G
s
may be considered to be
a normal subgroup of
¸
G
t
.
Exercise Let f : Z → Z
30
Z
100
be the homomorphism defined by f(m) =
([4m], [3m]). Find the kernel of f. Find the order of ([4], [3]) in Z
30
Z
100
.
Exercise Let f : Z → Z
90
Z
70
Z
42
be the group homomorphism defined by
f(m) = ([m], [m], [m]). Find the kernel of f and show that f is not surjective. Let
g : Z → Z
45
Z
35
Z
21
be defined by g(m) = ([m], [m], [m]). Find the kernel of
g and determine if g is surjective. Note that the gcd of ¦45, 35, 21¦ is 1. Now let
h : Z → Z
8
Z
9
Z
35
be defined by h(m) = ([m], [m], [m]). Find the kernel of h
and show that h is surjective. Finally suppose each of b, c, and d is greater than 1
and f : Z → Z
b
Z
c
Z
d
is defined by f(m) = ([m], [m], [m]). Find necessary and
sufficient conditions for f to be surjective (see the first exercise on page 18).
Exercise Suppose T is a non-void set, G is an additive group, and G
T
is the
collection of all functions f : T →G with addition defined by (f +g)(t) = f(t) +g(t).
Show G
T
is a group. For each t ∈ T, let G
t
= G. Note that G
T
is just another way
of writing
¸
t∈T
G
t
. Also note that if T = [0, 1] and G = R, the addition defined on
G
T
is just the usual addition of functions used in calculus. (For the ring and module
versions, see exercises on pages 44 and 69.)
Chapter 3
Rings
Rings are additive abelian groups with a second operation called multiplication. The
connection between the two operations is provided by the distributive law. Assuming
the results of Chapter 2, this chapter flows smoothly. This is because ideals are also
normal subgroups and ring homomorphisms are also group homomorphisms. We do
not show that the polynomial ring F[x] is a unique factorization domain, although
with the material at hand, it would be easy to do. Also there is no mention of prime
or maximal ideals, because these concepts are unnecessary for our development of
linear algebra. These concepts are developed in the Appendix. A section on Boolean
rings is included because of their importance in logic and computer science.
Suppose R is an additive abelian group, R = 0
¯
, and R has a second binary
operation (i.e., map from R R to R) which is denoted by multiplication. Consider
the following properties.
1) If a, b, c ∈ R, (a b) c = a (b c). (The associative property
of multiplication.)
2) If a, b, c ∈ R, a (b +c) = (a b) + (a c) and (b +c) a = (b a) + (c a).
(The distributive law, which connects addition and
multiplication.)
3) R has a multiplicative identity, i.e., there is an element
1
¯
= 1
¯
R
∈ R such that if a ∈ R, a 1
¯
= 1
¯
a = a.
4) If a, b ∈ R, a b = b a. (The commutative property for
multiplication.)
Definition If 1), 2), and 3) are satisfied, R is said to be a ring. If in addition 4)
is satisfied, R is said to be a commutative ring.
Examples The basic commutative rings in mathematics are the integers Z, the
37
38 Rings Chapter 3
rational numbers Q, the real numbers R, and the complex numbers C. It will be shown
later that Z
n
, the integers mod n, has a natural multiplication under which it is a
commutative ring. Also if R is any commutative ring, we will define R[x
1
, x
2
, . . . , x
n
],
a polynomical ring in n variables. Now suppose R is any ring, n ≥ 1, and R
n
is the
collection of all nn matrices over R. In the next chapter, operations of addition and
multiplication of matrices will be defined. Under these operations, R
n
is a ring. This
is a basic example of a non-commutative ring. If n > 1, R
n
is never commutative,
even if R is commutative.
The next two theorems show that ring multiplication behaves as you would wish
it to. They should be worked as exercises.
Theorem Suppose R is a ring and a, b ∈ R.
1) a 0
¯
= 0
¯
a = 0
¯
. Since R = 0
¯
, it follows that 1
¯
= 0
¯
.
2) (−a) b = a (−b) = −(a b).
Recall that, since R is an additive abelian group, it has a scalar multiplication
over Z (page 20). This scalar multiplication can be written on the right or left, i.e.,
na = an, and the next theorem shows it relates nicely to the ring multiplication.
Theorem Suppose a, b ∈ R and n, m ∈ Z.
1) (na) (mb) = (nm)(a b). (This follows from the distributive
law and the previous theorem.)
2) Let n
¯
= n1
¯
. For example, 2
¯
= 1
¯
+ 1
¯
. Then na = n
¯
a, that is, scalar
multiplication by n is the same as ring multiplication by n
¯
.
Of course, n
¯
may be 0
¯
even though n = 0.
Units
Definition An element a of a ring R is a unit provided ∃ an element a
−1
∈ R
with a a
−1
= a
−1
a = 1
¯
.
Theorem 0
¯
can never be a unit. 1
¯
is always a unit. If a is a unit, a
−1
is also a
unit with (a
−1
)
−1
= a. The product of units is a unit with (a b)
−1
= b
−1
a
−1
. More
Chapter 3 Rings 39
generally, if a
1
, a
2
, ..., a
n
are units, then their product is a unit with (a
1
a
2
a
n
)
−1
=
a
−1
n
a
−1
n−1
a
−1
1
. The set of all units of R forms a multiplicative group denoted by
R
∗
. Finally if a is a unit, (−a) is a unit and (−a)
−1
= −(a
−1
).
In order for a to be a unit, it must have a two-sided inverse. It suffices to require
a left inverse and a right inverse, as shown in the next theorem.
Theorem Suppose a ∈ R and ∃ elements b and c with b a = a c = 1
¯
. Then
b = c and so a is a unit with a
−1
= b = c.
Proof b = b 1
¯
= b (a c) = (b a) c = 1
¯
c = c.
Corollary Inverses are unique.
Domains and Fields In order to define these two types of rings, we first consider
the concept of zero divisor.
Definition Suppose R is a commutative ring. An element a ∈ R is called a zero
divisor provided it is non-zero and ∃ a non-zero element b with a b = 0
¯
. Note that
if a is a unit, it cannot be a zero divisor.
Theorem Suppose R is a commutative ring and a ∈ (R−0
¯
) is not a zero divisor.
Then (a b = a c) ⇒b = c. In other words, multiplication by a is an injective map
from R to R. It is surjective iff a is a unit.
Definition A domain (or integral domain) is a commutative ring such that, if
a = 0
¯
, a is not a zero divisor. A field is a commutative ring such that, if a = 0
¯
, a is
a unit. In other words, R is a field if it is commutative and its non-zero elements
form a group under multiplication.
Theorem A field is a domain. A finite domain is a field.
Proof A field is a domain because a unit cannot be a zero divisor. Suppose R is
a finite domain and a = 0
¯
. Then f : R → R defined by f(b) = a b is injective and,
by the pigeonhole principle, f is surjective. Thus a is a unit and so R is a field.
40 Rings Chapter 3
Exercise Let C be the additive abelian group R
2
. Define multiplication by
(a, b) (c, d) = (ac − bd, ad + bc). Show C is a commutative ring which is a field.
Note that 1
¯
= (1, 0) and if i = (0, 1), then i
2
= −1
¯
.
Examples Z is a domain. Q, R, and C are fields.
The Integers Mod n
The concept of integers mod n is fundamental in mathematics. It leads to a neat
little theory, as seen by the theorems below. However, the basic theory cannot be
completed until the product of rings is defined. (See the Chinese Remainder Theorem
on page 50.) We know from page 27 that Z
n
is an additive abelian group.
Theorem Suppose n > 1. Define a multiplication on Z
n
by [a] [b] = [ab]. This
is a well defined binary operation which makes Z
n
into a commutative ring.
Proof Since [a +kn] [b +l n] = [ab +n(al +bk +kl n)] = [ab], the multiplication
is well-defined. The ring axioms are easily verified.
Theorem Suppose n > 1 and a ∈ Z. Then the following are equivalent.
1) [a] is a generator of the additive group Z
n
.
2) (a, n) = 1.
3) [a] is a unit of the ring Z
n
.
Proof We already know from page 27 that 1) and 2) are equivalent. Recall that
if b is an integer, [a]b = [a] [b] = [ab]. Thus 1) and 3) are equivalent, because each
says ∃ an integer b with [a]b = [1].
Corollary If n > 1, the following are equivalent.
1) Z
n
is a domain.
2) Z
n
is a field.
3) n is a prime.
Proof We already know 1) and 2) are equivalent, because Z
n
is finite. Suppose
3) is true. Then by the previous theorem, each of [1], [2],...,[n − 1] is a unit, and
thus 2) is true. Now suppose 3) is false. Then n = ab where 1 < a < n, 1 < b < n,
Chapter 3 Rings 41
[a][b] = [0], and thus [a] is a zero divisor and 1) is false.
Exercise List the units and their inverses for Z
7
and Z
12
. Show that (Z
7
)
∗
is
a cyclic group but (Z
12
)
∗
is not. Show that in Z
12
the equation x
2
= 1
¯
has four
solutions. Finally show that if R is a domain, x
2
= 1
¯
can have at most two solutions
in R (see the first theorem on page 46).
Subrings Suppose S is a subset of a ring R. The statement that S is a subring
of R means that S is a subgroup of the group R, 1
¯
∈ S , and (a, b ∈ S ⇒a b ∈ S).
Then clearly S is a ring and has the same multiplicative identity as R. Note that Z
is a subring of Q, Q is a subring of R, and R is a subring of C. Subrings do not play
a role analogous to subgroups. That role is played by ideals, and an ideal is never a
subring (unless it is the entire ring). Note that if S is a subring of R and s ∈ S, then
s may be a unit in R but not in S. Note also that Z and Z
n
have no proper subrings,
and thus occupy a special place in ring theory, as well as in group theory.
Ideals and Quotient Rings
Ideals in ring theory play a role analagous to normal subgroups in group theory.
Definition A subset I of a ring R is a
left
right
2−sided
ideal provided it is a subgroup
of the additive group R and if a ∈ R and b ∈ I, then
a b ∈ I
b a ∈ I
a b and b a ∈ I
. The
word "ideal " means "2-sided ideal". Of course, if R is commutative, every right or
left ideal is an ideal.
Theorem Suppose R is a ring.
1) R and 0
¯
are ideals of R. These are called the improper ideals.
2) If ¦I
t
¦
t∈T
is a collection of right (left, 2-sided) ideals of R, then
¸
t∈T
I
t
is a
right (left, 2-sided) ideal of R. (See page 22.)
42 Rings Chapter 3
3) Furthermore, if the collection is monotonic, then
¸
t∈T
I
t
is a right (left, 2-sided)
ideal of R.
4) If a ∈ R, I = aR is a right ideal. Thus if R is commutative, aR is an ideal,
called a principal ideal. Thus every subgroup of Z is a principal ideal,
because it is of the form nZ.
5) If R is a commutative ring and I ⊂ R is an ideal, then the following are
equivalent.
i) I = R.
ii) I contains some unit u.
iii) I contains 1
¯
.
Exercise Suppose R is a commutative ring. Show that R is a field iff R contains
no proper ideals.
The following theorem is just an observation, but it is in some sense the beginning
of ring theory.
Theorem Suppose R is a ring and I ⊂ R is an ideal, I = R. Since I is a normal
subgroup of the additive group R, R/I is an additive abelian group. Multiplication
of cosets defined by (a +I) (b +I) = (ab +I) is well-defined and makes R/I a ring.
Proof (a + I) (b + I) = a b + aI + Ib + II ⊂ a b + I. Thus multiplication
is well defined, and the ring axioms are easily verified. The multiplicative identity is
(1
¯
+I).
Observation If R = Z, n > 1, and I = nZ, the ring structure on Z
n
= Z/nZ
is the same as the one previously defined.
Homomorphisms
Definition Suppose R and
¯
R are rings. A function f : R →
¯
R is a ring homo-
morphism provided
1) f is a group homomorphism
2) f(1
¯
R
) = 1
¯
¯
R
and
3) if a, b ∈ R then f(a b) = f(a) f(b). (On the left, multiplication
Chapter 3 Rings 43
is in R, while on the right multiplication is in
¯
R.)
The kernel of f is the kernel of f considered as a group homomorphism, namely
ker(f) = f
−1
(0
¯
).
Here is a list of the basic properties of ring homomorphisms. Much of this
work has already been done by the theorem in group theory on page 28.
Theorem Suppose each of R and
¯
R is a ring.
1) The identity map I
R
: R →R is a ring homomorphism.
2) The zero map from R to
¯
R is not a ring homomorphism
(because it does not send 1
¯
R
to 1
¯
¯
R
).
3) The composition of ring homomorphisms is a ring homomorphism.
4) If f : R →
¯
R is a bijection which is a ring homomorphism,
then f
−1
:
¯
R →R is a ring homomorphism. Such an f is called
a ring isomorphism. In the case R =
¯
R, f is also called a
ring automorphism.
5) The image of a ring homomorphism is a subring of the range.
6) The kernel of a ring homomorphism is an ideal of the domain.
In fact, if f : R →
¯
R is a homomorphism and I ⊂
¯
R is an ideal,
then f
−1
(I) is an ideal of R.
7) Suppose I is an ideal of R, I = R, and π : R →R/I is the
natural projection, π(a) = (a +I). Then π is a surjective ring
homomorphism with kernel I. Furthermore, if f : R →
¯
R is a surjective
ring homomorphism with kernel I, then R/I ≈
¯
R (see below).
8) From now on the word "homomorphism" means "ring homomorphism".
Suppose f : R →
¯
R is a homomorphism and I is an ideal of R, I = R.
If I ⊂ ker(f), then
¯
f : R/I →
¯
R defined by
¯
f(a +I) = f(a)
44 Rings Chapter 3
is a well-defined homomorphism making the following diagram commute.
R
¯
R
R/I
f
·
π
¯
f
Thus defining a homomorphism on a quotient ring is the same as
defining a homomorphism on the numerator which sends the
denominator to zero. The image of
¯
f is the image of f, and
the kernel of
¯
f is ker(f)/I. Thus if I = ker(f),
¯
f is
injective, and so R/I ≈ image (f).
Proof We know all this on the group level, and it is only necessary
to check that
¯
f is a ring homomorphism, which is obvious.
9) Given any ring homomorphism f, domain(f)/ker(f) ≈ image(f).
Exercise Find a ring R with a proper ideal I and an element b such that b is not
a unit in R but (b +I) is a unit in R/I.
Exercise Show that if u is a unit in a ring R, then conjugation by u is an
automorphism on R. That is, show that f : R →R defined by f(a) = u
−1
a u is
a ring homomorphism which is an isomorphism.
Exercise Suppose T is a non-void set, R is a ring, and R
T
is the collection of
all functions f : T → R. Define addition and multiplication on R
T
point-wise. This
means if f and g are functions from T to R, then (f + g)(t) = f(t) + g(t) and
(f g)(t) = f(t)g(t). Show that under these operations R
T
is a ring. Suppose S is a
non-void set and α : S →T is a function. If f : T →R is a function, define a function
α
∗
(f) : S →R by α
∗
(f) = f ◦ α. Show α
∗
: R
T
→R
S
is a ring homomorphism.
Exercise Now consider the case T = [0, 1] and R = R. Let A ⊂ R
[0,1]
be the
collection of all C
∞
functions, i.e., A =¦f : [0, 1] →R : f has an infinite number of
derivatives¦. Show A is a ring. Notice that much of the work has been done in the
previous exercise. It is only necessary to show that A is a subring of the ring R
[0,1]
.
Chapter 3 Rings 45
Polynomial Rings
In calculus, we consider real functions f which are polynomials, f(x) = a
0
+a
1
x +
+a
n
x
n
. The sum and product of polynomials are again polynomials, and it is easy
to see that the collection of polynomial functions forms a commutative ring. We can
do the same thing formally in a purely algebraic setting.
Definition Suppose R is a commutative ring and x is a "variable" or "symbol".
The polynomial ring R[x] is the collection of all polynomials f = a
0
+a
1
x + +a
n
x
n
where a
i
∈ R. Under the obvious addition and multiplication, R[x] is a commutative
ring. The degree of a non-zero polynomial f is the largest integer n such that a
n
= 0
¯
,
and is denoted by n = deg(f). If the top term a
n
= 1
¯
, then f is said to be monic.
To be more formal, think of a polynomial a
0
+ a
1
x + as an infinite sequence
(a
0
, a
1
, ...) such that each a
i
∈ R and only a finite number are non-zero. Then
(a
0
, a
1
, ...) + (b
0
, b
1
, ...) = (a
0
+b
0
, a
1
+b
1
, ...) and
(a
0
, a
1
, ...) (b
0
, b
1
, ...) = (a
0
b
0
, a
0
b
1
+a
1
b
0
, a
0
b
2
+a
1
b
1
+a
2
b
0
, ...).
Note that on the right, the ring multiplication a b is written simply as ab, as is
often done for convenience.
Theorem If R is a domain, R[x] is also a domain.
Proof Suppose f and g are non-zero polynomials. Then deg(f)+deg(g) = deg(fg)
and thus fg is not 0
¯
. Another way to prove this theorem is to look at the bottom
terms instead of the top terms. Let a
i
x
i
and b
j
x
j
be the first non-zero terms of f
and g. Then a
i
b
j
x
i+j
is the first non-zero term of fg.
Theorem (The Division Algorithm) Suppose R is a commutative ring, f ∈
R[x] has degree ≥ 1 and its top coefficient is a unit in R. (If R is a field, the
top coefficient of f will always be a unit.) Then for any g ∈ R[x], ∃! h, r ∈ R[x]
such that g = fh +r with r = 0
¯
or deg(r) < deg(f).
Proof This theorem states the existence and uniqueness of polynomials h and
r. We outline the proof of existence and leave uniqueness as an exercise. Suppose
f = a
0
+ a
1
x + +a
m
x
m
where m ≥ 1 and a
m
is a unit in R. For any g with
deg(g) < m, set h = 0
¯
and r = g. For the general case, the idea is to divide f into g
until the remainder has degree less than m. The proof is by induction on the degree
of g. Suppose n ≥ m and the result holds for any polynomial of degree less than
46 Rings Chapter 3
n. Suppose g is a polynomial of degree n. Now ∃ a monomial bx
t
with t = n − m
and deg(g − fbx
t
) < n. By induction, ∃ h
1
and r with fh
1
+ r = (g − fbx
t
) and
deg(r) < m. The result follows from the equation f(h
1
+bx
t
) +r = g.
Note If r = 0
¯
we say that f divides g. Note that f = x − c divides g iff c is
a root of g, i.e., g(c) = 0
¯
. More generally, x −c divides g with remainder g(c).
Theorem Suppose R is a domain, n > 0, and g(x) = a
0
+ a
1
x + + a
n
x
n
is a
polynomial of degree n with at least one root in R. Then g has at most n roots. Let
c
1
, c
2
, .., c
k
be the distinct roots of g in the ring R. Then ∃ a unique sequence of
positive integers n
1
, n
2
, .., n
k
and a unique polynomial h with no root in R so that
g(x) = (x − c
1
)
n
1
(x − c
k
)
n
k
h(x). (If h has degree 0, i.e., if h = a
n
, then we say
"all the roots of g belong to R". If g = a
n
x
n
, we say "all the roots of g are 0
¯
".)
Proof Uniqueness is easy so let's prove existence. The theorem is clearly true
for n = 1. Suppose n > 1 and the theorem is true for any polynomial of degree less
than n. Now suppose g is a polynomial of degree n and c
1
is a root of g. Then ∃
a polynomial h
1
with g(x) = (x − c
1
)h
1
. Since h
1
has degree less than n, the result
follows by induction.
Note If g is any non-constant polynomial in C[x], all the roots of g belong to C,
i.e., C is an algebraically closed field. This is called The Fundamental Theorem of
Algebra, and it is assumed without proof for this textbook.
Exercise Suppose g is a non-constant polynomial in R[x]. Show that if g has
odd degree then it has a real root. Also show that if g(x) = x
2
+ bx + c, then it has
a real root iff b
2
≥ 4c, and in that case both roots belong to R.
Definition A domain T is a principal ideal domain (PID) if, given any ideal I,
∃ t ∈ T such that I = tT. Note that Z is a PID and any field is PID.
Theorem Suppose F is a field, I is a proper ideal of F[x], and n is the smallest
positive integer such that I contains a polynomial of degree n. Then I contains a
unique polynomial of the form f = a
0
+ a
1
x + +a
n−1
x
n−1
+ x
n
and it has the
property that I = fF[x]. Thus F[x] is a PID. Furthermore, each coset of I can be
written uniquely in the form (c
0
+c
1
x + +c
n−1
x
n−1
+I).
Proof. This is a good exercise in the use of the division algorithm. Note this is
similar to showing that a subgroup of Z is generated by one element (see page 15).
Chapter 3 Rings 47
Theorem. Suppose R is a subring of a commutative ring C and c ∈ C. Then
∃! homomorphism h : R[x] → C with h(x) = c and h(r) = r for all r ∈ R. It is
defined by h(a
0
+ a
1
x + +a
n
x
n
) = a
0
+ a
1
c + +a
n
c
n
, i.e., h sends f(x) to f(c).
The image of h is the smallest subring of C containing R and c.
This map h is called an evaluation map. The theorem says that adding two
polynomials in R[x] and evaluating is the same as evaluating and then adding in C.
Also multiplying two polynomials in R[x] and evaluating is the same as evaluating
and then multiplying in C. In street language the theorem says you are free to send
x wherever you wish and extend to a ring homomorphism on R[x].
Exercise Let C = ¦a + bi : a, b ∈ R¦. Since R is a subring of C, there exists a
homomorphism h : R[x] → C which sends x to i, and this h is surjective. Show
ker(h) = (x
2
+ 1)R[x] and thus R[x]/(x
2
+ 1) ≈ C. This is a good way to look
at the complex numbers, i.e., to obtain C, adjoin x to R and set x
2
= −1.
Exercise Z
2
[x]/(x
2
+ x + 1) has 4 elements. Write out the multiplication table
for this ring and show that it is a field.
Exercise Show that, if R is a domain, the units of R[x] are just the units of R.
Thus if F is a field, the units of F[x] are the non-zero constants. Show that [1] +[2]x
is a unit in Z
4
[x].
In this chapter we do not prove F[x] is a unique factorization domain, nor do
we even define unique factorization domain. The next definition and theorem are
included merely for reference, and should not be studied at this stage.
Definition Suppose F is a field and f ∈ F[x] has degree ≥ 1. The statement
that g is an associate of f means ∃ a unit u ∈ F[x] such that g = uf. The statement
that f is irreducible means that if h is a non-constant polynomial which divides f,
then h is an associate of f.
We do not develop the theory of F[x] here. However, the development is easy
because it corresponds to the development of Z in Chapter 1. The Division Algo-
rithm corresponds to the Euclidean Algorithm. Irreducible polynomials correspond
to prime integers. The degree function corresponds to the absolute value function.
One difference is that the units of F[x] are non-zero constants, while the units of Z
48 Rings Chapter 3
are just ±1. Thus the associates of f are all cf with c = 0
¯
while the associates of an
integer n are just ±n. Here is the basic theorem. (This theory is developed in full in
the Appendix under the topic of Euclidean domains.)
Theorem Suppose F is a field and f ∈ F[x] has degree ≥ 1. Then f factors as the
product of irreducibles, and this factorization is unique up to order and associates.
Also the following are equivalent.
1) F[x]/(f) is a domain.
2) F[x]/(f) is a field.
3) f is irreducible.
Definition Now suppose x and y are "variables". If a ∈ R and n, m ≥ 0, then
ax
n
y
m
= ay
m
x
n
is called a monomial. Define an element of R[x, y] to be any finite
sum of monomials.
Theorem R[x, y] is a commutative ring and (R[x])[y] ≈ R[x, y] ≈ (R[y])[x]. In
other words, any polynomial in x and y with coefficients in R may be written as a
polynomial in y with coefficients in R[x], or as a polynomial in x with coefficients in
R[y].
Side Comment It is true that if F is a field, each f ∈ F[x, y] factors as the
product of irreducibles. However F[x, y] is not a PID. For example, the ideal
I = xF[x, y] +yF[x, y] = ¦f ∈ F[x, y] : f(0
¯
, 0
¯
) = 0
¯
¦ is not principal.
If R is a commutative ring and n ≥ 2, the concept of a polynomial ring in
n variables works fine without a hitch. If a ∈ R and v
1
, v
2
, ..., v
n
are non-negative
integers, then ax
v
1
1
x
v
2
2
x
vn
n
is called a monomial. Order does not matter here.
Define an element of R[x
1
, x
2
, ..., x
n
] to be any finite sum of monomials. This
gives a commutative ring and there is canonical isomorphism R[x
1
, x
2
, ..., x
n
] ≈
(R[x
1
, x
2
, ..., x
n−1
])[x
n
]. Using this and induction on n, it is easy to prove the fol-
lowing theorem.
Theorem If R is a domain, R[x
1
, x
2
, ..., x
n
] is a domain and its units are just the
units of R.
Chapter 3 Rings 49
Exercise Suppose R is a commutative ring and f : R[x, y] → R[x] is the eval-
uation map which sends y to 0
¯
. This means f(p(x, y)) = p(x, 0
¯
). Show f is a ring
homomorphism whose kernel is the ideal (y) = yR[x, y]. Use the fact that "the do-
main mod the kernel is isomorphic to the image" to show R[x, y]/(y) is isomorphic
to R[x]. That is, if you adjoin y to R[x] and then factor it out, you get R[x] back.
Product of Rings
The product of rings works fine, just as does the product of groups.
Theorem Suppose T is an index set and for each t ∈ T, R
t
is a ring. On the
additive abelian group
¸
t∈T
R
t
=
¸
R
t
, define multiplication by ¦r
t
¦ ¦s
t
¦ = ¦r
t
s
t
¦.
Then
¸
R
t
is a ring and each projection π
s
:
¸
R
t
→ R
s
is a ring homomorphism.
Suppose R is a ring. Under the natural bijection from ¦functions f : R →
¸
R
t
¦
to ¦sequences of functions ¦f
t
¦
t∈T
where f
t
: R → R
t
¦, f is a ring homomorphism
iff each f
t
is a ring homomorphism.
Proof We already know f is a group homomorphism iff each f
t
is a group homo-
morphism (see page 36). Note that ¦1
¯
t
¦ is the multiplicative identity of
¸
R
t
, and
f(1
¯
R
) = ¦1
¯
t
¦ iff f
t
(1
¯
R
) = 1
¯
t
for each t ∈ T. Finally, since multiplication is defined
coordinatewise, f is a ring homomorphism iff each f
t
is a ring homomorphism.
Exercise Suppose R and S are rings. Note that R 0 is not a subring of R S
because it does not contain (1
¯
R
, 1
¯
S
). Show R0
¯
is an ideal and (RS/R0
¯
) ≈ S.
Suppose I ⊂ R and J ⊂ S are ideals. Show I J is an ideal of RS and every
ideal of R S is of this form.
Exercise Suppose R and S are commutative rings. Show T = R S is not a
domain. Let e = (1, 0) ∈ RS and show e
2
= e, (1 −e)
2
= (1 −e), R0 = eT,
and 0 S = (1 −e)T.
Exercise If T is any ring, an element e of T is called an idempotent provided
e
2
= e. The elements 0 and 1 are idempotents called the trivial idempotents. Suppose
T is a commutative ring and e ∈ T is an idempotent with 0 = e = 1. Let R = eT
and S = (1 − e)T. Show each of the ideals R and S is a ring with identity, and
f : T →RS defined by f(t) = (et, (1−e)t) is a ring isomorphism. This shows that
a commutative ring T splits as the product of two rings iff it contains a non-trivial
idempotent.
50 Rings Chapter 3
The Chinese Remainder Theorem
The natural map from Z to Z
m
Z
n
is a group homomorphism and also a ring
homomorphism. If m and n are relatively prime, this map is surjective with kernel
mnZ, and thus Z
mn
and Z
m
Z
n
are isomorphic as groups and as rings. The next
theorem is a classical generalization of this. (See exercise three on page 35.)
Theorem Suppose n
1
, ..., n
t
are integers, each n
i
> 1, and (n
i
, n
j
) = 1 for all
i = j. Let f
i
: Z → Z
n
i
be defined by f
i
(a) = [a]. (Note that the bracket symbol is
used ambiguously.) Then the ring homomorphism f = (f
1
, .., f
t
) : Z →Z
n
1
Z
nt
is surjective. Furthermore, the kernel of f is nZ, where n = n
1
n
2
n
t
. Thus Z
n
and Z
n
1
Z
nt
are isomorphic as rings, and thus also as groups.
Proof We wish to show that the order of f(1) is n, and thus f(1) is a group
generator, and thus f is surjective. The element f(1)m = ([1], .., [1])m = ([m], .., [m])
is zero iff m is a multiple of each of n
1
, .., n
t
. Since their least common multiple is n,
the order of f(1) is n. (See the fourth exercise on page 36 for the case t = 3.)
Exercise Show that if a is an integer and p is a prime, then [a] = [a
p
] in Z
p
(Fermat's Little Theorem). Use this and the Chinese Remainder Theorem to show
that if b is a positive integer, it has the same last digit as b
5
.
Characteristic
The following theorem is just an observation, but it shows that in ring theory, the
ring of integers is a "cornerstone".
Theorem If R is a ring, there is one and only one ring homomorphism f : Z →R.
It is given by f(m) = m1
¯
= m
¯
. Thus the subgroup of R generated by 1
¯
is a subring
of R isomorphic to Z or isomorphic to Z
n
for some positive integer n.
Definition Suppose R is a ring and f : Z → R is the natural ring homomor-
phism f(m) = m1
¯
= m
¯
. The non-negative integer n with ker(f) = nZ is called the
characteristic of R. Thus f is injective iff R has characteristic 0 iff 1
¯
has infinite
order. If f is not injective, the characteristic of R is the order of 1
¯
.
It is an interesting fact that, if R is a domain, all the non-zero elements of R
have the same order. (See page 23 for the definition of order.)
Chapter 3 Rings 51
Theorem Suppose R is a domain. If R has characteristic 0, then each non-zero
a ∈ R has infinite order. If R has finite characteristic n, then n is a prime and each
non-zero a ∈ R has order n.
Proof Suppose R has characteristic 0, a is a non-zero element of R, and m is a
positive integer. Then ma = m
¯
a cannot be 0
¯
because m
¯
, a = 0
¯
and R is a domain.
Thus o(a) = ∞. Now suppose R has characteristic n. Then R contains Z
n
as a
subring, and thus Z
n
is a domain and n is a prime. If a is a non-zero element of R,
na = n
¯
a = 0
¯
a = 0
¯
and thus o(a)[n and thus o(a) = n.
Exercise Show that if F is a field of characteristic 0, F contains Q as a subring.
That is, show that the injective homomorphism f : Z → F extends to an injective
homomorphism
¯
f : Q →F.
Boolean Rings
This section is not used elsewhere in this book. However it fits easily here, and is
included for reference.
Definition A ring R is a Boolean ring if for each a ∈ R, a
2
= a, i.e., each
element of R is an idempotent.
Theorem Suppose R is a Boolean ring.
1) R has characteristic 2. If a ∈ R, 2a = a +a = 0
¯
, and so a = −a.
Proof (a +a) = (a +a)
2
= a
2
+ 2a
2
+a
2
= 4a. Thus 2a = 0
¯
.
2) R is commutative.
Proof (a +b) = (a +b)
2
= a
2
+ (a b) + (b a) +b
2
= a + (a b) −(b a) +b. Thus a b = b a.
3) If R is a domain, R ≈ Z
2
.
Proof Suppose a = 0
¯
. Then a (1
¯
−a) = 0
¯
and so a = 1
¯
.
4) The image of a Boolean ring is a Boolean ring. That is, if I is an ideal
of R with I = R, then every element of R/I is idempotent and thus
R/I is a Boolean ring. It follows from 3) that R/I is a domain iff R/I
is a field iff R/I ≈ Z
2
. (In the language of Chapter 6, I is a prime
ideal iff I is a maximal ideal iff R/I ≈ Z
2
).
52 Rings Chapter 3
Suppose X is a non-void set. If a is a subset of X, let a
= (X−a) be a complement
of a in X. Now suppose R is a non-void collection of subsets of X. Consider the
following properties which the collection R may possess.
1) a ∈ R ⇒ a
belongs to
R and so 3) is true. Since R is non-void, it contains some element a. Then ∅ = a ∩a
and X = a ∪ a
belong to R, and so 4) is true.
Theorem Suppose R is a Boolean algebra of sets. Define an addition on R by
a +b = (a ∪ b) −(a ∩ b). Under this addition, R is an abelian group with 0
¯
= ∅ and
a = −a. Define a multiplication on R by a b = a ∩ b. Under this multiplication R
becomes a Boolean ring with 1
¯
= X.
Exercise Let X = ¦1, 2, ..., n¦ and let R be the Boolean ring of all subsets of
X. Note that o(R) = 2
n
. Define f
i
: R → Z
2
by f
i
(a) = [1] iff i ∈ a. Show each
f
i
is a homomorphism and thus f = (f
1
, ..., f
n
) : R → Z
2
Z
2
Z
2
is a ring
homomorphism. Show f is an isomorphism. (See exercises 1) and 4) on page 12.)
Exercise Use the last exercise on page 49 to show that any finite Boolean ring is
isomorphic to Z
2
Z
2
Z
2
, and thus also to the Boolean ring of subsets above.
Note Suppose R is a Boolean ring. It is a classical theorem that ∃ a Boolean
algebra of sets whose Boolean ring is isomorphic to R. So let's just suppose R is
a Boolean algebra of sets which is a Boolean ring with addition and multiplication
defined as above. Now define a ∨ b = a ∪ b and a ∧ b = a ∩ b. These operations cup
and cap are associative, commutative, have identity elements, and each distributes
over the other. With these two operations (along with complement), R is called a
Boolean algebra. R is not a group under cup or cap. Anyway, it is a classical fact
that, if you have a Boolean ring (algebra), you have a Boolean algebra (ring). The
advantage of the algebra is that it is symmetric in cup and cap. The advantage of
the ring viewpoint is that you can draw from the rich theory of commutative rings.
Chapter 4
Matrices and Matrix Rings
We first consider matrices in full generality, i.e., over an arbitrary ring R. However,
after the first few pages, it will be assumed that R is commutative. The topics,
such as invertible matrices, transpose, elementary matrices, systems of equations,
and determinant, are all classical. The highlight of the chapter is the theorem that a
square matrix is a unit in the matrix ring iff its determinant is a unit in the ring.
This chapter concludes with the theorem that similar matrices have the same deter-
minant, trace, and characteristic polynomial. This will be used in the next chapter
to show that an endomorphism on a finitely generated vector space has a well-defined
determinant, trace, and characteristic polynomial.
Definition Suppose R is a ring and m and n are positive integers. Let R
m,n
be
the collection of all mn matrices
A = (a
i,j
) =
¸
¸
¸
a
1,1
. . . a
1,n
.
.
.
.
.
.
a
m,1
. . . a
m,n
¸
where each entry a
i,j
∈ R.
A matrix may be viewed as m n-dimensional row vectors or as n m-dimensional
column vectors. A matrix is said to be square if it has the same number of rows
as columns. Square matrices are so important that they have a special notation,
R
n
= R
n,n
. R
n
is defined to be the additive abelian group R R R.
To emphasize that R
n
does not have a ring structure, we use the "sum" notation,
R
n
= R⊕R⊕ ⊕R. Our convention is to write elements of R
n
as column vectors,
i.e., to identify R
n
with R
n,1
. If the elements of R
n
are written as row vectors, R
n
is
identified with R
1,n
.
53
54 Matrices Chapter 4
Addition of matrices To "add" two matrices, they must have the same number
of rows and the same number of columns, i.e., addition is a binary operation R
m,n
R
m,n
→R
m,n
. The addition is defined by (a
i,j
) +(b
i,j
) = (a
i,j
+b
i,j
), i.e., the i, j term
of the sum is the sum of the i, j terms. The following theorem is just an observation.
Theorem R
m,n
is an additive abelian group. Its "zero" is the matrix 0 = 0
m,n
all of whose terms are zero. Also −(a
i,j
) = (−a
i,j
). Furthermore, as additive groups,
R
m,n
≈ R
mn
.
Scalar multiplication An element of R is called a scalar. A matrix may be
"multiplied" on the right or left by a scalar. Right scalar multiplication is defined
by (a
i,j
)c = (a
i,j
c). It is a function R
m,n
R → R
m,n
. Note in particular that
scalar multiplication is defined on R
n
. Of course, if R is commutative, there is no
distinction between right and left scalar multiplication.
Theorem Suppose A, B ∈ R
m,n
and c, d ∈ R. Then
(A+B)c = Ac +Bc
A(c +d) = Ac +Ad
A(cd) = (Ac)d
and A1 = A
This theorem is entirely transparent. In the language of the next chapter, it merely
states that R
m,n
is a right module over the ring R.
Multiplication of Matrices The matrix product AB is defined iff the number
of columns of A is equal to the number of rows of B. The matrix AB will have the
same number of rows as A and the same number of columns as B, i.e., multiplication
is a function R
m,n
R
n,p
→R
m,p
. The product (a
i,j
)(b
i,j
) is defined to be the matrix
whose (s, t) term is a
s,1
b
1,t
+ + a
s,n
b
n,t
, i.e., the dot product of row s of A
with column t of B.
Exercise Consider real matrices A =
=
¸
j
a
s,j
y
j,t
which is the (s, t) term of A(BC).
Theorem For each ring R and integer n ≥ 1, R
n
is a ring.
Proof This elegant little theorem is immediate from the theorems above. The
units of R
n
are called invertible or non-singular matrices. They form a group under
multiplication called the general linear group and denoted by GL
n
(R) = (R
n
)
∗
.
Exercise Recall that if A is a ring and a ∈ A, then aA is right ideal of A. Let
A = R
2
and a = (a
i,j
) where a
1,1
= 1 and the other entries are 0. Find aR
2
and R
2
a.
Show that the only ideal of R
2
containing a is R
2
itself.
Multiplication by blocks Suppose A, E ∈ R
n
, B, F ∈ R
n,m
, C, G ∈ R
m,n
, and
D, H ∈ R
m
. Then multiplication in R
n+m
is given by
A B
C D
E F
G H
=
AE +BG AF +BH
CE +DG CF +DH
.
56 Matrices Chapter 4
Transpose
Notation For the remainder of this chapter on matrices, suppose R is a commu-
tative ring. Of course, for n > 1, R
n
is non-commutative.
Transpose is a function from R
m,n
to R
n,m
. If A ∈ R
m,n
, A
t
∈ R
n,m
is the matrix
whose (i, j) term is the (j, i) term of A. So row i (column i) of A becomes column
i (row i) of A
t
. If A is an n-dimensional row vector, then A
t
is an n-dimensional
column vector. If A is a square matrix, A
t
is also square.
Theorem 1) (A
t
)
t
= A
2) (A+B)
t
= A
t
+B
t
3) If c ∈ R, (Ac)
t
= A
t
c
4) (AB)
t
= B
t
A
t
5) If A ∈ R
n
, then A is invertible iff A
t
is invertible.
In this case (A
−1
)
t
= (A
t
)
−1
.
Proof of 5) Suppose A is invertible. Then I = I
t
= (AA
−1
)
t
= (A
−1
)
t
A
t
.
Exercise Characterize those invertible matrices A ∈ R
2
which have A
−1
= A
t
.
Show that they form a subgroup of GL
2
(R).
Triangular Matrices
If A ∈ R
n
, then A is upper (lower) triangular provided a
i,j
= 0 for all i > j (all
j > i). A is strictly upper (lower) triangular provided a
i,j
= 0 for all i ≥ j (all j ≥ i).
A is diagonal if it is upper and lower triangular, i.e., a
i,j
= 0 for all i = j. Note
that if A is upper (lower) triangular, then A
t
is lower (upper) triangular.
Theorem If A ∈ R
n
is strictly upper (or lower) triangular, then A
n
= 0.
Proof The way to understand this is just multiply it out for n = 2 and n = 3.
The geometry of this theorem will become transparent later in Chapter 5 when the
matrix A defines an R-module endomorphism on R
n
(see page 93).
Definition If T is any ring, an element t ∈ T is said to be nilpotent provided ∃n
such that t
n
= 0. In this case, (1 − t) is a unit with inverse 1 + t + t
2
+ + t
n−1
.
Thus if T = R
n
and B is a nilpotent matrix, I −B is invertible.
Chapter 4 Matrices 57
Exercise Let R = Z. Find the inverse of
¸
¸
1 2 −3
0 1 4
0 0 1
¸
.
Exercise Suppose A =
¸
¸
¸
¸
¸
¸
¸
a
1
a
2
0
0
a
n
¸
is a diagonal matrix, B ∈ R
m,n
,
and C ∈ R
n,p
. Show that BA is obtained from B by multiplying column i of B
by a
i
. Show AC is obtained from C by multiplying row i of C by a
i
. Show A is a
unit in R
n
iff each a
i
is a unit in R.
Scalar matrices A scalar matrix is a diagonal matrix for which all the diagonal
terms are equal, i.e., a matrix of the form cI
n
. The map R → R
n
which sends c to
cI
n
is an injective ring homomorphism, and thus we may consider R to be a subring
of R
n
. Multiplying by a scalar is the same as multiplying by a scalar matrix, and
thus scalar matrices commute with everything, i.e., if B ∈ R
n
, (cI
n
)B = cB = Bc =
B(cI
n
). Recall we are assuming R is a commutative ring.
Exercise Suppose A ∈ R
n
and for each B ∈ R
n
, AB = BA. Show A is a scalar
matrix. For n > 1, this shows how non-commutative R
n
is.
Elementary Operations and Elementary Matrices
Suppose R is a commutative ring and A is a matrix over R. There are 3 types of
elementary row and column operations on the matrix A. A need not be square.
Type 1 Multiply row i by some Multiply column i by some
unit a ∈ R. unit a ∈ R.
Type 2 Interchange row i and row j. Interchange column i and column j.
Type 3 Add a times row j Add a times column i
to row i where i = j and a to column j where i = j and a
is any element of R. is any element of R.
58 Matrices Chapter 4
Elementary Matrices Elementary matrices are square and invertible. There
are three types. They are obtained by performing row or column operations on the
identity matrix.
Type 1 B =
¸
¸
¸
¸
¸
¸
¸
¸
¸
1
1 0
a
1
0 1
1
¸
where a is a unit in R.
Type 2 B =
¸
¸
¸
¸
¸
¸
¸
¸
¸
1
0 1
1
1
1 0
1
¸
Type 3 B =
¸
¸
¸
¸
¸
¸
¸
¸
¸
1
1 a
i,j
1
1
0 1
1
¸
where i = j and a
i,j
is
any element of R.
In type 1, all the off-diagonal elements are zero. In type 2, there are two non-zero
off-diagonal elements. In type 3, there is at most one non-zero off-diagonal element,
and it may be above or below the diagonal.
Exercise Show that if B is an elementary matrix of type 1,2, or 3, then B is
invertible and B
−1
is an elementary matrix of the same type.
The following theorem is handy when working with matrices.
Theorem Suppose A is a matrix. It need not be square. To perform an elemen-
tary row (column) operation on A, perform the operation on an identity matrix to
obtain an elementary matrix B, and multiply on the left (right). That is, BA = row
operation on A and AB = column operation on A. (See the exercise on page 54.)
Chapter 4 Matrices 59
Exercise Suppose F is a field and A ∈ F
m,n
.
1) Show ∃ invertible matrices B ∈ F
m
and C ∈ F
n
such that BAC = (d
i,j
)
where d
1,1
= = d
t,t
= 1 and all other entries are 0. The integer t is
called the rank of A. (See page 89 of Chapter 5.)
2) Suppose A ∈ F
n
is invertible. Show A is the product of elementary
matrices.
3) A matrix T is said to be in row echelon form if, for each 1 ≤ i < m, the
first non-zero term of row (i + 1) is to the right of the first non-zero
term of row i. Show ∃ an invertible matrix B ∈ F
m
such that BA is in
row echelon form.
4) Let A =
3 11
0 4
and D =
3 11
1 4
. Write A and D as products
of elementary matrices over Q. Is it possible to write them as products
of elementary matrices over Z?
For 1), perform row and column operations on A to reach the desired form. This
shows the matrices B and C may be selected as products of elementary matrices.
Part 2) also follows from this procedure. For part 3), use only row operations. Notice
that if T is in row-echelon form, the number of non-zero rows is the rank of T.
Systems of Equations
Suppose A = (a
i,j
) ∈ R
m,n
and C =
or AX = C.
60 Matrices Chapter 4
Define f : R
n
→R
m
by f(D) = AD. Then f is a group homomorphism and also
f(Dc) = f(D)c for any c ∈ R. In the language of the next chapter, this says that
f is an R-module homomorphism. The next theorem summarizes what we already
know about solutions of linear equations in this setting.
Theorem
1) AX = 0 is called the homogeneous equation. Its solution set is ker(f).
2) AX = C has a solution iff C ∈ image(f). If D ∈ R
n
is one
solution, the solution set f
−1
(C) is the coset D + ker(f) in R
n
.
(See part 7 of the theorem on homomorphisms in Chapter 2, page 28.)
3) Suppose B ∈ R
m
is invertible. Then AX = C and (BA)X = BC have
the same set of solutions. Thus we may perform any row operation
on both sides of the equation and not change the solution set.
4) If m = n and A ∈ R
m
is invertible, then AX = C has the unique
solution X = A
−1
C.
The geometry of systems of equations over a field will not become really trans-
parent until the development of linear algebra in Chapter 5.
Determinants
The concept of determinant is one of the most amazing in all of mathematics.
The proper development of this concept requires a study of multilinear forms, which
is given in Chapter 6. In this section we simply present the basic properties.
For each n ≥ 1 and each commutative ring R, determinant is a function from R
n
to R. For n = 1, [ (a) [ = a. For n = 2,
a b
c d
= ad −bc.
Definition Let A = (a
i,j
) ∈ R
n
. If σ is a permutation on ¦1, 2, ..., n¦, let sign(σ) =
1 if σ is an even permutation, and sign(σ) = −1 if σ is an odd permutation. The
determinant is defined by [ A [=
¸
all σ
sign(σ) a
1,σ(1)
a
2,σ(2)
a
n,σ(n)
. Check that for
n = 2, this agrees with the definition above. (Note that here we are writing the
permutation functions as σ(i) and not as (i)σ.)
Chapter 4 Matrices 61
For each σ, a
1,σ(1)
a
2,σ(2)
a
n,σ(n)
contains exactly one factor from each row and
one factor from each column. Since R is commutative, we may rearrange the factors
so that the first comes from the first column, the second from the second column, etc.
This means that there is a permutation τ on ¦1, 2, . . . , n¦ such that a
1,σ(1)
a
n,σ(n)
=
a
τ(1),1
a
τ(n),n
. We wish to show that τ = σ
−1
and thus sign(σ) = sign(τ). To
reduce the abstraction, suppose σ(2) = 5. Then the first expression will contain
the factor a
2,5
. In the second expression, it will appear as a
τ(5),5
, and so τ(5) = 2.
Anyway, τ is the inverse of σ and thus there are two ways to define determinant. It
follows that the determinant of a matrix is equal to the determinant of its transpose.
Theorem [A[ =
¸
all σ
sign(σ)a
1,σ(1)
a
2,σ(2)
a
n,σ(n)
=
¸
all τ
sign(τ)a
τ(1),1
a
τ(2),2
a
τ(n),n
.
Corollary [A[ = [A
t
[.
You may view an n n matrix A as a sequence of n column vectors or as a
sequence of n row vectors. Here we will use column vectors. This means we write the
matrix A as A = (A
1
, A
2
, . . . , A
n
) where each A
i
∈ R
n,1
= R
n
.
Theorem If two columns of A are equal, then [A[ = 0
¯
.
Proof For simplicity, assume the first two columns are equal, i.e., A
1
= A
2
.
Now [A[ =
¸
all τ
sign(τ)a
τ(1),1
a
τ(2),2
a
τ(n),n
and this summation has n! terms and
n! is an even number. Let γ be the transposition which interchanges one and two.
Then for any τ, a
τ(1),1
a
τ(2),2
a
τ(n),n
= a
τγ(1),1
a
τγ(2),2
a
τγ(n),n
. This pairs up
the n! terms of the summation, and since sign(τ)=−sign(τγ), these pairs cancel
in the summation. Therefore [A[ = 0
¯
.
Theorem Suppose 1 ≤ r ≤ n, C
r
∈ R
n,1
, and a, c ∈ R. Then [(A
1
, . . . , A
r−1
,
aA
r
+cC
r
, A
r+1
, . . . , A
n
)[ = a[(A
1
, . . . , A
n
)[ + c[(A
1
, . . . , A
r−1
, C
r
, A
r+1
, . . . , A
n
)[
Proof This is immediate from the definition of determinant and the distributive
law of multiplication in the ring R.
Summary Determinant is a function d : R
n
→ R. In the language used in the
Appendix, the two previous theorems say that d is an alternating multilinear form.
The next two theorems show that alternating implies skew-symmetric (see page 129).
62 Matrices Chapter 4
Theorem Interchanging two columns of A multiplies the determinant by minus
one.
Proof For simplicity, show that [(A
2
, A
1
, A
3
, . . . , A
n
)[ = −[A[. We know 0
¯
=
[(A
1
+ A
2
, A
1
+ A
2
, A
3
, . . . , A
n
)[ = [(A
1
, A
1
, A
3
, . . . , A
n
)[ + [(A
1
, A
2
, A
3
, . . . , A
n
)[ +
[(A
2
, A
1
, A
3
, . . . , A
n
)[ + [(A
2
, A
2
, A
3
, . . . , A
n
)[. Since the first and last of these four
terms are zero, the result follows.
Theorem If τ is a permutation of (1, 2, . . . , n), then
[A[ = sign(τ)[(A
τ(1)
, A
τ(2)
, . . . , A
τ(n)
)[.
Proof The permutation τ is the finite product of transpositions.
Exercise Rewrite the four preceding theorems using rows instead of columns.
The following theorem is just a summary of some of the work done so far.
Theorem Multiplying any row or column of matrix by a scalar c ∈ R, multiplies
the determinant by c. Interchanging two rows or two columns multiplies the determi-
nant by −1. Adding c times one row to another row, or adding c times one column
to another column, does not change the determinant. If a matrix has two rows equal
or two columns equal, its determinant is zero. More generally, if one row is c times
another row, or one column is c times another column, then the determinant is zero.
There are 2n ways to compute [ A[; expansion by any row or expansion by any
column. Let M
i,j
be the determinant of the (n − 1) (n − 1) matrix obtained by
removing row i and column j from A. Let C
i,j
= (−1)
i+j
M
i,j
. M
i,j
and C
i,j
are
called the (i, j) minor and cofactor of A. The following theorem is useful but the
proof is a little tedious and should not be done as an exercise.
Theorem For any 1 ≤ i ≤ n, [ A[= a
i,1
C
i,1
+ a
i,2
C
i,2
+ + a
i,n
C
i,n
. For any
1 ≤ j ≤ n, [ A[= a
1,j
C
1,j
+a
2,j
C
2,j
+ +a
n,j
C
n,j
. Thus if any row or any column is
zero, the determinant is zero.
Exercise Let A =
¸
¸
a
1
a
2
a
3
b
1
b
2
b
3
c
1
c
2
c
3
¸
. The determinant of A is the sum of six terms.
Chapter 4 Matrices 63
Write out the determinant of A expanding by the first column and also expanding by
the second row.
Theorem If A is an upper or lower triangular matrix, [ A[ is the product of the
diagonal elements. If A is an elementary matrix of type 2, [ A [= −1. If A is an
elementary matrix of type 3, [ A[= 1.
Proof We will prove the first statement for upper triangular matrices. If A ∈ R
2
is an upper triangular matrix, then its determinant is the product of the diagonal
elements. Suppose n > 2 and the theorem is true for matrices in R
n−1
. Suppose
A ∈ R
n
is upper triangular. The result follows by expanding by the first column.
An elementary matrix of type 3 is a special type of upper or lower triangular
matrix, so its determinant is 1. An elementary matrix of type 2 is obtained from the
identity matrix by interchanging two rows or columns, and thus has determinant −1.
Theorem (Determinant by blocks) Suppose A ∈ R
n
, B ∈ R
n,m
, and D ∈ R
m
.
Then the determinant of
A B
O D
is [ A[[ D[.
Proof Expand by the first column and use induction on n.
The following remarkable theorem takes some work to prove. We assume it here
without proof. (For the proof, see page 130 of the Appendix.)
Theorem The determinant of the product is the product of the determinants,
i.e., if A, B ∈ R
n
, [ AB[ = [ A[[ B[. Thus [ AB[ = [ BA[ and if C is invertible,
[ C
−1
AC [ = [ACC
−1
[ = [ A[.
Corollary If A is a unit in R
n
, then [ A[ is a unit in R and [ A
−1
[ = [ A[
−1
.
Proof 1 = [ I [ = [ AA
−1
[ = [ A[[ A
−1
[ .
One of the major goals of this chapter is to prove the converse of the preceding
corollary.
Classical adjoint Suppose R is a commutative ring and A ∈ R
n
. The classical
adjoint of A is (C
i,j
)
t
, i.e., the matrix whose (j, i) term is the (i, j) cofactor. Before
64 Matrices Chapter 4
we consider the general case, let's examine 2 2 matrices.
If A =
.
Here is the general case.
Theorem If R is commutative and A ∈ R
n
, then A(C
i,j
)
t
= (C
i,j
)
t
A = [ A[ I.
Proof We must show that the diagonal elements of the product A(C
i,j
)
t
are all
[ A[ and the other elements are 0. The (s, s) term is the dot product of row s of A
with row s of (C
i,j
) and is thus [ A[ (computed by expansion by row s). For s = t,
the (s, t) term is the dot product of row s of A with row t of (C
i,j
). Since this is the
determinant of a matrix with row s = row t, the (s, t) term is 0. The proof that
(C
i,j
)
t
A = [A[I is similar and is left as an exercise.
We are now ready for one of the most beautiful and useful theorems in all of
mathematics.
Theorem Suppose R is a commutative ring and A ∈ R
n
. Then A is a unit in
R
n
iff [ A[ is a unit in R. (Thus if R is a field, A is invertible iff [ A[ = 0.) If A is
invertible, then A
−1
= [ A[
−1
(C
i,j
)
t
. Thus if [ A[ = 1, A
−1
= (C
i,j
)
t
, the classical
adjoint of A.
Proof This follows immediately from the preceding theorem.
Exercise Show that any right inverse of A is also a left inverse. That is, suppose
A, B ∈ R
n
and AB = I. Show A is invertible with A
−1
= B, and thus BA = I.
Similarity
Suppose A, B ∈ R
n
. B is said to be similar to A if ∃ an invertible C ∈ R
n
such
that B = C
−1
AC, i.e., B is similar to A iff B is a conjugate of A.
Theorem B is similar to B.
Chapter 4 Matrices 65
B is similar to A iff A is similar to B.
If D is similar to B and B is similar to A, then D is similar to A.
"Similarity" is an equivalence relation on R
n
.
Proof This is a good exercise using the definition.
Theorem Suppose A and B are similar. Then [ A[ = [ B[ and thus A is invertible
iff B is invertible.
Proof Suppose B = C
−1
AC. Then [ B[ = [ C
−1
AC [ = [ACC
−1
[ = [ A[.
Trace Suppose A = (a
i,j
) ∈ R
n
. Then the trace is defined by trace(A) = a
1,1
+
a
2,2
+ +a
n,n
. That is, the trace of A is the sum of its diagonal terms.
One of the most useful properties of trace is trace(AB) = trace(BA) whenever AB
and BA are defined. For example, suppose A = (a
1
, a
2
, ..., a
n
) and B = (b
1
, b
2
, ..., b
n
)
t
.
Then AB is the scalar a
1
b
1
+ + a
n
b
n
while BA is the n n matrix (b
i
a
j
). Note
that trace(AB) = trace(BA). Here is the theorem in full generality.
Theorem Suppose A ∈ R
m,n
and B ∈ R
n,m
. Then AB and BA are square
matrices with trace(AB) = trace(BA).
Proof This proof involves a change in the order of summation. By definition,
trace(AB) =
¸
1≤i≤m
a
i,1
b
1,i
+ +a
i,n
b
n,i
=
¸
1≤i≤m
1≤j≤n
a
i,j
b
j,i
=
¸
1≤j≤n
b
j,1
a
1,j
+ +b
j,m
a
m,j
=
trace(BA).
Theorem If A, B ∈ R
n
, trace(A+B) = trace(A) + trace(B) and
trace(AB) = trace(BA).
Proof The first part of the theorem is immediate, and the second part is a special
case of the previous theorem.
Theorem If A and B are similar, then trace(A) = trace(B).
Proof trace(B) = trace(C
−1
AC) = trace(ACC
−1
) = trace(A).
66 Matrices Chapter 4
Summary Determinant and trace are functions from R
n
to R. Determinant is a
multiplicative homomorphism and trace is an additive homomorphism. Furthermore
[ AB[ = [ BA[ and trace(AB) = trace(BA). If A and B are similar, [ A[ = [ B[ and
trace(A) = trace(B).
Exercise Suppose A ∈ R
n
and a ∈ R. Find [aA[ and trace(aA).
Characteristic polynomials If A ∈ R
n
, the characteristic polynomial CP
A
(x) ∈
R[x] is defined by CP
A
(x) = [ (xI − A) [. Any λ ∈ R which is a root of CP
A
(x) is
called a characteristic root of A.
Theorem CP
A
(x) = a
0
+ a
1
x + + a
n−1
x
n−1
+ x
n
where trace(A) = −a
n−1
and [ A[ = (−1)
n
a
0
.
Proof This follows from a direct computation of the determinant.
Theorem If A and B are similar, then they have the same characteristic polyno-
mials.
Proof Suppose B = C
−1
AC. CP
B
(x) = [ (xI −C
−1
AC) [ = [ C
−1
(xI −A)C[ =
[ (xI −A)[ = CP
A
(x).
Exercise Suppose R is a commutative ring, A =
a b
c d
is a matrix in R
2
, and
CP
A
(x) = a
0
+ a
1
x + x
2
. Find a
0
and a
1
and show that a
0
I + a
1
A + A
2
= 0, i.e.,
show A satisfies its characteristic polynomial. In other words, CP
A
(A) = 0.
Exercise Suppose F is a field and A ∈ F
2
. Show the following are equivalent.
1) A
2
= 0.
2) [ A [= trace(A) = 0.
3) CP
A
(x) = x
2
.
4) ∃ an elementary matrix C such that C
−1
AC is strictly upper triangular.
Note This exercise is a special case of a more general theorem. A square matrix
over a field is nilpotent iff all its characteristic roots are 0
¯
iff it is similar to a strictly
upper triangular matrix. This remarkable result cannot be proved by matrix theory
alone, but depends on linear algebra (see pages 93, 94, and 98).
Chapter 5
Linear Algebra
The exalted position held by linear algebra is based upon the subject's ubiquitous
utility and ease of application. The basic theory is developed here in full generality,
i.e., modules are defined over an arbitrary ring R and not just over a field. The
elementary facts about cosets, quotients, and homomorphisms follow the same pat-
tern as in the chapters on groups and rings. We give a simple proof that if R is a
commutative ring and f : R
n
→ R
n
is a surjective R-module homomorphism, then
f is an isomorphism. This shows that finitely generated free R-modules have a well
defined dimension, and simplifies some of the development of linear algebra. It is in
this chapter that the concepts about functions, solutions of equations, matrices, and
generating sets come together in one unified theory.
After the general theory, we restrict our attention to vector spaces, i.e., modules
over a field. The key theorem is that any vector space V has a free basis, and thus
if V is finitely generated, it has a well defined dimension, and incredible as it may
seem, this single integer determines V up to isomorphism. Also any endomorphism
f : V →V may be represented by a matrix, and any change of basis corresponds to
conjugation of that matrix. One of the goals in linear algebra is to select a basis so
that the matrix representing f has a simple form. For example, if f is not injective,
then f may be represented by a matrix whose first column is zero. As another
example, if f is nilpotent, then f may be represented by a strictly upper triangular
matrix. The theorem on Jordan canonical form is not proved in this chapter, and
should not be considered part of this chapter. It is stated here in full generality only
for reference and completeness. The proof is given in the Appendix. This chapter
concludes with the study of real inner product spaces, and with the beautiful theory
relating orthogonal matrices and symmetric matrices.
67
68 Linear Algebra Chapter 5
Definition Suppose R is a ring and M is an additive abelian group. The state-
ment that M is a right R-module means there is a scalar multiplication
M R → M satisfying (a
1
+a
2
)r = a
1
r +a
2
r
(m, r) → mr a(r
1
+r
2
) = ar
1
+ar
2
a(r
1
r
2
) = (ar
1
)r
2
a1
¯
= a
for all a, a
1
, a
2
∈ M and r, r
1
, r
2
∈ R.
The statement that M is a left R-module means there is a scalar multiplication
R M → M satisfying r(a
1
+a
2
) = ra
1
+ra
2
(r, m) → rm (r
1
+r
2
)a = r
1
a +r
2
a
(r
1
r
2
)a = r
1
(r
2
a)
1
¯
a = a
Note that the plus sign is used ambiguously, as addition in M and as addition in R.
Notation The fact that M is a right (left) R-module will be denoted by M = M
R
(M =
R
M). If R is commutative and M = M
R
then left scalar multiplication defined
by ra = ar makes M into a left R-module. Thus for commutative rings, we may write
the scalars on either side. In this text we stick to right R-modules.
Convention Unless otherwise stated, it is assumed that R is a ring and the word
"R-module" (or sometimes just "module") means "right R-module".
Theorem Suppose M is an R-module.
1) If r ∈ R, then f : M →M defined by f(a) = ar is a homomorphism of
additive groups. In particular (0
¯
M
)r = 0
¯
M
.
2) If a ∈ M, a0
¯
R
= 0
¯
M
.
3) If a ∈ M and r ∈ R, then (−a)r = −(ar) = a(−r).
Proof This is a good exercise in using the axioms for an R-module.
Chapter 5 Linear Algebra 69
Submodules If M is an R-module, the statement that a subset N ⊂ M is a
submodule means it is a subgroup which is closed under scalar multiplication, i.e., if
a ∈ N and r ∈ R, then ar ∈ N. In this case N will be an R-module because the
axioms will automatically be satisfied. Note that 0
¯
and M are submodules, called the
improper submodules of M.
Theorem Suppose M is an R-module, T is an index set, and for each t ∈ T,
N
t
is a submodule of M.
1)
¸
t∈T
N
t
is a submodule of M.
2) If ¦N
t
¦ is a monotonic collection,
¸
t∈T
N
t
is a submodule.
3) +
t∈T
N
t
= ¦all finite sums a
1
+ +a
m
: each a
i
belongs
to some N
t
¦ is a submodule. If T = ¦1, 2, .., n¦,
then this submodule may be written as
N
1
+N
2
+ +N
n
= ¦a
1
+a
2
+ +a
n
: each a
i
∈ N
i
¦.
Proof We know from page 22 that versions of 1) and 2) hold for subgroups, and
in particular for subgroups of additive abelian groups. To finish the proofs it is only
necessary to check scalar multiplication, which is immediate. Also the proof of 3) is
immediate. Note that if N
1
and N
2
are submodules of M, N
1
+ N
2
is the smallest
submodule of M containing N
1
∪ N
2
.
Exercise Suppose T is a non-void set, N is an R-module, and N
T
is the collection
of all functions f : T →N with addition defined by (f +g)(t) = f(t)+g(t), and scalar
multiplication defined by (fr)(t) = f(t)r. Show N
T
is an R-module. (We know from
the last exercise in Chapter 2 that N
T
is a group, and so it is only necessary to check
scalar multiplication.) This simple fact is quite useful in linear algebra. For example,
in 5) of the theorem below, it is stated that Hom
R
(M, N) forms an abelian group.
So it is only necessary to show that Hom
R
(M, N) is a subgroup of N
M
. Also in 8) it
is only necessary to show that Hom
R
(M, N) is a submodule of N
M
.
Homomorphisms
Suppose M and N are R-modules. A function f : M → N is a homomorphism
(i.e., an R-module homomorphism) provided it is a group homomorphism and if
a ∈ M and r ∈ R, f(ar) = f(a)r. On the left, scalar multiplication is in M and on
the right it is in N. The basic facts about homomorphisms are listed below. Much
70 Linear Algebra Chapter 5
of this work has already been done in the chapter on groups (see page 28).
Theorem
1) The zero map M →N is a homomorphism.
2) The identity map I : M →M is a homomorphism.
3) The composition of homomorphisms is a homomorphism.
4) The sum of homomorphisms is a homomorphism. If f, g : M →N are
homomorphisms, define (f +g) : M →N by (f +g)(a) = f(a) +g(a).
Then f +g is a homomorphism. Also (−f) defined by (−f)(a) = −f(a)
is a homomorphism. If h : N →P is a homomorphism,
h ◦ (f +g) = (h ◦ f) + (h ◦ g). If k : P →M is a homomorphism,
(f +g ) ◦ k = (f ◦ k) + (g ◦ k).
5) Hom
R
(M, N) = Hom(M
R
, N
R
), the set of all homomorphisms from M
to N, forms an abelian group under addition. Hom
R
(M, M), with
multiplication defined to be composition, is a ring.
6) If a bijection f : M →N is a homomorphism, then f
−1
: N →M is also
a homomorphism. In this case f and f
−1
are called isomorphisms. A
homomorphism f : M →M is called an endomorphism. An isomorphism
f : M →M is called an automorphism. The units of the endomorphism
ring Hom
R
(M, M) are the automorphisms. Thus the automorphisms on
M form a group under composition. We will see later that if M = R
n
,
Hom
R
(R
n
, R
n
) is just the matrix ring R
n
and the automorphisms
are merely the invertible matrices.
7) If R is commutative and r ∈ R, then g : M →M defined by g(a) = ar
is a homomorphism. Furthermore, if f : M →N is a homomorphism,
fr defined by (fr)(a) = f(ar) = f(a)r is a homomorphism.
8) If R is commutative, Hom
R
(M, N) is an R-module.
9) Suppose f : M →N is a homomorphism, G ⊂ M is a submodule,
and H ⊂ N is a submodule. Then f(G) is a submodule of N
and f
−1
(H) is a submodule of M. In particular, image(f) is a
submodule of N and ker(f) = f
−1
(0
¯
) is a submodule of M.
Proof This is just a series of observations.
Chapter 5 Linear Algebra 71
Abelian groups are Z-modules On page 21, it is shown that any additive
group M admits a scalar multiplication by integers, and if M is abelian, the properties
are satisfied to make M a Z-module. Note that this is the only way M can be a Z-
module, because a1 = a, a2 = a + a, etc. Furthermore, if f : M → N is a group
homomorphism of abelian groups, then f is also a Z-module homomorphism.
Summary Additive abelian groups are "the same things" as Z-modules. While
group theory in general is quite separate from linear algebra, the study of additive
abelian groups is a special case of the study of R-modules.
Exercise R-modules are also Z-modules and R-module homomorphisms are also
Z-module homomorphisms. If M and N are Q-modules and f : M → N is a
Z-module homomorphism, must it also be a Q-module homomorphism?
Homomorphisms on R
n
R
n
as an R-module On page 54 it was shown that the additive abelian
group R
m,n
admits a scalar multiplication by elements in R. The properties listed
there were exactly those needed to make R
m,n
an R-module. Of particular importance
is the case R
n
= R ⊕ ⊕R = R
n,1
(see page 53). We begin with the case n = 1.
R as a right R-module Let M = R and define scalar multiplication on the right
by ar = a r. That is, scalar multiplication is just ring multiplication. This makes
R a right R-module denoted by R
R
(or just R). This is the same as the definition
before for R
n
when n = 1.
Theorem Suppose R is a ring and N is a subset of R. Then N is a submodule
of R
R
(
R
R) iff N is a right (left) ideal of R.
Proof The definitions are the same except expressed in different language.
Theorem Suppose M = M
R
and f, g : R →M are homomorphisms with f(1
¯
) =
g(1
¯
). Then f = g. Furthermore, if m ∈ M, ∃! homomorphism h : R → M with
h(1
¯
) = m. In other words, Hom
R
(R, M) ≈ M.
Proof Suppose f(1
¯
) = g(1
¯
). Then f(r) = f(1
¯
r) = f(1
¯
)r = g(1
¯
)r = g(1
¯
r) =
g(r). Given m ∈ M, h : R → M defined by h(r) = mr is a homomorphism. Thus
72 Linear Algebra Chapter 5
evaluation at 1
¯
gives a bijection from Hom
R
(R, M) to M, and this bijection is clearly
a group isomorphism. If R is commutative, it is an isomorphism of R-modules.
In the case M = R, the above theorem states that multiplication on left by some
m ∈ R defines a right R-module homomorphism from R to R, and every module
homomorphism is of this form. The element m should be thought of as a 1 1
matrix. We now consider the case where the domain is R
n
.
Homomorphisms on R
n
Define e
i
∈ R
n
by e
i
=
¸
¸
¸
¸
¸
¸
¸
0
¯
1
¯
i
0
¯
¸
. Note that any
¸
¸
¸
¸
¸
¸
¸
r
1
r
n
¸
can be written uniquely as e
1
r
1
+ +e
n
r
n
. The sequence ¦e
1
, .., e
n
¦ is called the
canonical free basis or standard basis for R
n
.
Theorem Suppose M = M
R
and f, g : R
n
→ M are homomorphisms with
f(e
i
) = g(e
i
) for 1 ≤ i ≤ n. Then f = g. Furthermore, if m
1
, m
2
, ..., m
n
∈ M, ∃!
homomorphism h : R
n
→ M with h(e
i
) = m
i
for 1 ≤ i ≤ m. The homomorphism
h is defined by h(e
1
r
1
+ +e
n
r
n
) = m
1
r
1
+ +m
n
r
n
.
Proof The proof is straightforward. Note this theorem gives a bijection from
Hom
R
(R
n
, M) to M
n
= M M M and this bijection is a group isomorphism.
We will see later that the product M
n
is an R-module with scalar multiplication
defined by (m
1
, m
2
, .., m
n
)r = (m
1
r, m
2
r, .., m
n
r). If R is commutative so that
Hom
R
(R
n
, M) is an R-module, this theorem gives an R-module isomorphism from
Hom
R
(R
n
, M) to M
n
.
This theorem reveals some of the great simplicity of linear algebra. It does not
matter how complicated the ring R is, or which R-module M is selected. Any
R-module homomorphism from R
n
to M is determined by its values on the basis,
and any function from that basis to M extends uniquely to a homomorphism from
R
n
to M.
Exercise Suppose R is a field and f : R
R
→ M is a non-zero homomorphism.
Show f is injective.
Chapter 5 Linear Algebra 73
Now let's examine the special case M = R
m
and show Hom
R
(R
n
, R
m
) ≈ R
m,n
.
Theorem Suppose A = (a
i,j
) ∈ R
m,n
. Then f : R
n
→R
m
defined by f(B) = AB
is a homomorphism with f(e
i
) = column i of A. Conversely, if v
1
, . . . , v
n
∈ R
m
, define
A ∈ R
m,n
to be the matrix with column i = v
i
. Then f defined by f(B) = AB is
the unique homomorphism from R
n
to R
m
with f(e
i
) = v
i
.
Even though this follows easily from the previous theorem and properties of ma-
trices, it is one of the great classical facts of linear algebra. Matrices over R give
R-module homomorphisms! Furthermore, addition of matrices corresponds to addi-
tion of homomorphisms, and multiplication of matrices corresponds to composition
of homomorphisms. These properties are made explicit in the next two theorems.
Theorem If f, g : R
n
→ R
m
are given by matrices A, C ∈ R
m,n
, then f + g is
given by the matrix A+C. Thus Hom
R
(R
n
, R
m
) and R
m,n
are isomorphic as additive
groups. If R is commutative, they are isomorphic as R-modules.
Theorem If f : R
n
→ R
m
is the homomorphism given by A ∈ R
m,n
and g :
R
m
→R
p
is the homomorphism given by C ∈ R
p,m
, then g ◦ f : R
n
→R
p
is given by
CA ∈ R
p,n
. That is, composition of homomorphisms corresponds to multiplication
of matrices.
Proof This is just the associative law of matrix multiplication, C(AB) = (CA)B.
The previous theorem reveals where matrix multiplication comes from. It is the
matrix which represents the composition of the functions. In the case where the
domain and range are the same, we have the following elegant corollary.
Corollary Hom
R
(R
n
, R
n
) and R
n
are isomorphic as rings. The automorphisms
correspond to the invertible matrices.
This corollary shows one way non-commutative rings arise, namely as endomor-
phism rings. Even if R is commutative, R
n
is never commutative unless n = 1.
We now return to the general theory of modules (over some given ring R).
74 Linear Algebra Chapter 5
Cosets and Quotient Modules
After seeing quotient groups and quotient rings, quotient modules go through
without a hitch. As before, R is a ring and module means R-module.
Theorem Suppose M is a module and N ⊂ M is a submodule. Since N is a
normal subgroup of M, the additive abelian quotient group M/N is defined. Scalar
multiplication defined by (a + N)r = (ar + N) is well-defined and gives M/N the
structure of an R-module. The natural projection π : M → M/N is a surjective
homomorphism with kernel N. Furthermore, if f : M →
¯
M is a surjective homomor-
phism with ker(f) = N, then M/N ≈
¯
M (see below).
Proof On the group level, this is all known from Chapter 2 (see pages 27 and 29).
It is only necessary to check the scalar multiplication, which is obvious.
The relationship between quotients and homomorphisms for modules is the same
as for groups and rings, as shown by the next theorem.
Theorem Suppose f : M →
¯
M is a homomorphism and N is a submodule of M.
If N ⊂ ker(f), then
¯
f : (M/N) →
¯
M defined by
¯
f(a + N) = f(a) is a well-defined
homomorphism making the following diagram commute.
M
¯
M
M/N
f
·
π
¯
f
Thus defining a homomorphism on a quotient module is the same as defining a homo-
morphism on the numerator that sends the denominator to 0
¯
. The image of
¯
f is the
image of f, and the kernel of
¯
f is ker(f)/N. Thus if N = ker(f),
¯
f is injective, and
thus (M/N) ≈image(f). Therefore for any homomorphism f, (domain(f)/ker(f)) ≈
image(f).
Proof On the group level this is all known from Chapter 2 (see page 29). It is
only necessary to check that
¯
f is a module homomorphism, and this is immediate.
Chapter 5 Linear Algebra 75
Theorem Suppose M is an R-module and K and L are submodules of M.
i) The natural homomorphism K →(K + L)/L is surjective with kernel
K ∩ L. Thus (K/K ∩ L)
≈
→(K +L)/L is an isomorphism.
ii) Suppose K ⊂ L. The natural homomorphism M/K →M/L is surjective
with kernel L/K. Thus (M/K)/(L/K)
≈
→M/L is an isomorphism.
Examples These two examples are for the case R = Z, i.e., for abelian groups.
1) M = Z K = 3Z L = 5Z K ∩ L = 15Z K +L = Z
K/K ∩ L = 3Z/15Z ≈ Z/5Z = (K +L)/L
2) M = Z K = 6Z L = 3Z (K ⊂ L)
(M/K)/(L/K) = (Z/6Z)/(3Z/6Z) ≈ Z/3Z = M/L
Products and Coproducts
Infinite products work fine for modules, just as they do for groups and rings.
This is stated below in full generality, although the student should think of the finite
case. In the finite case something important holds for modules that does not hold
for non-abelian groups or rings – namely, the finite product is also a coproduct. This
makes the structure of module homomorphisms much more simple. For the finite
case we may use either the product or sum notation, i.e., M
1
M
2
M
n
=
M
1
⊕M
2
⊕ ⊕M
n
.
Theorem Suppose T is an index set and for each t ∈ T, M
t
is an R-module. On
the additive abelian group
¸
t∈T
M
t
=
¸
M
t
define scalar multiplication by ¦m
t
¦r =
¦m
t
r¦. Then
¸
M
t
is an R-module and, for each s ∈ T, the natural projection
π
s
:
¸
M
t
→M
s
is a homomorphism. Suppose M is a module. Under the natural 1-1
correspondence from ¦functions f : M →
¸
M
t
¦ to ¦sequence of functions ¦f
t
¦
t∈T
where f
t
: M →M
t
¦, f is a homomorphism iff each f
t
is a homomorphism.
Proof We already know from Chapter 2 that f is a group homomorphism iff each
f
t
is a group homomorphism. Since scalar multiplication is defined coordinatewise,
f is a module homomorphism iff each f
t
is a module homomorphism.
76 Linear Algebra Chapter 5
Definition If T is finite, the coproduct and product are the same module. If T
is infinite, the coproduct or sum
¸
t∈T
M
t
=
´
Theorem For finite T, the 1-1 correspondences in the above theorems actually
produce group isomorphisms. If R is commutative, they give isomorphisms of R-
modules.
Hom
R
(M, M
1
⊕ ⊕M
n
) ≈ Hom
R
(M, M
1
) ⊕ ⊕Hom
R
(M, M
n
) and
Hom
R
(M
1
⊕ ⊕M
n
, M) ≈ Hom
R
(M
1
, M) ⊕ ⊕Hom
R
(M
n
, M)
Proof Let's look at this theorem for products with n = 2. All it says is that if f =
(f
1
, f
2
) and h = (h
1
, h
2
), then f +h = (f
1
+h
1
, f
2
+h
2
). If R is commutative, so that
the objects are R-modules and not merely additive groups, then the isomorphisms
are module isomorphisms. This says merely that fr = (f
1
, f
2
)r = (f
1
r, f
2
r).
Chapter 5 Linear Algebra 77
Exercise Suppose M and N are R-modules. Show that M ⊕N is isomorphic to
N ⊕ M. Now suppose A ⊂ M, B ⊂ N are submodules and show (M ⊕ N)/(A ⊕B)
is isomorphic to (M/A) ⊕ (N/B). In particular, if a ∈ R and b ∈ R, then
(R ⊕ R)/(aR ⊕ bR) is isomorphic to (R/aR) ⊕ (R/bR). For example, the abelian
group (Z ⊕ Z)/(2Z ⊕ 3Z) is isomorphic to Z
2
⊕ Z
3
. These isomorphisms are trans-
parent and are used routinely in algebra without comment (see Th 4, page 118).
Exercise Suppose R is a commutative ring, M is an R-module, and n ≥ 1. Define
a function α : Hom
R
(R
n
, M) →M
n
which is a R-module isomorphism.
Summands
One basic question in algebra is "When does a module split as the sum of two
modules?". Before defining summand, here are two theorems for background.
Theorem Consider M
1
= M
1
⊕0
¯
as a submodule of M
1
⊕M
2
. Then the projection
map π
2
: M
1
⊕ M
2
→ M
2
is a surjective homomorphism with kernel M
1
. Thus
(M
1
⊕M
2
)/M
1
is isomorphic to M
2
. (See page 35 for the group version.)
This is exactly what you would expect, and the next theorem is almost as intuitive.
Theorem Suppose K and L are submodules of M and f : K ⊕ L → M is the
natural homomorphism, f(k, l) = k + l. Then the image of f is K + L and the
kernel of f is ¦(a, −a) : a ∈ K ∩ L¦. Thus f is an isomorphism iff K + L = M and
K ∩ L = 0
¯
. In this case we write K ⊕ L = M. This abuse of notation allows us to
avoid talking about "internal" and "external" direct sums.
Definition Suppose K is a submodule of M. The statement that K is a summand
of M means ∃ a submodule L of M with K ⊕ L = M. According to the previous
theorem, this is the same as there exists a submodule L with K + L = M and
K ∩ L = 0
¯
. If such an L exists, it need not be unique, but it will be unique up to
isomorphism, because L ≈ M/K. Of course, M and 0
¯
are always summands of M.
Exercise Suppose M is a module and K = ¦(m, m) : m ∈ M¦ ⊂ M ⊕M. Show
K is a submodule of M ⊕M which is a summand.
Exercise R is a module over Q, and Q ⊂ R is a submodule. Is Q a summand of
R? With the material at hand, this is not an easy question. Later on, it will be easy.
78 Linear Algebra Chapter 5
Exercise Answer the following questions about abelian groups, i.e., Z-modules.
(See the third exercise on page 35.)
1) Is 2Z a summand of Z?
2) Is 2Z
4
a summand of Z
4
?
3) Is 3Z
12
a summand of Z
12
?
4) Suppose m, n > 1. When is nZ
mn
a summand of Z
mn
?
Exercise If T is a ring, define the center of T to be the subring ¦t : ts =
st for all s ∈ T¦. Let R be a commutative ring and T = R
n
. There is a exercise
on page 57 to show that the center of T is the subring of scalar matrices. Show R
n
is a left T-module and find Hom
T
(R
n
, R
n
).
Independence, Generating Sets, and Free Basis
This section is a generalization and abstraction of the brief section Homomor-
phisms on R
n
. These concepts work fine for an infinite index set T because linear
combination means finite linear combination. However, to avoid dizziness, the student
should first consider the case where T is finite.
Definition Suppose M is an R-module, T is an index set, and for each t ∈ T,
s
t
∈ M. Let S be the sequence ¦s
t
¦
t∈T
= ¦s
t
¦. The statement that S is dependent
means ∃ a finite number of distinct elements t
1
, ..., t
n
in T, and elements r
1
, .., r
n
in
R, not all zero, such that the linear combination s
t
1
r
1
+ +s
tn
r
n
= 0
¯
. Otherwise,
S is independent. Note that if some s
t
= 0
¯
, then S is dependent. Also if ∃ distinct
elements t
1
and t
2
in T with s
t
1
= s
t
2
, then S is dependent.
Let SR be the set of all linear combinations s
t
1
r
1
+ +s
tn
r
n
. SR is a submodule
of M called the submodule generated by S. If S is independent and generates M,
then S is said to be a basis or free basis for M. In this case any v ∈ M can be written
uniquely as a linear combination of elements in S. An R-module M is said to be a
free R-module if it is zero or if it has a basis. The next two theorems are obvious,
except for the confusing notation. You might try first the case T = ¦1, 2, ..., n¦ and
⊕R
t
= R
n
(see p 72).
Theorem For each t ∈ T, let R
t
= R
R
and for each c ∈ T, let e
c
∈ ⊕R
t
=
t∈T
R
t
be e
c
= ¦r
t
¦ where r
c
= l
¯
and r
t
= 0
¯
if t = c. Then ¦e
c
¦
c∈T
is a basis for ⊕R
t
called
the canonical basis or standard basis.
Chapter 5 Linear Algebra 79
Theorem Suppose N is an R-module and M is a free R-module with a basis
¦s
t
¦. Then ∃ a 1-1 correspondence from the set of all functions g : ¦s
t
¦ →N and the
set of all homomorphisms f : M → N. Given g, define f by f(s
t
1
r
1
+ +s
tn
r
n
) =
g(s
t
1
)r
1
+ +g(s
tn
)r
n
. Given f, define g by g(s
t
) = f(s
t
). In other words, f is
completely determined by what it does on the basis S, and you are "free" to send the
basis any place you wish and extend to a homomorphism.
Recall that we have already had the preceding theorem in the case S is the canon-
ical basis for M = R
n
(p 72). The next theorem is so basic in linear algebra that it
is used without comment. Although the proof is easy, it should be worked carefully.
Theorem Suppose N is a module, M is a free module with basis S = ¦s
t
¦, and
f : M →N is a homomorphism. Let f(S) be the sequence ¦f(s
t
)¦ in N.
1) f(S) generates N iff f is surjective.
2) f(S) is independent in N iff f is injective.
3) f(S) is a basis for N iff f is an isomorphism.
4) If h : M →N is a homomorphism, then f = h iff f [ S = h [ S.
Exercise Let (A
1
, .., A
n
) be a sequence of n vectors with each A
i
∈ Z
n
.
Show this sequence is linearly independent over Z iff it is linearly independent over Q.
Is it true the sequence is linearly independent over Z iff it is linearly independent
over R? This question is difficult until we learn more linear algebra.
Characterization of Free Modules
Any free R-module is isomorphic to one of the canonical free R-modules ⊕R
t
.
This is just an observation, but it is a central fact in linear algebra.
Theorem A non-zero R-module M is free iff ∃ an index set T such that
M ≈
t∈T
R
t
. In particular, M has a finite free basis of n elements iff M ≈ R
n
.
Proof If M is isomorphic to ⊕R
t
then M is certainly free. So now suppose M
has a free basis ¦s
t
¦. Then the homomorphism f : M → ⊕R
t
with f(s
t
) = e
t
sends
the basis for M to the canonical basis for ⊕R
t
. By 3) in the preceding theorem, f is
an isomorphism.
80 Linear Algebra Chapter 5
Exercise Suppose R is a commutative ring, A ∈ R
n
, and the homomorphism
f : R
n
→ R
n
defined by f(B) = AB is surjective. Show f is an isomorphism, i.e.,
show A is invertible. This is a key theorem in linear algebra, although it is usually
stated only for the case where R is a field. Use the fact that ¦e
1
, .., e
n
¦ is a free basis
for R
n
.
The next exercise is routine, but still informative.
Exercise Let R = Z, A =
2 1 0
3 2 −5
and f: Z
3
→ Z
2
be the group homo-
morphism defined by A. Find a non-trivial linear combination of the columns of A
which is 0
¯
. Also find a non-zero element of kernel(f).
If R is a commutative ring, you can relate properties of R as an R-module to
properties of R as a ring.
Exercise Suppose R is a commutative ring and v ∈ R, v = 0
¯
.
1) v is independent iff v is .
2) v is a basis for R iff v generates R iff v is .
Note that 2) here is essentially the first exercise for the case n = 1. That is, if
f : R →R is a surjective R-module homomorphism, then f is an isomorphism.
Relating these concepts to matrices
The theorem stated below gives a summary of results we have already had. It
shows that certain concepts about matrices, linear independence, injective homo-
morphisms, and solutions of equations, are all the same — they are merely stated in
different language. Suppose A ∈ R
m,n
and f : R
n
→R
m
is the homomorphism associ-
ated with A, i.e., f(B) = AB. Let v
1
, .., v
n
∈ R
m
be the columns of A, i.e., f(e
i
) = v
i
= column i of A. Let B =
¸
¸
b
1
.
b
n
¸
represent an element of R
n
and C =
¸
¸
c
1
.
c
m
¸
Chapter 5 Linear Algebra 81
represent an element of R
m
.
Theorem
1) The element f(B) is a linear combination of the columns of A, that is
f(B) = f(e
1
b
1
+ +e
n
b
n
) = v
1
b
1
+ +v
n
b
n
. Thus the image of f
is generated by the columns of A. (See bottom of page 89.)
2) ¦v
1
, .., v
n
¦ generates R
m
iff f is surjective iff (for any C ∈ R
m
, AX = C
has a solution).
3) ¦v
1
, .., v
n
¦ is independent iff f is injective iff AX = 0
¯
has a unique
solution iff (∃ C ∈ R
m
such that AX = C has a unique solution).
4) ¦v
1
, .., v
n
¦ is a basis for R
m
iff f is an isomorphism iff (for any C ∈ R
m
,
AX = C has a unique solution).
Relating these concepts to square matrices
We now look at the preceding theorem in the special case where n = m and R
is a commutative ring. So far in this chapter we have just been cataloging. Now we
prove something more substantial, namely that if f : R
n
→ R
n
is surjective, then f
is injective. Later on we will prove that if R is a field, injective implies surjective.
Theorem Suppose R is a commutative ring, A ∈ R
n
, and f : R
n
→R
n
is defined
by f(B) = AB. Let v
1
, .., v
n
∈ R
n
be the columns of A, and w
1
, .., w
n
∈ R
n
= R
1,n
be the rows of A. Then the following are equivalent.
1) f is an automorphism.
2) A is invertible, i.e., [ A [ is a unit in R.
3) ¦v
1
, .., v
n
¦ is a basis for R
n
.
4) ¦v
1
, .., v
n
¦ generates R
n
.
5) f is surjective.
2
t
) A
t
is invertible, i.e., [ A
t
[ is a unit in R.
3
t
) ¦w
1
, .., w
n
¦ is a basis for R
n
.
82 Linear Algebra Chapter 5
4
t
) ¦w
1
, .., w
n
¦ generates R
n
.
Proof Suppose 5) is true and show 2). Since f is onto, ∃ u
1
, ..., u
n
∈ R
n
with
f(u
i
) = e
i
. Let g : R
n
→R
n
be the homomorphism satisfying g(e
i
) = u
i
. Then f ◦ g
is the identity. Now g comes from some matrix D and thus AD = I. This shows that
A has a right inverse and is thus invertible. Recall that the proof of this fact uses
determinant, which requires that R be commutative (see the exercise on page 64).
We already know the first three properties are equivalent, 4) and 5) are equivalent,
and 3) implies 4). Thus the first five are equivalent. Furthermore, applying this
result to A
t
shows that the last three properties are equivalent to each other. Since
[ A [=[ A
t
[, 2) and 2
t
) are equivalent.
Uniqueness of Dimension
There exists a ring R with R
2
≈ R
3
as R-modules, but this is of little interest. If
R is commutative, this is impossible, as shown below. First we make a convention.
Convention For the remainder of this chapter, R will be a commutative ring.
Theorem If f : R
m
→R
n
is a surjective R-module homomorphism, then m ≥ n.
Proof Suppose k = n − m is positive. Define h : (R
m
⊕ R
k
= R
n
) → R
n
by
h(u, v) = f(u). Then h is a surjective homomorphism, and by the previous section,
also injective. This is a contradiction and thus m ≥ n.
Corollary If f : R
m
→R
n
is an isomorphism, then m = n.
Proof Each of f and f
−1
is surjective, so m = n by the previous theorem.
Corollary If ¦v
1
, .., v
m
¦ generates R
n
, then m ≥ n.
Proof The hypothesis implies there is a surjective homomorphism R
m
→R
n
. So
this follows from the first theorem.
Lemma Suppose M is a f.g. module (i.e., a finite generated R-module). Then
if M has a basis, that basis is finite.
Chapter 5 Linear Algebra 83
Proof Suppose U ⊂ M is a finite generating set and S is a basis. Then any
element of U is a finite linear combination of elements of S, and thus S is finite.
Theorem Suppose M is a f.g. module. If M has a basis, that basis is finite
and any other basis has the same number of elements. This number is denoted by
dim(M), the dimension of M. (By convention, 0
¯
is a free module of dimension 0.)
Proof By the previous lemma, any basis for M must be finite. M has a basis of
n elements iff M ≈ R
n
. The result follows because R
n
≈ R
m
iff n = m.
Change of Basis
Before changing basis, we recall what a basis is. Previously we defined generat-
ing, independence, and basis for sequences, not for collections. For the concept of
generating it matters not whether you use sequences or collections, but for indepen-
dence and basis, you must use sequences. Consider the columns of the real matrix
A =
2 3 2
1 4 1
. If we consider the column vectors of A as a collection, there are
only two of them, yet we certainly don't wish to say the columns of A form a basis for
R
2
. In a set or collection, there is no concept of repetition. In order to make sense,
we must consider the columns of A as an ordered triple of vectors, and this sequence
is dependent. In the definition of basis on page 78, basis is defined for sequences, not
for sets or collections.
Two sequences cannot begin to be equal unless they have the same index set.
Here we follow the classical convention that an index set with n elements will be
¦1, 2, .., n¦, and thus a basis for M with n elements is a sequence S = ¦u
1
, .., u
n
¦
or if you wish, S = (u
1
, .., u
n
) ∈ M
n
. Suppose M is an R-module with a basis of
n elements. Recall there is a bijection α : Hom
R
(R
n
, M) → M
n
defined by α(h) =
(h(e
1
), .., h(e
n
)). Now h : R
n
→M is an isomorphism iff α(h) is a basis for M.
Summary The point of all this is that selecting a basis of n elements for M
is the same as selecting an isomorphism from R
n
to M, and from this viewpoint,
change of basis can be displayed by the diagram below.
Endomorphisms on R
n
are represented by square matrices, and thus have a de-
terminant and trace. Now suppose M is a f.g. free module and f : M → M is a
homomorphism. In order to represent f by a matrix, we must select a basis for M
(i.e., an isomorphism with R
n
). We will show that this matrix is well defined up to
similarity, and thus the determinant, trace, and characteristic polynomial of f are
well-defined.
84 Linear Algebra Chapter 5
Definition Suppose M is a free module, S = ¦u
1
, .., u
n
¦ is a basis for M, and
f : M → M is a homomorphism. The matrix A = (a
i,j
) ∈ R
n
of f w.r.t. the basis
S is defined by f(u
i
) = u
1
a
1,i
+ +u
n
a
n,i
. (Note that if M = R
n
and u
i
= e
i
, A is
the usual matrix associated with f).
Theorem Suppose T = ¦v
1
, .., v
n
¦ is another basis for M and B ∈ R
n
is the
matrix of f w.r.t. T. Define C = (c
i,j
) ∈ R
n
by v
i
= u
1
c
1,i
+ +u
n
c
n,i
. Then C is
invertible and B = C
−1
AC, i.e., A and B are similar. Therefore [A[ = [B[,
trace(A)=trace(B), and A and B have the same characteristic polynomial (see page
66 of chapter 4).
Conversely, suppose C = (c
i,j
) ∈ R
n
is invertible. Define T = ¦v
1
, .., v
n
¦ by
v
i
= u
1
c
1,i
+ +u
n
c
n,i
. Then T is a basis for M and the matrix of f w.r.t. T is
B = C
−1
AC. In other words, conjugation of matrices corresponds to change of basis.
Proof The proof follows by seeing that the following diagram is commutative.
R
n
R
n
R
n
R
n
M M C C
A
B
≈ ≈
≈ ≈
≈ ≈
e
i
v
i
e
i
u
i
v
i
e
i
u
i
e
i
f
· ·
´
»
»
´
»
´
´
»
The diagram also explains what it means for A to be the matrix of f w.r.t. the
basis S. Let h : R
n
→ M be the isomorphism with h(e
i
) = u
i
for 1 ≤ i ≤ n. Then
the matrix A ∈ R
n
is the one determined by the endomorphism h
−1
◦f ◦h : R
n
→R
n
.
In other words, column i of A is h
−1
(f(h(e
i
))).
An important special case is where M = R
n
and f : R
n
→ R
n
is given by some
matrix W. Then h is given by the matrix U whose i
th
column is u
i
and A =
U
−1
WU. In other words, W represents f w.r.t. the standard basis, and U
−1
WU
represents f w.r.t. the basis ¦u
1
, .., u
n
¦.
Definition Suppose M is a f.g. free module and f : M →M is a homomorphism.
Define [f[ to be [A[, trace(f) to be trace(A), and CP
f
(x) to be CP
A
(x), where A is
Chapter 5 Linear Algebra 85
the matrix of f w.r.t. some basis. By the previous theorem, all three are well-defined,
i.e., do not depend upon the choice of basis.
Exercise Let R = Z and f : Z
2
→ Z
2
be defined by f(D) =
3 3
0 −1
D.
Find the matrix of f w.r.t. the basis
2
1
,
3
1
¸
.
Exercise Let L ⊂ R
2
be the line L = ¦(r, 2r)
t
: r ∈ R¦. Show there is one
and only one homomorphism f : R
2
→ R
2
which is the identity on L and has
f((−1, 1)
t
) = (1, −1)
t
. Find the matrix A ∈ R
2
which represents f with respect
to the basis ¦(1, 2)
t
, (−1, 1)
t
¦. Find the determinant, trace, and characteristic
polynomial of f. Also find the matrix B ∈ R
2
which represents f with respect to
the standard basis. Finally, find an invertible matrix C ∈ R
2
with B = C
−1
AC.
Vector Spaces
So far in this chapter we have been developing the theory of linear algebra in
general. The previous theorem, for example, holds for any commutative ring R, but
it must be assumed that the module M is free. Endomorphisms in general will not
have a determinant, trace, or characteristic polynomial. We now focus on the case
where R is a field F, and show that in this case, every F-module is free. Thus any
finitely generated F-module will have a well-defined dimension, and endomorphisms
on it will have well-defined determinant, trace, and characteristic polynomial.
In this section, F is a field. F-modules may also be called vector spaces and
F-module homomorphisms may also be called linear transformations.
Theorem Suppose M is an F-module and v ∈ M. Then v = 0
¯
iff v is independent.
That is, if v ∈ V and r ∈ F, vr = 0
¯
implies v = 0
¯
in M or r = 0
¯
in F.
Proof Suppose vr = 0
¯
and r = 0
¯
. Then 0
¯
= (vr)r
−1
= v1
¯
= v.
Theorem Suppose M = 0
¯
is an F-module and v ∈ M. Then v generates M iff v
is a basis for M. Furthermore, if these conditions hold, then M ≈ F
F
, any non-zero
element of M is a basis, and any two elements of M are dependent.
86 Linear Algebra Chapter 5
Proof Suppose v generates M. Then v = 0
¯
and is thus independent by the
previous theorem. In this case M ≈ F, and any non-zero element of F is a basis, and
any two elements of F are dependent.
Theorem Suppose M = 0
¯
is a finitely generated F-module. If S = ¦v
1
, .., v
m
¦
generates M, then any maximal independent subsequence of S is a basis for M. Thus
any finite independent sequence can be extended to a basis. In particular, M has a
finite free basis, and thus is a free F-module.
Proof Suppose, for notational convenience, that ¦v
1
, .., v
n
¦ is a maximal inde-
pendent subsequence of S, and n < i ≤ m. It must be shown that v
i
is a linear
combination of ¦v
1
, .., v
n
¦. Since ¦v
1
, .., v
n
, v
i
¦ is dependent, ∃ r
1
, ..., r
n
, r
i
not all
zero, such that v
1
r
1
+ +v
n
r
n
+v
i
r
i
= 0
¯
. Then r
i
= 0
¯
and v
i
= −(v
1
r
1
+ +v
n
r
n
)r
−1
i
.
Thus ¦v
1
, .., v
n
¦ generates S and thus all of M. Now suppose T is a finite indepen-
dent sequence. T may be extended to a finite generating sequence, and inside that
sequence it may be extended to a maximal independent sequence. Thus T extends
to a basis.
After so many routine theorems, it is nice to have one with real power. It not
only says any finite independent sequence can be extended to a basis, but it can be
extended to a basis inside any finite generating set containing it. This is one of the
theorems that makes linear algebra tick. The key hypothesis here is that the ring
is a field. If R = Z, then Z is a free module over itself, and the element 2 of Z is
independent. However it certainly cannot be extended to a basis. Also the finiteness
hypothesis in this theorem is only for convenience, as will be seen momentarily.
Since F is a commutative ring, any two bases of M must have the same number
of elements, and thus the dimension of M is well defined (see theorem on page 83).
Theorem Suppose M is an F-module of dimension n, and ¦v
1
, ..., v
m
¦ is an
independent sequence in M. Then m ≤ n and if m = n, ¦v
1
, .., v
m
¦ is a basis.
Proof ¦v
1
, .., v
m
¦ extends to a basis with n elements.
The next theorem is just a collection of observations.
Theorem Suppose M and N are finitely generated F-modules.
Chapter 5 Linear Algebra 87
1) M ≈ F
n
iff dim(M) = n.
2) M ≈ N iff dim(M) = dim(N).
3) F
m
≈ F
n
iff n = m.
4) dim(M ⊕N) = dim(M) + dim(N).
Here is the basic theorem for vector spaces in full generality.
Theorem Suppose M = 0
¯
is an F-module and S = ¦v
t
¦
t∈T
generates M.
1) Any maximal independent subsequence of S is a basis for M.
2) Any independent subsequence of S may be extended to a maximal
independent subsequence of S, and thus to a basis for M.
3) Any independent subsequence of M can be extended to a basis for M.
In particular, M has a free basis, and thus is a free F-module.
Proof The proof of 1) is the same as in the case where S is finite. Part 2) will
follow from the Hausdorff Maximality Principle. An independent subsequence of S is
contained in a maximal monotonic tower of independent subsequences. The union of
these independent subsequences is still independent, and so the result follows. Part
3) follows from 2) because an independent sequence can always be extended to a
generating sequence.
Theorem Suppose M is an F-module and K ⊂ M is a submodule.
1) K is a summand of M, i.e., ∃ a submodule L of M with K ⊕L = M.
2) If M is f.g., then dim(K) ≤ dim(M) and K = M iff dim(K) = dim(M).
Proof Let T be a basis for K. Extend T to a basis S for M. Then S−T generates
a submodule L with K ⊕L = M. Part 2) follows from 1).
Corollary Q is a summand of R. In other words, ∃ a Q-submodule V ⊂ R
with Q⊕V = R as Q-modules. (See exercise on page 77.)
Proof Q is a field, R is a Q-module, and Q is a submodule of R.
Corollary Suppose M is a f.g. F-module, N is an F-module, and f : M → N
is a homomorphism. Then dim(M) = dim(ker(f)) + dim(image(f)).
88 Linear Algebra Chapter 5
Proof Let K = ker(f) and L ⊂ M be a submodule with K ⊕ L = M. Then
f [ L : L →image(f) is an isomorphism.
Exercise Suppose R is a domain with the property that, for R-modules, every
submodule is a summand. Show R is a field.
Exercise Find a free Z-module which has a generating set containing no basis.
Exercise The real vector space R
2
is generated by the sequence S =
¦(π, 0), (2, 1), (3, 2)¦. Show there are three maximal independent subsequences of
S, and each is a basis for R
2
. (Row vectors are used here just for convenience.)
The real vector space R
3
is generated by S = ¦(1, 1, 2), (1, 2, 1), (3, 4, 5), (1, 2, 0)¦.
Show there are three maximal independent subsequences of S and each is a basis
for R
3
. You may use determinant.
Square matrices over fields
This theorem is just a summary of what we have for square matrices over fields.
Theorem Suppose A ∈ F
n
and f : F
n
→ F
n
is defined by f(B) = AB. Let
v
1
, .., v
n
∈ F
n
be the columns of A, and w
1
, .., w
n
∈ F
n
= F
1,n
be the rows of A.
Then the following are equivalent.
1) ¦v
1
, .., v
n
¦ is independent, i.e., f is injective.
2) ¦v
1
, .., v
n
¦ is a basis for F
n
, i.e., f is an automorphism, i.e., A is
invertible, i.e., [ A [ = 0
¯
.
3) ¦v
1
, .., v
n
¦ generates F
n
, i.e., f is surjective.
1
t
) ¦w
1
, .., w
n
¦ is independent.
2
t
) ¦w
1
, .., w
n
¦ is a basis for F
n
, i.e., A
t
is invertible, i.e., [ A
t
[ = 0
¯
.
3
t
) ¦w
1
, .., w
n
¦ generates F
n
.
Chapter 5 Linear Algebra 89
Proof Except for 1) and 1
t
), this theorem holds for any commutative ring R.
(See the section Relating these concepts to square matrices, pages 81 and 82.)
Parts 1) and 1
t
) follow from the preceding section.
Exercise Add to this theorem more equivalent statements in terms of solutions
of n equations in n unknowns.
Overview Suppose each of X and Y is a set with n elements and f : X →Y is a
function. By the pigeonhole principle, f is injective iff f is bijective iff f is surjective.
Now suppose each of U and V is a vector space of dimension n and f : U → V is
a linear transformation. It follows from the work done so far that f is injective iff
f is bijective iff f is surjective. This shows some of the simple and definitive nature
of linear algebra.
Exercise Let A = (A
1
, .., A
n
) be an nn matrix over Z with column i = A
i
∈
Z
n
. Let f : Z
n
→ Z
n
be defined by f(B) = AB and
¯
f : R
n
→ R
n
be defined by
¯
f(C) = AC. Show the following are equivalent. (See the exercise on page 79.)
1) f : Z
n
→Z
n
is injective.
2) The sequence (A
1
, .., A
n
) is linearly independent over Z.
3) [A[ = 0.
4)
¯
f : R
n
→R
n
is injective.
5) The sequence (A
1
, .., A
n
) is linearly independent over R.
Rank of a matrix Suppose A ∈ F
m,n
. The row (column) rank of A is defined
to be the dimension of the submodule of F
n
(F
m
) generated by the rows (columns)
of A.
Theorem If C ∈ F
m
and D ∈ F
n
are invertible, then the row (column) rank of
A is the same as the row (column) rank of CAD.
Proof Suppose f : F
n
→ F
m
is defined by f(B) = AB. Each column of A
is a vector in the range F
m
, and we know from page 81 that each f(B) is a linear
90 Linear Algebra Chapter 5
combination of those vectors. Thus the image of f is the submodule of F
m
generated
by the columns of A, and its dimension is the column rank of A. This dimension
is the same as the dimension of the image of g ◦ f ◦ h : F
n
→ F
m
, where h is any
automorphism on F
n
and g is any automorphism on F
m
. This proves the theorem
for column rank. The theorem for row rank follows using transpose.
Theorem If A ∈ F
m,n
, the row rank and the column rank of A are equal. This
number is called the rank of A and is ≤ min¦m, n¦.
Proof By the theorem above, elementary row and column operations change
neither the row rank nor the column rank. By row and column operations, A may be
changed to a matrix H where h
1,1
= = h
t,t
= 1
¯
and all other entries are 0
¯
(see the
first exercise on page 59). Thus row rank = t = column rank.
Exercise Suppose A has rank t. Show that it is possible to select t rows and t
columns of A such that the determined t t matrix is invertible. Show that the rank
of A is the largest integer t such that this is possible.
Exercise Suppose A ∈ F
m,n
has rank t. What is the dimension of the solution
set of AX = 0
¯
?
Definition If N and M are finite dimensional vector spaces and f : N →M is a
linear transformation, the rank of f is the dimension of the image of f. If f : F
n
→F
m
is given by a matrix A, then the rank of f is the same as the rank of the matrix A.
Geometric Interpretation of Determinant
Suppose V ⊂ R
n
is some nice subset. For example, if n = 2, V might be the
interior of a square or circle. There is a concept of the n-dimensional volume of V .
For n = 1, it is length. For n = 2, it is area, and for n = 3 it is "ordinary volume".
Suppose A ∈ R
n
and f : R
n
→R
n
is the homomorphism given by A. The volume of
V does not change under translation, i.e., V and V +p have the same volume. Thus
f(V ) and f(V +p) = f(V ) +f(p) have the same volume. In street language, the next
theorem says that "f multiplies volume by the absolute value of its determinant".
Theorem The n-dimensional volume of f(V ) is ±[A[(the n-dimensional volume
of V ). Thus if [A[ = ±1, f preserves volume.
Chapter 5 Linear Algebra 91
Proof If [A[ = 0, image(f) has dimension < n and thus f(V ) has n-dimensional
volume 0. If [A[ = 0 then A is the product of elementary matrices (see page 59)
and for elementary matrices, the theorem is obvious. The result follows because the
determinant of the composition is the product of the determinants.
Corollary If P is the n-dimensional parallelepiped determined by the columns
v
1
, .. , v
n
of A, then the n-dimensional volume of P is ±[A[.
Proof Let V = [0, 1] [0, 1] = ¦e
1
t
1
+ +e
n
t
n
: 0 ≤ t
i
≤ 1¦. Then
P = f(V ) = ¦v
1
t
1
+ +v
n
t
n
: 0 ≤ t
i
≤ 1¦.
Linear functions approximate differentiable functions locally
We continue with the special case F = R. Linear functions arise naturally in
business, science, and mathematics. However this is not the only reason that linear
algebra is so useful. It is a central fact that smooth phenomena may be approx-
imated locally by linear phenomena. Without this great simplification, the world
of technology as we know it today would not exist. Of course, linear transforma-
tions send the origin to the origin, so they must be adjusted by a translation. As
a simple example, suppose h : R → R is differentiable and p is a real number. Let
f : R →R be the linear transformation f(x) = h
V
1dxdy. From the previous section we know that
any homomorphism f multiplies area by [ f [. The student may now understand
the following theorem from calculus. (Note that if h is the restriction of a linear
transformation from R
2
to R
2
, this theorem is immediate from the previous section.)
Theorem Suppose the determinant of J(h)(x, y) is non-negative for each
(x, y) ∈ V . Then the area of h(V ) is
V
[ J(h) [ dxdy.
92 Linear Algebra Chapter 5
The Transpose Principle
We now return to the case where F is a field (of arbitrary characteristic). F-
modules may also be called vector spaces and submodules may be called subspaces.
The study of R-modules in general is important and complex. However the study of
F-modules is short and simple – every vector space is free and every subspace is a
summand. The core of classical linear algebra is not the study of vector spaces, but
the study of homomorphisms, and in particular, of endomorphisms. One goal is to
show that if f : V → V is a homomorphism with some given property, there exists
a basis of V so that the matrix representing f displays that property in a prominent
manner. The next theorem is an illustration of this.
Theorem Let F be a field and n be a positive integer.
1) Suppose V is an n-dimensional vector space and f : V →V is a
homomorphism with [f[ = 0
¯
. Then ∃ a basis of V such that the matrix
representing f has its first row zero.
2) Suppose A ∈ F
n
has [A[ = 0
¯
. Then ∃ an invertible matrix C such that
C
−1
AC has its first row zero.
3) Suppose V is an n-dimensional vector space and f : V →V is a
homomorphism with [f[ = 0. Then ∃ a basis of V such that the matrix
representing f has its first column zero.
4) Suppose A ∈ F
n
has [A[ = 0
¯
. Then ∃ an invertible matrix D such that
D
−1
AD has its first column zero.
We first wish to show that these 4 statements are equivalent. We know that
1) and 2) are equivalent and also that 3) and 4) are equivalent because change of
basis corresponds to conjugation of the matrix. Now suppose 2) is true and show
4) is true. Suppose [A[ = 0
¯
. Then [A
t
[ = 0
¯
and by 2) ∃ C such that C
−1
A
t
C has
first row zero. Thus (C
−1
A
t
C)
t
= C
t
A(C
t
)
−1
has first row column zero. The result
follows by defining D = (C
t
)
−1
. Also 4) implies 2).
This is an example of the transpose principle. Loosely stated, it is that theorems
about change of basis correspond to theorems about conjugation of matrices and
theorems about the rows of a matrix correspond to theorems about the columns of a
matrix, using transpose. In the remainder of this chapter, this will be used without
further comment.
Chapter 5 Linear Algebra 93
Proof of the theorem We are free to select any of the 4 parts, and we select
part 3). Since [ f [= 0, f is not injective and ∃ a non-zero v
1
∈ V with f(v
1
) = 0
¯
.
Now v
1
is independent and extends to a basis ¦v
1
, .., v
n
¦. Then the matrix of f w.r.t
this basis has first column zero.
Exercise Let A =
and find an invertible matrix D ∈ R
3
so that D
−1
AD has first column zero.
Exercise Suppose M is an n-dimensional vector space over a field F, k is an
integer with 0 < k < n, and f : M → M is an endomorphism of rank k. Show
there is a basis for M so that the matrix representing f has its first n −k rows zero.
Also show there is a basis for M so that the matrix representing f has its first n −k
columns zero. Work these out directly without using the transpose principle.
Nilpotent Homomorphisms
In this section it is shown that an endomorphism f is nilpotent iff all of its char-
acteristic roots are 0
¯
iff it may be represented by a strictly upper triangular matrix.
Definition An endomorphism f : V →V is nilpotent if ∃ m with f
m
= 0
¯
. Any
f represented by a strictly upper triangular matrix is nilpotent (see page 56).
Theorem Suppose V is an n-dimensional vector space and f : V → V is a
nilpotent homomorphism. Then f
n
= 0
¯
and ∃ a basis of V such that the matrix
representing f w.r.t. this basis is strictly upper triangular. Thus the characteristic
polynomial of f is CP
f
(x) = x
n
.
Proof Suppose f = 0
¯
is nilpotent. Let t be the largest positive integer with
f
t
= 0
¯
. Then f
t
(V ) ⊂ f
t−1
(V ) ⊂ ⊂ f(V ) ⊂ V . Since f is nilpotent, all of these
inclusions are proper. Therefore t < n and f
n
= 0
¯
. Construct a basis for V by
starting with a basis for f
t
(V ), extending it to a basis for f
t−1
(V ), etc. Then the
matrix of f w.r.t. this basis is strictly upper triangular.
Note To obtain a matrix which is strictly lower triangular, reverse the order of
the basis.
94 Linear Algebra Chapter 5
Exercise Use the transpose principle to write 3 other versions of this theorem.
Theorem Suppose V is an n-dimensional vector space and f : V → V is a
homomorphism. Then f is nilpotent iff CP
f
(x) = x
n
. (See the exercise at the end
of Chapter 4 for the case n = 2.)
Proof Suppose CP
f
(x) = x
n
. For n = 1 this implies f = 0
¯
, so suppose n > 1.
Since the constant term of CP
f
(x) is 0
¯
, the determinant of f is 0
¯
. Thus ∃ a basis
of V such that the matrix A representing f has its first column zero. Let B ∈ F
n−1
be the matrix obtained from A by removing its first row and first column. Now
CP
A
(x) = x
n
= xCP
B
(x). Thus CP
B
(x) = x
n−1
and by induction on n, B is
nilpotent and so ∃ C such that C
−1
BC is strictly upper triangular. Then
¸
¸
¸
¸
¸
¸
¸
1 0 0
0
C
−1
0
¸
¸
¸
¸
¸
¸
¸
¸
0 ∗ ∗
B
0
¸
¸
¸
¸
¸
¸
¸
¸
1 0 0
0
C
0
¸
=
¸
¸
¸
¸
¸
¸
¸
0 ∗ ∗
0
C
−1
BC
0
¸
is strictly upper triangular.
Exercise Suppose F is a field, A ∈ F
3
is a strictly lower triangular matrix of
rank 2, and B =
¸
¸
0 0 0
1 0 0
0 1 0
¸
. Using conjugation by elementary matrices, show there
is an invertible matrix C so that C
−1
AC = B. Now suppose V is a 3-dimensional
vector space and f : V →V is a nilpotent endomorphism of rank 2. We know f can
be represented by a strictly lower triangular matrix. Show there is a basis ¦v
1
, v
2
, v
3
¦
for V so that B is the matrix representing f. Also show that f(v
1
) = v
2
, f(v
2
) = v
3
,
and f(v
3
) = 0
¯
. In other words, there is a basis for V of the form ¦v, f(v), f
2
(v)¦
with f
3
(v) = 0
¯
.
Exercise Suppose V is a 3-dimensional vector space and f : V →V is a nilpotent
endomorphism of rank 1. Show there is a basis for V so that the matrix representing
f is
¸
¸
0 0 0
1 0 0
0 0 0
¸
.
Chapter 5 Linear Algebra 95
Eigenvalues
Our standing hypothesis is that V is an n-dimensional vector space over a field F
and f : V →V is a homomorphism.
Definition An element λ ∈ F is an eigenvalue of f if ∃ a non-zero v ∈ V with
f(v) = λv. Any such v is called an eigenvector. E
λ
⊂ V is defined to be the set of
all eigenvectors for λ (plus 0
¯
). Note that E
λ
= ker(λI − f) is a subspace of V . The
next theorem shows the eigenvalues of f are just the characteristic roots of f.
Theorem If λ ∈ F then the following are equivalent.
1) λ is an eigenvalue of f, i.e., (λI −f) : V →V is not injective.
2) [ (λI −f) [= 0
¯
.
3) λ is a characteristic root of f, i.e., a root of the characteristic
polynomial CP
f
(x) =[ (xI −A) [, where A is any matrix representing f.
Proof It is immediate that 1) and 2) are equivalent, so let's show 2) and 3)
are equivalent. The evaluation map F[x] → F which sends h(x) to h(λ) is a ring
homomorphism (see theorem on page 47). So evaluating (xI − A) at x = λ and
taking determinant gives the same result as taking the determinant of (xI − A) and
evaluating at x = λ. Thus 2) and 3) are equivalent.
The nicest thing you can say about a matrix is that it is similar to a diagonal
matrix. Here is one case where that happens.
Theorem Suppose λ
1
, .., λ
k
are distinct eigenvalues of f, and v
i
is an eigenvector
of λ
i
for 1 ≤ i ≤ k. Then the following hold.
1) ¦v
1
, .., v
k
¦ is independent.
2) If k = n, i.e., if CP
f
(x) = (x −λ
1
) (x −λ
n
), then ¦v
1
, .., v
n
¦ is a
basis for V . The matrix of f w.r.t. this basis is the diagonal matrix whose
(i, i) term is λ
i
.
Proof Suppose ¦v
1
, .., v
k
¦ is dependent. Suppose t is the smallest positive integer
such that ¦v
1
, .., v
t
¦ is dependent, and v
1
r
1
+ +v
t
r
t
= 0
¯
is a non-trivial linear
combination. Note that at least two of the coefficients must be non-zero. Now
(f −λ
t
)(v
1
r
1
+ +v
t
r
t
) = v
1
(λ
1
−λ
t
)r
1
+ +v
t−1
(λ
t−1
−λ
t
)r
t−1
+0
¯
= 0
¯
is a shorter
96 Linear Algebra Chapter 5
non-trivial linear combination. This is a contradiction and proves 1). Part 2) follows
from 1) because dim(V ) = n.
Exercise Let A =
0 1
−1 0
∈ R
2
. Find an invertible C ∈ C
2
such that
C
−1
AC is diagonal. Show that C cannot be selected in R
2
. Find the characteristic
polynomial of A.
Exercise Suppose V is a 3-dimensional vector space and f : V →V is an endo-
morphism with CP
f
(x) = (x−λ)
3
. Show that (f −λI) has characteristic polynomial
x
3
and is thus a nilpotent endomorphism. Show there is a basis for V so that the
matrix representing f is
¸
¸
λ 0 0
1 λ 0
0 1 λ
¸
,
¸
¸
λ 0 0
1 λ 0
0 0 λ
¸
or
¸
¸
λ 0 0
0 λ 0
0 0 λ
¸
.
We could continue and finally give an ad hoc proof of the Jordan canonical form,
but in this chapter we prefer to press on to inner product spaces. The Jordan form
will be developed in Chapter 6 as part of the general theory of finitely generated
modules over Euclidean domains. The next section is included only as a convenient
reference.
Jordan Canonical Form
This section should be just skimmed or omitted entirely. It is unnecessary for the
rest of this chapter, and is not properly part of the flow of the chapter. The basic
facts of Jordan form are summarized here simply for reference.
The statement that a square matrix B over a field F is a Jordan block means that
∃ λ ∈ F such that B is a lower triangular matrix of the form
B =
∈ R
n
has f(C) = b. Then f
−1
(b) is the set of all solutions to a
1
x
1
+ +a
n
x
n
= b which
is the coset L+C, and this the set of all solutions to a
1
(x
1
−c
1
) + +a
n
(x
n
−c
n
) = 0.
Gram-Schmidt orthonormalization
Theorem (Fourier series) Suppose W is an IPS with an orthonormal basis
¦w
1
, .., w
n
¦. Then if v ∈ W, v = w
1
(v w
1
) + +w
n
(v w
n
).
Proof v = w
1
r
1
+ +w
n
r
n
and v w
i
= (w
1
r
1
+ +w
n
r
n
) w
i
= r
i
Theorem Suppose W is an IPS, Y ⊂ W is a subspace with an orthonormal basis
¦w
1
, .., w
k
¦, and v ∈ W−Y . Define the projection of v onto Y by p(v) = w
1
(v w
1
)+
+w
k
(vw
k
), and let w = v−p(v). Then (ww
i
) = (v−w
1
(vw
1
)−w
k
(vw
k
))w
i
= 0.
Thus if w
k+1
=
w
|w|
, then ¦w
1
, .., w
k+1
¦ is an orthonormal basis for the subspace
generated by ¦w
1
, .., w
k
, v¦. If ¦w
1
, .., w
k
, v¦ is already orthonormal, w
k+1
= v.
Theorem (Gram-Schmidt) Suppose W is an IPS with a basis ¦v
1
, .., v
n
¦.
Then W has an orthonormal basis ¦w
1
, .., w
n
¦. Moreover, any orthonormal sequence
in W extends to an orthonormal basis of W.
Proof Let w
1
=
v
1
|v
1
|
. Suppose inductively that ¦w
1
, .., w
k
¦ is an orthonormal
basis for Y , the subspace generated by ¦v
1
, .., v
k
¦. Let w = v
k+1
− p(v
k+1
) and
Chapter 5 Linear Algebra 101
w
k+1
=
w
|w|
. Then by the previous theorem, ¦w
1
, .., w
k+1
¦ is an orthonormal basis
for the subspace generated by ¦w
1
, .., w
k
, v
k+1
¦. In this manner an orthonormal basis
for W is constructed. Notice that this construction defines a function h which sends
a basis for W to an orthonormal basis for W (see topology exercise on page 103).
Now suppose W has dimension n and ¦w
1
, .., w
k
¦ is an orthonormal sequence in
W. Since this sequence is independent, it extends to a basis ¦w
1
, .., w
k
, v
k+1
, .., v
n
¦.
The process above may be used to modify this to an orthonormal basis ¦w
1
, .., w
n
¦.
Exercise Let f : R
3
→ R be the homomorphism defined by the matrix (2,1,3).
Find an orthonormal basis for the kernel of f. Find the projection of (e
1
+ e
2
) onto
ker(f). Find the angle between e
1
+e
2
and the plane ker(f).
Exercise Let W = R
3
have the standard inner product and Y ⊂ W be the
subspace generated by ¦w
1
, w
2
¦ where w
1
= (1, 0, 0)
t
and w
2
= (0, 1, 0)
t
. W is
generated by the sequence ¦w
1
, w
2
, v¦ where v = (1, 2, 3)
t
. As in the first theorem
of this section, let w = v − p(v), where p(v) is the projection of v onto Y , and set
w
3
=
w
|w|
. Find w
3
and show that for any t with 0 ≤ t ≤ 1, ¦w
1
, w
2
, (1 −t)v +tw
3
¦
is a basis for W. This is a key observation for an exercise on page 103 showing O(n)
is a deformation retract of GL
n
(R).
Isometries Suppose each of U and V is an IPS. A homomorphism f : U → V
is said to be an isometry provided it is an isomorphism and for any u
1
, u
2
in U,
(u
1
u
2
)
U
= (f(u
1
) f(u
2
))
V
.
Theorem Suppose each of U and V is an n-dimensional IPS, ¦u
1
, .., u
n
¦ is an
orthonormal basis for U, and f : U →V is a homomorphism. Then f is an isometry
iff ¦f(u
1
), .., f(u
n
)¦ is an orthonormal sequence in V .
Proof Isometries certainly preserve orthonormal sequences. So suppose T =
¦f(u
1
), .., f(u
n
)¦ is an orthonormal sequence in V . Then T is independent and thus
T is a basis for V and thus f is an isomorphism (see the second theorem on page 79).
It is easy to check that f preserves inner products.
We now come to one of the definitive theorems in linear algebra. It is that, up to
isometry, there is only one inner product space for each dimension.
102 Linear Algebra Chapter 5
Theorem Suppose each of U and V is an n-dimensional IPS. Then ∃ an isometry
f : U →V. In particular, U is isometric to R
n
with its standard inner product.
Proof There exist orthonormal bases ¦u
1
, .., u
n
¦ for U and ¦v
1
, .., v
n
¦ for V .
By the first theorem on page 79, there exists a homomorphism f : U → V with
f(u
i
) = v
i
, and by the previous theorem, f is an isometry.
Exercise Let f : R
3
→ R be the homomorphism defined by the matrix (2,1,3).
Find a linear transformation h : R
2
→R
3
which gives an isometry from R
2
to ker(f).
Orthogonal Matrices
As noted earlier, linear algebra is not so much the study of vector spaces as it is
the study of endomorphisms. We now wish to study isometries from R
n
to R
n
.
We know from a theorem on page 90 that an endomorphism preserves volume iff
its determinant is ±1. Isometries preserve inner product, and thus preserve angle and
distance, and so certainly preserve volume.
Theorem Suppose A ∈ R
n
and f : R
n
→ R
n
is the homomorphism defined by
f(B) = AB. Then the following are equivalent.
1) The columns of A form an orthonormal basis for R
n
, i.e., A
t
A = I.
2) The rows of A form an orthonormal basis for R
n
, i.e., AA
t
= I.
3) f is an isometry.
Proof A left inverse of a matrix is also a right inverse (see the exercise on
page 64). Thus 1) and 2) are equivalent because each of them says A is invert-
ible with A
−1
= A
t
. Now ¦e
1
, .., e
n
¦ is the canonical orthonormal basis for R
n
, and
f(e
i
) is column i of A. Thus by the previous section, 1) and 3) are equivalent.
Definition If A ∈ R
n
satisfies these three conditions, A is said to be orthogonal.
The set of all such A is denoted by O(n), and is called the orthogonal group.
Theorem
1) If A is orthogonal, [ A [= ±1.
2) If A is orthogonal, A
−1
is orthogonal. If A and C are orthogonal, AC is
orthogonal. Thus O(n) is a multiplicative subgroup of GL
n
(R).
Chapter 5 Linear Algebra 103
3) Suppose A is orthogonal and f is defined by f(B) = AB. Then f preserves
distances and angles. This means that if u, v ∈ R
n
, |u −v| =
|f(u)−f(v)| and the angle between u and v is equal to the angle between
f(u) and f(v).
Proof Part 1) follows from [A[
2
= [A[ [A
t
[ = [I[ = 1. Part 2) is imme-
diate, because isometries clearly form a subgroup of the multiplicative group of
all automorphisms. For part 3) assume f : R
n
→ R
n
is an isometry. Then
|u − v|
2
= (u − v) (u − v) = f(u − v) f(u − v) = |f(u − v)|
2
= |f(u) − f(v)|
2
.
The proof that f preserves angles follows from u v = |u||v|cosΘ.
Exercise Show that if A ∈ O(2) has [A[ = 1, then A =
cosΘ −sinΘ
sinΘ cosΘ
for
some number Θ. (See the exercise on page 56.)
Exercise (topology) Let R
n
≈ R
n
2
have its usual metric topology. This means
a sequence of matrices ¦A
i
¦ converges to A iff it converges coordinatewise. Show
GL
n
(R) is an open subset and O(n) is closed and compact. Let h : GL
n
(R) →
O(n) be defined by Gram-Schmidt. Show H : GL
n
(R) [0, 1] → GL
n
(R) defined
by H(A, t) = (1 −t)A+th(A) is a deformation retract of GL
n
(R) to O(n).
Diagonalization of Symmetric Matrices
We continue with the case F = R. Our goals are to prove that, if A is a symmetric
matrix, all of its eigenvalues are real and that ∃ an orthogonal matrix C such that
C
−1
AC is diagonal. As background, we first note that symmetric is the same as
self-adjoint.
Theorem Suppose A ∈ R
n
and u, v ∈ R
n
. Then (A
t
u) v = u (Av).
Proof If y, z ∈ R
n
, then the dot product y z, is the matrix product y
t
z, and
matrix multiplication is associative. Thus (A
t
u) v = (u
t
A)v = u
t
(Av) = u (Av).
Definition Suppose A ∈ R
n
. A is said to be symmetric provided A
t
= A. Note
that any diagonal matrix is symmetric. A is said to be self-adjoint if (Au)v = u(Av)
for all u, v ∈ R
n
. The next theorem is just an exercise using the previous theorem.
Theorem A is symmetric iff A is self-adjoint.
104 Linear Algebra Chapter 5
Theorem Suppose A ∈ R
n
is symmetric. Then ∃ real numbers λ
1
, .., λ
n
(not
necessarily distinct) such that CP
A
(x) = (x − λ
1
)(x − λ
2
) (x − λ
n
). That is, all
the eigenvalues of A are real.
Proof We know CP
A
(x) factors into linears over C. If µ = a + bi is a complex
number, its conjugate is defined by ¯ µ = a −bi. If h : C →C is defined by h(µ) = ¯ µ,
then h is a ring isomorphism which is the identity on R. If w = (a
i,j
) is a complex
matrix or vector, its conjugate is defined by ¯ w = (¯ a
i,j
). Since A ∈ R
n
is a real
symmetric matrix, A = A
t
=
¯
A
t
. Now suppose λ is a complex eigenvalue of A
and v ∈ C
n
is an eigenvector with Av = λv. Then λ(v
t
¯ v) = (λv)
t
¯ v = (Av)
t
¯ v =
(v
t
A)¯ v = v
t
(A¯ v) = v
t
(Av) = v
t
(λv) =
¯
λ(v
t
¯ v). Thus λ =
¯
λ and λ ∈ R. Or
you can define a complex inner product on C
n
by (w v) = w
t
¯ v. The proof then
reads as λ(v v) = (λv v) = (Av v) = (v Av) = (v λv) =
¯
λ(v v). Either way,
λ is a real number.
We know that eigenvectors belonging to distinct eigenvalues are linearly indepen-
dent. For symmetric matrices, we show more, namely that they are perpendicular.
Theorem Suppose A is symmetric, λ
1
, λ
2
∈ R are distinct eigenvalues of A, and
Au = λ
1
u and Av = λ
2
v. Then u v = 0.
Proof λ
1
(u v) = (Au) v = u (Av) = λ
2
(u v).
Review Suppose A ∈ R
n
and f : R
n
→ R
n
is defined by f(B) = AB. Then A
represents f w.r.t. the canonical orthonormal basis. Let S = ¦v
1
, .., v
n
¦ be another
basis and C ∈ R
n
be the matrix with v
i
as column i. Then C
−1
AC is the matrix
representing f w.r.t. S. Now S is an orthonormal basis iff C is an orthogonal matrix.
Summary Representing f w.r.t. an orthonormal basis is the same as conjugating
A by an orthogonal matrix.
Theorem Suppose A ∈ R
n
and C ∈ O(n). Then A is symmetric iff C
−1
AC
is symmetric.
Proof Suppose A is symmetric. Then (C
−1
AC)
t
= C
t
A(C
−1
)
t
= C
−1
AC.
The next theorem has geometric and physical implications, but for us, just the
incredibility of it all will suffice.
Chapter 5 Linear Algebra 105
Theorem If A ∈ R
n
, the following are equivalent.
1) A is symmetric.
2) ∃ C ∈ O(n) such that C
−1
AC is diagonal.
Proof By the previous theorem, 2) ⇒ 1). Show 1) ⇒ 2). Suppose A is a
symmetric 22 matrix. Let λ be an eigenvalue for A and ¦v
1
, v
2
¦ be an orthonormal
basis for R
2
with Av
1
= λv
1
. Then w.r.t this basis, the transformation determined
by A is represented by
λ b
0 d
. Since this matrix is symmetric, b = 0.
Now suppose by induction that the theorem is true for symmetric matrices in
R
t
for t < n, and suppose A is a symmetric n n matrix. Denote by λ
1
, .., λ
k
the
distinct eigenvalues of A, k ≤ n. If k = n, the proof is immediate, because then there
is a basis of eigenvectors of length 1, and they must form an orthonormal basis. So
suppose k < n. Let v
1
, .., v
k
be eigenvectors for λ
1
, .., λ
k
with each | v
i
|= 1. They
may be extended to an orthonormal basis v
1
, .., v
n
. With respect to this basis, the
transformation determined by A is represented by
¸
¸
¸
¸
¸
¸
¸
¸
¸
¸
¸
¸
¸
λ
1
λ
k
¸
(B)
(0) (D)
¸
.
Since this is a symmetric matrix, B = 0 and D is a symmetric matrix of smaller
size. By induction, ∃ an orthogonal C such that C
−1
DC is diagonal. Thus conjugating
by
I 0
0 C
makes the entire matrix diagonal.
This theorem is so basic we state it again in different terminology. If V is an IPS, a
linear transformation f : V →V is said to be self-adjoint provided (uf(v)) = (f(u)v)
for all u, v ∈ V .
Theorem If V is an n-dimensional IPS and f : V →V is a linear transformation,
then the following are equivalent.
1) f is self-adjoint.
2) ∃ an orthonormal basis ¦v
1
, ..., v
n
¦ for V with each
v
i
an eigenvector of f.
106 Linear Algebra Chapter 5
Exercise Let A =
2 2
2 2
. Find an orthogonal C such that C
−1
AC is diagonal.
Do the same for A =
2 1
1 2
.
Exercise Suppose A, D ∈ R
n
are symmetric. Under what conditions are A and D
similar? Show that, if A and D are similar, ∃ an orthogonal C such that D = C
−1
AC.
Exercise Suppose V is an n-dimensional real vector space. We know that V is
isomorphic to R
n
. Suppose f and g are isomorphisms from V to R
n
and A is a subset
of V . Show that f(A) is an open subset of R
n
iff g(A) is an open subset of R
n
. This
shows that V , an algebraic object, has a god-given topology. Of course, if V has
an inner product, it automatically has a metric, and this metric will determine that
same topology. Finally, suppose V and W are finite-dimensional real vector spaces
and h : V →W is a linear transformation. Show that h is continuous.
Exercise Define E : C
n
→C
n
by E(A) = e
A
= I +A+(1/2!)A
2
+. This series
converges and thus E is a well defined function. If AB = BA, then E(A + B) =
E(A)E(B). Since A and −A commute, I = E(0
¯
) = E(A − A) = E(A)E(−A), and
thus E(A) is invertible with E(A)
−1
= E(−A). Furthermore E(A
t
) = E(A)
t
, and
if C is invertible, E(C
−1
AC) = C
−1
E(A)C. Now use the results of this section to
prove the statements below. (For part 1, assume the Jordan form, i.e., assume any
A ∈ C
n
is similar to a lower triangular matrix.)
1) If A ∈ C
n
, then [ e
A
[= e
trace(A)
. Thus if A ∈ R
n
, [ e
A
[= 1
iff trace(A) = 0.
2) ∃ a non-zero matrix N ∈ R
2
with e
N
= I.
3) If N ∈ R
n
is symmetric, then e
N
= I iff N = 0
¯
.
4) If A ∈ R
n
and A
t
= −A, then e
A
∈ O(n).
Chapter 6
Appendix
The five previous chapters were designed for a year undergraduate course in algebra.
In this appendix, enough material is added to form a basic first year graduate course.
Two of the main goals are to characterize finitely generated abelian groups and to
prove the Jordan canonical form. The style is the same as before, i.e., everything is
right down to the nub. The organization is mostly a linearly ordered sequence except
for the last two sections on determinants and dual spaces. These are independent
sections added on at the end.
Suppose R is a commutative ring. An R-module M is said to be cyclic if it can
be generated by one element, i.e., M ≈ R/I where I is an ideal of R. The basic
theorem of this chapter is that if R is a Euclidean domain and M is a finitely generated
R-module, then M is the sum of cyclic modules. Thus if M is torsion free, it is a
free R-module. Since Z is a Euclidean domain, finitely generated abelian groups
are the sums of cyclic groups – one of the jewels of abstract algebra.
Now suppose F is a field and V is a finitely generated F-module. If T : V →V is
a linear transformation, then V becomes an F[x]-module by defining vx = T(v). Now
F[x] is a Euclidean domain and so V
F[x]
is the sum of cyclic modules. This classical
and very powerful technique allows an easy proof of the canonical forms. There is a
basis for V so that the matrix representing T is in Rational canonical form. If the
characteristic polynomial of T factors into the product of linear polynomials, then
there is a basis for V so that the matrix representing T is in Jordan canonical form.
This always holds if F = C. A matrix in Jordan form is a lower triangular matrix
with the eigenvalues of T displayed on the diagonal, so this is a powerful concept.
In the chapter on matrices, it is stated without proof that the determinant of the
product is the product of the determinants. A proof of this, which depends upon the
classification of certain types of alternating multilinear forms, is given in this chapter.
The final section gives the fundamentals of dual spaces.
107
108 Appendix Chapter 6
The Chinese Remainder Theorem
On page 50 in the chapter on rings, the Chinese Remainder Theorem was proved
for the ring of integers. In this section this classical topic is presented in full generality.
Surprisingly, the theorem holds even for non-commutative rings.
Definition Suppose R is a ring and A
1
, A
2
, ..., A
m
are ideals of R. Then the sum
A
1
+ A
2
+ + A
m
is the set of all a
1
+ a
2
+ + a
m
with a
i
∈ A
i
. The product
A
1
A
2
A
m
is the set of all finite sums of elements a
1
a
2
a
m
with a
i
∈ A
i
. Note
that the sum and product of ideals are ideals and A
1
A
2
A
m
⊂ (A
1
∩A
2
∩ ∩A
m
).
Definition Ideals A and B of R are said to be comaximal if A+B = R.
Theorem If A and B are ideals of a ring R, then the following are equivalent.
1) A and B are comaximal.
2) ∃ a ∈ A and b ∈ B with a +b = 1
¯
.
3) π(A) = R/B where π : R →R/B is the projection.
Theorem If A
1
, A
2
, ..., A
m
and B are ideals of R with A
i
and B comaximal for
each i, then A
1
A
2
A
m
and B are comaximal. Thus A
1
∩ A
2
∩ ∩ A
m
and B
are comaximal.
Proof Consider π : R →R/B. Then π(A
1
A
2
A
m
) = π(A
1
)π(A
2
) π(A
m
) =
(R/B)(R/B) (R/B) = R/B.
Chinese Remainder Theorem Suppose A
1
, A
2
, ..., A
n
are pairwise comaximal
ideals of R, with each A
i
= R. Then the natural map π : R →R/A
1
R/A
2
R/A
n
is a surjective ring homomorphism with kernel A
1
∩ A
2
∩ ∩ A
n
.
Proof There exists a
i
∈ A
i
and b
i
∈ A
1
A
2
A
i−1
A
i+1
A
n
with a
i
+b
i
= 1
¯
. Note
that π(b
i
) = (0, .., 0, 1
¯
i
, 0, .., 0). If (r
1
+ A
1
, r
2
+ A
2
, ..., r
n
+ A
n
) is an element of the
range, it is the image of r
1
b
1
+r
2
b
2
++r
n
b
n
= r
1
(1
¯
−a
1
)+r
2
(1
¯
−a
2
)++r
n
(1
¯
−a
n
).
Theorem If R is commutative and A
1
, A
2
, ..., A
n
are pairwise comaximal ideals
of R, then A
1
A
2
A
n
= A
1
∩ A
2
∩ ∩ A
n
.
Proof for n = 2. Show A
1
∩A
2
⊂ A
1
A
2
. ∃ a
1
∈ A
1
and a
2
∈ A
2
with a
1
+a
2
= 1
¯
.
If c ∈ A
1
∩ A
2
, then c = c(a
1
+a
2
) ∈ A
1
A
2
.
Chapter 6 Appendix 109
Prime and Maximal Ideals and UFD
s
In the first chapter on background material, it was shown that Z is a unique
factorization domain. Here it will be shown that this property holds for any principle
ideal domain. Later on it will be shown that every Euclidean domain is a principle
ideal domain. Thus every Euclidean domain is a unique factorization domain.
Definition Suppose R is a commutative ring and I ⊂ R is an ideal.
I is prime means I = R and if a, b ∈ R have ab ∈ I, then a or b ∈ I.
I is maximal means I = R and there are no ideals properly between I and R.
Theorem 0
¯
is a prime ideal of R iff R is
0
¯
is a maximal ideal of R iff R is
Theorem Suppose J ⊂ R is an ideal, J = R.
J is a prime ideal iff R/J is
J is a maximal ideal iff R/J is
Corollary Maximal ideals are prime.
Proof Every field is a domain.
Theorem If a ∈ R is not a unit, then ∃ a maximal ideal I of R with a ∈ I.
Proof This is a classical application of the Hausdorff Maximality Principle. Con-
sider ¦J : J is an ideal of R containing a with J = R¦. This collection contains a
maximal monotonic collection ¦V
t
¦
t∈T
. The ideal V =
¸
t∈T
V
t
does not contain 1
¯
and
thus is not equal to R. Therefore V is equal to some V
t
and is a maximal ideal
containing a.
Note To properly appreciate this proof, the student should work the exercise in
group theory at the end of this section (see page 114).
Definition Suppose R is a domain and a, b ∈ R. Then we say a ∼ b iff there
exists a unit u with au = b. Note that ∼ is an equivalence relation. If a ∼ b, then a
110 Appendix Chapter 6
and b are said to be associates.
Examples If R is a domain, the associates of 1
¯
are the units of R, while the only
associate of 0
¯
is 0
¯
itself. If n ∈ Z is not zero, then its associates are n and −n.
If F is a field and g ∈ F[x] is a non-zero polynomial, then the associates of g are
all cg where c is a non-zero constant.
The following theorem is elementary, but it shows how associates fit into the
scheme of things. An element a divides b (a[b) if ∃! c ∈ R with ac = b.
Theorem Suppose R is a domain and a, b ∈ (R − 0
¯
). Then the following are
equivalent.
1) a ∼ b.
2) a[b and b[a.
3) aR = bR.
Parts 1) and 3) above show there is a bijection from the associate classes of R to
the principal ideals of R. Thus if R is a PID, there is a bijection from the associate
classes of R to the ideals of R. If an element of a domain generates a non-zero prime
ideal, it is called a prime element.
Definition Suppose R is a domain and a ∈ R is a non-zero non-unit.
1) a is irreducible if it does not factor, i.e., a = bc ⇒ b or c is a unit.
2) a is prime if it generates a prime ideal, i.e., a[bc ⇒ a[b or a[c.
Note If a is a prime and a[c
1
c
2
c
n
, then a[c
i
for some i. This follows from the
definition and induction on n. If each c
j
is irreducible, then a ∼ c
i
for some i.
Note If a ∼ b, then a is irreducible (prime) iff b is irreducible (prime). In other
words, if a is irreducible (prime) and u is a unit, then au is irreducible (prime).
Note a is prime ⇒a is irreducible. This is immediate from the definitions.
Theorem Factorization into primes is unique up to order and associates, i.e., if
d = b
1
b
2
b
n
= c
1
c
2
c
m
with each b
i
and each c
i
prime, then n = m and for some
permutation σ of the indices, b
i
and c
σ(i)
are associates for every i. Note also ∃ a unit
u and primes p
1
, p
2
, . . . , p
t
where no two are associates and du = p
s
1
1
p
s
2
2
p
st
t
.
Chapter 6 Appendix 111
Proof This follows from the notes above.
Definition R is a factorization domain (FD) means that R is a domain and if a is
a non-zero non-unit element of R, then a factors into a finite product of irreducibles.
Definition R is a unique factorization domain (UFD) means R is a FD in which
factorization is unique (up to order and associates).
Theorem If R is a UFD and a is a non-zero non-unit of R, then a is irreducible
⇔ a is prime. Thus in a UFD, elements factor as the product of primes.
Proof Suppose R is a UFD, a is an irreducible element of R, and a[bc. If either
b or c is a unit or is zero, then a divides one of them, so suppose each of b and c is
a non-zero non-unit element of R. There exists an element d with ad = bc. Each of
b and c factors as the product of irreducibles and the product of these products is
the factorization of bc. It follows from the uniqueness of the factorization of ad = bc,
that one of these irreducibles is an associate of a, and thus a[b or a[c. Therefore
the element a is a prime.
Theorem Suppose R is a FD. Then the following are equivalent.
1) R is a UFD.
2) Every irreducible element of R is prime, i.e., a irreducible ⇔ a is prime.
Proof We already know 1) ⇒ 2). Part 2) ⇒ 1) because factorization into primes
is always unique.
This is a revealing and useful theorem. If R is a FD, then R is a UFD iff each
irreducible element generates a prime ideal. Fortunately, principal ideal domains
have this property, as seen in the next theorem.
Theorem Suppose R is a PID and a ∈ R is non-zero non-unit. Then the following
are equivalent.
1) aR is a maximal ideal.
2) aR is a prime ideal, i.e., a is a prime element.
3) a is irreducible.
Proof Every maximal ideal is a prime ideal, so 1) ⇒ 2). Every prime element is
an irreducible element, so 2) ⇒ 3). Now suppose a is irreducible and show aR is a
maximal ideal. If I is an ideal containing aR, ∃ b ∈ R with I = bR. Since b divides
a, the element b is a unit or an associate of a. This means I = R or I = aR.
112 Appendix Chapter 6
Our goal is to prove that a PID is a UFD. Using the two theorems above, it
only remains to show that a PID is a FD. The proof will not require that ideals be
principally generated, but only that they be finitely generated. This turns out to
be equivalent to the property that any collection of ideals has a "maximal" element.
We shall see below that this is a useful concept which fits naturally into the study of
unique factorization domains.
Theorem Suppose R is a commutative ring. Then the following are equivalent.
1) If I ⊂ R is an ideal, ∃ a finite set ¦a
1
, a
2
, ..., a
n
¦ ⊂ R such that I =
a
1
R +a
2
R + +a
n
R, i.e., each ideal of R is finitely generated.
2) Any non-void collection of ideals of R contains an ideal I which is maximal in
the collection. This means if J is an ideal in the collection with J ⊃ I, then
J = I. (The ideal I is maximal only in the sense described. It need not contain
all the ideals of the collection, nor need it be a maximal ideal of the ring R.)
3) If I
1
⊂ I
2
⊂ I
3
⊂ ... is a monotonic sequence of ideals, ∃ t
0
≥ 1 such that I
t
= I
t
0
for all t ≥ t
0
.
Proof Suppose 1) is true and show 3). The ideal I = I
1
∪ I
2
∪ . . . is finitely
generated and ∃ t
0
≥ 1 such that I
t
0
contains those generators. Thus 3) is true. Now
suppose 2) is true and show 1). Let I be an ideal of R, and consider the collection
of all finitely generated ideals contained in I. By 2) there is a maximal one, and it
must be I itself, and thus 1) is true. We now have 2)⇒1)⇒3), so suppose 2) is false
and show 3) is false. So there is a collection of ideals of R such that any ideal in the
collection is properly contained in another ideal of the collection. Thus it is possible
to construct a sequence of ideals I
1
⊂ I
2
⊂ I
3
. . . with each properly contained in
the next, and therefore 3) is false. (Actually this construction requires the Hausdorff
Maximality Principle or some form of the Axiom of Choice, but we slide over that.)
Definition If R satisfies these properties, R is said to be Noetherian, or it is said
to satisfy the ascending chain condition. This property is satisfied by many of the
classical rings in mathematics. Having three definitions makes this property useful
and easy to use. For example, see the next theorem.
Theorem A Noetherian domain is a FD. In particular, a PID is a FD.
Proof Suppose there is a non-zero non-unit element that does not factor as the
finite product of irreducibles. Consider all ideals dR where d does not factor. Since
R is Noetherian, ∃ a maximal one cR. The element c must be reducible, i.e., c = ab
where neither a nor b is a unit. Each of aR and bR properly contains cR, and so each
Chapter 6 Appendix 113
of a and b factors as a finite product of irreducibles. This gives a finite factorization
of c into irreducibles, which is a contradiction.
Corollary A PID is a UFD. So Z is a UFD and if F is a field, F[x] is a UFD.
You see the basic structure of UFD
s
is quite easy. It takes more work to prove
the following theorems, which are stated here only for reference.
Theorem If R is a UFD then R[x
1
, ..., x
n
] is a UFD. Thus if F is a field,
F[x
1
, ..., x
n
] is a UFD. (This theorem goes all the way back to Gauss.)
If R is a PID, then the formal power series R[[x
1
, ..., x
n
]] is a UFD. Thus if F
is a field, F[[x
1
, ..., x
n
]] is a UFD. (There is a UFD R where R[[x]] is not a UFD.
See page 566 of Commutative Algebra by N. Bourbaki.)
Theorem Germs of analytic functions on C
n
form a UFD.
Proof See Theorem 6.6.2 of An Introduction to Complex Analysis in Several Vari-
ables by L. H¨ ormander.
Theorem Suppose R is a commutative ring. Then R is Noetherian ⇒R[x
1
, ..., x
n
]
and R[[x
1
, ..., x
n
]] are Noetherian. (This is the famous Hilbert Basis Theorem.)
Theorem If R is Noetherian and I ⊂ R is a proper ideal, then R/I is Noetherian.
(This follows immediately from the definition. This and the previous theorem show
that Noetherian is a ubiquitous property in ring theory.)
Domains With Non-unique Factorizations Next are presented two of the
standard examples of Noetherian domains that are not unique factorization domains.
Exercise Let R = Z(
√
5) = ¦n +m
√
5 : n, m ∈ Z¦. Show that R is a subring of
R which is not a UFD. In particular 2 2 = (1 −
√
5) (−1 −
√
5) are two distinct
irreducible factorizations of 4. Show R is isomorphic to Z[x]/(x
2
−5), where (x
2
−5)
represents the ideal (x
2
− 5)Z[x], and R/(2) is isomorphic to Z
2
[x]/(x
2
− [5]) =
Z
2
[x]/(x
2
+ [1]), which is not a domain.
114 Appendix Chapter 6
Exercise Let R = R[x, y, z]/(x
2
− yz). Show x
2
− yz is irreducible and thus
prime in R[x, y, z]. If u ∈ R[x, y, z], let ¯ u ∈ R be the coset containing u. Show R
is not a UFD. In particular ¯ x ¯ x = ¯ y ¯ z are two distinct irreducible factorizations
of ¯ x
2
. Show R/(¯ x) is isomorphic to R[y, z]/(yz), which is not a domain. An easier
approach is to let f : R[x, y, z] → R[x, y] be the ring homomorphism defined by
f(x) = xy, f(y) = x
2
, and f(z) = y
2
. Then S = R[xy, x
2
, y
2
] is the image of
f and S is isomorphic to R. Note that xy, x
2
, and y
2
are irreducible in S and
(xy)(xy) = (x
2
)(y
2
) are two distinct irreducible factorizations of (xy)
2
in S.
Exercise In Group Theory If G is an additive abelian group, a subgroup H
of G is said to be maximal if H = G and there are no subgroups properly between
H and G. Show that H is maximal iff G/H ≈ Z
p
for some prime p. For simplicity,
consider the case G = Q. Which one of the following is true?
1) If a ∈ Q, then there is a maximal subgroup H of Q which contains a.
2) Q contains no maximal subgroups.
Splitting Short Exact Sequences
Suppose B is an R-module and K is a submodule of B. As defined in the chapter
on linear algebra, K is a summand of B provided ∃ a submodule L of B with
K+L = B and K∩L = 0
¯
. In this case we write K⊕L = B. When is K a summand
of B? It turns out that K is a summand of B iff there is a splitting map from
B/K to B. In particular, if B/K is free, K must be a summand of B. This is used
below to show that if R is a PID, then every submodule of R
n
is free.
Theorem 1 Suppose R is a ring, B and C are R-modules, and g : B → C is a
surjective homomorphism with kernel K. Then the following are equivalent.
1) K is a summand of B.
2) g has a right inverse, i.e., ∃ a homomorphism h : C →B with g ◦h = I : C →C.
(h is called a splitting map.)
Proof Suppose 1) is true, i.e., suppose ∃ a submodule L of B with K ⊕ L = B.
Then (g[L) : L → C is an isomorphism. If i : L → B is inclusion, then h defined
by h = i ◦ (g[L)
−1
is a right inverse of g. Now suppose 2) is true and h : C → B
is a right inverse of g. Then h is injective, K + h(C) = B and K ∩ h(C) = 0
¯
.
Thus K ⊕h(C) = B.
Chapter 6 Appendix 115
Definition Suppose f : A → B and g : B → C are R-module homomorphisms.
The statement that 0 →A
f
→B
g
→C →0 is a short exact sequence (s.e.s) means
f is injective, g is surjective and f(A) = ker(g). The canonical split s.e.s. is A →
A ⊕ C → C where f = i
1
and g = π
2
. A short exact sequence is said to split if ∃
an isomorphism B
≈
→A⊕C such that the following diagram commutes.
0 → A B C →0
A⊕C
≈
f
g
i
1
π
2
`
`
`
`
`
`
·
We now restate the previous theorem in this terminology.
Theorem 1.1 A short exact sequence 0 → A → B → C → 0 splits iff f(A) is
a summand of B, iff B → C has a splitting map. If C is a free R-module, there is
a splitting map and thus the sequence splits.
Proof We know from the previous theorem f(A) is a summand of B iff B → C
has a splitting map. Showing these properties are equivalent to the splitting of the
sequence is a good exercise in the art of diagram chasing. Now suppose C has a free
basis T ⊂ C, and g : B → C is surjective. There exists a function h : T → B such
that g ◦ h(c) = c for each c ∈ T. The function h extends to a homomorphism from
C to B which is a right inverse of g.
Theorem 2 If R is a domain, then the following are equivalent.
1) R is a PID.
2) Every submodule of R
R
is a free R-module of dimension ≤ 1.
This theorem restates the ring property of PID as a module property. Although
this theorem is transparent, 1)⇒2) is a precursor to the following classical result.
Theorem 3 If R is a PID and A ⊂ R
n
is a submodule, then A is a free R-module
of dimension ≤ n. Thus subgroups of Z
n
are free Z-modules of dimension ≤ n.
Proof From the previous theorem we know this is true for n = 1. Suppose n > 1
and the theorem is true for submodules of R
n−1
. Suppose A ⊂ R
n
is a submodule.
116 Appendix Chapter 6
Consider the following short exact sequences, where f : R
n−1
→R
n−1
⊕R is inclusion
and g = π : R
n−1
⊕R →R is the projection.
0 −→R
n−1
f
−→R
n−1
⊕R
π
−→R −→0
0 −→A∩ R
n−1
−→A −→π(A) −→0
By induction, A∩ R
n−1
is free of dimension ≤ n −1. If π(A) = 0
¯
, then A ⊂ R
n−1
.
If π(A) = 0
¯
, it is free of dimension 1 and thus the sequence splits by Theorem 1.1.
In either case, A is a free submodule of dimension ≤ n.
Exercise Let A ⊂ Z
2
be the subgroup generated by ¦(6, 24), (16, 64)¦. Show A
is a free Z-module of dimension 1. Also show the s.e.s. Z
4
×3
−→ Z
12
−→ Z
3
splits
but Z
×2
−→Z −→Z
2
and Z
2
×2
−→Z
4
−→Z
2
do not (see top of page 78).
Euclidean Domains
The ring Z possesses the Euclidean algorithm and the polynomial ring F[x] has
the division algorithm (pages 14 and 45). The concept of Euclidean domain is an
abstraction of these properties, and the efficiency of this abstraction is displayed in
this section. Furthermore the first axiom, φ(a) ≤ φ(ab), is used only in Theorem
2, and is sometimes omitted from the definition. Anyway it is possible to just play
around with matrices and get some deep results. If R is a Euclidean domain and M
is a finitely generated R-module, then M is the sum of cyclic modules. This is one of
the great classical theorems of abstract algebra, and you don't have to worry about
it becoming obsolete. Here N will denote the set of all non-negative integers, not
just the set of positive integers.
Definition A domain R is a Euclidean domain provided ∃ φ : (R−0
¯
) −→N such
that if a, b ∈ (R −0
¯
), then
1) φ(a) ≤ φ(ab).
2) ∃ q, r ∈ R such that a = bq +r with r = 0
¯
or φ(r) < φ(b).
Examples of Euclidean Domains
Z with φ(n) = [n[.
A field F with φ(a) = 1 ∀ a = 0
¯
or with φ(a) = 0 ∀ a = 0
¯
.
F[x] where F is a field with φ(f = a
0
+a
1
x + +a
n
x
n
) = deg(f).
Z[i] = ¦a +bi : a, b ∈ Z¦ = Gaussian integers with φ(a +bi) = a
2
+b
2
.
Chapter 6 Appendix 117
Theorem 1 If R is a Euclidean domain, then R is a PID and thus a UFD.
Proof If I is a non-zero ideal, then ∃ b ∈ I −0
¯
satisfying φ(b) ≤ φ(a) ∀ a ∈ I −0
¯
.
Then b generates I because if a ∈ I − 0
¯
, ∃ q, r with a = bq + r. Now r ∈ I and
r = 0
¯
⇒φ(r) < φ(b) which is impossible. Thus r = 0
¯
and a ∈ bR so I = bR.
Theorem 2 If R is a Euclidean domain and a, b ∈ R −0
¯
, then
φ(1
¯
) is the smallest integer in the image of φ.
a is a unit in R iff φ(a) = φ(1
¯
).
a and b are associates ⇒ φ(a) = φ(b).
Proof This is a good exercise. However it is unnecessary for Theorem 3 below.
The following remarkable theorem is the foundation for the results of this section.
Theorem 3 If R is a Euclidean domain and (a
i,j
) ∈ R
n,t
is a non-zero matrix,
then by elementary row and column operations (a
i,j
) can be transformed to
¸
¸
¸
¸
¸
¸
¸
¸
¸
¸
¸
¸
d
1
0 0
0 d
2
.
.
.
.
.
.
d
m
0
0 0
¸
where each d
i
= 0
¯
, and d
i
[d
i+1
for 1 ≤ i < m. Also d
1
generates the ideal of R
generated by the entries of (a
i,j
).
Proof Let I ⊂ R be the ideal generated by the elements of the matrix A = (a
i,j
).
If E ∈ R
n
, then the ideal J generated by the elements of EA has J ⊂ I. If E is
invertible, then J = I. In the same manner, if E ∈ R
t
is invertible and J is the ideal
generated by the elements of AE, then J = I. This means that row and column
operations on A do not change the ideal I. Since R is a PID, there is an element
d
1
with I = d
1
R, and this will turn out to be the d
1
displayed in the theorem.
The matrix (a
i,j
) has at least one non-zero element d with φ(d) a miminum.
However, row and column operations on (a
i,j
) may produce elements with smaller
118 Appendix Chapter 6
φ values. To consolidate this approach, consider matrices obtained from (a
i,j
) by a
finite number of row and column operations. Among these, let (b
i,j
) be one which
has an entry d
1
= 0 with φ(d
1
) a minimum. By elementary operations of type 2, the
entry d
1
may be moved to the (1, 1) place in the matrix. Then d
1
will divide the other
entries in the first row, else we could obtain an entry with a smaller φ value. Thus
by column operations of type 3, the other entries of the first row may be made zero.
In a similar manner, by row operations of type 3, the matrix may be changed to the
following form.
¸
¸
¸
¸
¸
¸
¸
¸
¸
¸
d
1
0 0
0
.
.
. c
ij
0
¸
Note that d
1
divides each c
i,j
, and thus I = d
1
R. The proof now follows by induction
on the size of the matrix.
This is an example of a theorem that is easy to prove playing around at the
blackboard. Yet it must be a deep theorem because the next two theorems are easy
consequences.
Theorem 4 Suppose R is a Euclidean domain, B is a finitely generated free R-
module and A ⊂ B is a non-zero submodule. Then ∃ free bases ¦a
1
, a
2
, ..., a
t
¦ for A
and ¦b
1
, b
2
, ..., b
n
¦ for B, with t ≤ n, and such that each a
i
= d
i
b
i
, where each d
i
= 0
¯
,
and d
i
[d
i+1
for 1 ≤ i < t. Thus B/A ≈ R/d
1
⊕R/d
2
⊕ ⊕R/d
t
⊕R
n−t
.
Proof By Theorem 3 in the section Splitting Short Exact Sequences, A has a
free basis ¦v
1
, v
2
, ..., v
t
¦. Let ¦w
1
, w
2
, ..., w
n
¦ be a free basis for B, where n ≥ t. The
composition
R
t
≈
−→A
⊂
−→B
≈
−→R
n
e
i
−→v
i
w
i
−→e
i
is represented by a matrix (a
i,j
) ∈ R
n,t
where v
i
= a
1,i
w
1
+a
2,i
w
2
+ +a
n,i
w
n
. By
the previous theorem, ∃ invertible matrixes U ∈ R
n
and V ∈ R
t
such that
Chapter 6 Appendix 119
U(a
i,j
)V =
¸
¸
¸
¸
¸
¸
¸
¸
¸
¸
d
1
0 0
0 d
2
0
.
.
. 0
.
.
.
d
t
0 0
¸
with d
i
[d
i+1
. Since changing the isomorphisms R
t
≈
−→A and B
≈
−→R
n
corresponds
to changing the bases ¦v
1
, v
2
, ..., v
t
¦ and ¦w
1
, w
2
, ..., w
n
¦, the theorem follows.
Theorem 5 If R is a Euclidean domain and M is a finitely generated R-module,
then M ≈ R/d
1
⊕R/d
2
⊕ ⊕R/d
t
⊕R
m
where each d
i
= 0
¯
, and d
i
[d
i+1
for 1 ≤ i < t.
Proof By hypothesis ∃ a finitely generated free module B and a surjective homo-
morphism B −→M −→0. Let A be the kernel, so 0 −→A
⊂
−→B −→M −→0 is
a s.e.s. and B/A ≈ M. The result now follows from the previous theorem.
The way Theorem 5 is stated, some or all of the elements d
i
may be units, and for
such d
i
, R/d
i
= 0
¯
. If we assume that no d
i
is a unit, then the elements d
1
, d
2
, ..., d
t
are
called invariant factors. They are unique up to associates, but we do not bother with
that here. If R = Z and we select the d
i
to be positive, they are unique. If R = F[x]
and we select the d
i
to be monic, then they are unique. The splitting in Theorem 5
is not the ultimate because the modules R/d
i
may split into the sum of other cyclic
modules. To prove this we need the following Lemma.
Lemma Suppose R is a PID and b and c are non-zero non-unit elements of R.
Suppose b and c are relatively prime, i.e., there is no prime common to their prime
factorizations. Then bR and cR are comaximal ideals. (See p 108 for comaximal.)
Proof There exists an a ∈ R with aR = bR + cR. Since a[b and a[c, a is a
unit, so R = bR +cR.
Theorem 6 Suppose R is a PID and d is a non-zero non-unit element of R.
Assume d = p
s
1
1
p
s
2
2
p
st
t
is the prime factorization of d (see bottom of p 110). Then
the natural map R/d
≈
−→R/p
s
1
1
⊕ ⊕ R/p
st
t
is an isomorphism of R-modules.
(The elements p
s
i
i
are called elementary divisors of R/d.)
Proof If i = j, p
s
i
i
and p
s
j
j
are relatively prime. By the Lemma above, they are
120 Appendix Chapter 6
comaximal and thus by the Chinese Remainder Theorem, the natural map is a ring
isomorphism (page 108). Since the natural map is also an R-module homomorphism,
it is an R-module isomorphism.
This theorem carries the splitting as far as it can go, as seen by the next exercise.
Exercise Suppose R is a PID, p ∈ R is a prime element, and s ≥ 1. Then the
R-module R/p
s
has no proper submodule which is a summand.
Torsion Submodules This will give a little more perspective to this section.
Definition Suppose M is a module over a domain R. An element m ∈ M is said
to be a torsion element if ∃ r ∈ R with r = 0
¯
and mr = 0
¯
. This is the same as
saying m is dependent. If R = Z, it is the same as saying m has finite order. Denote
by T(M) the set of all torsion elements of M. If T(M) = 0
¯
, we say that M is torsion
free.
Theorem 7 Suppose M is a module over a domain R. Then T(M) is a submodule
of M and M/T(M) is torsion free.
Proof This is a simple exercise.
Theorem 8 Suppose R is a Euclidean domain and M is a finitely generated
R-module which is torsion free. Then M is a free R-module, i.e., M ≈ R
m
.
Proof This follows immediately from Theorem 5.
Theorem 9 Suppose R is a Euclidean domain and M is a finitely generated
R-module. Then the following s.e.s. splits.
0 −→T(M) −→M −→M/T(M) −→0
Proof By Theorem 7, M/T(M) is torsion free. By Theorem 8, M/T(M) is a free
R-module, and thus there is a splitting map. Of course this theorem is transparent
anyway, because Theorem 5 gives a splitting of M into a torsion part and a free part.
Chapter 6 Appendix 121
Note It follows from Theorem 9 that ∃ a free submodule V of M such that T(M)⊕
V = M. The first summand T(M) is unique, but the complementary summand V is
not unique. V depends upon the splitting map and is unique only up to isomorphism.
To complete this section, here are two more theorems that follow from the work
we have done.
Theorem 10 Suppose T is a domain and T
∗
is the multiplicative group of units
of T. If G is a finite subgroup of T
∗
, then G is a cyclic group. Thus if F is a finite
field, the multiplicative group F
∗
is cyclic. Thus if p is a prime, (Z
p
)
∗
is cyclic.
Proof This is a corollary to Theorem 5 with R = Z. The multiplicative group G
is isomorphic to an additive group Z/d
1
⊕Z/d
2
⊕ ⊕ Z/d
t
where each d
i
> 1 and
d
i
[d
i+1
for 1 ≤ i < t. Every u in the additive group has the property that ud
t
= 0
¯
.
So every g ∈ G is a solution to x
dt
− 1
¯
= 0
¯
. If t > 1, the equation will have degree
less than the number of roots, which is impossible. Thus t = 1 and so G is cyclic.
Exercise For which primes p and q is the group of units (Z
p
Z
q
)
∗
a cyclic group?
We know from Exercise 2) on page 59 that an invertible matrix over a field is the
product of elementary matrices. This result also holds for any invertible matrix over
a Euclidean domain.
Theorem 11 Suppose R is a Euclidean domain and A ∈ R
n
is a matrix with
non-zero determinant. Then by elementary row and column operations, A may be
transformed to a diagonal matrix
¸
¸
¸
¸
¸
d
1
0
d
2
.
.
.
0 d
n
¸
where each d
i
= 0
¯
and d
i
[d
i+1
for 1 ≤ i < n. Also d
1
generates the ideal generated
by the entries of A. Furthermore A is invertible iff each d
i
is a unit. Thus if A is
invertible, A is the product of elementary matrices.
122 Appendix Chapter 6
Proof It follows from Theorem 3 that A may be transformed to a diagonal matrix
with d
i
[d
i+1
. Since the determinant of A is not zero, it follows that each d
i
= 0
¯
.
Furthermore, the matrix A is invertible iff the diagonal matrix is invertible, which is
true iff each d
i
is a unit. If each d
i
is a unit, then the diagonal matrix is the product
of elementary matrices of type 1. Therefore if A is invertible, it is the product of
elementary matrices.
Exercise Let R = Z, A =
3 11
0 4
and D =
3 11
1 4
. Perform elementary
operations on A and D to obtain diagonal matrices where the first diagonal element
divides the second diagonal element. Write D as the product of elementary matri-
ces. Find the characteristic polynomials of A and D. Find an elementary matrix B
over Z such that B
−1
AB is diagonal. Find an invertible matrix C in R
2
such that
C
−1
DC is diagonal. Show C cannot be selected in Q
2
.
Jordan Blocks
In this section, we define the two special types of square matrices used in the
Rational and Jordan canonical forms. Note that the Jordan block B(q) is the sum
of a scalar matrix and a nilpotent matrix. A Jordan block displays its eigenvalue
on the diagonal, and is more interesting than the companion matrix C(q). But as
we shall see later, the Rational canonical form will always exist, while the Jordan
canonical form will exist iff the characteristic polynomial factors as the product of
linear polynomials.
Suppose R is a commutative ring, q = a
0
+ a
1
x + + a
n−1
x
n−1
+ x
n
∈ R[x]
is a monic polynomial of degree n ≥ 1, and V is the R[x]-module V = R[x]/q.
V is a torsion module over the ring R[x], but as an R-module, V has a free basis
¦1, x, x
2
, . . . , x
n−1
¦. (See the last part of the last theorem on page 46.) Multipli-
cation by x defines an R-module endomorphism on V , and C(q) will be the ma-
trix of this endomorphism with respect to this basis. Let T : V → V be defined
by T(v) = vx. If h(x) ∈ R[x], h(T) is the R-module homomorphism given by
multiplication by h(x). The homomorphism from R[x]/q to R[x]/q given by
multiplication by h(x), is zero iff h(x) ∈ qR[x]. That is to say q(T) = a
0
I + a
1
T+
+ T
n
is the zero homomorphism, and h(T) is the zero homomorphism iff
h(x) ∈ qR[x]. All of this is supposed to make the next theorem transparent.
Theorem Let V have the free basis ¦1, x, x
2
, ..., x
n−1
¦. The companion matrix
Chapter 6 Appendix 123
representing T is
C(q) =
The characteristic polynomial of B(q) is q, and [B(q)[ = λ
n
= (−1)
n
a
0
. Finally, if
h(x) ∈ R[x], h(B(q)) is zero iff h(x) ∈ qR[x].
Note For n = 1, C(a
0
+ x) = B(a
0
+x) = (−a
0
). This is the only case where a
block matrix may be the zero matrix.
Note In B(q), if you wish to have the 1
s
above the diagonal, reverse the order of
the basis for V .
Jordan Canonical Form
We are finally ready to prove the Rational and Jordan forms. Using the previous
sections, all that's left to do is to put the pieces together. (For an overview of Jordan
form, read first the section in Chapter 5, page 96.)
124 Appendix Chapter 6
Suppose R is a commutative ring, V is an R-module, and T : V → V is an
R-module homomorphism. Define a scalar multiplication V R[x] → V by
v(a
0
+a
1
x + +a
r
x
r
) = va
0
+T(v)a
1
+ +T
r
(v)a
r
.
Theorem 1 Under this scalar multiplication, V is an R[x]-module.
This is just an observation, but it is one of the great tricks in mathematics.
Questions about the transformation T are transferred to questions about the module
V over the ring R[x]. And in the case R is a field, R[x] is a Euclidean domain and so
we know almost everything about V as an R[x]-module.
Now in this section, we suppose R is a field F, V is a finitely generated F-module,
T : V →V is a linear transformation and V is an F[x]-module with vx = T(v). Our
goal is to select a basis for V such that the matrix representing T is in some simple
form. A submodule of V
F[x]
is a submodule of V
F
which is invariant under T. We
know V
F[x]
is the sum of cyclic modules from Theorems 5 and 6 in the section on
Euclidean Domains. Since V is finitely generated as an F-module, the free part of
this decomposition will be zero. In the section on Jordan Blocks, a basis is selected
for these cyclic modules and the matrix representing T is described. This gives the
Rational Canonical Form and that is all there is to it. If all the eigenvalues for T are
in F, we pick another basis for each of the cyclic modules (see the second theorem in
the section on Jordan Blocks). Then the matrix representing T is called the Jordan
Canonical Form. Now we say all this again with a little more detail.
From Theorem 5 in the section on Euclidean Domains, it follows that
V
F[x]
≈ F[x]/d
1
⊕F[x]/d
2
⊕ ⊕F[x]/d
t
where each d
i
is a monic polynomial of degree ≥ 1, and d
i
[d
i+1
. Pick ¦1, x, x
2
, . . . , x
m−1
¦
as the F-basis for F[x]/d
i
where m is the degree of the polynomial d
i
.
Theorem 2 With respect to this basis, the matrix representing T is
¸
¸
¸
¸
¸
¸
¸
¸
¸
¸
C(d
1
)
C(d
2
)
.
.
.
C(d
t
)
¸
Chapter 6 Appendix 125
The characteristic polynomial of T is p = d
1
d
2
d
t
and p(T) = 0
¯
. This is a type
of canonical form but it does not seem to have a name.
Now we apply Theorem 6 to each F[x]/d
i
. This gives V
F[x]
≈ F[x]/p
s
1
1
⊕ ⊕
F[x]/p
sr
r
where the p
i
are irreducible monic polynomials of degree at least 1. The p
i
need not be distinct. Pick an F-basis for each F[x]/p
s
i
i
as before.
Theorem 3 With respect to this basis, the matrix representing T is
¸
¸
¸
¸
¸
¸
¸
¸
C(p
s
1
1
)
C(p
s
2
2
) 0
0
.
.
.
C(p
sr
r
)
¸
The characteristic polynomial of T is p = p
s
1
1
p
sr
r
and p(T) = 0
¯
. This is called
the Rational canonical form for T.
Now suppose the characteristic polynomial of T factors in F[x] as the product of
linear polynomials. Thus in the Theorem above, p
i
= x −λ
i
and
V
F[x]
≈ F[x]/(x −λ
1
)
s
1
⊕ ⊕F[x]/(x −λ
r
)
sr
is an isomorphism of F[x]-modules. Pick ¦1, (x − λ
i
), (x − λ
i
)
2
, . . . , (x − λ
i
)
m−1
¦ as
the F-basis for F[x]/(x −λ
i
)
s
i
where m is s
i
.
Theorem 4 With respect to this basis, the matrix representing T is
126 Appendix Chapter 6
The characteristic polynomial of T is p = (x−λ
1
)
s
1
(x−λ
r
)
sr
and p(T) = 0
¯
. This
is called the Jordan canonical form for T. Note that the λ
i
need not be distinct.
Note A diagonal matrix is in Rational canonical form and in Jordan canonical
form. This is the case where each block is one by one. Of course a diagonal matrix
is about as canonical as you can get. Note also that if a matrix is in Jordan form,
its trace is the sum of the eigenvalues and its determinant is the product of the
eigenvalues. Finally, this section is loosely written, so it is important to use the
transpose principle to write three other versions of the last two theorems.
Exercise Suppose F is a field of characteristic 0 and T ∈ F
n
has trace(T
i
) = 0
¯
for 0 < i ≤ n. Show T is nilpotent. Let p ∈ F[x] be the characteristic polynomial of
T. The polynomial p may not factor into linears in F[x], and thus T may have no
conjugate in F
n
which is in Jordan form. However this exercise can still be worked
using Jordan form. This is based on the fact that there exists a field
¯
F containing F
as a subfield, such that p factors into linears in
¯
F[x]. This fact is not proved in this
book, but it is assumed for this exercise. So ∃ an invertible matrix U ∈
¯
F
n
so that
U
−1
TU is in Jordan form, and of course, T is nilpotent iff U
−1
TU is nilpotent. The
point is that it sufficies to consider the case where T is in Jordan form, and to show
the diagonal elements are all zero.
So suppose T is in Jordan form and trace (T
i
) = 0
¯
for 1 ≤ i ≤ n. Thus trace
(p(T)) = a
0
n where a
0
is the constant term of p(x). We know p(T) = 0
¯
and thus
trace (p(T)) = 0
¯
, and thus a
0
n = 0
¯
. Since the field has characteristic 0, a
0
= 0
¯
and so 0
¯
is an eigenvalue of T. This means that one block of T is a strictly lower
triangular matrix. Removing this block leaves a smaller matrix which still satisfies
the hypothesis, and the result follows by induction on the size of T. This exercise
illustrates the power and facility of Jordan form. It also has a cute corollary.
Corollary Suppose F is a field of characteristic 0, n ≥ 1, and (λ
1
, λ
2
, .., λ
n
) ∈ F
n
satisfies λ
i
1
+λ
i
2
+ +λ
i
n
= 0
¯
for each 1 ≤ i ≤ n. Then λ
i
= 0
¯
for 1 ≤ i ≤ n.
Minimal polynomials To conclude this section here are a few comments on the
minimal polynomial of a linear transformation. This part should be studied only if
you need it. Suppose V is an n-dimensional vector space over a field F and T : V →V
is a linear transformation. As before we make V a module over F[x] with T(v) = vx.
Chapter 6 Appendix 127
Definition Ann(V
F[x]
) is the set of all h ∈ F[x] which annihilate V , i.e., which
satisfy V h = 0
¯
. This is a non-zero ideal of F[x] and is thus generated by a unique
monic polynomial u(x) ∈ F(x), Ann(V
F[x]
) = uF[x]. The polynomial u is called the
minimal polynomial of T. Note that u(T) = 0
¯
and if h(x) ∈ F[x], h(T) = 0
¯
iff
h is a multiple of u in F[x]. If p(x) ∈ F[x] is the characteristic polynomial of T,
p(T) = 0
¯
and thus p is a multiple of u.
Now we state this again in terms of matrices. Suppose A ∈ F
n
is a matrix
representing T. Then u(A) = 0
¯
and if h(x) ∈ F[x], h(A) = 0
¯
iff h is a multiple of
u in F[x]. If p(x) ∈ F[x] is the characteristic polynomial of A, then p(A) = 0
¯
and
thus p is a multiple of u. The polynomial u is also called the minimal polynomial of
A. Note that these properties hold for any matrix representing T, and thus similar
matrices have the same minimal polynomial. If A is given to start with, use the linear
transformation T : F
n
→F
n
determined by A to define the polynomial u.
Now suppose q ∈ F[x] is a monic polynomial and C(q) ∈ F
n
is the compan-
ion matrix defined in the section Jordan Blocks. Whenever q(x) = (x − λ)
n
, let
B(q) ∈ F
n
be the Jordan block matrix also defined in that section. Recall that q is
the characteristic polynomial and the minimal polynomial of each of these matrices.
This together with the rational form and the Jordan form will allow us to understand
the relation of the minimal polynomial to the characteristic polynomial.
Exercise Suppose A
i
∈ F
n
i
has q
i
as its characteristic polynomial and its minimal
polynomial, and A =
¸
¸
¸
¸
¸
A
1
0
A
2
.
.
.
0 A
r
¸
. Find the characteristic polynomial
and the minimal polynomial of A.
Exercise Suppose A ∈ F
n
.
1) Suppose A is the matrix displayed in Theorem 2 above. Find the characteristic
and minimal polynomials of A.
2) Suppose A is the matrix displayed in Theorem 3 above. Find the characteristic
and minimal polynomials of A.
3) Suppose A is the matrix displayed in Theorem 4 above. Find the characteristic
and minimal polynomials of A.
128 Appendix Chapter 6
4) Suppose λ ∈ F. Show λ is a root of the characteristic polynomial of A iff λ
is a root of the minimal polynomial of A. Show that if λ is a root, its order
in the characteristic polynomial is at least as large as its order in the minimal
polynomial.
5) Suppose
¯
F is a field containing F as a subfield. Show that the minimal poly-
nomial of A ∈ F
n
is the same as the minimal polynomial of A considered as a
matrix in
¯
F
n
. (This funny looking exercise is a little delicate.)
6) Let F = R and A =
Introduction
In 1965 I first taught an undergraduate course in abstract algebra. It was fun to teach because the material was interesting and the class was outstanding. Five of those students later earned a Ph.D. in mathematics. Since then I have taught the course about a dozen times from various texts. Over the years I developed a set of lecture notes and in 1985 I had them typed so they could be used as a text. They now appear (in modified form) as the first five chapters of this book. Here were some of my motives at the time. 1) To have something as short and inexpensive as possible. In my experience, students like short books. 2) To avoid all innovation. To organize the material in the most simple-minded straightforward manner. 3) To order the material linearly. To the extent possible, each section should use the previous sections and be used in the following sections. 4) To omit as many topics as possible. This is a foundational course, not a topics course. If a topic is not used later, it should not be included. There are three good reasons for this. First, linear algebra has top priority. It is better to go forward and do more linear algebra than to stop and do more group and ring theory. Second, it is more important that students learn to organize and write proofs themselves than to cover more subject matter. Algebra is a perfect place to get started because there are many "easy" theorems to prove. There are many routine theorems stated here without proofs, and they may be considered as exercises for the students. Third, the material should be so fundamental that it be appropriate for students in the physical sciences and in computer science. Zillions of students take calculus and cookbook linear algebra, but few take abstract algebra courses. Something is wrong here, and one thing wrong is that the courses try to do too much group and ring theory and not enough matrix theory and linear algebra. 5) To offer an alternative for computer science majors to the standard discrete mathematics courses. Most of the material in the first four chapters of this text is covered in various discrete mathematics courses. Computer science majors might benefit by seeing this material organized from a purely mathematical viewpoint.
Chapter 6 continues the material to complete a first year graduate course. Indeed. but still somewhat informal. These were independent topics stuck on at the end.e. Finishing the chapter on linear algebra gives a basic one year undergraduate course in abstract algebra. As bare as the first four chapters are. It hung together pretty well except for the last two sections on determinants and dual spaces. More advanced classes can do four chapters the first semester and chapters 5 and 6 the second semester. Chapter 2 is the most difficult part of the book because groups are written in additive and multiplicative notation. Dmitry Gokhman. and many of my friends have contributed to this text. and chapters 4 and 5 in the second semester. Finally. The proof is contained in Chapter 6. you still have to truck right along to finish them in one semester. but as an additional chapter for more advanced courses. For the proper flow of the course. Huseyin Kocak. This is the personal background of how this book came about.iv Over the years I used the five chapters that were typed as a base for my algebra courses. In 1996 I wrote a sixth chapter. giving enough material for a full first year graduate course. These proofs are to be provided by the professor in class or assigned as homework exercises. My sincere gratitude goes especially to Marilyn Gonzalez. After Chapter 2 the book gets easier as you go along. There is a non-trivial theorem stated without proof in Chapter 4. The Jordan form should not be considered part of Chapter 5. This book is a survey of abstract algebra with emphasis on linear algebra. Lourdes Robles. This chapter was written in the same "style" as the previous chapters. everything was right down to the nub. Brian Coomes. The first three or four chapters can stand alone as a one semester course in abstract algebra. this theorem should be assumed there without proof. i. Chapter 6 is not written primarily for reference. namely the determinant of the product is the product of the determinants. John Zweibel. Classes with little background can do the first three chapters in the first semester. this book is fondly dedicated.
. In the academic year 1997-98 I revised all six chapters and had them typed in LaTeX. supplementing them as I saw fit. Marta Alpar. the linear algebra follows easily.. However they are structured to provide the background for the chapter on linear algebra. and Shulim Kaliman. and the physical sciences. It is difficult to do anything in life without help from friends. The proofs of many of the elementary theorems are omitted. after the first four chapters. and the concept of coset is confusing at first. To these and all who contributed. It is intended for students in mathematics. computer science. The presentation is compact and tightly organized. It is stated there only as a reference for undergraduate courses.
it is learned. but it requires a lot of organization. you still have to build it.edu
." Mathematics is not taught. Basic algebra is a subject of incredible elegance and utility. Teaching abstract algebra and linear algebra as distinct courses results in a loss of synergy and a loss of momentum. I would have made the book shorter. Because after you extract it. E. The purpose of class is to learn. not just back. FL 33124 ec@math. I am convinced it is easier to build a course from a base than to extract it from a big book. The bare bones nature of this book adds to its flexibility. they participate little and learn little. Also with this text the professor does not extract the course from the text. This book is my attempt at that organization. enjoy it. the student already has the outline of the next lecture. Professors should give more direction in that regard. Every effort has been extended to make the subject move rapidly and to make the flow from one topic to the next as seamless as possible. The professor picks which topics to assign for serious study and which ones to "wave arms at". This book works best when viewed lightly and read as a story. The style and approach of this book is to make it a little lighter. Unfortunately mathematics is a difficult and heavy subject. I am here. The student has limited time during the semester for serious study. The goal is to stay focused and go forward. A few minutes of preparation does wonders to leverage classroom learning. Connell Department of Mathematics University of Miami Coral Gables. and this book is intended to be used in that manner. and this time should be allocated with care. but I did not have any more time. I hope the students and professors who try it. but rather builds the course upon it. because you can build whatever course you want around it. H. Study forward. When students come to class cold and spend the period taking notes. When using this text.v This text is written with the conviction that it is more effective to teach abstract and linear algebra as one coherent discipline rather than as two separate ones. This leads to a dead class and also to the bad psychology of "O K. and each assignment should include the study of the next few pages.miami. so teach me the subject. because mathematics is learned in hindsight. and many students never learn how to learn. not to do transcription work.
And before beginning the course. not on sleight of hand. This trick is based. In this puzzle. Is it possible to slide the tiles around to get them all in order. a frame has 12 spaces.viii
1 5 9
2 6 11
3 7 10
4 8
Abstract algebra is not only a major subject of science. The last two tiles are out of order. but it is also magic and fun. and end again with the last space vacant? After giving up on this. for example. but rather on a theorem in abstract algebra. the first 11 with numbered tiles and the last vacant. and it is certainly not a dull boy. the neat card trick on page 18. but to understand it you need some group theory. Anyone can do it. you can study permutation groups and learn the answer!
. Abstract algebra is not all work and no play. See. you might first try your skills on the famous (some would say infamous) tile puzzle.
.} Q = the field of rational numbers = {a/b : a. injective. b = 0} R = the field of real numbers C = the field of complex numbers = {a + bi : a. . and most properties of functions can be stated in terms of solutions of equations.. An equivalence relation on a set A is shown to be simply a partition of A into disjoint subsets. 2. and bijective.Chapter 1
Background and Fundamentals of Mathematics
This chapter is fundamental. The symbol ∀ means "for each" and ⇒ means "implies". and the properties of surjective. C. 3. but for all fields related to mathematics. The basic concepts are products of sets. We use the standard notation for intersection and union.. not just for algebra.. are sets. 1. and the integers. Notation Mathematics has its own universally accepted shorthand. −1... b ∈ R} (i2 = −1) Sets Suppose A.. 2. 0. −2. . The symbol ∃ means "there exists" and ∃! means "there exists a unique". A ∩ B = {x : x ∈ A and x ∈ B} = the set of all x which are elements 1
. In elementary courses the section on the Hausdorff Maximality Principle should be ignored.. N = Z+ = the set of positive integers = {1. equivalence relations. Five of these are listed below. B.. partial orderings.} Z = the ring of integers = {. The notion of a solution of an equation is central in mathematics.. Some sets (or collections) are so basic they have their own proprietary symbols. The final section gives a proof of the unique factorization theorem for the integers. functions. There is an emphasis on the concept of function. b ∈ Z.
x ∈ At }
t∈T
Let ∅ be the null set. At is a set. x ∈ At } = {x : ∀t ∈ T. the complement of C in S. or B contains A. Definition Suppose each of A and B is a set.e. a ∈ A ⇒ a ∈ B. then a is an element of B.
. be defined by C = S − C = {x ∈ S : x ∈ C}. If A ⊂ B we may say A is contained in B. The statement that A is a subset of B (A ⊂ B) means that if a is an element of A. Suppose T is an index set and for each t ∈ T . let C . That is. Exercise Suppose each of A and B is a set.. y) : x ∈ X and y ∈ Y }. if C is a subset of S). Any set called an index set is assumed to be non-void.2 of A and B. Theorem (De Morgan's laws) Suppose S is a set.
t∈T
At = {x : ∃ t ∈ T with x ∈ At } At = {x : if t ∈ T. The statement that A is not a subset of B means . B ⊂ S. If C ⊂ S (i. In other words. then A and B are said to be disjoint. X × Y = {(x. (A ∩ B) = A ∪ B (A ∪ B) = A ∩ B and
Cartesian Products If X and Y are sets. Then for any A.
Background
Chapter 1
A ∪ B = {x : x ∈ A or x ∈ B} = the set of all x which are elements of A or B. the Cartesian product of X and Y is defined to be the set of all ordered pairs whose first term is in X and whose second term is in Y . Example R × R = R2 = the plane. If A ∩ B = ∅.
b ∈ A. .. and we write this fact by the expression a ∼ b..Chapter 1
Background
3
Definition If each of X1 . 2). If a ∼ b. then (reflexive) (symmetric) a = b. Example Example A = R with the ordinary ordering. X1 × · · · × Xn = {(x1 . (anti-symmetric) a ∼ c. and 3) is called a partial ordering. then a ≤ a. Example Question R × · · · × R = Rn = real n-space. is a partial ordering.. Here are several properties which a relation may possess. xn ) : xi ∈ Xi for 1 ≤ i ≤ n} = the set of all ordered n-tuples whose i-th term is in Xi . the relation also satisfies these properties when restricted to S. 2 ) If a ≤ b and b ≤ a. Then 1) If a ∈ A. Definition A linear ordering is a partial ordering with the additional property that. a partial ordering on A defines a partial ordering
. then b ∼ a. or 3) on A. Is (R × R2 ) = (R2 × R) = R3 ? Relations If A is a non-void set. then a ≤ c. In particular. a non-void subset R ⊂ A × A is called a relation on A. 2 ). If a ∼ b and b ∼ a. then a ≤ b or b ≤ a. In this case we write a ∼ b as a ≤ b. then a = b. then a ∼ a. 1) 2) 2) 3) If a ∈ A. If (a.. If the relation satisfies any of the properties 1).
Hausdorff Maximality Principle (HMP) Suppose S is a non-void subset of A and ∼ is a relation on A. if a. . 2 ). This defines a relation on S.. (transitive)
Definition A relation which satisfies 1). b) ∈ R we say that a is related to b. 3) If a ≤ b and b ≤ c. with a ≤ b defined by a ⊂ b. A = all subsets of R2 . then If a ∼ b and b ∼ c. Xn is a set.. is a linear ordering.
one is contained in the other. and thus the HMP may be ignored. Show this is a partial ordering which is linear on S = {(a. Definition A collection of sets is said to be monotonic if. we define the equivalence class containing a by cl(a) = {x ∈ A : a ∼ x}. One of the most useful applications of the HMP is to obtain maximal monotonic collections of subsets. Corollary to HMP Suppose X is a non-void set and A is some non-void collection of subsets of X. to show that infinitely generated vector spaces have free bases. given any two sets of the collection. First. In elementary courses. Proof Define a partial ordering on A by V ≤ W iff V ⊂ W. the maximal monotonic subcollection will have a maximal element. Exercise Define a relation on A = R2 by (a. a) : a < 0}. and 3) is called
Exercise Define a relation on A = Z by n ∼ m iff n − m is a multiple of 3.
A relation satisfying properties 1). these results may be assumed. in the Appendix. and second. d) provided a ≤ c and b ≤ d. Definition If ∼ is an equivalence relation on A and a ∈ A. In each of these applications.
Equivalence Relations an equivalence relation.4
Background
Chapter 1
on S. and apply HMP. Show this is an equivalence relation. b) ∼ (c. The HMP is that any linearly ordered subset of a partially ordered set is contained in a maximal linearly ordered subset.
The HMP is used twice in this book. and S is a subcollection of A which is monotonic. However the ordering may be linear on S but not linear on A. Then ∃ a maximal monotonic subcollection of A which contains S.
. 2). to show that rings have maximal ideals (see pages 87 and 109). Find at least two maximal linearly ordered subsets of R2 which contain S.
If each of U. then U = V. In other words.Chapter 1 Theorem 1)
Background
5
2) 3)
If b ∈ cl(a) then cl(b) = cl(a).
Definition A partition of A is a collection of disjoint non-void subsets whose union is A.e. Then ∼ is an equivalence relation. and the other is the "graph" or "ordered pairs" definition. Consider the collection of all translates of H. 2).. In either case. Define a relation on A by a ∼ b iff a and b belong to the same subset of the partition. and the other is as a partition of A into disjoint subsets. Find the equivalence relation on R2 defined by this partition of R2 . there are two ways of defining a function. Summary There are two ways of viewing an equivalence relation — one is as a relation on A satisfying 1). What are the equivalence classes? Exercise Is there a relation on R satisfying 1). Thus we may speak of a subset of A being an equivalence class with no mention of any element contained in it. and the equivalence classes are just the subsets of the partition. Note that if A has an equivalence relation. a collection of non-void subsets of A is a partition of A provided any a ∈ A is an element of one and only one subset of the collection. Functions Just as there are two ways of viewing an equivalence relation. 2a) : a ∈ R}. domain and range are inherent parts of the definition. We use the "intuitive" definition because everyone thinks that way. V ⊂ A is an equivalence class and U ∩ V = ∅. i. Each element of A is an element of one and only one equivalence class. is there
Exercise Let H ⊂ R2 be the line H = {(a. 2 ) and 3) ? an equivalence relation on R which is also a partial ordering? That is. and 3). the equivalence classes form a partition of A. Exercise Define an equivalence relation on Z by n ∼ m iff n − m is a multiple of 3. Theorem Suppose A is a non-void set with a partition.
. all lines in the plane with slope 2. One is the "intuitive" definition. 2).
The function is defined by "f (x) is the second term of the ordered pair in Γ whose first term is x. if Γ is a subset of X × Y with the property that each x ∈ X is the first term of one and only ordered pair in Γ. f ) where f assigns to each x ∈ X a well defined element f (x) ∈ Y . Y.
. Y.6
Background
Chapter 1
Definition If X and Y are (non-void) sets. then the graph Γ ⊂ X × Y has the property that each x ∈ X is the first term of one and only one ordered pair in Γ. a function or mapping or map with domain X and range Y .
f g
define g ◦ f : W → Y by
f g h
Theorem (The associative law of composition) If V → W → X → Y . define f | S : S → Y by (f | S)(s) = f (s) for all s ∈ S. Definition The graph of a function (X. Composition Given W → X → Y (g ◦ f )(x) = g(f (x)). define the inclusion i : S → X by i(s) = s for all s ∈ S. Suppose y0 ∈ Y . f ) is a function is written f as f : X → Y or X → Y ."
Example Identity functions Here X = Y and f : X → X is defined by f (x) = x for all x ∈ X. then ∃! f : X → Y whose graph is Γ. The statement that (X. Example Constant functions y0 for all x ∈ X. f (x)) : x ∈ X}. Inclusion If S is a non-void subset of X. This may be written as h ◦ g ◦ f . Conversely. Y. The identity on X is denoted by IX or just I : X → X. The connection between the "intuitive" and "graph" viewpoints is given in the next theorem. then h ◦ (g ◦ f ) = (h ◦ g) ◦ f. Define f : X → Y by f (x) =
Restriction Given f : X → Y and a non-void subset S of X. Theorem If f : X → Y . f ) is the subset Γ ⊂ X × Y defined by Γ = {(x. is an ordered triple (X. Note that inclusion is a restriction of the identity.
. Find f −1 (f ([1. Exercise Suppose f : [−2. y0 ) : x ∈ X} = (X × y0 ) is called a horizontal strip. then f is not injective. Of course. 2] → R is defined by f (x) = x2 . If m > n. .
Pigeonhole Principle Suppose X is a finite set with m elements. then you have placed 2 pigeons in one hole. 5])). Also find the relationship between T and f (f −1 (T )).
If x0 ∈ X. as can be seen by the following exercise.8
Background
Chapter 1
Exercise Show there are natural bijections from (R × R2 ) to (R2 × R) and from (R2 × R) to R × R × R. S ⊂ X and T ⊂ Y . T = f (f −1 (T )). Exercise Show there is a function f : Z+ → Z+ which is injective but not surjective. Exercise 1) 2) 3) Suppose X is a set with 6 elements and Y is a finite set with n elements. 1) 2) 3) If m = n. Find the relationship between S and f −1 (f (S)).
If you are placing 6 pigeons in 6 holes. If m < n. if f is not surjective then f is not injective. 2])). then f is injective iff f is surjective iff f is bijective.
Strips
We now define the vertical and horizontal strips of X × Y . and f : X → Y is a function. y) : y ∈ Y } = (x0 × Y ) is called a vertical strip. in part 1) for m = n = 6. There exists an injective f : X → Y iff n There exists a surjective f : X → Y iff n There exists a bijective f : X → Y iff n . {(x. {(x0 . Also show there is one which is surjective but not injective. Y is a finite set with n elements.
. the pigeonhole principle does not hold for infinite sets. These three sets are disjoint. Show that if f is injective. then f is not surjective. Exercise Suppose f : X → Y is a function. In other words. but the bijections between them are so natural that we sometimes identify them. and you run out of pigeons before you fill the holes. If y0 ∈ Y. S = f −1 (f (S)). Show that if f is surjective. Also find f (f −1 ([3.
which are defined after the next theorem. The equation f (x) = y0 has at least one solution for each y0 ∈ Y iff f is . The purpose of the next theorem is to restate properties of functions in terms of horizontal strips. Consider the equation f (x) = y0 . The subset S is the graph of a function with domain X and range Y iff each vertical strip intersects S in exactly one point.
Right and Left Inverses One way to understand functions is to study right and left inverses. . Note that the set of all solutions to f (x) = y0 is f −1 (y0 ). A solution to this equation is any x0 ∈ X with f (x0 ) = y0 . Also f (x) = y0 has a solution iff y0 ∈ image(f ) iff f −1 (y0 ) is non-void.
Solutions of Equations Now we restate these properties in terms of solutions of equations. Here y0 is given and x is considered to be a "variable". Theorem 1) Suppose X → Y → W are functions.
Theorem 1) 2) 3)
Suppose f : X → Y has graph Γ.Chapter 1
Background
9
Theorem Suppose S ⊂ X × Y . . Theorem 1) 2) 3) Suppose f : X → Y . This is just a restatement of the property of a graph of a function. then f is injective. The equation f (x) = y0 has at most one solution for each y0 ∈ Y iff .
f g
. Suppose f : X → Y and y0 ∈ Y . f is The equation f (x) = y0 has a unique solution for each y0 ∈ Y iff f is . Then Each horizontal strip intersects Γ in at least one point iff f is Each horizontal strip intersects Γ in at most one point iff f is Each horizontal strip intersects Γ in exactly one point iff f is . If g ◦ f is injective.
f (p) = p. Here g ◦ f is the identity. However in this text we do not go that deeply into set theory. Suppose f : X → Y is a function. then f has a right inverse h. The Axiom of Choice If f : X → Y is surjective. q}. Also a function from X to Y is bijective iff it has a left inverse and a right inverse iff it has a left and right inverse. Definition Suppose f : X → Y is a function. it is possible to choose an x ∈ f −1 (y) and thus to define h(y) = x. Note It is a classical theorem in set theory that the Axiom of Choice and the Hausdorff Maximality Principle are equivalent.
Corollary Suppose each of X and Y is a non-void set. If y belongs to the image of f .
. Theorem 1) 2) Suppose f : X → Y is a function. In the next chapter where f is a group homomorphism. Show this is an equivalence relation. A left inverse of f is a function g : Y → X such that g ◦ f = IX : X → X. f has a right inverse iff f is surjective. Then ∃ an injective f : X → Y iff ∃ a surjective g : Y → X. but f is not surjective and g is not injective. Y = {p.10 2) 3)
Background If g ◦ f is surjective. Any such right inverse must be injective. then f −1 (y) is an equivalence class and every equivalence class is of this form. However. f has a left inverse iff f is injective. Note The Axiom of Choice is not discussed in this book. If g ◦ f is bijective. for each y ∈ Y .
Chapter 1
Example X = W = {p}. For our purposes it is assumed that the Axiom of Choice and the HMP are true. Any such left inverse must be surjective. these equivalence classes will be called cosets. then f is injective and g is surjective. For completeness. then g is surjective. and g(p) = g(q) = p. That is. A right inverse of f is a function h : Y → X such that f ◦ h = IY : Y → Y . you unknowingly used one version of it. if you worked 1) of the theorem above. we state this part of 1) again. Define a relation on X by a ∼ b if Exercise f (a) = f (b).
4) If S is any subset of T . let Yt be a copy of Y for each t ∈ T. to be the collection of all subsets of T (including the null set). . then Y T has nm elements. (This is the fundamental property of Cartesian products presented in the two previous theorems. Y T = Y × Y × Y has n3 elements. 1}. when T = {1. 1}T → P(T ) by β(f ) = f −1 (1). and be 0 when t ∈ S. 1}T 5) Suppose γ : T → {0.2. and if f : T → {0. It is included here for students who wish to do a little more set theory. P(T ) ←→ {0. Define f : T → {0. Use the mean value theorem to show that D is injective. 1) If Y is a non-void set. . Show that f is not in the image of γ and thus γ cannot be surjective. Thus α is a bijection and β = α−1 . Let A0 ⊂ A be the subcollection of those functions f with f (0) = 0.) 3) Define P(T ). 1} then α ◦ β(f ) = f . Show that if S ⊂ T then β ◦ α(S) = S. 1} by f (t) = 0 if ft (t) = 1. then the set {0. define its characteristic function χS : T → {0. This shows that if T is an infinite set. Show that if n ≥ 3. 1}T is a function and show that it cannot be surjective. Y N is the set of all infinite sequences (y1 . Show that if T and Y are finite sets with m and n elements.12
Background
Chapter 1
A Calculus Exercise Let A be the collection of all functions f : [0. 1] → R which have an infinite number of derivatives. In particular. denote γ(t) by γ(t) = ft : T → {0. P(T ) has 2m elements. T . Suppose T is a non-void set. Then Y T =
t∈T
2) Suppose each of Y1 and Y2 is a non-void set. . define Y T to be the collection of all functions with domain T and range Y .3} of all injective functions has n(n − 1)(n − 2) elements. 1}T represents a "higher order of infinity than T ". If t ∈ T . 6) An infinite set Y is said to be countable if there is a bijection from the positive
.) where each yi ∈ Y . the subset of Y {1. Exercise This exercise is not used elsewhere in this text and may be omitted. Define D : A0 → A by D(f ) = df /dx. That is. Use the fundamental theorem of calculus to show that D is surjective. Define β : {0. y2 . Define α : P(T ) → {0. then Y T is the infinite product Y × Y × · · · . 1} by letting χS (t) be 1 when t ∈ S. the power set of T . Show there is a natural bijection from (Y1 × Y2 )T to Y1T × Y2T . For any Y and Yt . and f (t) = 1 if ft (t) = 0. Show that if T is a finite set with m elements. 2. If T = N. 1}T by | α(S) = χS . These injective functions are called permutations on Y taken 3 at a time. 3}.
We know that ii) and iii) are equal and there is a natural bijection between i) and ii). iii) The collection of all sequences (y1 . "x ∈ B " means "x is an element of the set B. ii) {0. (This is not so easy. Mathematical symbols are shorthand for phrases and sentences in the English language. To start with. i) P(N).) where each yi is 0 or 1. then B is true. Conclusion Statement B.
Mathematical Induction is based upon the fact that if S ⊂ Z+ is a non-void subset.. 1}N is uncountable. If A is true. 1}N to the real numbers R. the collection of all subsets of N. show there is a bijection from {0.) Notation for the Logic of Mathematics Each of the words "Lemma". 1}N . Show Q is countable but the following three collections are not. . and "Corollary" means "true statement". The important thing to remember is that thoughts and expressions flow through the language. or to suppose B is false and show A is false. then "A ⇒ B"means "If x is a positive integer. {0. y2 .Chapter 1
Background
13
integers N to Y. A ⇒ B (A implies B ).
. the collection of all functions f : N → {0.e. "Theorem"." If A is the statement "x ∈ Z+ " and B is the statement "x2 ∈ Z+ ". "A is equivalent to B". then x2 is a positive integer". i. you have to decide what the real numbers are. Suppose A and B are statements. then S contains a smallest element. Suppose A is true. that A ⇒ B and B ⇒ A). 1}. We also know there is no surjective map from N to {0. 1}N . Finally. . and "A is true iff B is true " have the same meaning (namely. For example. The expressions "A ⇔ B".
There are two ways to prove the theorem — to suppose A is true and show B is true. . Then B is true. A theorem may be stated in any of the following ways: Theorem Theorem Theorem Theorem Hypothesis Statement A.
± 1 divides anything . we say that b divides a or a is a multiple of b.
3)
All of this will follow from long division. Since P (m − 1) is true. . then ∃ a smallest positive integer m such that P (m) is false. Then for each n ≥ 1.e. 2. .. If n ≥ 2. 1) 2) If G is a subgroup of Z.
. Note that b | a ⇔ the rational number a/b is an integer ⇔ ∃! m such that a = bm ⇔ a ∈ bZ. Euclidean Algorithm Given a. Proof If the theorem is false. for each n ≥ 1. Exercise Use induction to show that. P (n) is true.. For example. b. g2 ∈ G ⇒ (g1 + g2 ) ∈ G). Here we will establish the following three basic properties of the integers.
Definition A non-void subset G ⊂ Z is a subgroup provided (g ∈ G ⇒ −g ∈ G) and (g1 ... . i. In other words. 0 does not divide anything. and its positive generator is the greatest common divisor of a and b... If a and b are integers.. . then n factors uniquely as the product of primes. −n. The Integers In this section. Also n divides a and b with the same remainder iff n divides (a − b). b divides a "m times with a remainder of r".14
Background
Chapter 1
Theorem Suppose P (n) is a statement for each n = 1. and G is the collection of all linear combinations of a and b. 2n. P (n) ⇒ P (n + 1). will represent integers. 0.}. 1 + 2 + · · · + n = n(n + 1)/2. which we now state formally. then ∃ n ≥ 0 such that G = nZ. Definition If r = 0. then m = −4 and r = 3. n. the set of integers which n divides is nZ = {nm : m ∈ Z} = {. b with b = 0. then G is a subgroup of Z. this is impossible. Suppose P (1) is true and for each n ≥ 1. ∃! m and r with 0 ≤ r <|b| and a = bm + r. −2n. If n = 0. −17 = 5(−4) + 3.. not both zero. if a = −17 and b = 5. We say that G is closed under negation and closed under addition. Note Anything (except 0) divides 0. This fact is written as b | a. lower case letters a.. elements of Z.. c.
Chapter 1
Background
15
Theorem If n ∈ Z then nZ is a subgroup. Since r ∈ G. In fact. Let n be the smallest positive integer in G. it is the smallest subgroup containing a and b.. b). n ∈ Z}. By the previous theorem. and so 1) is true.. Since G is the smallest subgroup containing a and b. if G = {0} and n is the smallest positive integer in G. b) is called the greatest common divisor of a and b. and g ∈ nZ.
4)
Proof of 4) Suppose n | a and n | b i. b) | b. Now (−g) ∈ G and thus 0 = g + (−g) belongs to G. Part 2) is straightforward.
. G is a subgroup. the set of integers which n divides is a subgroup of Z. It is called the subgroup generated by a and b. Thus if n = 0. Then 1) 2) 3) G contains a and b. so consider 3). it must be 0. then n also divides (a.e. Denote by (a. b)Z. If G = 0. m2 . suppose a. G = {ma + nb : m.e. b). Also note that ∃ m. and thus n | (a.e. If n is an integer which divides a and b. If g1 and g2 ∈ G. it must contain a positive element. ∃ g ∈ G. b).
Now suppose a. The integer (a. b ∈ nZ. b)Z. i. then G = nZ. b) | a and (a. ∃! non-negative integer n such that G = nZ. The next theorem states that every subgroup of Z is of this form. Then 0 ∈ G. b ∈ Z and at least one of a and b is non-zero. Corollary 1) The following are equivalent. g = nm + r where 0 ≤ r < n. (n | a and n | b) ⇒ n = ±1. If g ∈ G.
Proof Since G is non-void. b) the smallest positive integer in G. n such that ma + nb = (a. i. nZ ⊃ (a. then (m1 g1 + m2 g2 ) ∈ G for all integers m1 . G = (a. Theorem Let G be the set of all linear combinations of a and b..
a and b have no common divisors. and thus (a. Theorem 1) 2) 3) Suppose G ⊂ Z is a subgroup. In fact.
then a|bc ⇒ a|c.
. ∃ m. Part 3) follows from 2) and induction on n.1.
This next theorem is the basis for unique factorization.. the subgroup generated by a and b is all of Z. pk and positive integers s1 . Theorem 1) 2) 3) Suppose p is a prime.e. if a is any integer.. (p..
Proof Part 1) follows immediately from the definition of prime. If p does not divide a.. Thus a | (mac + nbc) and so a | c. sk such that a = ±ps1 ps2 · · · psk . this factorization is unique. 1 2 k Proof Factorization into primes is obvious. Thus 2) is true. then by 1). and uniqueness follows from 3) in the theorem above. or -1. In other words. a) = 1.. except for order.. s2 . Then there exist m.. then p is equal to some ai . a) = p or (p. . If p | ab then p | a or p | b. we say that a and b are relatively prime. then (p. The first few primes are 2. a) = 1 and by the previous theorem. Then a may be factored into the product of primes and. If p | a1 a2 · · · an then p divides some ai . if p = ab then a = ±1 or a = ±p. n ∈Z with ma + nb = 1. 5. 13. . p2 . 7. n with ma + nb = 1. c ∈ Z and a | bc. p must divide b. The power of this theorem is uniqueness.
Proof Suppose a and b are relatively prime.. Now a | mac and a | nbc. 11.
The Unique Factorization Theorem Suppose a is an integer which is not 0. (p. 3. i. That is. not existence.
If a is an integer which is not a multiple of p.. b) = 1. Thus if each ai is a prime. a) = 1. Now suppose p | ab.. . Theorem If a and b are relatively prime with a not zero..
Chapter 1
Definition If any one of these three conditions is satisfied. 17. and thus mac + nbc = c.e. Definition A prime is an integer p > 1 which does not factor.16 2) 3)
Background (a. ∃ a unique collection of distinct primes p1 . i.
b).)
2)
3)
3)
4)
5)
Exercise Find (180. (Proof: If c is √ rational..) (See the fifth exercise below. then their least common multiple is just their product.28}. find the greatest common divisor of 180 and 28. 28).
. Then c = pv1 · · · pvk is the least 1 k (positive) common multiple of a and b. 1 k Let vi be the maximum of si and ti . n with ma + nb = 1.. Then 1 k (a. Finally. k. Thus if there is no common prime in their factorizations. this prime will also appear in the prime factorization of a.. Thus a = ±ps1 · · · psk where 0 ≤ si and 1 k b = ±pt1 · · · ptk where 0 ≤ ti . i. . then (a. Find integers m and n such that 180m + 28n = (180. if a and b are positive.. and if in addition a and b are relatively prime. then it is divisible by some prime. For example (23 · 5 · 11.) √ √ Suppose c is an integer √ greater than 1.Chapter 1
Background
17
Now that we have unique factorization and part 3) above. Let ui be the minimum of si and ti . pk .e. . and also (a2 . Then the only 1 k t divisors of a are of the form ±pt1 · · · pkk where 0 ≤ ti ≤ si for i = 1. . and show that it is equal to (180 · 28)/(180. 22 · 54 · 7) = 22 · 5. ∃ positive integers a and b with c = a/b and (a. ∃ m.. There is an infinite number of primes. 28). (Proof: Suppose there were only a finite number of primes p1 . b) = 1. If b > 1. In particular.. Suppose | a |> 1 and | b |> 1. This is a √ contradiction and thus b = 1 and c is an integer. and if n is a multiple of a and b. p2 . pk } be the union of the distinct primes of their factorizations. . 1 If | a |> 1 and | b |> 1. their least common multiple is c = ab/(a. the picture becomes transparent. i. and since cb2 = a2 . Here are some of the basic properties of the integers in this light.28). b) = 1 iff there is no common prime in their factorizations. b) = pu1 · · · puk .. . b2 ) = 1. Then no prime would divide (p1 p2 · · · pk + 1).e. Note that c is a multiple of a and b. . Then c is rational iff c is an √ √ integer. Theorem (Summary) 1) Suppose | a |> 1 has prime factorization a = ±ps1 · · · psk . then n is a multiple of c. Let {p1 . Find the least common multiple of 180 and 28.. 2 and 3 are irrational. find the positive generator of the subgroup generated by {180.
. let a = an an−1 . but they are not to know where. 42} is 2. its nth root is an irrational number. leaving it face up on top. Exercise Show that if each of G1 .. Exercise Show that if the nth root of an integer is a rational number. Also find the lcm of the elements of S. Now let b = an + an−1 + · · · + a0 . Now suppose n ≥ 2 and S = {a1 . Thus if p is a prime. Suppose they say three. a2 . then G1 ∩ G2 ∩ · · · ∩ Gm is also a subgroup of Z. That is. When is the lcm of S equal to the product a1 a2 · · · an ? Show that the set of all linear combinations of the elements of S is a subgroup of Z.
. Exercise Show that the gcd of S = {90. and find integers n1 . suppose c and n are integers greater than 1. There is a unique positive real number x with xn = c.. n3 such that 90n1 + 70n2 + 42n3 = 2. Then repeat the process. . Exercise Show that a positive integer is divisible by 3 iff the sum of its digits is divisible by 3. Define the gcd and the lcm of the elements of S and develop their properties. where it is shown that any non-zero element of Z7 has order 7. Magic? Stay tuned for Chapter 2. . and bring the seven cards in front in your left hand. and show that 3 divides a and b with the same remainder.. and then you turn over the third card. Now let G = (90Z) ∩ (70Z) ∩ (42Z) and find the positive integer n with G = nZ.18
Background
Chapter 1
Exercise We have defined the greatest common divisor (gcd) and the least common multiple (lcm) of a pair of integers. it will be more transparent later on. Card Trick Ask friends to pick out seven cards from a deck and then to select one to look at without showing it to you. moving the top two cards to the bottom and turning the third card face up on top. Take the six cards face down in your left hand and the selected card in your right hand. n2 . Continue until there is only one card face down. 70. an } is a finite collection of integers with |ai | > 1 for 1 ≤ i ≤ n. Show that if x is rational. Gm is a subgroup of Z. . Ask your friends to give you a number between one and seven (not allowing one). a0 = an 10n + an−1 10n−1 + · · · + a0 where 0 ≤ ai ≤ 9. and this will be the selected card. it says that [a] = [b] in Z3 . then the second card to the bottom. then it is an integer. G2 . and announce you will place the selected card in with the other six. In the language of the next chapter. You move the top card to the bottom. More generally. Although this is a straightforward exercise in long division.. . and its positive generator is the gcd of the elements of S. Express the gcd and the lcm in terms of the prime factorizations of the ai . then it itself is an integer. Put your hands behind your back and place the selected card on top.
b) = a + b. φ) is said to be a group. c ∈ G then a · (b · c) = (a · b) · c. it leads to a rich and amazing theory. Consider the following properties. we say it is a multiplicative group. by far and above. rings. If a. c ∈ G then a + (b + c) = (a + b) + c. φ is called a binary operation. 19
. and also the concept of coset is confusing at first. 2). b) = a+b. b ∈ G. The approach is to do quickly the fundamentals of groups. b ∈ G. ∃b ∈ G with a · b = b · a = e If a ∈ G. If we write φ(a. 2) ∃ e = eG ∈ G such that if a ∈ G e · a = a · e = a. Everything presented here is standard. 1) If a. the most difficult chapter in the book. b. then a · b = b · a. If a. If we write φ(a. Definition Suppose G is a non-void set and φ : G × G → G is a function. b) = a·b or φ(a. ∃ 0=0G ∈ G such that if a ∈ G ¯ ¯ 0+a = a+0= a. b. and we will write φ(a. because group operations may be written as addition or multiplication. This chapter is. Also ring homomorphisms and module homomorphisms are special cases of group homomorphisms.
Definition If properties 1). and matrices.Chapter 2
Groups
Groups are the central objects of algebra. then a + b = b + a. ¯ ¯
3) If a ∈ G. and 3) hold. This is the notation used in later chapters for the products of rings and modules. and to push forward to the chapter on linear algebra. 4) If a. This chapter and the next two chapters are restricted to the most basic topics. (b is written as b = −a). b) = a · b. ∃b ∈ G with a + b = b + a = 0 ¯ (b is written as b = a−1 ). except that the product of groups is given in the additive notation. (G. Even though the definition of group is simple. In later chapters we will define rings and modules and see that they are special cases of groups.
Theorem (i) Let (G.. If in addition. Recall that b is an identity in G provided it is a right and left identity for any a in G. . ¯ let an = (a + · · +a) where the sum is n times. However. ¯ ¯ ¯ Also c · a = c · a ⇒ c = c. b ∈ G. i. ¯ ¯ if a. if f : G → G is defined by f (c) = a · c. then e = e. Then a · c = a · c ⇒ c = c. φ) be a multiplicative group. Thus inverses are unique. and a(−n) = (−a) + (−a) · · + (−a). If a ∈ G. This means that if a is in G and n is an integer. group structure is so rigid that if ∃ a ∈ G such that b is a right identity for a. Write out the above theorem where G is an additive group. i. b ∈ G. if G is abelian and a. c ∈ G. c. e is unique. In general. then (a · b)n = an · bn .. In fact.e. there is defined an element an in G.. if a · b = e then b = a−1 . If n1 .20
Groups
Chapter 2
we say it is an additive group.e. Also (an )m = anm . This is so basic. Also f is bijective with f −1 given by f −1 (c) = a−1 · c. this is just a special case of the cancellation law in (i). then f is injective.
. let a0 =0 and if n > 0. Let a0 = e and if n > 0. 1 n−1 n Suppose a ∈ G. if e ∈ G satisfies 2). Finally. ¯ ¯ In other words. Also if b · a = e then b = a−1 . an = a · · · a (n times) and a−n = a−1 · · · a−1 (n times).. Suppose a. Every right inverse is an inverse. Also (a1 · a2 · · · an )−1 = a−1 · a−1 · · · a−1 . Suppose G is an additive group. then b = e.
(ii)
(iii)
(iv) (v)
(vi)
(vii)
Exercise. The multiplication a1 · a2 · a3 = a1 · (a2 · a3 ) = (a1 · a2 ) · a3 is well-defined. Note that part (vii) states that G has a scalar multiplication over Z. If a. we say the group is abelian or commutative. property 4) holds. a1 · a2 · · · an is well defined. If a ∈ G. n2 . Theorem. b ∈ G then (a · b = a) ⇒ (b = e) and (a · b = b) ⇒ (a = e). (a · b)−1 = b−1 · a−1 . then (a−1 )−1 = a. Of course.. nt ∈ Z then an1 · an2 · · · ant = an1 +···+nt . that we state it explicitly.
and f has a left inverse in G iff f is injective (see page 10). Then the following properties hold in general. and thus G is not a group. b) = ab is a multiplicative abelian group. ∃b ∈ G with a · b = e]. Show (G. b) = a·b which satisfies 1). or G = Z with φ(a. In the language used in Chapter 5. 2) and [ 3 ) If a ∈ G. b ∈ H then a · b ∈ H 2) if a ∈ H then a−1 ∈ H. Exercise Suppose G is the set of all functions from Z to Z with multiplication defined by composition. except the first requires that G be abelian. i. If every element has a right inverse. G = Z − 0 with φ(a. In other words.) Exercise Suppose G is a non-void set with a binary operation φ(a.
. Show that f has a right inverse in G iff f is surjective. b) = a + b is an additive abelian group. G = R − 0 or G = Q − 0 with φ(a.Chapter 2
Groups
21
which we write as (−a − a · · − a). Note that G satisfies 1) and 2) but not 3).e. Examples G = R. (a + b)n a(n + m) a(nm) a1 = = = = an + bn an + am (an)m a
Note that the plus sign is used ambiguously — sometimes for addition in G and sometimes for addition in Z. φ) is a group. G = Q. (See page 71. Subgroups Theorem satisfying and Suppose G is a multiplicative group and H ⊂ G is a non-void subset
Examples
1) if a.. f · g = f ◦ g. G = R+ = {r ∈ R : r > 0} with φ(a.e. b) = ab is a multiplicative abelian group. i. Also show that the set of all bijections from Z to Z is a group under composition. this theorem states that any additive abelian group is a Z-module. show b · a = e.. b) = ab is not a group. the group axioms are stronger than necessary. then every element has a two sided inverse.
Example G is a subgroup of G and e is a subgroup of G. The associative law is immediate and so H is a group.e. a−1 ∈ H and so by 1). 1] → R}. Let H be the subset of G composed of all continuous functions. Let K be the subset of G composed of all differentiable functions. By 2). These are called the improper subgroups of G. Suppose H1 and H2 are subgroups of G. then
Ht is a subgroup of G. with neither H1 nor H2 contained in the other. Show H1 ∪ H2 is not a subgroup of G. H is called a subgroup of G. What theorems in calculus show that H and K are subgroups of G? What theorem shows that K is a subset (and thus subgroup) of H?
Order
Suppose G is a multiplicative group. Define an addition on G by (f + g)(t) = f (t) + g(t) for all t ∈ [0. Suppose T is an index set and for each t ∈ T . e ∈ H. i. This is a key property of the integers. Show H1 ∩ H2 is a subgroup of G. Show Ht is a subgroup of G. Proof Since H is non-void. ∃a ∈ H. then H = nZ is a subgroup of Z. This makes G into an abelian group. 1]. every subgroup of Z is of this form (see page 15).
Let H be the center of G. Suppose H1 and H2 are subgroups of G. H = {h ∈ G : g · h = h · g for all g ∈ G}. Example If G = Z under addition.
t∈T
Suppose G= {all functions f : [0. Show H is a subgroup of G.
Exercises 1)
Suppose G is a multiplicative group.. By a theorem in the section on the integers in Chapter 1. if {Ht } is a monotonic collection.22
Groups
Chapter 2
Then e ∈ H and H is a group under multiplication. Ht is a subgroup of G.
t∈T
2) 3)
4) 5) 6)
Furthermore. and n ∈ Z. If G has an infinite number of
.
Note that in the case G has finite order n. then by the Euclidean algorithm. Exercise Show that if G is a finite group of even order. and H = {ai : i ∈ Z}.. and i H = {a : i ∈ Z}. If H = e.
Proof Suppose G = {ai : i ∈ Z} is a cyclic group and H is a subgroup of G. Suppose a ∈ G and H = {ai : i ∈ Z}. We define the order of the element a to be the order of H. Also note that this theorem was proved on page 15 for the additive group Z. Then ai−j = e and thus ∃ a smallest positive integer n with an = e. m divides t. then G has an odd number of elements of order 2. and we must show they are all of H. the order of a is the smallest positive integer n with an = e. then a has some finite order n. i. and thus the positive integer m divides n. . a1 . a1 . Proof Suppose j < i and ai = aj . then o(G) = n. . an−1 }. We come now to the first real theorem in group theory. .Chapter 2
Groups
23
elements. Let f : Z → H be the surjective function defined by f (m) = am .. Note that f (k + l) = f (k) · f (l) where the addition is in Z and the multiplication is in the group H. Note that e is the only element of order 1. . . then H is cyclic. If m ∈ Z. so suppose H = e.. we say that o(G). In either case. and so H = {a0 . and in this case. suppose a is an element of an additive group G. Definition Theorem A group G is cyclic if ∃ an element of G which generates G...e. To begin. H is isomorphic to Z or Zn . H is an abelian subgroup of G called the subgroup generated by a.. It says that the element a has finite order iff f is not injective. i. H = {a0 . in additive notation.. Later in this chapter we will see that f is a homomorphism from an additive group to a multiplicative group and that. . and am = e iff n|m. Thus am = anq · ar = ar . If G is cyclic and H is a subgroup of G. an−1 }. then an = e ∈ H.
. In particular. Now there is a smallest positive integer m with am ∈ H. is infinite. This implies that the elements of {a0 . and f −1 (e) = nZ. and am = e iff n|m. If G has n elements. G = {a0 . .e. the Euclidean algorithm states that ∃ integers q and r with 0 ≤ r < n and m = nq + r. an−1 }. and thus am generates H. an−1 } are distinct.. the order of a is the smallest positive integer n with an = e. we have a clear picture of the subgroups of G. Theorem Suppose a is an element of a multiplicative group G. a1 . In this case H has n distinct elements. . If t is an integer with at ∈ H. the order of G. If ∃ distinct integers i and j with ai = aj . the order of the subgroup generated by a. Exercise Write out this theorem for G an additive group. a1 . then H is cyclic. . .
24
Groups
Chapter 2
Cosets Suppose H is a subgroup of a group G. It will be shown below that H partitions G into right cosets. It also partitions G into left cosets, and in general these partitions are distinct. If H is a subgroup of a multiplicative group G, then a ∼ b defined by Theorem −1 a ∼ b iff a · b ∈ H is an equivalence relation. If a ∈ G, cl(a) = {b ∈ G : a ∼ b} = {h · a : h ∈ H} = Ha. Note that a · b−1 ∈ H iff b · a−1 ∈ H. If H is a subgroup of an additive group G, then a ∼ b defined by a ∼ b iff (a − b) ∈ H is an equivalence relation. If a ∈ G, cl(a) = {b ∈ G : a ∼ b} = {h + a : h ∈ H} = H + a. Note that (a − b) ∈ H iff (b − a) ∈ H. Definition These equivalence classes are called right cosets. If the relation is defined by a ∼ b iff b−1 · a ∈ H, then the equivalence classes are cl(a) = aH and they are called left cosets. H is a left and right coset. If G is abelian, there is no distinction between right and left cosets. Note that b−1 · a ∈ H iff a−1 · b ∈ H. In the theorem above, H is used to define an equivalence relation on G, and thus a partition of G. We now do the same thing a different way. We define the right cosets directly and show they form a partition of G. You might find this easier. Theorem Suppose H is a subgroup of a multiplicative group G. If a ∈ G, define the right coset containing a to be Ha = {h · a : h ∈ H}. Then the following hold. Ha = H iff a ∈ H. If b ∈ Ha, then Hb = Ha, i.e., if h ∈ H, then H(h · a) = (Hh)a = Ha. If Hc ∩ Ha = ∅, then Hc = Ha. The right cosets form a partition of G, i.e., each a in G belongs to one and only one right coset. 5) Elements a and b belong to the same right coset iff a · b−1 ∈ H iff b · a−1 ∈ H. 1) 2) 3) 4) Proof There is no better way to develop facility with cosets than to prove this theorem. Also write this theorem for G an additive group.
Theorem
Suppose H is a subgroup of a multiplicative group G.
Chapter 2 1)
Groups
25
Any two right cosets have the same number of elements. That is, if a, b ∈ G, f : Ha → Hb defined by f (h · a) = h · b is a bijection. Also any two left cosets have the same number of elements. Since H is a right and left coset, any two cosets have the same number of elements. G has the same number of right cosets as left cosets. The function F defined by F (Ha) = a−1 H is a bijection from the collection of right cosets to the left cosets. The number of right (or left) cosets is called the index of H in G. If G is finite, o(H) (index of H) = o(G) and so o(H) | o(G). In other words, o(G)/o(H) = the number of right cosets = the number of left cosets. If G is finite, and a ∈ G, then o(a) | o(G). (Proof: The order of a is the order of the subgroup generated by a, and by 3) this divides the order of G.) If G has prime order, then G is cyclic, and any element (except e) is a generator. (Proof: Suppose o(G) = p and a ∈ G, a = e. Then o(a) | p and thus o(a) = p.) If o(G) = n and a ∈ G, then an = e. (Proof: ao(a) = e and n = o(a) (o(G)/o(a)) .)
2)
3)
4)
5)
6)
Exercises i) Suppose G is a cyclic group of order 4, G = {e, a, a2 , a3 } with a4 = e. Find the order of each element of G. Find all the subgroups of G. Suppose G is the additive group Z and H = 3Z. Find the cosets of H. Think of a circle as the interval [0, 1] with end points identified. Suppose G = R under addition and H = Z. Show that the collection of all the cosets of H can be thought of as a circle. Let G = R2 under addition, and H be the subgroup defined by H = {(a, 2a) : a ∈ R}. Find the cosets of H. (See the last exercise on p 5.) Normal Subgroups We would like to make a group out of the collection of cosets of a subgroup H. In
ii) iii)
iv)
26
Groups
Chapter 2
general, there is no natural way to do that. However, it is easy to do in case H is a normal subgroup, which is described below. Theorem 1) 2) 3) 4) If H is a subgroup of a group G, then the following are equivalent. If a ∈ G, then aHa−1 = H If a ∈ G, then aHa−1 ⊂ H If a ∈ G, then aH = Ha Every right coset is a left coset, i.e., if a ∈ G, ∃ b ∈ G with Ha = bH.
Proof 1) ⇒ 2) is obvious. Suppose 2) is true and show 3). We have (aHa−1 )a ⊂ Ha so aH ⊂ Ha. Also a(a−1 Ha) ⊂ aH so Ha ⊂ aH. Thus aH = Ha. 3) ⇒ 4) is obvious. Suppose 4) is true and show 3). Ha = bH contains a, so bH = aH because a coset is an equivalence class. Thus aH = Ha. Finally, suppose 3) is true and show 1). Multiply aH = Ha on the right by a−1 . Definition If H satisfies any of the four conditions above, then H is said to be a normal subgroup of G. (This concept goes back to Evariste Galois in 1831.) Note For any group G, G and e are normal subgroups. If G is an abelian group, then every subgroup of G is normal. Exercise Show that if H is a subgroup of G with index 2, then H is normal.
Exercise Show the intersection of a collection of normal subgroups of G is a normal subgroup of G. Show the union of a monotonic collection of normal subgroups of G is a normal subgroup of G. Exercise Let A ⊂ R2 be the square with vertices (−1, 1), (1, 1), (1, −1), and (−1, −1), and G be the collection of all "isometries" of A onto itself. These are bijections of A onto itself which preserve distance and angles, i.e., which preserve dot product. Show that with multiplication defined as composition, G is a multiplicative group. Show that G has four rotations, two reflections about the axes, and two reflections about the diagonals, for a total of eight elements. Show the collection of rotations is a cyclic subgroup of order four which is a normal subgroup of G. Show that the reflection about the x-axis together with the identity form a cyclic subgroup of order two which is not a normal subgroup of G. Find the four right cosets of this subgroup. Finally, find the four left cosets of this subgroup.
[n − 1]. Proof Multiplication of elements in G/N is multiplication of subsets in G. Also Zn is cyclic because each of [1] and [−1] = [n − 1] is a generator.
Proof The element [a] is a generator iff the subgroup generated by [a] contains [1] iff ∃ an integer k such that [a]k = [1] iff ∃ integers k and l such that ak + nl = 1. . It follows that they honor identities and inverses. and G/N is the collection of all cosets. The coset E does not depend upon the choice of c and d. We already know that if p is a prime. Theorem Suppose G is a multiplicative group. i. n > 1. Furthermore. Then (N a) · (N b) = N (a · b) is a well defined multiplication (binary operation) on G/N . n) = 1. if G is finite.. In this section we list
.Chapter 2
Groups
27
Quotient Groups Suppose N is a normal subgroup of G. Zn . This is made precise in the next theorem. and C and D are cosets.) Homomorphisms Homomorphisms are functions between groups that commute with the group operations.. Note that [a] = [r] where r is the remainder of a divided by n. [1]. (N a) · (N b) = N (aN )b = N (N a)b = N (a · b). N is a normal subgroup. determine those elements of finite order. If a is an integer. Note that [a] + [b] = [a + b]. Exercise Show that a positive integer is divisible by 3 iff the sum of its digits is divisible by 3. Once multiplication is well defined. and thus the distinct elements of Zn are [0]. In the additive abelian group R/Z. and [a] = [a + nl] for any integer l. Its identity is N and (N a)−1 = (N a−1 ). Exercise Write out the above theorem for G an additive group. then [a] is a generator of Zn iff (a. and N = nZ. the group axioms are immediate.. Example Suppose G = Z under +. (See the fifth exercise on page 18. the group of integers mod n is defined by Zn = Z/nZ. Note that [10] = [1] in Z3 .e. G/N is a group. Theorem If n > 1 and a is any integer. −[a] = [−a]. We wish to define a coset E which is the product of C and D. and with this multiplication. o(G/N ) = o(G)/o(N ).. E = N (c · d). define E to be the coset containing c · d. and in this case it is just [a]m = [am]. the coset a + nZ is denoted by [a]. because Zp has p elements. which is quite easy. any non-zero element of Zp is a generator. Any additive abelian group has a scalar multiplication over Z. If c ∈ C and d ∈ D.
if H is ¯ ¯ normal in G. if the equation f (x) = g has a ¯ ¯ Suppose G and G are groups and f : G → G is a homomorphism. and the second is in G. the kernel is the set of e ¯ ¯ solutions to the equation f (x) = e. if f (g) = g then ¯ ¯ g ¯ −1 g ¯ f (¯) = N g where N = ker(f ). f (a · b) = f (a) · f (b). the student should rewrite the material in additive notation. a function f : G → G is a homomorphism if. 12).e. for all a. Furthermore. (If G is an additive group. and should be considered as the cornerstones of abstract algebra. ¯ If H is a subgroup of G. In other words. ¯ ¯ ¯ ¯ If H is a subgroup of G. f is injective ⇔ ker(f ) = e. In particular. f −1 (H) is a subgroup of G. In other words. the inclusion i : H → G is a homomorphism. while on the right side it is in G. f −1 (¯) is void or is a coset of ker(f ). ker(f ) = f −1 (0). the group ¯ operation is in G. ¯ ¯ f (a−1 ) = f (a)−1 . ¯ ¯ Definition If G and G are multiplicative groups. The function f : Z → Z defined by f (t) = 2t is a homomorphism of additive groups.
5)
6) 7)
.
We now catalog the basic properties of homomorphisms.. then f −1 (H) is normal in G. The function h : Z → R − 0 defined by h(t) = 2t is a homomorphism from an additive group to a multiplicative group. b ∈ G. If g ∈ G. As always.28
Groups
Chapter 2
the basic properties. while the function defined by f (t) = t + 2 is not a homomorphism. f (H) is a subgroup of G. image(f ) is ¯ a subgroup of G. The first inverse is in G. Properties 11). The kernel of f is a normal subgroup of G. These will be helpful later on in the study of ring homomorphisms and module homomorphisms. and 13) show the connections between coset groups and homomorphisms.) ¯ ¯ ¯ Examples The constant map f : G → G defined by f (a) = e is a homomorphism. The kernel of f is defined by −1 ker(f ) = f (¯) = {a ∈ G : f (a) = e}. ¯ If H is a subgroup of G. Theorem 1) 2) 3) 4) f (e) = e. i. On the left side.
=
9)
10) Isomorphisms preserve all algebraic properties. f is called an isomorphism. then H is a subgroup of G ¯ ¯ iff f (H) is a subgroup of G. ¯ f is injective. ¯ In this case. by definition. The ¯ ¯ ¯ image of f is the image of f and the kernel of f is ker(f )/H. and we write G ≈ G. then G/H ≈ G (see below). Then π : G → G/H defined by π(a) = Ha is a surjective homomorphism with kernel H. ¯ ¯ If f : G → G is a bijection. i. For example. then f : G/H → G ¯(Ha) = f (a) is a well-defined homomorphism making defined by f the following diagram commute. This is a key fact which is used routinely in topics such as systems of equations and linear differential equations. if h : G →G is = a homomorphism. then the set of all solutions is a coset of N = ker(f ). then the function f −1 : G → G is a homomorphism. G π G/H
c& &
f
& & ¯
E b & &
¯ G
f
Thus defining a homomorphism on a quotient group is the same as defining a homomorphism on the numerator which sends the denominator to e. domain(f )/ker(f ) ≈ image(f ). G is ¯ is cyclic. and thus G/H ≈ image(f )..
. If H ⊂ ker(f ). Of course. because an cyclic iff G algebraic property is one that. H is normal in G iff f (H) is normal in G. 13) Given any group homomorphism f . f is also called an automorphism. 11) Suppose H is a normal subgroup of G.e. This is the fundamental connection between quotient groups and homomorphisms. this is somewhat of a cop-out. Furthermore.Chapter 2
Groups
29
solution. ¯ ¯ 12) Suppose H is a normal subgroup of G. if f is an isomorphism and H ⊂ G is a subset. is preserved under isomorphisms. 8) ¯ The composition of homomorphisms is a homomorphism. Thus if H = ker(f ). if ¯ ¯ f : G → G is a surjective homomorphism with kernel H. In the case ¯ G = G. then h ◦ f : G →G is a homomorphism. etc.
Under what conditions on g is there a homomorphism f : Z7 → G with f ([1]) = g ? Under what conditions on g is there a homomorphism f : Z15 → G with f ([1]) = g ? Under what conditions on G is there an injective homomorphism f : Z15 → G ? Under what conditions on G is there a surjective homomorphism f : Z15 → G ?
2)
3) 4)
Exercise We know every finite group of prime order is cyclic and thus abelian. Then G is a group under addition. Show that every group of order four is abelian. Find f −1 (g) and show it is a coset of ker(f ). Then f is a group homomorphism and the differential equation h +5h +6t2 h = g has a solution iff g lies in the image of f . Then f : Z → K defined by f (m) = am is a homomorphism from an additive group to a multiplicative group. 1] → R be defined by g(t) = t3 − 3t + 4. If o(a) = n. Exercise If a is an element of a group G. Define f : G → G by f (h) = dh = h . Then K is an infinite cycle group iff K is isomorphic to the integers under addition. When is there a homomorphism from Zn to G which sends [1] to a? What are the homomorphisms from Z2 to Z6 ? What are the homomorphisms from Z4 to Z8 ? Exercise 1) Suppose G is a group and g is an element of G. Exercise Let G be as above and g ∈ G. K is a cyclic group of order n iff K ≈ Zn . Exercise Let G = {h : [0.. Now suppose this equation has a solution and S ⊂ G is the set of all solutions. If o(a) is infinite.30
Groups
Chapter 2
14) Suppose K is a group. there is always a homomorphism from Z to G which sends 1 to a. ¯ Proof of 14) Suppose G = K is generated by some element a. For which subgroup H of G is S an H-coset?
. Let g : [0. Define f : G → G by f (h) = h + 5h + 6t2 h. ker(f ) = nZ and ¯ f : Zn → K is an isomorphism. 1] → R : h has an infinite number of derivatives}. g = e.e. Show f dt is a homomorphism and find its kernel and image. f is an isomorphism. K ≈ Z. i.
Define f : G → G to be conjugation by a. Also show f is an automorphism and find its inverse. ha·b = ha ◦ hb .. Theorem (Cayley's Theorem) Suppose G is a multiplicative group with n elements and Sn is the group of all permutations on the set G. They are all denoted by the one symbol Sn . i. 2. Exercise Show that o(Sn ) = n!. and H = {f ∈ Sn : (n)f = n}. i.e. Let X = {1. The proof follows from the following observations.. variables are written on the left. i. A bijection f : X → X is called a permutation on X.. there is an isomorphism between S(X) and S(Y ). n}. Let g be any permutation on X with (n)g = 1... 2. f = (x)f . Therefore the composition f ◦ g means "f followed by g". S(X) forms a multiplicative group under composition. Permutations Suppose X is a (non-void) set. Then G is isomorphic to a subgroup of Sn . Thus if each of X and Y has n elements. and the collection of all these permutations is denoted by S = S(X). Proof Let h : G → Sn be the function which sends a to the bijection ha : G → G defined by (g)ha = g · a.Chapter 2
Groups
31
Exercise Suppose G is a multiplicative group and a ∈ G.. n}. f (g) = a−1 · g · a.
The Symmetric Groups Now let n ≥ 2 and let Sn be the group of all permutations on {1. h is a homomorphism. Show H is a subgroup of Sn which is isomorphic to Sn−1 .e. and these groups are called the symmetric groups on n elements.. Sn = S(X).e. . .. Show that f is a homomorphism. ha is a bijection from G to G. Exercise Show that if there is a bijection between X and Y . The following definition shows that each element of Sn may
.. In this setting. The next theorem shows that the symmetric groups are incredibly rich and complex. h is injective and thus G is isomorphic to image(h) ⊂ Sn . S(X) ≈ S(Y ). 1) 2) 3) For each given a. Find g −1 Hg.
. .32 be represented by a matrix... b2 . (This is obvious. . a2 )(a1 . In this case. Listed here are eight basic properties of permutations. c ) are disjoint provided ai = cj for all 1 ≤ i ≤ k and 1 ≤ j ≤ . a1 a2 . bk } is the same collection in some different order.. (This is easy. then g ◦ f is
3)
4)
5)
6)
...) Every nonidentity permutation can be written uniquely (except for order) as the product of disjoint cycles... which takes a little work.. bk and (a)f = a for all other a. . a3 ) · · · (a1 .. and is denoted by (a1 .ak−1 ak is called a k-cycle. 2) and (a1 . then p is even iff q is even.... ak } is a collection of distinct integers with 1 ≤ ai ≤ n.. Suppose f. {a1 . .) Every permutation can be written (non-uniquely) as the product of transpositions. f is an odd permutation... f is said to be an even permutation. Definition a1 a2 . 2)(1. a2 a3 . 2)(1. The cycles (a1 . ak ). 3) = (1... 2. For example (1. ak ).. Properties 9) and 10) are listed solely for reference. a2 . ak ) and (c1 . . g ∈ Sn . Theorem 1) 2) Disjoint cycles commute. They are all easy except 4). b1 b2 .. For these we have a special notation. and {b1 .. If one of f and g is even and the other is odd. This means that if f is the product of p transpositions and also of q transpositions. ) The parity of the number of these transpositions is unique..... A k-cycle is even (odd) iff k is odd (even). a2 . In the other case.ak a1 A 2-cycle is called a transposition.
Groups
Chapter 2
Definition Suppose 1 < k ≤ n. 3) is an even permutation. ak ) = (a1 ... There is a special type of permutation called a cycle. The composition of two permutations is computed by applying the matrix on the left first and the matrix on the right second. ak Then the matrix represents f ∈ Sn defined by (ai )f = bi for 1 ≤ i ≤ k. (Proof: I = (1.
b)(b. 2. . c. where t is the smallest odd integer this is possible. If f and g are both even or both odd. then (a. n). and (a.1) as the product of disjoint cycles. n)(b. (1.5. . b)(a. 6 5 4 3 1 7 2 Write (1. c) = (a. . {(1. b) and (a.
8)
The following parts are not included in this course. then (a. n}. This can be done by inductively "pushing n to the right" using the equations below.4) as the product of disjoint cycles. b). 9) For any n = 4.. . c) = (b. n)(a.3. Since I = (1. Since t is odd. n) = I. If a.7. Which of these permutations are odd and which are even? Write
. . b. (Of course there are subgroups of Sn which cannot be generated by two elements).. n). c)(c. n)} generates Sn . has no proper normal subgroups. it is at least 3. Proof of 4) It suffices to prove if the product of t transpositions is the identity I on {1. 2. b)(c. c)(a.4)(3. every even permutation is the product of 3-cycles. n). n)(a. . 3)3 . n)(b. c. . Thus An is a normal subgroup of index 2. 2. They are presented here merely for reference. (a. (a.. i.6.Chapter 2
Groups
33
odd. then g ◦ f is even. Its kernel (the subgroup of even permutations) is denoted by An and is called the alternating group.3. 2. An is simple. 2. (Now you may solve the tile puzzle on page viii. b. Note that (a. and Sn /An ≈ Z2 . n) cannot occur here because it would result in a shorter odd product. ..6.1)(1. . (Obvious.7)(2.
10) Sn can be generated by two elements. d)(a. c and d are distinct integers in {1. . then t is even. If a. . c.5.) 7) The map h : Sn → Z2 defined by h(even)= [0] and h(odd)= [1] is a homomorphism from a multiplicative group to an additive group. . Suppose this is false and I is written as t transpositions. We will rewrite I as a product of transpositions σ1 σ2 · · · σt where (n)σi = (n) for 1 ≤ i < t and (n)σt = n. n − 1}. and c are distinct integers in {1. which will be a contradiction. In fact. Suppose for convenience the first transposition is (a. d) = (a.7. n).) Exercise 1) 1 2 3 4 5 6 7 as the product of disjoint cycles. . Write (3. c) = (a.e. n) = (a. it follows that for n ≥ 3.7)(2. n)(a. n}. 2).
Define an addition on G1 × G2 by (a1 . op Show that an element of Sn is a permutation on {1. . . −1). a2 + b2 ). Let A ⊂ R2 be the square with vertices (−1. Show that it has the same identity and the same inverses as G. and G be the collection of all isometries of A onto itself. n} with the variable op written on the right. this section should be rewritten using multiplicative notation. −a2 ). Theorem Suppose G1 and G2 are additive groups. . 2. and that f : G → Gop defined by f (a) = a−1 is a group isomorphism. either Sn or Sn may be called the symmetric group. a2 ) + (b1 . Its "zero" is (01 . Also let τ = (4. (3)σ). We know from a previous exercise that G is a group with eight elements. a2 ) = (−a1 . 1).) Product of Groups
3)
4)
5)
6)
The product of groups is usually presented for multiplicative groups. 5. . Now consider the special case G = Sn . For simplicity we first consider the product of two groups. define a new multiplication on the set G by a ◦ b = b · a. This operation makes G1 × G2 into a group. 6) and find τ −1 (1. This defines a new group denoted by Gop . 2. depending on personal preference or context. (1. b2 ) = (a1 + b1 . The two theorems below are transparent and easy. although the case of infinite products is only slightly more difficult. n} with the variable written on the left. 3. . This shows that conjugation by σ is just a type of relabeling. the new multiplication is the old multiplication in the opposite order. . . . In other words. It is presented here for additive groups because this is the form that occurs in later chapters. . What is the order of their product? Suppose σ ∈ Sn . For background. It follows from Cayley's theorem that G is isomorphic to a subgroup of S8 . 4. and (−1. (1. . Show that σ −1 (1. . 2. read first the two theorems on page 11. 5)τ . but quite useful. . The convention used in this section is that an element of Sn is a permutation on {1. 2. Show that G is isomorphic to a subgroup of S4 . . c ) are disjoint cycles. . . . (Of course.34 2)
Groups
Chapter 2
Suppose (a1 . The projections π1 : G1 × G2 → G1 ¯ ¯
. −1). (2)σ. ak ) and (c1 . Show that H = {σ ∈ S6 : (6)σ = 6} is a subgroup of S6 and find its right cosets and its left cosets. If G is a multiplicative group. 02 ) and −(a1 . As an exercise. 3)σ = ((1)σ. the opposite group. 1).
Suppose G. a2 ) in G1 × G2 . so let's prove the last part. Exercise Suppose G1 and G2 are groups. Exercise Let R be the reals under addition. f2 ) where f1 : G → G1 and f2 : G → G2 }. Proof It is transparent that the product of groups is a group. Finally. Under this bijection. Exercise Suppose G1 and G2 are groups and i1 : G1 → G1 × G2 is defined by i1 (g1 ) = (g1 . Exercise Suppose n > 2. f2 (a) + f2 (b)). f2 (a)) + (f1 (b). Show that the addition in the product R × R is just the usual addition in analytic geometry. Show that G1 × G2 and G2 × G1 are If o(a1 ) = m and o(a2 ) = n. f2 (a + b)) and f (a) + f (b) = (f1 (a). f2 (b)) = (f1 (a) + f1 (b). show that Zmn is isomorphic to Zm × Zn iff (m. Is Sn isomorphic to An × G where G is a multiplicative group of order 2 ? One nice thing about the product of groups is that it works fine for any finite number. n) = 1.
. or even any infinite number. An examination of these two equations shows that f is a group homomorphism iff each of f1 and f2 is a group homomorphism. Show Z12 is isomorphic to Z4 × Z3 . and G2 are groups and f = (f1 . Suppose G is an additive group. Show Z4 is not isomorphic to Z2 × Z2 . f2 ) is a function from G to G1 × G2 . Exercise isomorphic. Now f (a + b) = (f1 (a + b). find the order of (a1 .
Exercise Show that if G is any group of order 4. G1 . f is a group homomorphism iff each of f1 and f2 is a group homomorphism. Let π2 : G1 × G2 → G2 be the projection map defined in the Background chapter. We know there is a bijection from {functions f : G → G1 × G2 } to {ordered pairs of functions (f1 . Usually G1 is identified with its image under i1 . Show i1 is an injective group homomorphism and its image is a ¯ normal subgroup of G1 × G2 . so G1 may be considered to be a normal subgroup of G1 × G2 . Show π2 is a surjective homomorphism with kernel G1 . The next theorem is stated in full generality. G is isomorphic to Z4 or Z2 ×Z2 .Chapter 2
Groups
35
and π2 : G1 × G2 → G2 are group homomorphisms. Therefore (G1 × G2 )/G1 ≈ G2 as you would expect. 02 ).
c. Its "zero" is {0t } and −{at } = {−at }. Show GT is a group. [m]. {at }n = {at n}. (For the ring and module versions. see exercises on pages 44 and 69. Note that the gcd of {45. [m]). Now let h : Z → Z8 × Z9 × Z35 be defined by h(m) = ([m]. [m]). Exercise Let f : Z → Z30 × Z100 be the homomorphism defined by f (m) = ([4m]. Note that GT is just another way of writing Gt .e.. [3]) in Z30 × Z100 . Exercise Suppose T is a non-void set.)
. This opt∈T
eration makes the product into a group. Find the kernel of f. the addition defined on
t∈T
GT is just the usual addition of functions used in calculus. Thus each Gs may be considered to be a normal subgroup of Gt . Under the natural bijection from {functions f : G → Gt } to {sequences of functions {ft }t∈T where ft : G → Gt }. i. the scalar multiplication on Gt by integers is given coordinatewise. Exercise Let f : Z → Z90 × Z70 × Z42 be the group homomorphism defined by f (m) = ([m]. Gt is an additive group. [m]. [m]. ¯ Each projection πs : Gt → Gs is a group homomorphism. let Gt = G. and d is greater than 1 and f : Z → Zb × Zc × Zd is defined by f (m) = ([m]. Let g : Z → Z45 × Z35 × Z21 be defined by g(m) = ([m]. Define an addition on Gt = Gt by {at } + {bt } = {at + bt }. 1] and G = R. Find the kernel of f and show that f is not surjective. Suppose G is an additive group. Finally suppose each of b. and GT is the collection of all functions f : T → G with addition defined by (f + g)(t) = f (t) + g(t). [m]). 35. f is a group homomorphism iff each ft is a group homomorphism. [3m]). G is an additive group. Find the kernel of g and determine if g is surjective. Proof The addition on Gt is coordinatewise. and for any t ∈ T .36
Groups
Chapter 2
Theorem Suppose T is an index set. [m]). Also note that if T = [0.
Exercise Suppose s is an element of T and πs : Gt → Gs is the projection map defined in the Background chapter. 21} is 1. Exercise Suppose s is an element of T and is : Gs → Gt is defined by is (a) = {at } where at = 0 if t = s and as = a. Find necessary and sufficient conditions for f to be surjective (see the first exercise on page 18). [m]. Find the order of ([4]. Finally. Show πs is a surjective homomorphism and find its kernel. Show is is an injective homomorphism ¯ and its image is a normal subgroup of Gt . Find the kernel of h and show that h is surjective. For each t ∈ T .
A section on Boolean rings is included because of their importance in logic and computer science.Chapter 3
Rings
Rings are additive abelian groups with a second operation called multiplication. there is an element 1 = 1R ∈ R such that if a ∈ R. If in addition 4) is satisfied. R is said to be a ring. and 3) are satisfied. Assuming the results of Chapter 2. this chapter flows smoothly. (The distributive law. (The associative property of multiplication. which connects addition and multiplication. 1) 2) If a. These concepts are developed in the Appendix. a · b = b · a. the 37
. c ∈ R.. it would be easy to do. ¯ ¯ ¯ ¯ If a. and R has a second binary ¯ operation (i. This is because ideals are also normal subgroups and ring homomorphisms are also group homomorphisms. b. i. map from R × R to R) which is denoted by multiplication.e. c ∈ R.e.. (a · b) · c = a · (b · c).) R has a multiplicative identity. b ∈ R. We do not show that the polynomial ring F [x] is a unique factorization domain. Also there is no mention of prime or maximal ideals. Consider the following properties. a · 1 = 1 · a = a.)
3) 4)
Definition If 1). (The commutative property for multiplication. b. R = 0. 2). The connection between the two operations is provided by the distributive law. although with the material at hand. Suppose R is an additive abelian group. because these concepts are unnecessary for our development of linear algebra. Examples The basic commutative rings in mathematics are the integers Z. R is said to be a commutative ring.) If a. a · (b + c) = (a · b) + (a · c) and (b + c) · a = (b · a) + (c · a).
Rn is never commutative. na = an.
Recall that. . we will define R[x1 . It will be shown later that Zn . ¯ Theorem 0 can never be a unit. b ∈ R. The product of units is a unit with (a · b)−1 = b−1 · a−1 . ¯ ¯ Units Definition An element a of a ring R is a unit provided ∃ an element a−1 ∈ R with a · a−1 = a−1 · a = 1. Rn is a ring. This is a basic example of a non-commutative ring. ¯ Of course. . scalar ¯ ¯ ¯ ¯ ¯ ¯ multiplication by n is the same as ring multiplication by n. This scalar multiplication can be written on the right or left. The next two theorems show that ring multiplication behaves as you would wish it to. a · 0 = 0 · a = 0. Since R = 0. Theorem 1) Suppose a. More
2)
. Then na = n · a. even if R is commutative. They should be worked as exercises. Under these operations.. n may be 0 even though n = 0. x2 . a polynomical ring in n variables. m ∈ Z. xn ]. that is.e. and the next theorem shows it relates nicely to the ring multiplication. and Rn is the collection of all n×n matrices over R. i. . In the next chapter. For example. Theorem 1) 2) Suppose R is a ring and a. the integers mod n. has a natural multiplication under which it is a commutative ring. b ∈ R and n.) Let n = n1. operations of addition and multiplication of matrices will be defined. 2 = 1 + 1. n ≥ 1. 1 is always a unit. the real numbers R. If n > 1. If a is a unit. . a−1 is also a ¯ −1 ¯ unit with (a )−1 = a. (This follows from the distributive law and the previous theorem. Now suppose R is any ring. and the complex numbers C. since R is an additive abelian group. it follows that 1 = 0.38
Rings
Chapter 3
rational numbers Q. ¯ ¯ ¯ ¯ ¯ ¯ (−a) · b = a · (−b) = −(a · b). Also if R is any commutative ring. it has a scalar multiplication over Z (page 20). (na) · (mb) = (nm)(a · b).
an are units. as shown in the next theorem. (−a) is a unit and (−a)−1 = −(a−1 ). Suppose R is a finite domain and a = 0. we first consider the concept of zero divisor. Then ¯ b = c and so a is a unit with a−1 = b = c. Proof b = b · 1 = b · (a · c) = (b · a) · c = 1 · c = c. R is a field if it is commutative and its non-zero elements form a group under multiplication.. . a is not a zero divisor. It suffices to require a left inverse and a right inverse. if a1 . Finally if a is a unit. Thus a is a unit and so R is a field. Definition A domain (or integral domain) is a commutative ring such that. a is ¯ ¯ a unit. Theorem Suppose a ∈ R and ∃ elements b and c with b · a = a · c = 1. f is surjective. it must have a two-sided inverse. An element a ∈ R is called a zero divisor provided it is non-zero and ∃ a non-zero element b with a · b = 0. a2 . then their product is a unit with (a1 · a2 · · · an )−1 = a−1 · a−1 · · · a−1 . A finite domain is a field. ¯ Then (a · b = a · c) ⇒ b = c. Theorem A field is a domain. ¯ ¯ Inverses are unique.
Proof A field is a domain because a unit cannot be a zero divisor. it cannot be a zero divisor. Theorem Suppose R is a commutative ring and a ∈ (R − 0) is not a zero divisor. if a = 0. ¯ by the pigeonhole principle. In other words. A field is a commutative ring such that. if a = 0.Chapter 3
Rings
39
generally. Definition Suppose R is a commutative ring.. In other words. multiplication by a is an injective map from R to R. The set of all units of R forms a multiplicative group denoted by n−1 1 n R∗ .. Then f : R → R defined by f (b) = a · b is injective and. Note that ¯ if a is a unit.
Corollary
Domains and Fields In order to define these two types of rings. In order for a to be a unit. It is surjective iff a is a unit.
.
[n − 1] is a unit. Suppose 3) is true. 1).. Now suppose 3) is false. Zn is a field. and C are fields.40
Rings
Chapter 3
Exercise Let C be the additive abelian group R2 .. [a]b = [a] · [b] = [ab].
Proof We already know 1) and 2) are equivalent. each of [1]. It leads to a neat little theory. Define multiplication by (a. the following are equivalent. because each says ∃ an integer b with [a]b = [1]. b) · (c. Show C is a commutative ring which is a field. because Zn is finite. (See the Chinese Remainder Theorem on page 50. [2]. [a] is a unit of the ring Zn . The ring axioms are easily verified. However. d) = (ac − bd. [a] is a generator of the additive group Zn . Then the following are equivalent. Note that 1 = (1.) We know from page 27 that Zn is an additive abelian group. 0) and if i = (0. R. Theorem 1) 2) 3) Suppose n > 1 and a ∈ Z.
Proof We already know from page 27 that 1) and 2) are equivalent. ad + bc). Thus 1) and 3) are equivalent. Corollary 1) 2) 3) If n > 1. Then by the previous theorem. as seen by the theorems below. Recall that if b is an integer. ¯ ¯ Examples Z is a domain. (a.
. the basic theory cannot be completed until the product of rings is defined. Then n = ab where 1 < a < n. The Integers Mod n The concept of integers mod n is fundamental in mathematics. Theorem Suppose n > 1. Since [a + kn] · [b + l n] = [ab + n(al + bk + kl n)] = [ab]. the multiplication Proof is well-defined. then i2 = −1. Zn is a domain. n is a prime. Q. n) = 1. This is a well defined binary operation which makes Zn into a commutative ring. Define a multiplication on Zn by [a] · [b] = [ab].. 1 < b < n.. and thus 2) is true.
R and 0 are ideals of R. Note that Z is a subring of Q. x2 = 1 can have at most two solutions ¯ in R (see the first theorem on page 46). Exercise List the units and their inverses for Z7 and Z12 . and (a. then .
Subrings Suppose S is a subset of a ring R. and thus occupy a special place in ring theory. Of course. That role is played by ideals. and R is a subring of C. then s may be a unit in R but not in S. every right or left ideal is an ideal. Show that (Z7 )∗ is a cyclic group but (Z12 )∗ is not. and thus [a] is a zero divisor and 1) is false. as well as in group theory. 2-sided) ideals of R. 1 ∈ S . and an ideal is never a subring (unless it is the entire ring). Note that if S is a subring of R and s ∈ S. ¯ If {It }t∈T is a collection of right (left. The a · b and b · a ∈ I word "ideal " means "2-sided ideal". then right (left. ¯ Then clearly S is a ring and has the same multiplicative identity as R. Finally show that if R is a domain. Ideals and Quotient Rings Ideals in ring theory play a role analagous to normal subgroups in group theory. Show that in Z12 the equation x2 = 1 has four ¯ solutions.
Theorem 1) 2)
Suppose R is a ring. Subrings do not play a role analogous to subgroups. These are called the improper ideals. Q is a subring of R. 2-sided) ideal of R. Note also that Z and Zn have no proper subrings. The statement that S is a subring of R means that S is a subgroup of the group R.Chapter 3
Rings
41
[a][b] = [0]. if R is commutative.) It is a
t∈T
.
left Definition A subset I of a ring R is a right ideal provided it is a subgroup 2−sided a·b ∈ I b·a∈I of the additive group R and if a ∈ R and b ∈ I. (See page 22. b ∈ S ⇒ a · b ∈ S).
I = R. called a principal ideal. Thus every subgroup of Z is a principal ideal. then the following are equivalent.42 3) Furthermore. and the ring axioms are easily verified. (On the left. iii) I contains 1. i) I = R. 2-sided)
ideal of R. Show that R is a field iff R contains no proper ideals. R/I is an additive abelian group. Homomorphisms ¯ ¯ Definition Suppose R and R are rings. if the collection is monotonic. and I = nZ. The following theorem is just an observation. Multiplication of cosets defined by (a + I) · (b + I) = (ab + I) is well-defined and makes R/I a ring. 4) If a ∈ R. Thus if R is commutative. but it is in some sense the beginning of ring theory. b ∈ R then f (a · b) = f (a) · f (b). Thus multiplication is well defined. ¯ Observation If R = Z. ii) I contains some unit u. because it is of the form nZ. If R is a commutative ring and I ⊂ R is an ideal. Proof (a + I) · (b + I) = a · b + aI + Ib + II ⊂ a · b + I. aR is an ideal. The multiplicative identity is (1 + I). ¯
5)
Exercise Suppose R is a commutative ring. Since I is a normal subgroup of the additive group R. Theorem Suppose R is a ring and I ⊂ R is an ideal. then
t∈T
Rings
Chapter 3
It is a right (left. multiplication
. I = aR is a right ideal. A function f : R → R is a ring homomorphism provided 1) 2) 3) f is a group homomorphism f (1R ) = 1R and ¯ ¯ ¯ if a. n > 1. the ring structure on Zn = Z/nZ is the same as the one previously defined.
¯
Here is a list of the basic properties of ring homomorphisms.)
43
The kernel of f is the kernel of f considered as a group homomorphism. The kernel of a ring homomorphism is an ideal of the domain.
The identity map IR : R → R is a ring homomorphism. while on the right multiplication is in R. f is also called a ring automorphism. Theorem 1) 2) ¯ Suppose each of R and R is a ring. Suppose I is an ideal of R. ¯ Suppose f : R → R is a homomorphism and I is an ideal of R. From now on the word "homomorphism" means "ring homomorphism". π(a) = (a + I). if f : R → R is a surjective ¯ ring homomorphism with kernel I. ¯ ¯ ¯ The composition of ring homomorphisms is a ring homomorphism. Much of this work has already been done by the theorem in group theory on page 28. ¯ ¯ ¯ If I ⊂ ker(f ).Chapter 3
Rings ¯ is in R. Then π is a surjective ring ¯ homomorphism with kernel I. then f : R/I → R defined by f (a + I) = f (a)
3) 4)
5) 6)
7)
8)
. namely ker(f ) = f −1 (0). then f −1 (I) is an ideal of R. ¯ ¯ In fact. and π : R → R/I is the natural projection. I = R. The image of a ring homomorphism is a subring of the range. ¯ The zero map from R to R is not a ring homomorphism (because it does not send 1R to 1R ). ¯ If f : R → R is a bijection which is a ring homomorphism. −1 ¯ → R is a ring homomorphism. Furthermore. I = R. if f : R → R is a homomorphism and I ⊂ R is an ideal. Such an f is called then f : R ¯ a ring isomorphism. In the case R = R. then R/I ≈ R (see below).
Notice that much of the work has been done in the previous exercise. Suppose S is a non-void set and α : S → T is a function. domain(f )/ker(f ) ≈ image(f ). 1] and R = R. i. 9) Given any ring homomorphism f. It is only necessary to show that A is a subring of the ring R[0..
Exercise Find a ring R with a proper ideal I and an element b such that b is not a unit in R but (b + I) is a unit in R/I. Define addition and multiplication on RT point-wise. Exercise Now consider the case T = [0. Proof We know all this on the group level. A ={f : [0.
.1] . Exercise Show that if u is a unit in a ring R. and it is only necessary ¯ to check that f is a ring homomorphism. Exercise Suppose T is a non-void set. which is obvious. The image of f is the image of f . f is injective. 1] → R : f has an infinite number of derivatives}.e. This means if f and g are functions from T to R. R π
c &&
f
&
E b & &
¯ R
&
&
&
¯ f
R/I Thus defining a homomorphism on a quotient ring is the same as defining a homomorphism on the numerator which sends the ¯ denominator to zero. show that f : R → R defined by f (a) = u−1 · a · u is a ring homomorphism which is an isomorphism. and RT is the collection of all functions f : T → R.44
Rings
Chapter 3
is a well-defined homomorphism making the following diagram commute. If f : T → R is a function. Show that under these operations R T is a ring. and so R/I ≈ image (f ). define a function α∗ (f ) : S → R by α∗ (f ) = f ◦ α. R is a ring.1] be the collection of all C ∞ functions. Show A is a ring. and ¯ ¯ the kernel of f is ker(f )/I. then conjugation by u is an automorphism on R. That is. Let A ⊂ R[0. Show α∗ : RT → RS is a ring homomorphism. then (f + g)(t) = f (t) + g(t) and (f · g)(t) = f (t)g(t). Thus if I = ker(f ).
.. . Suppose n ≥ m and the result holds for any polynomial of degree less than
.) + (b0 .). Note that on the right. R[x ] is a commutative ring.. We outline the proof of existence and leave uniqueness as an exercise. ∃! h.. b1 . the top coefficient of f will always be a unit.. and it is easy to see that the collection of polynomial functions forms a commutative ring. a1 . ¯ To be more formal. . f (x ) = a0 + a1 x + · · +an x n . Under the obvious addition and multiplication. a0 b1 + a1 b0 . Definition Suppose R is a commutative ring and x is a "variable" or "symbol".) such that each ai ∈ R and only a finite number are non-zero. Suppose f = a0 + a1 x + · · +am xm where m ≥ 1 and am is a unit in R. If the top term an = 1.
Proof Suppose f and g are non-zero polynomials. as is often done for convenience.) · (b0 .. ¯ Proof This theorem states the existence and uniqueness of polynomials h and r.. ¯ and is denoted by n = deg(f ). a1 . Then ai bj xi+j is the first non-zero term of f g. think of a polynomial a0 + a1 x + · · · as an infinite sequence (a0 . the idea is to divide f into g ¯ until the remainder has degree less than m. set h = 0 and r = g. a1 + b1 .) = (a0 b0 ...) and (a0 . a1 . a0 b2 + a1 b1 + a2 b0 . . Another way to prove this theorem is to look at the bottom ¯ terms instead of the top terms. we consider real functions f which are polynomials. The polynomial ring R[x ] is the collection of all polynomials f = a0 + a1 x + · · +an x n where ai ∈ R. Then (a0 . We can do the same thing formally in a purely algebraic setting.. the ring multiplication a · b is written simply as ab. For the general case. f ∈ R[x ] has degree ≥ 1 and its top coefficient is a unit in R.) Then for any g ∈ R[x ]. The proof is by induction on the degree of g.
Theorem
If R is a domain. For any g with deg(g) < m. r ∈ R[x ] such that g = f h + r with r = 0 or deg(r) < deg(f ). Theorem (The Division Algorithm) Suppose R is a commutative ring. . (If R is a field. . The degree of a non-zero polynomial f is the largest integer n such that an = 0. R[x ] is also a domain. .) = (a0 + b0 ... b1 . then f is said to be monic.. The sum and product of polynomials are again polynomials..Chapter 3
Rings Polynomial Rings
45
In calculus. . Let ai xi and bj xj be the first non-zero terms of f and g. Then deg(f )+deg(g) = deg(f g) and thus f g is not 0.
each coset of I can be written uniquely in the form (c0 + c1 x + · · +cn−1 x n−1 + I). (If h has degree 0. Also show that if g(x) = x2 + bx + c. n > 0. This is called The Fundamental Theorem of Algebra.. Theorem Suppose F is a field. Let c1 . The result follows from the equation f (h1 + bxt ) + r = g.e. c2 . i. and g(x) = a0 + a1 x + · · · + an xn is a Theorem polynomial of degree n with at least one root in R. given any ideal I. Definition A domain T is a principal ideal domain (PID) if. Note that Z is a PID and any field is PID.) ¯ Proof Uniqueness is easy so let's prove existence. Note that f = x − c divides g iff c is ¯ a root of g. C is an algebraically closed field. Furthermore. This is a good exercise in the use of the division algorithm.
. Note this is Proof.e. ck be the distinct roots of g in the ring R. we say "all the roots of g are 0". ∃ h1 and r with f h1 + r = (g − f bxt ) and deg(r) < m. Note If g is any non-constant polynomial in C[x].46
Rings
Chapter 3
n. and n is the smallest positive integer such that I contains a polynomial of degree n. x − c divides g with remainder g(c). I is a proper ideal of F [x ]. g(c) = 0. Show that if g has odd degree then it has a real root.. Exercise Suppose g is a non-constant polynomial in R[x]. Now suppose g is a polynomial of degree n and c1 is a root of g. Thus F [x ] is a PID. Now ∃ a monomial bxt with t = n − m and deg(g − f bxt ) < n.e. Suppose n > 1 and the theorem is true for any polynomial of degree less than n. n2 . Then ∃ a unique sequence of positive integers n1 . i. ∃ t ∈ T such that I = tT. similar to showing that a subgroup of Z is generated by one element (see page 15). Since h1 has degree less than n. if h = an . More generally... . and in that case both roots belong to R. then we say "all the roots of g belong to R". By induction. . Then I contains a unique polynomial of the form f = a0 + a1 x + · · +an−1 x n−1 + x n and it has the property that I = f F [x ]. If g = an xn . Suppose g is a polynomial of degree n. nk and a unique polynomial h with no root in R so that g(x) = (x − c1 )n1 · · · (x − ck )nk h(x).. then it has a real root iff b2 ≥ 4c.. and it is assumed without proof for this textbook. Then g has at most n roots. Note If r = 0 we say that f divides g. Then ∃ a polynomial h1 with g(x) = (x − c1 )h1 .. The theorem is clearly true for n = 1. ¯ Suppose R is a domain. the result follows by induction. all the roots of g belong to C. i.
i. Suppose R is a subring of a commutative ring C and c ∈ C. This is a good way to look at the complex numbers.. We do not develop the theory of F [x ] here. Write out the multiplication table for this ring and show that it is a field. The next definition and theorem are included merely for reference. The image of h is the smallest subring of C containing R and c. Thus if F is a field. then h is an associate of f . the units of R[x ] are just the units of R. Exercise Show that.e. One difference is that the units of F [x ] are non-zero constants. This map h is called an evaluation map. Also multiplying two polynomials in R[x ] and evaluating is the same as evaluating and then multiplying in C. adjoin x to R and set x2 = −1. and this h is surjective. The statement that f is irreducible means that if h is a non-constant polynomial which divides f . i. Then ∃! homomorphism h : R[x ] → C with h(x ) = c and h(r) = r for all r ∈ R. The Division Algorithm corresponds to the Euclidean Algorithm. Show ker(h) = (x2 + 1)R[x ] and thus R[x ]/(x 2 + 1) ≈ C. The statement that g is an associate of f means ∃ a unit u ∈ F [x] such that g = uf .Chapter 3
Rings
47
Theorem.
In this chapter we do not prove F [x] is a unique factorization domain. Exercise Z2 [x ]/(x 2 + x + 1) has 4 elements. b ∈ R}. It is defined by h(a0 + a1 x + · · +an x n ) = a0 + a1 c + · · +an cn . h sends f (x) to f (c). However. to obtain C. Definition Suppose F is a field and f ∈ F [x] has degree ≥ 1. there exists a homomorphism h : R[x] → C which sends x to i. The theorem says that adding two polynomials in R[x ] and evaluating is the same as evaluating and then adding in C. The degree function corresponds to the absolute value function. Exercise Let C = {a + bi : a. Irreducible polynomials correspond to prime integers. while the units of Z
.e. In street language the theorem says you are free to send x wherever you wish and extend to a ring homomorphism on R[x]. nor do we even define unique factorization domain.. the units of F [x ] are the non-zero constants. Show that [1] + [2]x is a unit in Z4 [x ]. Since R is a subring of C. the development is easy because it corresponds to the development of Z in Chapter 1. if R is a domain. and should not be studied at this stage.
y] is a commutative ring and (R[x ])[y] ≈ R[x . x2 . vn are non-negative v v v integers. xn ] is a domain and its units are just the
.. y] to be any finite sum of monomials. Using this and induction on n. x2 .. and this factorization is unique up to order and associates. x2 . Theorem R[x .. the ideal I = xF [x.. If R is a domain. then ax n y m = ay m x n is called a monomial... it is easy to prove the following theorem.. the concept of a polynomial ring in n variables works fine without a hitch. Define an element of R[x . xn ] to be any finite sum of monomials. y] : f (0.. y] = {f ∈ F [x. any polynomial in x and y with coefficients in R may be written as a polynomial in y with coefficients in R[x ]. However F [x . x2 .. Here is the basic theorem. y] is not a PID. This gives a commutative ring and there is canonical isomorphism R[x1 . 1) 2) 3) F [x ]/(f ) is a domain. Thus the associates of f are all cf with c = 0 while the associates of an ¯ integer n are just ±n. For example. or as a polynomial in x with coefficients in R[y]. y] factors as the product of irreducibles. Define an element of R[x1 ..48
Rings
Chapter 3
are just ±1. . 0) = 0} is not principal. m ≥ 0. Side Comment It is true that if F is a field. In other words. xn ] ≈ (R[x1 .. then ax1 1 x2 2 · · · xn n is called a monomial. f is irreducible. R[x1 . xn−1 ])[xn ].) Theorem Suppose F is a field and f ∈ F [x ] has degree ≥ 1. . . (This theory is developed in full in the Appendix under the topic of Euclidean domains.. Then f factors as the product of irreducibles. If a ∈ R and v1 . Theorem units of R. each f ∈ F [x .. . v2 . F [x ]/(f ) is a field. Also the following are equivalent. If a ∈ R and n. . y] ≈ (R[y])[x ].. y] + yF [x.. Order does not matter here.
Definition Now suppose x and y are "variables". ¯ ¯ ¯ If R is a commutative ring and n ≥ 2.
Show f is a ring ¯ ¯ homomorphism whose kernel is the ideal (y) = yR[x. Exercise Suppose R and S are commutative rings. since multiplication is defined ¯ ¯ ¯ ¯ coordinatewise. an element e of T is called an idempotent provided e2 = e. Suppose T is a commutative ring and e ∈ T is an idempotent with 0 = e = 1. Proof We already know f is a group homomorphism iff each ft is a group homomorphism (see page 36). Theorem Suppose T is an index set and for each t ∈ T . 1S ). Under the natural bijection from {functions f : R → Rt } to {sequences of functions {ft }t∈T where ft : R → Rt }. Rt is a ring. Show T = R × S is not a domain. just as does the product of groups. and ¯ f (1R ) = {1t } iff ft (1R ) = 1t for each t ∈ T. On the additive abelian group Rt = Rt . and 0 × S = (1 − e)T . Finally. That is. ¯ ¯ ¯ ¯ Suppose I ⊂ R and J ⊂ S are ideals. f is a ring homomorphism iff each ft is a ring homomorphism. Note that {1t } is the multiplicative identity of Rt . Let e = (1. define multiplication by {rt } · {st } = {rt · st }. y]. y)) = p(x.
t∈T
. Exercise If T is any ring. if you adjoin y to R[x] and then factor it out. and f : T → R × S defined by f (t) = (et. Show R × 0 is an ideal and (R × S/R × 0) ≈ S. (1 − e)t) is a ring isomorphism. Show I ×J is an ideal of R×S and every ideal of R × S is of this form. (1 − e)2 = (1 − e). Then Rt is a ring and each projection πs : Rt → Rs is a ring homomorphism. This means f (p(x. Let R = eT and S = (1 − e)T . This shows that a commutative ring T splits as the product of two rings iff it contains a non-trivial idempotent. 0) ∈ R × S and show e2 = e. R × 0 = eT . you get R[x] back. y] → R[x] is the evaluation map which sends y to 0. Exercise Suppose R and S are rings. 0). Product of Rings The product of rings works fine. Show each of the ideals R and S is a ring with identity. Use the fact that "the domain mod the kernel is isomorphic to the image" to show R[x. y]/(y) is isomorphic to R[x]. The elements 0 and 1 are idempotents called the trivial idempotents.Chapter 3
Rings
49
Exercise Suppose R is a commutative ring and f : R[x. Note that R × 0 is not a subring of R × S because it does not contain (1R . Suppose R is a ring. f is a ring homomorphism iff each ft is a ring homomorphism.
the ring of integers is a "cornerstone". Thus Zn and Zn1 × · · ×Znt are isomorphic as rings.. ft ) : Z → Zn1 × · · ×Znt is surjective. . The element f (1)m = ([1]. Proof We wish to show that the order of f (1) is n. Use this and the Chinese Remainder Theorem to show that if b is a positive integer. each ni > 1. Thus the subgroup of R generated by 1 is a subring ¯ ¯ ¯ of R isomorphic to Z or isomorphic to Zn for some positive integer n. nt are integers. and thus f (1) is a group generator. (See page 23 for the definition of order. The next theorem is a classical generalization of this. and thus f is surjective. but it shows that in ring theory. If m and n are relatively prime. Furthermore. if R is a domain. and thus Zmn and Zm × Zn are isomorphic as groups and as rings. [m]) is zero iff m is a multiple of each of n1 ... .. then [a] = [ap ] in Zp (Fermat's Little Theorem). nj ) = 1 for all i = j. Theorem If R is a ring. the kernel of f is nZ.. . this map is surjective with kernel mnZ. Characteristic The following theorem is just an observation. and (ni . there is one and only one ring homomorphism f : Z → R.) Exercise Show that if a is an integer and p is a prime.) Theorem Suppose n1 . where n = n1 n2 · · nt . the order of f (1) is n. . (See exercise three on page 35.. (Note that the bracket symbol is used ambiguously. (See the fourth exercise on page 36 for the case t = 3. all the non-zero elements of R have the same order..50 The Chinese Remainder Theorem
Rings
Chapter 3
The natural map from Z to Zm × Zn is a group homomorphism and also a ring homomorphism. nt . Let fi : Z → Zni be defined by fi (a) = [a]. The non-negative integer n with ker(f ) = nZ is called the ¯ ¯ characteristic of R. Thus f is injective iff R has characteristic 0 iff 1 has infinite ¯ order... If f is not injective. and thus also as groups. [1])m = ([m].. It is given by f (m) = m1 = m.)
.) Then the ring homomorphism f = (f1 . Since their least common multiple is n.. ¯ It is an interesting fact that. . Definition Suppose R is a ring and f : Z → R is the natural ring homomorphism f (m) = m1 = m. the characteristic of R is the order of 1. it has the same last digit as b5 .
then every element of R/I is idempotent and thus R/I is a Boolean ring. If a ∈ R. Now suppose R has characteristic n. However it fits easily here. 3) If R is a domain. then n is a prime and each non-zero a ∈ R has order n. ¯ 2 2 2 2 Proof (a + a) = (a + a) = a + 2a + a = 4a. and so a = −a. ¯ ¯ ¯ ¯ Thus o(a) = ∞.. Then ma = m · a cannot be 0 because m. (In the language of Chapter 6. a is a non-zero element of R. If R has characteristic 0. each element of R is an idempotent. and is included for reference. ¯ ¯ ¯ Exercise Show that if F is a field of characteristic 0.
. Thus 2a = 0. I is a prime ideal iff I is a maximal ideal iff R/I ≈ Z2 ). 2a = a + a = 0. ¯ ¯ ¯ ¯ 4) The image of a Boolean ring is a Boolean ring. a2 = a. If a is a non-zero element of R.Chapter 3
Rings
51
Theorem Suppose R is a domain. if I is an ideal of R with I = R. It follows from 3) that R/I is a domain iff R/I is a field iff R/I ≈ Z2 . That is. Then a · (1 − a) = 0 and so a = 1. If R has finite characteristic n. Thus a · b = b · a. and thus Zn is a domain and n is a prime. and m is a positive integer. then each non-zero a ∈ R has infinite order. Definition A ring R is a Boolean ring if for each a ∈ R. Theorem 1) Suppose R is a Boolean ring. ¯ R is commutative.
R has characteristic 2. Boolean Rings This section is not used elsewhere in this book. That is. F contains Q as a subring.
2)
Proof (a + b) = (a + b)2 = a2 + (a · b) + (b · a) + b2 = a + (a · b) − (b · a) + b. na = n · a = 0 · a = 0 and thus o(a)|n and thus o(a) = n. a = 0 and R is a domain. show that the injective homomorphism f : Z → F extends to an injective ¯ homomorphism f : Q → F . i.e. Proof Suppose R has characteristic 0.
Proof Suppose a = 0. Then R contains Zn as a subring. R ≈ Z2 .
it contains some element a. It is a classical theorem that ∃ a Boolean algebra of sets whose Boolean ring is isomorphic to R. b ∈ R ⇒ (a ∩ b) ∈ R. Show each fi is a homomorphism and thus f = (f1 . . Under this addition. Since R is non-void. let a = (X −a) be a complement of a in X. Then a ∪ b = (a ∩ b ) belongs to R and so 3) is true. These operations cup and cap are associative. Theorem Suppose R is a Boolean algebra of sets. R is not a group under cup or cap. If a is a subset of X. fn ) : R → Z2 × Z2 × · · ×Z2 is a ring homomorphism. In this case. The advantage of the ring viewpoint is that you can draw from the rich theory of commutative rings. and thus also to the Boolean ring of subsets above. Under this multiplication R becomes a Boolean ring with 1 = X. Anyway.. Now define a ∨ b = a ∪ b and a ∧ b = a ∩ b. Then ∅ = a ∩ a and X = a ∪ a belong to R. So let's just suppose R is a Boolean algebra of sets which is a Boolean ring with addition and multiplication defined as above.
Theorem If 1) and 2) are satisfied... n} and let R be the Boolean ring of all subsets of X. it is a classical fact that. R is called a Boolean algebra. (See exercises 1) and 4) on page 12. have identity elements. and each distributes over the other. 2. Define fi : R → Z2 by fi (a) = [1] iff i ∈ a. Show f is an isomorphism. a. Now suppose R is a non-void collection of subsets of X.. then 3) and 4) are satisfied. ¯ Exercise Let X = {1.) Exercise Use the last exercise on page 49 to show that any finite Boolean ring is isomorphic to Z2 × Z2 × · · ×Z2 . 1) 2) 3) 4) a ∈ R ⇒ a ∈ R. The advantage of the algebra is that it is symmetric in cup and cap. b ∈ R ⇒ (a ∪ b) ∈ R. Note Suppose R is a Boolean ring.. Define a multiplication on R by a · b = a ∩ b. and a. you have a Boolean algebra (ring). Define an addition on R by a + b = (a ∪ b) − (a ∩ b). . Note that o(R) = 2n .. commutative. if you have a Boolean ring (algebra). With these two operations (along with complement). a. b ∈ R. and so 4) is true. Proof Suppose 1) and 2) are true.52
Rings
Chapter 3
Suppose X is a non-void set.
. ∅ ∈ R and X ∈ R. Consider the following properties which the collection R may possess. R is called a Boolean algebra of sets. R is an abelian group with 0 = ∅ and ¯ a = −a.
A matrix is said to be square if it has the same number of rows as columns. Definition Suppose R is a ring and m and n are positive integers. A = (ai. we use the "sum" notation.Chapter 4
Matrices and Matrix Rings
We first consider matrices in full generality.1 . .1 . If the elements of Rn are written as row vectors. Rn is defined to be the additive abelian group R × R × · · · × R. Square matrices are so important that they have a special notation. such as invertible matrices. The highlight of the chapter is the theorem that a square matrix is a unit in the matrix ring iff its determinant is a unit in the ring. Let Rm.n be the collection of all m × n matrices a1. Rn is identified with R1. a1. .e. are all classical... .n . i. Rn = R ⊕ R ⊕ · · · ⊕ R.1 .j . and characteristic polynomial. The topics. Our convention is to write elements of R n as column vectors. where each entry a ∈ R. am. trace. and determinant. Rn = Rn. over an arbitrary ring R. . i. This chapter concludes with the theorem that similar matrices have the same determinant. i. elementary matrices.j ) = . However.
53
. This will be used in the next chapter to show that an endomorphism on a finitely generated vector space has a well-defined determinant. . am.e.n A matrix may be viewed as m n-dimensional row vectors or as n m-dimensional column vectors.n . it will be assumed that R is commutative. . To emphasize that Rn does not have a ring structure. to identify Rn with Rn. and characteristic polynomial.n . transpose. after the first few pages. systems of equations. . trace.
Note in particular that scalar multiplication is defined on Rn .n · bn.. The matrix AB will have the same number of rows as A and the same number of columns as B. Rm.p → Rm.t . In the language of the next chapter.n × R → Rm. VA. Exercise and W = Consider real matrices A = 1 2 0 1 a b c d .
Scalar multiplication An element of R is called a scalar.n all of whose terms are zero. AV. j term of the sum is the sum of the i.e.n × Rn.n and c. Theorem Rm. the dot product of row s of A with column t of B.n . there is no distinction between right and left scalar multiplication.n . addition is a binary operation Rm. Find the matrices AU.p .j ) is defined to be the matrix whose (s.j ). Then (A + B)c = Ac + Bc A(c + d) = Ac + Ad A(cd) = (Ac)d A1 = A
and
This theorem is entirely transparent.j ) + (bi. The addition is defined by (ai.V = 0 1 1 0 .. Theorem Suppose A.j )c = (ai.j ) = (−ai. d ∈ R.n is an additive abelian group. i.j + bi. j terms. multiplication is a function Rm. i.n ≈ Rmn . Of course.n is a right module over the ring R.
Multiplication of Matrices The matrix product AB is defined iff the number of columns of A is equal to the number of rows of B. it merely states that Rm. and W
.t + · · · + as. Its "zero" is the matrix 0 = 0m. A matrix may be "multiplied" on the right or left by a scalar. The product (ai.. t) term is as. . The following theorem is just an observation.e.e.n × Rm.U = 2 0 0 1 .1 · b1. Furthermore. UA.j ).n → Rm. i.j · c). as additive groups.54
Matrices
Chapter 4
Addition of matrices To "add" two matrices.j )(bi. they must have the same number of rows and the same number of columns. B ∈ Rm. It is a function Rm. i. Also −(ai. AW .
A.j ) = (ai. if R is commutative. Right scalar multiplication is defined by (ai. the i..e.
i ci.j ) where a1. Note that ABC ∈ Rm. t) term of (AB)C is as.j bj. They form a group under multiplication called the general linear group and denoted by GLn (R) = (Rn )∗ . Find aR2 and R2 a.p Im A = A = AIn (The distributive laws) (A + B)C = AC + BC C(A + B) = CA + CB and whenever the
operations are defined. Then (AB)C = A(BC).j ) = BC.n . Show that the only ideal of R2 containing a is R2 itself.t = as. B. Then multiplication in Rn+m is given by A C B D E G F H = AE + BG CE + DG AF + BH CF + DH .n . Exercise Recall that if A is a ring and a ∈ A. t) terms are equal. Proof We must show that the (s.
Theorem
For each ring R and integer n ≥ 1.
Proof This elegant little theorem is immediate from the theorems above. C. H ∈ Rm .Chapter 4
Matrices
55
Definition The identity matrix In ∈ Rn is the square matrix whose diagonal terms are 1 and whose off-diagonal terms are 0.q .m . 0p.t =
j
as.j bj.j
j i
bj.j
as.i ci. The units of Rn are called invertible or non-singular matrices. and C ∈ Rp. E ∈ Rn . and D.n A0n. Theorem 1) 2) Theorem Suppose A ∈ Rm.i ci. Rn is a ring.
Multiplication by blocks Suppose A.
.i ci. Let (xi. The proof involves writing it out and changing the order of summation. F ∈ Rn. t) term of A(BC). xs.p = 0m. G ∈ Rm. then aA is right ideal of A. Let A = R2 and a = (ai.t = Then the (s. B ∈ Rn.p .q .1 = 1 and the other entries are 0.j yj.m A = 0p.j ) = AB and (yi.t which is the (s.n . Theorem (The associative law for matrix multiplication) Suppose A ∈ Rm.t =
i i j i.
Then I = I t = (AA−1 )t = (A−1 )t At . If A ∈ Rm. The geometry of this theorem will become transparent later in Chapter 5 when the matrix A defines an R-module endomorphism on Rn (see page 93). So row i (column i) of A becomes column i (row i) of At .
Proof of 5)
Exercise Characterize those invertible matrices A ∈ R2 which have A−1 = At .j = 0 for all i ≥ j (all j ≥ i).56 Transpose Notation tative ring.
. In this case.m . then A is upper (lower) triangular provided ai.n .
Matrices
Chapter 4
For the remainder of this chapter on matrices.j = 0 for all i > j (all j > i). In this case (A−1 )t = (At )−1 .e. Triangular Matrices If A ∈ Rn . i) term of A.m is the matrix whose (i. At ∈ Rn.. Theorem 1) 2) 3) 4) 5) (At )t = A (A + B)t = At + B t If c ∈ R. Thus if T = Rn and B is a nilpotent matrix. an element t ∈ T is said to be nilpotent provided ∃n n such that t = 0.
Proof The way to understand this is just multiply it out for n = 2 and n = 3.j = 0 for all i = j. (1 − t) is a unit with inverse 1 + t + t2 + · · · + tn−1 . Suppose A is invertible. Show that they form a subgroup of GL2 (R). then At is an n-dimensional column vector. Definition If T is any ring. (Ac)t = At c (AB)t = B t At If A ∈ Rn . j) term is the (j. If A is a square matrix. At is also square. A is diagonal if it is upper and lower triangular.
Transpose is a function from Rm.n to Rn. i. A is strictly upper (lower) triangular provided ai. Note that if A is upper (lower) triangular. then An = 0. Theorem If A ∈ Rn is strictly upper (or lower) triangular. If A is an n-dimensional row vector. Rn is non-commutative. ai. then At is lower (upper) triangular. I − B is invertible. suppose R is a commuOf course. then A is invertible iff At is invertible. for n > 1.
Multiplying by a scalar is the same as multiplying by a scalar matrix. a matrix of the form cIn . i.n .Chapter 4
Matrices 1 2 −3 Find the inverse of 0 1 4 . There are 3 types of elementary row and column operations on the matrix A.. For n > 1. Multiply column i by some unit a ∈ R. Type 1 Multiply row i by some unit a ∈ R.
Scalar matrices A scalar matrix is a diagonal matrix for which all the diagonal terms are equal. 0 0 1
57
Exercise
Let R = Z. Show AC is obtained from C by multiplying row i of C by ai . B ∈ Rm. Show A is a unit in Rn iff each ai is a unit in R. Exercise Suppose A ∈ Rn and for each B ∈ Rn . Elementary Operations and Elementary Matrices Suppose R is a commutative ring and A is a matrix over R. i. this shows how non-commutative Rn is. AB = BA.p .
is a diagonal matrix.
Type 2 Type 3
.e. and thus we may consider R to be a subring of Rn . and thus scalar matrices commute with everything. Add a times row j to row i where i = j and a is any element of R. Show A is a scalar matrix. The map R → Rn which sends c to cIn is an injective ring homomorphism. if B ∈ Rn . (cIn )B = cB = Bc = B(cIn ).
a1 a2 0 · 0 ·
Exercise
Suppose A =
an and C ∈ Rn. Interchange row i and row j.. A need not be square. Show that BA is obtained from B by multiplying column i of B by ai . Recall we are assuming R is a commutative ring. Interchange column i and column j.e. Add a times column i to column j where i = j and a is any element of R.
In type 2. then B is invertible and B −1 is an elementary matrix of the same type.2. They are obtained by performing row or column operations on the identity matrix. (See the exercise on page 54. or 3. It need not be square. and multiply on the left (right). all the off-diagonal elements are zero. there are two non-zero off-diagonal elements. There are three types.
In type 1.)
. BA = row operation on A and AB = column operation on A. perform the operation on an identity matrix to obtain an elementary matrix B. Theorem Suppose A is a matrix.58
Matrices
Chapter 4
Elementary Matrices Elementary matrices are square and invertible. In type 3. there is at most one non-zero off-diagonal element.
Type 2
B=
Type 3
B=
where i = j and ai. and it may be above or below the diagonal. The following theorem is handy when working with matrices.j 1 0
Type 1
B=
where a is a unit in R. Exercise Show that if B is an elementary matrix of type 1.j is any element of R. To perform an elementary row (column) operation on A.
1 1 a 1 0 1 1 1 0 1 1 1 0 1 1 1 1 1 0 1 1 ai. That is.
59
where d1.1 x1 + · · · + a1.j )
A matrix T is said to be in row echelon form if. perform row and column operations on A to reach the desired form.n .j ) ∈ Rm. . . can be written as one . namely as (ai.
Suppose A = (ai.1 .n xn = cm c1 x1 · · or AX = C. Is it possible to write them as products matrices over Z?
4)
For 1). the first non-zero term of row (i + 1) is to the right of the first non-zero term of row i. = matrix equation in one unknown. 3 0 of elementary of elementary Let A = 11 3 11 and D = .1 x1 + · · · + am. Write A and D as products 1 4 4 matrices over Q.j ) · · cm xn
a1.) 2) 3) Suppose A ∈ Fn is invertible. . the number of non-zero rows is the rank of T . .n xn = . for each 1 ≤ i < m. Notice that if T is in row-echelon form. For part 3).n and C =
c1 · · cm
∈ Rm = Rm. . The integer t is called the rank of A.Chapter 4 Exercise 1)
Matrices Suppose F is a field and A ∈ Fm. The system
. am.
Show ∃ invertible matrices B ∈ Fm and C ∈ Fn such that BAC = (di. use only row operations. Show A is the product of elementary matrices. Show ∃ an invertible matrix B ∈ Fm such that BA is in row echelon form. Systems of Equations
c1 .t = 1 and all other entries are 0. This shows the matrices B and C may be selected as products of elementary matrices. of m equations in n unknowns. Part 2) also follows from this procedure.1 = · · · = dt. . (See page 89 of Chapter 5.
the solution set f −1 (C) is the coset D + ker(f ) in Rn . which is given in Chapter 6.. For n = 2.) Suppose B ∈ Rm is invertible. If σ is a permutation on {1. Then AX = C and (BA)X = BC have the same set of solutions. If D ∈ Rn is one solution. (See part 7 of the theorem on homomorphisms in Chapter 2.j ) ∈ Rn . 2. Its solution set is ker(f ). this agrees with the definition above.
3)
4)
The geometry of systems of equations over a field will not become really transparent until the development of linear algebra in Chapter 5. The determinant is defined by | A |= sign(σ) a1. Determinants The concept of determinant is one of the most amazing in all of mathematics. let sign(σ) = 1 if σ is an even permutation. For each n ≥ 1 and each commutative ring R. For n = 1.σ(1) · a2. The proper development of this concept requires a study of multilinear forms. and sign(σ) = −1 if σ is an odd permutation. In this section we simply present the basic properties. If m = n and A ∈ Rm is invertible. In the language of the next chapter. page 28. Thus we may perform any row operation on both sides of the equation and not change the solution set. | (a) | = a. The next theorem summarizes what we already know about solutions of linear equations in this setting.σ(2) · · · an.. = ad − bc. c d Definition Let A = (ai.60
Matrices
Chapter 4
Define f : Rn → Rm by f (D) = AD.)
. Check that for
all σ
n = 2. then AX = C has the unique solution X = A−1 C.. Then f is a group homomorphism and also f (Dc) = f (D)c for any c ∈ R. this says that f is an R-module homomorphism. . Theorem 1) 2) AX = 0 is called the homogeneous equation. n}.σ(n) . AX = C has a solution iff C ∈ image(f ). (Note that here we are writing the permutation functions as σ(i) and not as (i)σ. determinant is a function from Rn a b to R.
Ar+1 . .n . We wish to show that τ = σ −1 and thus sign(σ) = sign(τ ).2 · · · aτ (n). . An )| + c|(A1 . .1 = Rn . ¯ Proof For simplicity. n} such that a1. .σ(1) · a2. It follows that the determinant of a matrix is equal to the determinant of its transpose. In the second expression. . Let γ be the transposition which interchanges one and two. Summary Determinant is a function d : Rn → R. An ) where each Ai ∈ Rn. and so τ (5) = 2.5 . Here we will use column vectors. . . it will appear as aτ (5). Then the first expression will contain the factor a2..1 · aτ (2).σ(2) · · · an. . .2 · · · aτ (n). . .1 . ¯ Theorem Suppose 1 ≤ r ≤ n.σ(1) · · · an. Anyway. Theorem |A| = sign(σ)a1. . This means that there is a permutation τ on {1. c ∈ R. An )| = a|(A1 .1 · aτ (2). . these pairs cancel in the summation.e.n . . Ar−1 .
all τ
. .σ(1) · a2. Therefore |A| = 0.5 . Cr . i. 2. .n . Ar+1 . .
You may view an n × n matrix A as a sequence of n column vectors or as a sequence of n row vectors. .σ(n) = sign(τ )aτ (1). . A1 = A2 . If two columns of A are equal. and since sign(τ )=−sign(τ γ). assume the first two columns are equal.1 · aτ (2). we may rearrange the factors so that the first comes from the first column. suppose σ(2) = 5. In the language used in the Appendix. To reduce the abstraction. sign(τ )aτ (1).Chapter 4
Matrices
61
For each σ. a1. .1 · · · aτ (n). .n and this summation has n! terms and Now |A| = Theorem n! is an even number. . .2 · · · aτ (n).2 · · · aτ γ(n). . aAr + cCr . aτ (1). τ is the inverse of σ and thus there are two ways to define determinant. This means we write the matrix A as A = (A1 . . Then for any τ . the two previous theorems say that d is an alternating multilinear form. A2 . and a. etc. This pairs up the n! terms of the summation.
all σ
all τ
Corollary
|A| = |At |. The next two theorems show that alternating implies skew-symmetric (see page 129). Cr ∈ Rn. Then |(A1 .σ(n) = aτ (1). . then |A| = 0. the second from the second column.σ(n) contains exactly one factor from each row and one factor from each column. . . Ar−1 . Since R is commutative.n = aτ γ(1). An )| Proof This is immediate from the definition of determinant and the distributive law of multiplication in the ring R.1 · aτ γ(2).σ(2) · · · an.
j be the determinant of the (n − 1) × (n − 1) matrix obtained by removing row i and column j from A. More generally. Rewrite the four preceding theorems using rows instead of columns. . . An )| + |(A2 . .j Cn. If a matrix has two rows equal or two columns equal. An )| + |(A1 . . The permutation τ is the finite product of transpositions.j = (−1)i+j Mi. does not change the determinant. Interchanging two rows or two columns multiplies the determinant by −1.2 Ci. Since the first and last of these four terms are zero. Aτ (2) . We know 0 = ¯ |(A1 + A2 . or adding c times one column to another column. Let Mi. Mi. .
There are 2n ways to compute | A |. . multiplies the determinant by c.j C2. . . expansion by any row or expansion by any column. the determinant is zero. A3 . Theorem Multiplying any row or column of matrix by a scalar c ∈ R.j C1. . | A |= a1. . Thus if any row or any column is zero. 2.1 + ai. then |A| = sign(τ )|(Aτ (1) . A2 .j + a2.2 + · · · + ai. Let Ci.62 Theorem one. .n . An )| + |(A2 . j) minor and cofactor of A. | A |= ai. A1 + A2 . its determinant is zero.n Ci. A1 . . . . A3 . Aτ (n) )|. An )| = |(A1 . Theorem For any 1 ≤ i ≤ n. An )|. c1 c2 c3
Exercise
. then the determinant is zero. show that |(A2 . .
The following theorem is just a summary of some of the work done so far. An )| = −|A|. n). . For any 1 ≤ j ≤ n. A1 . . The following theorem is useful but the proof is a little tedious and should not be done as an exercise. A3 . A2 . A3 . Theorem Proof Exercise If τ is a permutation of (1. A1 .j are called the (i. . Adding c times one row to another row. . . .j and Ci. A3 . a1 a2 a3 Let A = b1 b2 b3 . A3 . The determinant of A is the sum of six terms. the result follows. .j . .1 Ci. .j + · · · + an. . . .
Matrices
Chapter 4
Interchanging two columns of A multiplies the determinant by minus
Proof For simplicity. .j . . . . or one column is c times another column. . if one row is c times another row.
(For the proof. the matrix whose (j. An elementary matrix of type 3 is a special type of upper or lower triangular matrix.
One of the major goals of this chapter is to prove the converse of the preceding corollary. B ∈ Rn . then its determinant is the product of the diagonal elements.. i. | A |= −1. We assume it here without proof. Before
. Suppose A ∈ Rn is upper triangular. j) cofactor. so its determinant is 1. If A is an elementary matrix of type 2.Chapter 4
Matrices
63
Write out the determinant of A expanding by the first column and also expanding by the second row. | C −1 AC | = |ACC −1 | = | A |. | A | is the product of the diagonal elements. We will prove the first statement for upper triangular matrices.
Classical adjoint Suppose R is a commutative ring and A ∈ Rn . (Determinant by blocks) Suppose A ∈ Rn .
The following remarkable theorem takes some work to prove.e. An elementary matrix of type 2 is obtained from the identity matrix by interchanging two rows or columns. Suppose n > 2 and the theorem is true for matrices in Rn−1 .
1 = | I | = | AA−1 | = | A || A−1 | . i.) Theorem The determinant of the product is the product of the determinants. then | A | is a unit in R and | A−1 | = | A |−1 . and D ∈ Rm . If A ∈ R2 Proof is an upper triangular matrix..j )t . Thus | AB | = | BA | and if C is invertible. The result follows by expanding by the first column. A B Then the determinant of is | A || D |. Corollary Proof If A is a unit in Rn . B ∈ Rn. Theorem If A is an upper or lower triangular matrix. if A.e.m . If A is an elementary matrix of type 3. i) term is the (i. | AB | = | A || B |. The classical adjoint of A is (Ci. and thus has determinant −1. O D Theorem Proof Expand by the first column and use induction on n. see page 130 of the Appendix. | A |= 1.
i.64 we consider the general case.j )t are all | A | and the other elements are 0. We are now ready for one of the most beautiful and useful theorems in all of mathematics.
Proof We must show that the diagonal elements of the product A(Ci.j )t =
d −b −c a
.
.j ) and is thus | A | (computed by expansion by row s).) If A is invertible. B ∈ Rn and AB = I.j )t . That is. A is invertible iff | A | = 0. Thus if | A | = 1. t) term is 0. (Thus if R is a field.e. Theorem B is similar to B.j )t A = | A | I. if | A | = 1.j )t = (Ci.j ).j )t . and thus BA = I. suppose A. If A = a b c d then (Ci. The proof that (Ci. the (s. A is d −b −c a . In particular. t) term is the dot product of row s of A with row t of (Ci.j )t = (Ci. then A(Ci. then A−1 = | A |−1 (Ci.
invertible and A−1 = | A |−1 (Ci.j )t .j ) = |A| 0 0 |A| d −c −b a
Matrices
Chapter 4
and so (Ci. Then A is a unit in Rn iff | A | is a unit in R.
Exercise Show that any right inverse of A is also a left inverse. let's examine 2 × 2 matrices. Theorem
If R is commutative and A ∈ Rn . Similarity Suppose A. B is similar to A iff B is a conjugate of A. A−1 = (Ci. Thus if | A | is a unit in R. Since this is the determinant of a matrix with row s = row t. Show A is invertible with A−1 = B. For s = t. Then
A(Ci. B is said to be similar to A if ∃ an invertible C ∈ Rn such that B = C −1 AC. A−1 = Here is the general case.j )t A = |A|I is similar and is left as an exercise. The (s. B ∈ Rn . s) term is the dot product of row s of A with row s of (Ci. Theorem Suppose R is a commutative ring and A ∈ Rn . the classical adjoint of A. the (s.j )t A =
= | A | I.. Proof This follows immediately from the preceding theorem.
. Theorem Suppose A ∈ Rm.
Theorem Suppose A and B are similar. then trace(A) = trace(B).m am. One of the most useful properties of trace is trace(AB) = trace(BA) whenever AB and BA are defined.j ) ∈ Rn .j + · · ·+ bj.n . "Similarity" is an equivalence relation on Rn . bn )t .j = ai.. the trace of A is the sum of its diagonal terms. trace(A + B) = trace(A) + trace(B) and trace(AB) = trace(BA). Note that trace(AB) = trace(BA).
.2 + · · · + an. bj. By definition. B ∈ Rn . Here is the theorem in full generality. then D is similar to A.
Trace Suppose A = (ai. Theorem If A.1 b1. For example. Proof Suppose B = C −1 AC.i = trace(AB) =
1≤i≤m
1≤i≤m
1≤j≤n
1≤j≤n
trace(BA). trace(B) = trace(C −1 AC) = trace(ACC −1 ) = trace(A).
65
Proof
This is a good exercise using the definition. If D is similar to B and B is similar to A. and the second part is a special case of the previous theorem. .Chapter 4
Matrices B is similar to A iff A is similar to B.1 + a2. Then the trace is defined by trace(A) = a1. Proof This proof involves a change in the order of summation. b2 .. That is.m ...1 a1..n and B ∈ Rn.i = ai. an ) and B = (b1 . a2 .i + · · ·+ ai.j bj..n bn. Then AB and BA are square matrices with trace(AB) = trace(BA). Theorem Proof If A and B are similar. suppose A = (a1 . Then | A | = | B | and thus A is invertible iff B is invertible. Then | B | = | C −1 AC | = |ACC −1 | = | A |. Then AB is the scalar a1 b1 + · · · + an bn while BA is the n × n matrix (bi aj ). Proof The first part of the theorem is immediate.
then they have the same characteristic polynoCPB (x) = | (xI − C −1 AC) | = | C −1 (xI − A)C | =
Proof Suppose B = C −1 AC. i. but depends on linear algebra (see pages 93. Any λ ∈ R which is a root of CPA (x) is called a characteristic root of A. If A and B are similar. Show the following are equivalent. | A |= trace(A) = 0. Exercise Suppose A ∈ Rn and a ∈ R.e. | A | = | B | and trace(A) = trace(B). Determinant is a multiplicative homomorphism and trace is an additive homomorphism..
. Theorem CPA (x) = a0 + a1 x + · · · + an−1 xn−1 + xn where trace(A) = −an−1 and | A | = (−1)n a0 . A = Exercise 1) 2) 3) 4) Suppose F is a field and A ∈ F2 . In other words. Suppose R is a commutative ring. and c d CPA (x) = a0 + a1 x + x2 . Exercise
a b is a matrix in R2 . 94.
Note This exercise is a special case of a more general theorem. the characteristic polynomial CPA (x) ∈ R[x] is defined by CPA (x) = | (xI − A) |. If A and B are similar.66
Matrices
Chapter 4
Summary Determinant and trace are functions from Rn to R. and 98). CPA (A) = 0.
Characteristic polynomials If A ∈ Rn . A square matrix over a field is nilpotent iff all its characteristic roots are 0 iff it is similar to a strictly ¯ upper triangular matrix. ∃ an elementary matrix C such that C −1 AC is strictly upper triangular. Furthermore | AB | = | BA | and trace(AB) = trace(BA). | (xI − A) | = CPA (x). This remarkable result cannot be proved by matrix theory alone. Proof Theorem mials. show A satisfies its characteristic polynomial. This follows from a direct computation of the determinant. Find |aA| and trace(aA). Find a0 and a1 and show that a0 I + a1 A + A2 = 0. A2 = 0. CPA (x) = x2 .
this single integer determines V up to isomorphism. It is stated here in full generality only for reference and completeness. we restrict our attention to vector spaces. quotients. and any change of basis corresponds to conjugation of that matrix. and generating sets come together in one unified theory. Also any endomorphism f : V → V may be represented by a matrix. One of the goals in linear algebra is to select a basis so that the matrix representing f has a simple form. if f is nilpotent. It is in this chapter that the concepts about functions. This shows that finitely generated free R-modules have a well defined dimension. We give a simple proof that if R is a commutative ring and f : Rn → Rn is a surjective R-module homomorphism. The elementary facts about cosets. then f may be represented by a strictly upper triangular matrix. The key theorem is that any vector space V has a free basis.. For example.e. it has a well defined dimension. The basic theory is developed here in full generality. The proof is given in the Appendix. and with the beautiful theory relating orthogonal matrices and symmetric matrices. and homomorphisms follow the same pattern as in the chapters on groups and rings.
67
. and simplifies some of the development of linear algebra.e. then f is an isomorphism. i. solutions of equations.. and thus if V is finitely generated. modules are defined over an arbitrary ring R and not just over a field. and incredible as it may seem. As another example. if f is not injective. matrices.Chapter 5
Linear Algebra
The exalted position held by linear algebra is based upon the subject's ubiquitous utility and ease of application. The theorem on Jordan canonical form is not proved in this chapter. and should not be considered part of this chapter. After the general theory. i. then f may be represented by a matrix whose first column is zero. modules over a field. This chapter concludes with the study of real inner product spaces.
Thus for commutative rings. a0R = 0M . a1 . then (−a)r = −(ar) = a(−r). it is assumed that R is a ring and the word "R-module" (or sometimes just "module") means "right R-module". ¯ ¯ If a ∈ M . a2 ∈ M and r. If R is commutative and M = MR then left scalar multiplication defined by ra = ar makes M into a left R-module. r1 . Theorem 1) 2) 3) Proof Suppose M is an R-module. m) → rm satisfying r(a1 + a2 ) (r1 + r2 )a (r1 · r2 )a 1a ¯ = = = = ra1 + ra2 r1 a + r2 a r1 (r2 a) a
Note that the plus sign is used ambiguously. we may write the scalars on either side.
. as addition in M and as addition in R.68
Linear Algebra
Chapter 5
Definition Suppose R is a ring and M is an additive abelian group. In this text we stick to right R-modules. Convention Unless otherwise stated.
Notation The fact that M is a right (left) R-module will be denoted by M = MR (M = R M ). ¯ ¯ If a ∈ M and r ∈ R. This is a good exercise in using the axioms for an R-module. then f : M → M defined by f (a) = ar is a homomorphism of additive groups. r2 ∈ R. The statement that M is a right R-module means there is a scalar multiplication M ×R → M (m. In particular (0M )r = 0M . r) → mr satisfying (a1 + a2 )r a(r1 + r2 ) a(r1 · r2 ) a1 ¯ = = = = a1 r + a2 r ar1 + ar2 (ar1 )r2 a
for all a. If r ∈ R. The statement that M is a left R-module means there is a scalar multiplication R×M → M (r.
. then this submodule may be written as N1 + N2 + · · +Nn = {a1 + a2 + · · +an : each ai ∈ Ni }. (We know from the last exercise in Chapter 2 that N T is a group. On the left. T is an index set. N ) is a subgroup of N M . f (ar) = f (a)r. i..
t∈T
2) 3)
+t∈T Nt = {all finite sums a1 + · · +am : each ai belongs to some Nt } is a submodule. In this case N will be an R-module because the axioms will automatically be satisfied. For example. Also in 8) it is only necessary to show that HomR (M. the statement that a subset N ⊂ M is a submodule means it is a subgroup which is closed under scalar multiplication. an R-module homomorphism) provided it is a group homomorphism and if a ∈ M and r ∈ R. and for each t ∈ T . Note that 0 and M are submodules. n}. 1)
t∈T
Nt is a submodule of M . Exercise Suppose T is a non-void set. Also the proof of 3) is immediate. if a ∈ N and r ∈ R. Nt is a submodule of M . 2.e. scalar multiplication is in M and on the right it is in N . which is immediate.. in 5) of the theorem below. N ) is a submodule of N M . and in particular for subgroups of additive abelian groups. N1 + N2 is the smallest submodule of M containing N1 ∪ N2 . N ) forms an abelian group.
Proof We know from page 22 that versions of 1) and 2) hold for subgroups.. If T = {1.Chapter 5
Linear Algebra
69
Submodules If M is an R-module. Much
. . then ar ∈ N. To finish the proofs it is only necessary to check scalar multiplication. it is stated that HomR (M. So it is only necessary to show that HomR (M. Homomorphisms Suppose M and N are R-modules. and scalar multiplication defined by (f r)(t) = f (t)r. Show N T is an R-module.) This simple fact is quite useful in linear algebra. Note that if N1 and N2 are submodules of M . Theorem Suppose M is an R-module. called the ¯ improper submodules of M . and N T is the collection of all functions f : T → N with addition defined by (f +g)(t) = f (t)+g(t). N is an R-module. A function f : M → N is a homomorphism (i. Nt is a submodule. If {Nt } is a monotonic collection. The basic facts about homomorphisms are listed below. and so it is only necessary to check scalar multiplication.e.
An isomorphism f : M → M is called an automorphism. g : M → N are homomorphisms. In particular. ¯ This is just a series of observations. G ⊂ M is a submodule. Also (−f ) defined by (−f )(a) = −f (a) is a homomorphism. Theorem 1) 2) 3) 4) The zero map M → N is a homomorphism. HomR (M. NR ). The units of the endomorphism ring HomR (M. with multiplication defined to be composition. if f : M → N is a homomorphism. N ) = Hom(MR . Rn ) is just the matrix ring Rn and the automorphisms are merely the invertible matrices. Suppose f : M → N is a homomorphism. is a ring. N ) is an R-module. M ). The identity map I : M → M is a homomorphism. We will see later that if M = R n . f r defined by (f r)(a) = f (ar) = f (a)r is a homomorphism. HomR (M.70
Linear Algebra
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of this work has already been done in the chapter on groups (see page 28). (f + g ) ◦ k = (f ◦ k) + (g ◦ k). If k : P → M is a homomorphism. If R is commutative and r ∈ R.
5)
6)
7)
8) 9)
Proof
. If h : N → P is a homomorphism. forms an abelian group under addition. h ◦ (f + g) = (h ◦ f ) + (h ◦ g). If a bijection f : M → N is a homomorphism. and H ⊂ N is a submodule. In this case f and f −1 are called isomorphisms. The composition of homomorphisms is a homomorphism. A homomorphism f : M → M is called an endomorphism. the set of all homomorphisms from M to N . HomR (Rn . Thus the automorphisms on M form a group under composition. If f. If R is commutative. then g : M → M defined by g(a) = ar is a homomorphism. Then f (G) is a submodule of N and f −1 (H) is a submodule of M. define (f + g) : M → N by (f + g)(a) = f (a) + g(a). Furthermore. The sum of homomorphisms is a homomorphism. M ) are the automorphisms. image(f ) is a submodule of N and ker(f ) = f −1 (0) is a submodule of M . then f −1 : N → M is also a homomorphism. HomR (M. Then f + g is a homomorphism.
That is. ¯ Proof Suppose f (1) = g(1). Of particular importance is the case Rn = R ⊕ · · ⊕R = Rn. it is shown that any additive group M admits a scalar multiplication by integers. Given m ∈ M . This is the same as the definition before for Rn when n = 1. Proof The definitions are the same except expressed in different language. and if M is abelian. Summary Additive abelian groups are "the same things" as Z-modules. Then f (r) = f (1 · r) = f (1)r = g(1)r = g(1 · r) = ¯ ¯ ¯ ¯ ¯ ¯ g(r). the properties are satisfied to make M a Z-module. Thus
. We begin with the case n = 1. if m ∈ M . Furthermore. Furthermore. If M and N are Q-modules and f : M → N is a Z-module homomorphism. HomR (R.n an R-module.
Theorem Suppose M = MR and f.n admits a scalar multiplication by elements in R. etc. the study of additive abelian groups is a special case of the study of R-modules.1 (see page 53). This makes R a right R-module denoted by RR (or just R). h : R → M defined by h(r) = mr is a homomorphism. a2 = a + a. Then f = g. because a1 = a. ∃! homomorphism h : R → M with ¯ h(1) = m. R as a right R-module Let M = R and define scalar multiplication on the right by ar = a · r. Theorem Suppose R is a ring and N is a subset of R. scalar multiplication is just ring multiplication. Note that this is the only way M can be a Zmodule. While group theory in general is quite separate from linear algebra.Chapter 5
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71
Abelian groups are Z-modules On page 21. Then N is a submodule of RR (R R) iff N is a right (left) ideal of R. R-modules are also Z-modules and R-module homomorphisms are also Exercise Z-module homomorphisms. must it also be a Q-module homomorphism? Homomorphisms on Rn Rn as an R-module On page 54 it was shown that the additive abelian group Rm. g : R → M are homomorphisms with f (1) = ¯ g(1). then f is also a Z-module homomorphism. if f : M → N is a group homomorphism of abelian groups. M ) ≈ M . In other words. The properties listed there were exactly those needed to make Rm.
. Furthermore.
0 r1 ¯ · · Homomorphisms on Rn Define ei ∈ Rn by ei = 1i . Any R-module homomorphism from Rn to M is determined by its values on the basis. M ) is an R-module. Show f is injective.. . ∃! homomorphism h : Rn → M with h(ei ) = mi for 1 ≤ i ≤ m. M ) to M . Theorem Suppose M = MR and f. We will see later that the product M n is an R-module with scalar multiplication defined by (m1 . If R is commutative so that HomR (Rn .72
Linear Algebra
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evaluation at 1 gives a bijection from HomR (R. this theorem gives an R-module isomorphism from HomR (Rn . the above theorem states that multiplication on left by some m ∈ R defines a right R-module homomorphism from R to R. and every module homomorphism is of this form. en } is called the canonical free basis or standard basis for R n . mn r). or which R-module M is selected. M ) to M n = M × M × · · ×M and this bijection is a group isomorphism. In the case M = R. mn ∈ M . Note that any ¯ · · 0 rn ¯ can be written uniquely as e1 r1 + · · +en rn .. m2 . Exercise Suppose R is a field and f : RR → M is a non-zero homomorphism. if m1 ... The homomorphism h is defined by h(e1 r1 + · · +en rn ) = m1 r1 + · · +mn rn . Then f = g. it is an isomorphism of R-modules. M ) to M n ..
. . g : Rn → M are homomorphisms with f (ei ) = g(ei ) for 1 ≤ i ≤ n. The element m should be thought of as a 1 × 1 matrix. mn )r = (m1 r. and any function from that basis to M extends uniquely to a homomorphism from Rn to M . and this bijection is clearly ¯ a group isomorphism. m2 . m2 r. This theorem reveals some of the great simplicity of linear algebra. We now consider the case where the domain is Rn . ... Proof The proof is straightforward. Note this theorem gives a bijection from HomR (Rn . If R is commutative. It does not matter how complicated the ring R is... The sequence {e1 .
These properties are made explicit in the next two theorems. Corollary HomR (Rn .Chapter 5
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Now let's examine the special case M = Rm and show HomR (Rn . Thus HomR (Rn .m . then f + g is given by the matrix A + C. . Theorem Suppose A = (ai. Matrices over R give R-module homomorphisms! Furthermore. Rm ) ≈ Rm. g : Rn → Rm are given by matrices A.
.n . composition of homomorphisms corresponds to multiplication of matrices. In the case where the domain and range are the same. we have the following elegant corollary. Conversely. C ∈ Rm. . Then f : Rn → Rm defined by f (B) = AB is a homomorphism with f (ei ) = column i of A. Even though this follows easily from the previous theorem and properties of matrices. Rm ) and Rm. they are isomorphic as R-modules. addition of matrices corresponds to addition of homomorphisms.
The previous theorem reveals where matrix multiplication comes from. . Proof This is just the associative law of matrix multiplication. Even if R is commutative.n and g : Rm → Rp is the homomorphism given by C ∈ Rp.n to be the matrix with column i = vi . If R is commutative. it is one of the great classical facts of linear algebra.j ) ∈ Rm.n are isomorphic as additive groups. This corollary shows one way non-commutative rings arise. Then f defined by f (B) = AB is the unique homomorphism from Rn to Rm with f (ei ) = vi . We now return to the general theory of modules (over some given ring R). C(AB) = (CA)B. then g ◦ f : Rn → Rp is given by CA ∈ Rp. namely as endomorphism rings. Rn ) and Rn are isomorphic as rings. The automorphisms correspond to the invertible matrices. Rn is never commutative unless n = 1. Theorem If f : Rn → Rm is the homomorphism given by A ∈ Rm. define A ∈ Rm.n . That is. It is the matrix which represents the composition of the functions. . and multiplication of matrices corresponds to composition of homomorphisms.n . vn ∈ Rm . Theorem If f.n . if v1 .
as shown by the next theorem. the additive abelian quotient group M/N is defined. The image of f is the ¯ ¯ ¯ image of f . Since N is a normal subgroup of M . quotient modules go through without a hitch. then M/N ≈ M (see below). Theorem Suppose M is a module and N ⊂ M is a submodule. Proof On the group level this is all known from Chapter 2 (see page 29). f is injective. Furthermore. Scalar multiplication defined by (a + N )r = (ar + N ) is well-defined and gives M/N the structure of an R-module. f M π
c && & & & E b & &
¯ M
&
¯ f
M/N Thus defining a homomorphism on a quotient module is the same as defining a homo¯ morphism on the numerator that sends the denominator to 0. ¯ : (M/N ) → M defined by f (a + N ) = f (a) is a well-defined ¯ ¯ If N ⊂ ker(f ). Thus if N = ker(f ). The natural projection π : M → M/N is a surjective ¯ homomorphism with kernel N . (domain(f )/ker(f )) ≈ image(f ).74
Linear Algebra Cosets and Quotient Modules
Chapter 5
After seeing quotient groups and quotient rings. this is all known from Chapter 2 (see pages 27 and 29). ¯ Theorem Suppose f : M → M is a homomorphism and N is a submodule of M . if f : M → M is a surjective homomor¯ phism with ker(f ) = N . It is ¯ only necessary to check that f is a module homomorphism. It is only necessary to check the scalar multiplication. and the kernel of f is ker(f )/N . As before. Therefore for any homomorphism f .
The relationship between quotients and homomorphisms for modules is the same as for groups and rings.
. then f homomorphism making the following diagram commute. R is a ring and module means R-module. and thus (M/N ) ≈ image(f ). which is obvious. and this is immediate. Proof On the group level.
.e.. ii) Suppose K ⊂ L. Under the natural 1-1 correspondence from {functions f : M → Mt } to {sequence of functions {ft }t∈T where ft : M → Mt }. This makes the structure of module homomorphisms much more simple. Suppose M is a module. These two examples are for the case R = Z. i. On the additive abelian group Mt = Mt define scalar multiplication by {mt }r =
t∈T
{mt r}. Since scalar multiplication is defined coordinatewise. Theorem Suppose T is an index set and for each t ∈ T . In the finite case something important holds for modules that does not hold for non-abelian groups or rings – namely. Proof We already know from Chapter 2 that f is a group homomorphism iff each ft is a group homomorphism. although the student should think of the finite case.. f is a homomorphism iff each ft is a homomorphism.
The natural homomorphism K → (K + L)/L is surjective with kernel ≈ K ∩ L. For the finite case we may use either the product or sum notation. The natural homomorphism M/K → M/L is surjective ≈ with kernel L/K. Then Mt is an R-module and. This is stated below in full generality. Thus (K/K ∩ L) → (K + L)/L is an isomorphism. M =Z K = 3Z L = 5Z K ∩ L = 15Z K/K ∩ L = 3Z/15Z ≈ Z/5Z = (K + L)/L K +L=Z
Examples 1) 2)
M =Z K = 6Z L = 3Z (K ⊂ L) (M/K)/(L/K) = (Z/6Z)/(3Z/6Z) ≈ Z/3Z = M/L Products and Coproducts
Infinite products work fine for modules. the natural projection πs : Mt → Ms is a homomorphism. just as they do for groups and rings. for abelian groups.Chapter 5
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75
Theorem i)
Suppose M is an R-module and K and L are submodules of M . i. the finite product is also a coproduct. Mt is an R-module.e. Thus (M/K)/(L/K) → M/L is an isomorphism. for each s ∈ T . M1 × M2 × · · ×Mn = M1 ⊕ M2 ⊕ · · ⊕Mn . f is a module homomorphism iff each ft is a module homomorphism.
The statement that K is a summand of M means ∃ a submodule L of M with K ⊕ L = M . Of course. −a) : a ∈ K ∩ L}. Then the projection ¯ map π2 : M1 ⊕ M2 → M2 is a surjective homomorphism with kernel M1 . M ) → M n which is a R-module isomorphism. In particular. Thus (M1 ⊕ M2 )/M1 is isomorphic to M2 . If such an L exists.
. These isomorphisms are transparent and are used routinely in algebra without comment (see Th 4. f (k. Show K is a submodule of M ⊕ M which is a summand. Theorem Consider M1 = M1 ⊕0 as a submodule of M1 ⊕M2 . because L ≈ M/K. but it will be unique up to ¯ isomorphism. M and 0 are always summands of M . and n ≥ 1. Exercise Suppose R is a commutative ring. Thus f is an isomorphism iff K + L = M and K ∩ L = 0. Is Q a summand of R? With the material at hand. if a ∈ R and b ∈ R. Show that M ⊕ N is isomorphic to N ⊕ M. Summands One basic question in algebra is "When does a module split as the sum of two modules?". and Q ⊂ R is a submodule. (See page 35 for the group version. Before defining summand. and the next theorem is almost as intuitive. M is an R-module. this is the same as there exists a submodule L with K + L = M and K ∩ L = 0. page 118).) This is exactly what you would expect. Define a function α : HomR (Rn . Exercise R is a module over Q. Later on. Now suppose A ⊂ M. In this case we write K ⊕ L = M . Suppose K and L are submodules of M and f : K ⊕ L → M is the Theorem natural homomorphism. For example. then (R ⊕ R)/(aR ⊕ bR) is isomorphic to (R/aR) ⊕ (R/bR). ¯ Exercise Suppose M is a module and K = {(m. According to the previous theorem. it will be easy. this is not an easy question. Then the image of f is K + L and the kernel of f is {(a. here are two theorems for background. the abelian group (Z ⊕ Z)/(2Z ⊕ 3Z) is isomorphic to Z2 ⊕ Z3 . m) : m ∈ M } ⊂ M ⊕ M . This abuse of notation allows us to ¯ avoid talking about "internal" and "external" direct sums.Chapter 5
Linear Algebra
77
Exercise Suppose M and N are R-modules. l) = k + l. Definition Suppose K is a submodule of M . B ⊂ N are submodules and show (M ⊕ N )/(A ⊕ B) is isomorphic to (M/A) ⊕ (N/B). it need not be unique.
and Free Basis This section is a generalization and abstraction of the brief section Homomorphisms on Rn ..78
Linear Algebra
Chapter 5
Exercise Answer the following questions about abelian groups.. to avoid dizziness. Definition Suppose M is an R-module. Theorem For each t ∈ T . then S is said to be a basis or free basis for M . (See the third exercise on page 35. then S is dependent. rn in R.
. st ∈ M . The statement that S is dependent means ∃ a finite number of distinct elements t1 . tn in T . . If S is independent and generates M . the student should first consider the case where T is finite. define the center of T to be the subring {t : ts = st for all s ∈ T }. .. then S is dependent.. These concepts work fine for an infinite index set T because linear combination means finite linear combination. When is nZmn a summand of Zmn ?
Exercise If T is a ring. Let S be the sequence {st }t∈T = {st }.. SR is a submodule of M called the submodule generated by S. T is an index set. and elements r1 . such that the linear combination st1 r1 + · · +stn rn = 0. let Rt = RR and for each c ∈ T ... In this case any v ∈ M can be written uniquely as a linear combination of elements in S.) 1) 2) 3) 4) Is 2Z a summand of Z? Is 2Z4 a summand of Z4 ? Is 3Z12 a summand of Z12 ? Suppose m. n > 1. However. Note that if some st = 0. 2. Show R n is a left T -module and find HomT (Rn .. let ec ∈ ⊕Rt = Rt
t∈T
be ec = {rt } where rc = l and rt = 0 if t = c. Independence. ¯ S is independent. n} and ⊕Rt = Rn (see p 72). Generating Sets. You might try first the case T = {1. . Otherwise. An R-module M is said to be a free R-module if it is zero or if it has a basis. except for the confusing notation. not all zero. Let R be a commutative ring and T = Rn . The next two theorems are obvious.e. There is a exercise on page 57 to show that the center of T is the subring of scalar matrices. and for each t ∈ T . i. Also if ∃ distinct ¯ elements t1 and t2 in T with st1 = st2 . Rn ).. Z-modules. Let SR be the set of all linear combinations st1 r1 + · · +stn rn . Then {ec }c∈T is a basis for ⊕Rt called ¯ ¯ the canonical basis or standard basis.
Chapter 5
Linear Algebra
79
Theorem Suppose N is an R-module and M is a free R-module with a basis {st }. M≈
t∈T
Proof If M is isomorphic to ⊕Rt then M is certainly free. f (S) is independent in N iff f is injective. Then ∃ a 1-1 correspondence from the set of all functions g : {st } → N and the set of all homomorphisms f : M → N . If h : M → N is a homomorphism.. Theorem Suppose N is a module. Given g. f is completely determined by what it does on the basis S. it should be worked carefully.. Let f (S) be the sequence {f (st )} in N . The next theorem is so basic in linear algebra that it is used without comment. M is a free module with basis S = {st }. This is just an observation. Theorem A non-zero R-module M is free iff ∃ an index set T such that Rt . 1) 2) 3) 4) f (S) generates N iff f is surjective.
. . and you are "free" to send the basis any place you wish and extend to a homomorphism. define f by f (st1 r1 + · · +stn rn ) = g(st1 )r1 + · · +g(stn )rn . f (S) is a basis for N iff f is an isomorphism. and f : M → N is a homomorphism. In other words. Although the proof is easy.
Exercise Let (A1 . define g by g(st ) = f (st ). In particular. So now suppose M has a free basis {st }. Show this sequence is linearly independent over Z iff it is linearly independent over Q. then f = h iff f | S = h | S. An ) be a sequence of n vectors with each Ai ∈ Zn . f is an isomorphism. Is it true the sequence is linearly independent over Z iff it is linearly independent over R? This question is difficult until we learn more linear algebra. Recall that we have already had the preceding theorem in the case S is the canonical basis for M = Rn (p 72). Given f . but it is a central fact in linear algebra. By 3) in the preceding theorem. Characterization of Free Modules Any free R-module is isomorphic to one of the canonical free R-modules ⊕Rt . Then the homomorphism f : M → ⊕Rt with f (st ) = et sends the basis for M to the canonical basis for ⊕Rt . M has a finite free basis of n elements iff M ≈ Rn .
en } is a free basis for Rn . ¯ v is independent iff v is v is a basis for R iff v generates R iff v is
. A = If R is a commutative ring. The next exercise is routine. and the homomorphism f : Rn → Rn defined by f (B) = AB is surjective. A ∈ Rn . Let B = .. represent an element of Rn and C = . i. Use the fact that {e1 .e. and solutions of equations. are all the same — they are merely stated in different language.. i.e.80
Linear Algebra
Chapter 5
Exercise Suppose R is a commutative ring. .
Relating these concepts to matrices The theorem stated below gives a summary of results we have already had. show A is invertible.. injective homomorphisms. . you can relate properties of R as an R-module to properties of R as a ring.Let v1 .. . but still informative. f (B) = AB.. Suppose A ∈ Rm. Exercise 1) 2) Suppose R is a commutative ring and v ∈ R. linear independence. bn cm
. Find a non-trivial linear combination of the columns of A which is 0.e. It shows that certain concepts about matrices. This is a key theorem in linear algebra.
Note that 2) here is essentially the first exercise for the case n = 1. v = 0. then f is an isomorphism. f (ei ) = vi b1 c1 = column i of A. Show f is an isomorphism. That is..n and f : Rn → Rm is the homomorphism associated with A. Also find a non-zero element of kernel(f ). 2 1 0 and f: Z3 → Z2 be the group homo3 2 −5 morphism defined by A. ¯ Exercise Let R = Z. vn ∈ Rm be the columns of A. although it is usually stated only for the case where R is a field.. i. if f : R → R is a surjective R-module homomorphism.
. .. . {v1 . wn } is a basis for Rn .Chapter 5
Linear Algebra
81
represent an element of Rm . vn } is a basis for Rm iff f is an isomorphism iff (for any C ∈ Rm . {v1 . Then the following are equivalent. At is invertible. So far in this chapter we have just been cataloging... A ∈ Rn . | At | is a unit in R. AX = C has a solution).. Later on we will prove that if R is a field. A is invertible. . | A | is a unit in R. (See bottom of page 89. Now we prove something more substantial.n be the rows of A.e. 1) 2) 3) 4) 5) 2t ) 3t ) f is an automorphism. i.. .. {v1 . {v1 . . vn ∈ Rn be the columns of A.. {w1 . Theorem 1) The element f (B) is a linear combination of the columns of A. and f : Rn → Rn is defined by f (B) = AB.. . .. . that is f (B) = f (e1 b1 + · · +en bn ) = v1 b1 + · · +vn bn . Let v1 . Thus the image of f is generated by the columns of A. injective implies surjective... and w1 . namely that if f : Rn → Rn is surjective. vn } generates Rn . Theorem Suppose R is a commutative ring.
. i. then f is injective... AX = C has a unique solution). wn ∈ Rn = R1. vn } is a basis for Rn . vn } generates Rm iff f is surjective iff (for any C ∈ Rm ...
2)
3)
4)
Relating these concepts to square matrices We now look at the preceding theorem in the special case where n = m and R is a commutative ring.. f is surjective.) {v1 . vn } is independent iff f is injective iff AX = 0 has a unique ¯ solution iff (∃ C ∈ Rm such that AX = C has a unique solution)..e.
g. Then h is a surjective homomorphism. Let g : Rn → Rn be the homomorphism satisfying g(ei ) = ui . as shown below. then m ≥ n. If R is commutative. R will be a commutative ring.82 4t ) {w1 . and by the previous section. a finite generated R-module)..
Proof Suppose k = n − m is positive. This is a contradiction and thus m ≥ n. which requires that R be commutative (see the exercise on page 64). but this is of little interest. Since f is onto.. Then if M has a basis. Define h : (Rm ⊕ Rk = Rn ) → Rn by h(u. Furthermore. 2) and 2t ) are equivalent. So this follows from the first theorem. then m ≥ n. Thus the first five are equivalent. Now g comes from some matrix D and thus AD = I. 4) and 5) are equivalent. First we make a convention.
Corollary
Proof The hypothesis implies there is a surjective homomorphism R m → Rn .e... wn } generates Rn . vm } generates Rn . ∃ u1 . module (i. Then f ◦ g is the identity.. If {v1 . so m = n by the previous theorem. . This shows that A has a right inverse and is thus invertible.. also injective. applying this result to At shows that the last three properties are equivalent to each other. this is impossible.
Linear Algebra
Chapter 5
Proof Suppose 5) is true and show 2). un ∈ Rn with f (ui ) = ei . that basis is finite.. Corollary Proof If f : Rm → Rn is an isomorphism. Convention Theorem For the remainder of this chapter. Lemma Suppose M is a f.. . Since | A |=| At |. then m = n.
. and 3) implies 4). We already know the first three properties are equivalent. v) = f (u). If f : Rm → Rn is a surjective R-module homomorphism.
Each of f and f −1 is surjective. . Recall that the proof of this fact uses determinant. Uniqueness of Dimension There exists a ring R with R2 ≈ R3 as R-modules.
h(en )). trace. change of basis can be displayed by the diagram below. Endomorphisms on Rn are represented by square matrices..
. The point of all this is that selecting a basis of n elements for M Summary is the same as selecting an isomorphism from Rn to M . . M has a basis of n elements iff M ≈ Rn .. In order to represent f by a matrix. un } or if you wish. We will show that this matrix is well defined up to similarity. an isomorphism with Rn ). independence. 2. (By convention.g. the dimension of M. you must use sequences.. basis is defined for sequences. n}.Chapter 5
Linear Algebra
83
Proof Suppose U ⊂ M is a finite generating set and S is a basis.. but for independence and basis. . Now suppose M is a f. and thus the determinant. Here we follow the classical convention that an index set with n elements will be {1. and thus S is finite. Two sequences cannot begin to be equal unless they have the same index set. Change of Basis Before changing basis. and characteristic polynomial of f are well-defined. If M has a basis. Now h : Rn → M is an isomorphism iff α(h) is a basis for M ... there is no concept of repetition. and thus a basis for M with n elements is a sequence S = {u1 . Then any element of U is a finite linear combination of elements of S. In order to make sense.) ¯ Proof By the previous lemma. not for collections. Consider the columns of the real matrix 2 3 2 A= . S = (u1 . we must select a basis for M (i. yet we certainly don't wish to say the columns of A form a basis for R2 . free module and f : M → M is a homomorphism. Theorem Suppose M is a f. Recall there is a bijection α : HomR (Rn .. not for sets or collections. M ) → M n defined by α(h) = (h(e1 ). . . we must consider the columns of A as an ordered triple of vectors. The result follows because Rn ≈ Rm iff n = m.. and from this viewpoint. Previously we defined generating. that basis is finite and any other basis has the same number of elements. there are 1 4 1 only two of them. and thus have a determinant and trace. module. Suppose M is an R-module with a basis of n elements. If we consider the column vectors of A as a collection. and this sequence is dependent. un ) ∈ M n . This number is denoted by dim(M ).. any basis for M must be finite. In the definition of basis on page 78. For the concept of generating it matters not whether you use sequences or collections. and basis for sequences.g. In a set or collection.e. 0 is a free module of dimension 0. we recall what a basis is.
Find the determinant. 1)t }. any non-zero element of M is a basis. The previous theorem. Then 0= (vr)r = v1 = v. and characteristic polynomial. ¯ That is. trace. and any two elements of M are dependent. In this section. Finally. Also find the matrix B ∈ R2 which represents f with respect to the standard basis. By the previous theorem. some basis. −1)t . and show that in this case. ¯ ¯ ¯ ¯ Theorem Suppose M = 0 is an F -module and v ∈ M . Find the matrix A ∈ R2 which represents f with respect to the basis {(1.
. or characteristic polynomial.Chapter 5
Linear Algebra
85
the matrix of f w. and characteristic polynomial of f . Theorem Suppose M is an F -module and v ∈ M . but it must be assumed that the module M is free. trace.r. Endomorphisms in general will not have a determinant. (−1. if these conditions hold. for example. 1)t ) = (1. every F -module is free.
3 3 0 −1
D. Vector Spaces So far in this chapter we have been developing the theory of linear algebra in general.. find an invertible matrix C ∈ R2 with B = C −1 AC. We now focus on the case where R is a field F . all three are well-defined. Show there is one and only one homomorphism f : R2 → R2 which is the identity on L and has f ((−1.t. if v ∈ V and r ∈ F . the basis
Exercise Let L ⊂ R2 be the line L = {(r. then M ≈ FF .r.e. Furthermore.t. vr = 0 implies v = 0 in M or r = 0 in F . F is a field. holds for any commutative ring R. and endomorphisms on it will have well-defined determinant. Then v = 0 iff v is independent. 2)t .
Find the matrix of f w. F -modules may also be called vector spaces and F -module homomorphisms may also be called linear transformations. Then v generates M iff v ¯ is a basis for M . 2r)t : r ∈ R}. i. trace. do not depend upon the choice of basis. 3 1 . Thus any finitely generated F -module will have a well-defined dimension. ¯ ¯ ¯ −1 Proof Suppose vr = 0 and r = 0.
Exercise
Let R = Z and f : Z2 → Z2 be defined by f (D) = 2 1 .
Theorem Suppose M = 0 is a finitely generated F -module. vm } is an independent sequence in M.. Thus T extends to a basis. Then v = 0 and is thus independent by the ¯ previous theorem.
The next theorem is just a collection of observations. .. and any two elements of F are dependent. Theorem Suppose M is an F -module of dimension n. Also the finiteness hypothesis in this theorem is only for convenience. This is one of the theorems that makes linear algebra tick. such that v1 r1 + · · +vn rn + vi ri = 0. vm } extends to a basis with n elements. It not only says any finite independent sequence can be extended to a basis.. ∃ r1 . vn }. Thus any finite independent sequence can be extended to a basis. and thus the dimension of M is well defined (see theorem on page 83). . and the element 2 of Z is independent. {v1 .. In this case M ≈ F . Proof Suppose. vm } ¯ generates M . Theorem Suppose M and N are finitely generated F -modules. vi } is dependent.. ri not all −1 zero... for notational convenience.
Since F is a commutative ring.. . then any maximal independent subsequence of S is a basis for M . vn } generates S and thus all of M . M has a finite free basis.
. After so many routine theorems. it is nice to have one with real power..86
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Proof Suppose v generates M . If S = {v1 . . The key hypothesis here is that the ring is a field. and {v1 . . It must be shown that vi is a linear combination of {v1 . vn } is a maximal independent subsequence of S. .. ¯ ¯ Thus {v1 . . Since {v1 ... then Z is a free module over itself. vn .. but it can be extended to a basis inside any finite generating set containing it. Then m ≤ n and if m = n. .. If R = Z. and n < i ≤ m. .. and thus is a free F -module.. and inside that sequence it may be extended to a maximal independent sequence. Then ri = 0 and vi = −(v1 r1 + · · +vn rn )ri .... rn . Proof {v1 . that {v1 . In particular. any two bases of M must have the same number of elements. T may be extended to a finite generating sequence. However it certainly cannot be extended to a basis. and any non-zero element of F is a basis. as will be seen momentarily.. Now suppose T is a finite independent sequence. vm } is a basis.
Then S −T generates a submodule L with K ⊕ L = M. K is a summand of M . Theorem 1) 2) Suppose M is an F -module and K ⊂ M is a submodule. Then dim(M ) = dim(ker(f )) + dim(image(f )).
87
Here is the basic theorem for vector spaces in full generality.e. Theorem 1) 2) 3) Suppose M = 0 is an F -module and S = {vt }t∈T generates M . Part 2) follows from 1). F -module. Any independent subsequence of S may be extended to a maximal independent subsequence of S. dim(M ⊕ N ) = dim(M ) + dim(N ).
Proof The proof of 1) is the same as in the case where S is finite. Any independent subsequence of M can be extended to a basis for M . ¯ Any maximal independent subsequence of S is a basis for M . (See exercise on page 77. i. and Q is a submodule of R. Extend T to a basis S for M .) Proof Q is a field. F m ≈ F n iff n = m.
Corollary Suppose M is a f.
. An independent subsequence of S is contained in a maximal monotonic tower of independent subsequences. Corollary Q is a summand of R. If M is f. R is a Q-module.. ∃ a submodule L of M with K ⊕ L = M . M ≈ N iff dim(M ) = dim(N ). In other words. ∃ a Q-submodule V ⊂ R with Q ⊕ V = R as Q-modules.g. Part 3) follows from 2) because an independent sequence can always be extended to a generating sequence.Chapter 5 1) 2) 3) 4)
Linear Algebra M ≈ F n iff dim(M ) = n. and thus to a basis for M . M has a free basis. The union of these independent subsequences is still independent.g. and thus is a free F -module. Part 2) will follow from the Hausdorff Maximality Principle. N is an F -module. and f : M → N is a homomorphism. and so the result follows. then dim(K) ≤ dim(M ) and K = M iff dim(K) = dim(M ). In particular.
Proof Let T be a basis for K..
every submodule is a summand. f is injective.e.. 1. for R-modules. f is an automorphism. . | At |= 0. wn } is independent. You may use determinant.. (3. .. 1).. A is invertible. (2. ¯ {v1 . Theorem Suppose A ∈ Fn and f : F n → F n is defined by f (B) = AB.e.. Show R is a field. Exercise Find a free Z-module which has a generating set containing no basis. {w1 .. i. wn ∈ F n = F1.
Square matrices over fields This theorem is just a summary of what we have for square matrices over fields.. wn } is a basis for F n . . 1). 0).e.e.. wn } generates F n . 1) 2) {v1 . vn ∈ F n be the columns of A. 2).... i. {w1 . vn } is independent. . and w1 .... vn } generates F n ... (3. (1.n be the rows of A. i.e.. Then f | L : L → image(f ) is an isomorphism. (1.. Let v1 . Show there are three maximal independent subsequences of S. vn } is a basis for F n . 0)}. . {v1 . (Row vectors are used here just for convenience. . | A |= 0. . i. Show there are three maximal independent subsequences of S and each is a basis for R3 .
3) 1t ) 2t ) 3t )
. 5). f is surjective.) The real vector space R3 is generated by S = {(1. i. i.
Exercise The real vector space R2 is generated by the sequence S = {(π. ¯ {w1 . 2)}. At is invertible. i.. Exercise Suppose R is a domain with the property that. 2...e. and each is a basis for R2 . Then the following are equivalent.e..88
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Proof Let K = ker(f ) and L ⊂ M be a submodule with K ⊕ L = M . 2.. . 4.
Show the following are equivalent. then the row (column) rank of A is the same as the row (column) rank of CAD. pages 81 and 82. The sequence (A1 . and we know from page 81 that each f (B) is a linear
. f is injective iff f is bijective iff f is surjective. An ) is linearly independent over Z.
Rank of a matrix Suppose A ∈ Fm. The sequence (A1 . This shows some of the simple and definitive nature of linear algebra. Exercise Add to this theorem more equivalent statements in terms of solutions of n equations in n unknowns.. . An ) be an n × n matrix over Z with column i = Ai ∈ n ¯ Z . By the pigeonhole principle. Now suppose each of U and V is a vector space of dimension n and f : U → V is a linear transformation. Let f : Zn → Zn be defined by f (B) = AB and f : Rn → Rn be defined by ¯(C) = AC. It follows from the work done so far that f is injective iff f is bijective iff f is surjective.) f 1) 2) 3) 4) 5) f : Zn → Zn is injective. Exercise Let A = (A1 . . Overview Suppose each of X and Y is a set with n elements and f : X → Y is a function.. . ¯ f : Rn → Rn is injective. this theorem holds for any commutative ring R. An ) is linearly independent over R.Chapter 5
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Proof Except for 1) and 1t ).. The row (column) rank of A is defined to be the dimension of the submodule of F n (F m ) generated by the rows (columns) of A. |A| = 0.) Parts 1) and 1t ) follow from the preceding section. (See the exercise on page 79.. Theorem If C ∈ Fm and D ∈ Fn are invertible.. Proof Suppose f : F n → F m is defined by f (B) = AB. (See the section Relating these concepts to square matrices. Each column of A is a vector in the range F m .n ..
By the theorem above. For n = 2. the row rank and the column rank of A are equal. Thus the image of f is the submodule of F m generated by the columns of A. the rank of f is the dimension of the image of f . There is a concept of the n-dimensional volume of V . and for n = 3 it is "ordinary volume". the next theorem says that "f multiplies volume by the absolute value of its determinant". and its dimension is the column rank of A. This proves the theorem for column rank.1 = ·· = ht. By row and column operations. f preserves volume. A may be changed to a matrix H where h1. n}. it is length. V and V + p have the same volume. Thus f (V ) and f (V + p) = f (V ) + f (p) have the same volume.n has rank t. Show that it is possible to select t rows and t columns of A such that the determined t × t matrix is invertible. it is area.e. Suppose A ∈ Rn and f : Rn → Rn is the homomorphism given by A. The volume of V does not change under translation. What is the dimension of the solution set of AX = 0? ¯ Definition If N and M are finite dimensional vector spaces and f : N → M is a linear transformation.t = 1 and all other entries are 0 (see the ¯ ¯ first exercise on page 59). Thus if |A| = ±1. i. If f : F n → F m is given by a matrix A. In street language. where h is any automorphism on F n and g is any automorphism on F m . For example. This number is called the rank of A and is ≤ min{m.90
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combination of those vectors. V might be the interior of a square or circle. The theorem for row rank follows using transpose. Theorem The n-dimensional volume of f (V ) is ±|A|(the n-dimensional volume of V ). Geometric Interpretation of Determinant Suppose V ⊂ Rn is some nice subset.. Theorem If A ∈ Fm. Exercise Suppose A has rank t. Thus row rank = t = column rank.n . then the rank of f is the same as the rank of the matrix A. Show that the rank of A is the largest integer t such that this is possible. For n = 1. This dimension is the same as the dimension of the image of g ◦ f ◦ h : F n → F m . Exercise Suppose A ∈ Fm.
. elementary row and column operations change Proof neither the row rank nor the column rank. if n = 2.
Without this great simplification. suppose h : R → R is differentiable and p is a real number. The area of V is any homomorphism f multiplies area by | f |. Of course.
. y) ∈ V . y) : R2 → R2 be the homomorphism defined by J(h)(x. science. 1] × · · ×[0.. vn of A. Then h is approximated near p by g(x) = h(p) + f (x − p) = h(p) + h (p)(x − p). Define the Jacobian by J(h)(x. p2 ). y). However this is not the only reason that linear algebra is so useful. The result follows because the determinant of the composition is the product of the determinants. the world of technology as we know it today would not exist. Corollary If P is the n-dimensional parallelepiped determined by the columns v1 . Proof Let V = [0. let f (x. It is a central fact that smooth phenomena may be approximated locally by linear phenomena. so they must be adjusted by a translation. then the n-dimensional volume of P is ±|A|. Linear functions approximate differentiable functions locally We continue with the special case F = R. and mathematics. y) ∈ V . As a simple example. Let f : R → R be the linear transformation f (x) = h (p)x. . Now suppose V ⊂ R2 is some nice subset and h = (h1 .) Theorem Suppose the determinant of J(h)(x. From the previous section we know that
(x. 1] = {e1 t1 + · · +en tn : 0 ≤ ti ≤ 1}.Chapter 5
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Proof If |A| = 0. p2 ) (after translation) by f (p1 . this theorem is immediate from the previous section. p2 ) ∈ V . the theorem is obvious. The student may now understand the following theorem from calculus. image(f ) has dimension < n and thus f (V ) has n-dimensional volume 0. y) is non-negative for each
V V ∂h1 ∂x ∂h2 ∂x ∂h1 ∂y ∂h2 ∂y
and for each
1dxdy. h2 ) : V → R2 is injective and differentiable. . Then the area of h(V ) is
| J(h) | dxdy. h is approximated near (p1 . Then for any (p1 . Linear functions arise naturally in business. linear transformations send the origin to the origin. Then P = f (V ) = {v1 t1 + · · +vn tn : 0 ≤ ti ≤ 1}. If |A| = 0 then A is the product of elementary matrices (see page 59) and for elementary matrices. (Note that if h is the restriction of a linear transformation from R2 to R2 . y) = (x.
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We now return to the case where F is a field (of arbitrary characteristic). F modules may also be called vector spaces and submodules may be called subspaces. The study of R-modules in general is important and complex. However the study of F -modules is short and simple – every vector space is free and every subspace is a summand. The core of classical linear algebra is not the study of vector spaces, but the study of homomorphisms, and in particular, of endomorphisms. One goal is to show that if f : V → V is a homomorphism with some given property, there exists a basis of V so that the matrix representing f displays that property in a prominent manner. The next theorem is an illustration of this. Theorem 1) Let F be a field and n be a positive integer. Suppose V is an n-dimensional vector space and f : V → V is a homomorphism with |f | = 0. Then ∃ a basis of V such that the matrix ¯ representing f has its first row zero. Suppose A ∈ Fn has |A| = 0. Then ∃ an invertible matrix C such that ¯ C −1 AC has its first row zero. Suppose V is an n-dimensional vector space and f : V → V is a homomorphism with |f | = 0. Then ∃ a basis of V such that the matrix representing f has its first column zero. Suppose A ∈ Fn has |A| = 0. Then ∃ an invertible matrix D such that ¯ D −1 AD has its first column zero.
2) 3)
4)
We first wish to show that these 4 statements are equivalent. We know that 1) and 2) are equivalent and also that 3) and 4) are equivalent because change of basis corresponds to conjugation of the matrix. Now suppose 2) is true and show 4) is true. Suppose |A| = 0. Then |At | = 0 and by 2) ∃ C such that C −1 At C has ¯ ¯ first row zero. Thus (C −1 At C)t = C t A(C t )−1 has first row column zero. The result follows by defining D = (C t )−1 . Also 4) implies 2). This is an example of the transpose principle. Loosely stated, it is that theorems about change of basis correspond to theorems about conjugation of matrices and theorems about the rows of a matrix correspond to theorems about the columns of a matrix, using transpose. In the remainder of this chapter, this will be used without further comment.
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Proof of the theorem We are free to select any of the 4 parts, and we select part 3). Since | f |= 0, f is not injective and ∃ a non-zero v1 ∈ V with f (v1 ) = 0. ¯ Now v1 is independent and extends to a basis {v1 , .., vn }. Then the matrix of f w.r.t this basis has first column zero. Exercise Let A = 3π 6 2π 4 . Find an invertible matrix C ∈ R2 so that C −1 AC
0 0 0 has first row zero. Also let A = 1 3 4 and find an invertible matrix D ∈ R3 2 1 4 so that D −1 AD has first column zero. Exercise Suppose M is an n-dimensional vector space over a field F , k is an integer with 0 < k < n, and f : M → M is an endomorphism of rank k. Show there is a basis for M so that the matrix representing f has its first n − k rows zero. Also show there is a basis for M so that the matrix representing f has its first n − k columns zero. Work these out directly without using the transpose principle. Nilpotent Homomorphisms In this section it is shown that an endomorphism f is nilpotent iff all of its characteristic roots are 0 iff it may be represented by a strictly upper triangular matrix. ¯ Definition An endomorphism f : V → V is nilpotent if ∃ m with f m = 0. Any ¯ f represented by a strictly upper triangular matrix is nilpotent (see page 56). Theorem Suppose V is an n-dimensional vector space and f : V → V is a nilpotent homomorphism. Then f n = 0 and ∃ a basis of V such that the matrix ¯ representing f w.r.t. this basis is strictly upper triangular. Thus the characteristic polynomial of f is CPf (x) = xn . Proof Suppose f = 0 is nilpotent. Let t be the largest positive integer with ¯ t f = 0. Then f t (V ) ⊂ f t−1 (V ) ⊂ ·· ⊂ f (V ) ⊂ V . Since f is nilpotent, all of these ¯ inclusions are proper. Therefore t < n and f n = 0. Construct a basis for V by ¯ starting with a basis for f t (V ), extending it to a basis for f t−1 (V ), etc. Then the matrix of f w.r.t. this basis is strictly upper triangular. Note To obtain a matrix which is strictly lower triangular, reverse the order of the basis.
94 Exercise
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Use the transpose principle to write 3 other versions of this theorem.
Theorem Suppose V is an n-dimensional vector space and f : V → V is a homomorphism. Then f is nilpotent iff CPf (x) = xn . (See the exercise at the end of Chapter 4 for the case n = 2.) Proof Suppose CPf (x) = xn . For n = 1 this implies f = 0, so suppose n > 1. ¯ Since the constant term of CPf (x) is 0, the determinant of f is 0. Thus ∃ a basis ¯ ¯ of V such that the matrix A representing f has its first column zero. Let B ∈ Fn−1 be the matrix obtained from A by removing its first row and first column. Now CPA (x) = xn = xCPB (x). Thus CPB (x) = xn−1 and by induction on n, B is nilpotent and so ∃ C such that C −1 BC is strictly upper triangular. Then
Suppose F is field, A ∈ F3 is a strictly lower triangular matrix of a 0 0 0 rank 2, and B = 1 0 0 . Using conjugation by elementary matrices, show there 0 1 0 is an invertible matrix C so that C −1 AC = B. Now suppose V is a 3-dimensional vector space and f : V → V is a nilpotent endomorphism of rank 2. We know f can be represented by a strictly lower triangular matrix. Show there is a basis {v1 , v2 , v3 } for V so that B is the matrix representing f . Also show that f (v1 ) = v2 , f (v2 ) = v3 , and f (v3 ) = 0. In other words, there is a basis for V of the form {v, f (v), f 2 (v)} ¯ with f 3 (v) = 0. ¯ Exercise Exercise Suppose V is a 3-dimensional vector space and f : V → V is a nilpotent endomorphism of rank 1. Show there is a basis for V so that the matrix representing 0 0 0 f is 1 0 0 . 0 0 0
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Our standing hypothesis is that V is an n-dimensional vector space over a field F and f : V → V is a homomorphism. Definition An element λ ∈ F is an eigenvalue of f if ∃ a non-zero v ∈ V with f (v) = λv. Any such v is called an eigenvector. Eλ ⊂ V is defined to be the set of all eigenvectors for λ (plus 0). Note that Eλ = ker(λI − f ) is a subspace of V . The ¯ next theorem shows the eigenvalues of f are just the characteristic roots of f . Theorem 1) 2) 3) If λ ∈ F then the following are equivalent. λ is an eigenvalue of f , i.e., (λI − f ) : V → V is not injective. | (λI − f ) |= 0. ¯ λ is a characteristic root of f , i.e., a root of the characteristic polynomial CPf (x) = | (xI − A) |, where A is any matrix representing f .
Proof It is immediate that 1) and 2) are equivalent, so let's show 2) and 3) are equivalent. The evaluation map F [x] → F which sends h(x) to h(λ) is a ring homomorphism (see theorem on page 47). So evaluating (xI − A) at x = λ and taking determinant gives the same result as taking the determinant of (xI − A) and evaluating at x = λ. Thus 2) and 3) are equivalent. The nicest thing you can say about a matrix is that it is similar to a diagonal matrix. Here is one case where that happens. Theorem Suppose λ1 , .., λk are distinct eigenvalues of f , and vi is an eigenvector of λi for 1 ≤ i ≤ k. Then the following hold. 1) 2) {v1 , .., vk } is independent. If k = n, i.e., if CPf (x) = (x − λ1 ) · · · (x − λn ), then {v1 , .., vn } is a basis for V . The matrix of f w.r.t. this basis is the diagonal matrix whose (i, i) term is λi .
non-trivial linear combination. This is a contradiction and proves 1). Part 2) follows from 1) because dim(V ) = n. Exercise 0 1 ∈ R2 . Find an invertible C ∈ C2 such that −1 0 C −1 AC is diagonal. Show that C cannot be selected in R2 . Find the characteristic polynomial of A. Let A =
Exercise Suppose V is a 3-dimensional vector space and f : V → V is an endomorphism with CPf (x) = (x − λ)3 . Show that (f − λI) has characteristic polynomial x3 and is thus a nilpotent endomorphism. Show there is a basis for V so that the λ 0 0 λ 0 0 λ 0 0 matrix representing f is 1 λ 0 , 1 λ 0 or 0 λ 0 . 0 1 λ 0 0 λ 0 0 λ We could continue and finally give an ad hoc proof of the Jordan canonical form, but in this chapter we prefer to press on to inner product spaces. The Jordan form will be developed in Chapter 6 as part of the general theory of finitely generated modules over Euclidean domains. The next section is included only as a convenient reference. Jordan Canonical Form This section should be just skimmed or omitted entirely. It is unnecessary for the rest of this chapter, and is not properly part of the flow of the chapter. The basic facts of Jordan form are summarized here simply for reference. The statement that a square matrix B over a field F is a Jordan block means that ∃ λ ∈ F such that B is a lower triangular matrix of the form
e.)
Theorem Jordan form (when it exists) is unique. This means that if A and D are similar matrices in Jordan form.e. iff each Jordan block is a 1 × 1 matrix. . . Theorem 1) 2) If A ∈ Fn . namely that if A has n distinct eigenvalues in F .. (x − 2)(x − 3)(x − 4). they have the same Jordan blocks. C may be selected to be in Rn iff all the eigenvalues of A are real. Then n1 + · · +nt = n and CPD (x) = (x − λ1 )n1 · ·(x − λt )nt .Chapter 5 Definition
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A matrix D ∈ Fn is in Jordan form if ∃ Jordan blocks B1 . (x − 2)2 (x − 3)2 . The complex numbers are algebraically closed. D is a diagonal matrix iff each ni = 1. Also note that we know one special case of this theorem. then A is similar to a diagonal matrix. Note that a diagonal matrix is a special case of Jordan form. i. ¯ ¯
. This means that CPA (x) will factor completely in C[x].. Exercise Find all real matrices in Jordan form that have the following characteristic polynomials: x(x − 2). (x − 2)(x − 3)2 . ∃ λ1 .. ∃ an invertible C ∈ Fn such that C −1 AC is in Jordan form. The reader should use the transpose principle to write three other versions of the first theorem.. i. Show a0 I + a1 D + · · +D n = 0. the following are equivalent.. . except possibly in different order. (In this case we say that all the eigenvalues of A belong to F . λn ∈ F (not necessarily distinct) such that CPA (x) = (x − λ1 ) · · (x − λn ).
that D =
Suppose D is of this form and Bi ∈ Fni has
eigenvalue λi . then A is similar to a diagonal matrix. show CPD (D) = 0. Let's look at the classical case A ∈ Rn . (x − 2)(x − 3)3 . and thus ∃ C ∈ Cn with C −1 AC in Jordan form. Exercise Suppose D ∈ Fn is in Jordan form and has characteristic polynomial a0 + a1 x + · · +xn . Bt such B1 B2 · 0 · Bt 0
. Later on it will be shown that if A is a symmetric real matrix. (x − 2)2 .
As in the first theorem of this section. So suppose T = {f (u1 ). .. vk+1 }. Then T is independent and thus T is a basis for V and thus f is an isomorphism (see the second theorem on page 79). w2 .
Isometries Suppose each of U and V is an IPS.. It is that. Find the projection of (e1 + e2 ) onto ker(f ). and f : U → V is a homomorphism. {w1 .. 2. . Then by the previous theorem. wn }. 1.. f (un )} is an orthonormal sequence in V . un } is an orthonormal basis for U .. .. Theorem Suppose each of U and V is an n-dimensional IPS. wk+1 } is an orthonormal basis w for the subspace generated by {w1 . It is easy to check that f preserves inner products. (u1 · u2 )U = (f (u1 ) · f (u2 ))V .3). We now come to one of the definitive theorems in linear algebra.... u2 in U . wk } is an orthonormal sequence in W . This is a key observation for an exercise on page 103 showing O(n) is a deformation retract of GLn (R). 0. Proof Isometries certainly preserve orthonormal sequences. where p(v) is the projection of v onto Y . (1 − t)v + tw3 } w is a basis for W . 0)t . Exercise Find an orthonormal basis for the kernel of f . wk . {u1 . Now suppose W has dimension n and {w1 ..1. there is only one inner product space for each dimension.. . In this manner an orthonormal basis for W is constructed.. Then f is an isometry iff {f (u1 ). and set w w3 = .. vk+1 . Find w3 and show that for any t with 0 ≤ t ≤ 1. The process above may be used to modify this to an orthonormal basis {w1 . . 3)t . .. v} where v = (1.. W is generated by the sequence {w1 .. {w1 .. 0)t and w2 = (0. Find the angle between e1 + e2 and the plane ker(f ). w2 .Chapter 5 wk+1 =
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w . . . Since this sequence is independent. Notice that this construction defines a function h which sends a basis for W to an orthonormal basis for W (see topology exercise on page 103). Exercise Let W = R3 have the standard inner product and Y ⊂ W be the subspace generated by {w1 ..
Let f : R3 → R be the homomorphism defined by the matrix (2. w2 } where w1 = (1. wk . up to isometry. f (un )} is an orthonormal sequence in V . A homomorphism f : U → V is said to be an isometry provided it is an isomorphism and for any u1 . it extends to a basis {w1 .
. vn }. let w = v − p(v). .
f is an isometry. Thus O(n) is a multiplicative subgroup of GLn (R). Proof There exist orthonormal bases {u1 . In particular. Orthogonal Matrices As noted earlier. We now wish to study isometries from Rn to Rn .1. and so certainly preserve volume.e. vn } for V .
Proof A left inverse of a matrix is also a right inverse (see the exercise on page 64). and is called the orthogonal group. At A = I. Exercise Let f : R3 → R be the homomorphism defined by the matrix (2..... Theorem 1) 2) If A is orthogonal. and thus preserve angle and distance. AAt = I. AC is orthogonal. i.102
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Theorem Suppose each of U and V is an n-dimensional IPS. i. Thus 1) and 2) are equivalent because each of them says A is invertible with A−1 = At . If A is orthogonal. 1) and 3) are equivalent. The set of all such A is denoted by O(n). Theorem Suppose A ∈ Rn and f : Rn → Rn is the homomorphism defined by f (B) = AB. and f (ei ) is column i of A. Now {e1 . Isometries preserve inner product. and by the previous theorem.. . linear algebra is not so much the study of vector spaces as it is the study of endomorphisms. Thus by the previous section. Then the following are equivalent. U is isometric to Rn with its standard inner product. .e.3). . en } is the canonical orthonormal basis for Rn . Definition If A ∈ Rn satisfies these three conditions.. un } for U and {v1 . 1) 2) 3) The columns of A form an orthonormal basis for Rn . there exists a homomorphism f : U → V with f (ui ) = vi .. A−1 is orthogonal. A is said to be orthogonal.. Find a linear transformation h : R2 → R3 which gives an isometry from R2 to ker(f ). f is an isometry.
. By the first theorem on page 79. | A |= ±1. If A and C are orthogonal. We know from a theorem on page 90 that an endomorphism preserves volume iff its determinant is ±1. Then ∃ an isometry f : U → V. The rows of A form an orthonormal basis for Rn .
v ∈ Rn . (See the exercise on page 56. u − v = f (u)−f (v) and the angle between u and v is equal to the angle between f (u) and f (v). The next theorem is just an exercise using the previous theorem. This means that if u. z ∈ Rn . we first note that symmetric is the same as self-adjoint.
Proof Part 1) follows from |A|2 = |A| |At | = |I| = 1. Our goals are to prove that. and matrix multiplication is associative. t) = (1 − t)A + th(A) is a deformation retract of GLn (R) to O(n). Theorem A is symmetric iff A is self-adjoint. then A =
2
cosΘ −sinΘ sinΘ cosΘ
for
some number Θ. Let h : GLn (R) → O(n) be defined by Gram-Schmidt. This means a sequence of matrices {Ai } converges to A iff it converges coordinatewise. As background.
. Then u − v 2 = (u − v) · (u − v) = f (u − v) · f (u − v) = f (u − v) 2 = f (u) − f (v) 2 . Diagonalization of Symmetric Matrices We continue with the case F = R. Then f preserves distances and angles. For part 3) assume f : Rn → Rn is an isometry. if A is a symmetric matrix. Show GLn (R) is an open subset and O(n) is closed and compact. is the matrix product y t z. A is said to be self-adjoint if (Au)·v = u·(Av) for all u. 1] → GLn (R) defined by H(A. Then (At u) · v = u · (Av). Show H : GLn (R) × [0. A is said to be symmetric provided At = A.
Proof If y. Definition Suppose A ∈ Rn . all of its eigenvalues are real and that ∃ an orthogonal matrix C such that C −1 AC is diagonal.Chapter 5 3)
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Suppose A is orthogonal and f is defined by f (B) = AB. The proof that f preserves angles follows from u · v = u v cosΘ. because isometries clearly form a subgroup of the multiplicative group of all automorphisms. Thus (At u) · v = (ut A)v = ut (Av) = u · (Av). Exercise Show that if A ∈ O(2) has |A| = 1. Part 2) is immediate. v ∈ Rn . v ∈ Rn . then the dot product y · z. Note that any diagonal matrix is symmetric.) Exercise (topology) Let Rn ≈ Rn have its usual metric topology. Theorem Suppose A ∈ Rn and u.
and suppose A is a symmetric n × n matrix.. the
λ1 · ·
transformation determined by A is represented by
Since this is a symmetric matrix.. 1) 2) f is self-adjoint.Chapter 5 Theorem 1) 2)
Linear Algebra If A ∈ Rn .. Theorem If V is an n-dimensional IPS and f : V → V is a linear transformation. v2 } be an orthonormal basis for R2 with Av1 = λv1 . Let λ be an eigenvalue for A and {v1 . then the following are equivalent. . because then there is a basis of eigenvectors of length 1.
λk
(B)
. the transformation determined λ b by A is represented by . . Suppose A is a symmetric 2 × 2 matrix.t this basis. They may be extended to an orthonormal basis v1 . . b = 0. Show 1) ⇒ 2)... k ≤ n. vn . A is symmetric. and they must form an orthonormal basis. λk with each vi = 1. If k = n. . So suppose k < n. Thus conjugating I 0 makes the entire matrix diagonal. a linear transformation f : V → V is said to be self-adjoint provided (u·f (v)) = (f (u)·v) for all u..
105
Proof By the previous theorem. vk be eigenvectors for λ1 .
(0)
(D)
. Let v1 .. λk the distinct eigenvalues of A... 2) ⇒ 1). ∃ C ∈ O(n) such that C −1 AC is diagonal. With respect to this basis..r. By induction.. Denote by λ1 . . ∃ an orthonormal basis {v1 . by 0 C This theorem is so basic we state it again in different terminology. the proof is immediate. Then w. 0 d Now suppose by induction that the theorem is true for symmetric matrices in Rt for t < n. vn } for V with each vi an eigenvector of f . ∃ an orthogonal C such that C −1 DC is diagonal. Since this matrix is symmetric. If V is an IPS. B = 0 and D is a symmetric matrix of smaller size. v ∈ V . the following are equivalent.
i. 2
Exercise Suppose A. Thus if A ∈ Rn . Exercise Define E : Cn → Cn by E(A) = eA = I + A + (1/2!)A2 + ··. ∃ a non-zero matrix N ∈ R2 with eN = I. Furthermore E(At ) = E(A)t . an algebraic object. Find an orthogonal C such that C −1 AC is diagonal. This shows that V . and this metric will determine that same topology. | eA |= 1 iff trace(A) = 0. and if C is invertible. if V has an inner product. ∃ an orthogonal C such that D = C −1 AC. then | eA |= etrace(A) . ¯ If A ∈ Rn and At = −A. then E(A + B) = E(A)E(B). We know that V is isomorphic to Rn . Show that h is continuous. Suppose f and g are isomorphisms from V to Rn and A is a subset of V . then eN = I iff N = 0. D ∈ Rn are symmetric. E(C −1 AC) = C −1 E(A)C. assume the Jordan form. if A and D are similar. If N ∈ Rn is symmetric. assume any A ∈ Cn is similar to a lower triangular matrix. Now use the results of this section to prove the statements below. I = E(0) = E(A − A) = E(A)E(−A). Show that f (A) is an open subset of Rn iff g(A) is an open subset of Rn . Exercise Suppose V is an n-dimensional real vector space. This series converges and thus E is a well defined function.
2) 3) 4)
. If AB = BA. suppose V and W are finite-dimensional real vector spaces and h : V → W is a linear transformation. then eA ∈ O(n). it automatically has a metric.106
Linear Algebra
Chapter 5
Exercise
Let A =
Do the same for A =
2 2 2 1
2 . Of course. Since A and −A commute. 2 1 . (For part 1. and ¯ thus E(A) is invertible with E(A)−1 = E(−A). Under what conditions are A and D similar? Show that. Finally.) 1) If A ∈ Cn .. has a god-given topology.e.
This always holds if F = C. then M is the sum of cyclic modules. i. Thus if M is torsion free.. i. enough material is added to form a basic first year graduate course. Suppose R is a commutative ring. 107
. The final section gives the fundamentals of dual spaces. everything is right down to the nub.Chapter 6
Appendix
The five previous chapters were designed for a year undergraduate course in algebra. Now F [x] is a Euclidean domain and so VF [x] is the sum of cyclic modules. An R-module M is said to be cyclic if it can be generated by one element. If the characteristic polynomial of T factors into the product of linear polynomials. so this is a powerful concept. then there is a basis for V so that the matrix representing T is in Jordan canonical form. The basic theorem of this chapter is that if R is a Euclidean domain and M is a finitely generated R-module. M ≈ R/I where I is an ideal of R. The style is the same as before.. In the chapter on matrices. finitely generated abelian groups are the sums of cyclic groups – one of the jewels of abstract algebra.e. In this appendix. There is a basis for V so that the matrix representing T is in Rational canonical form. A matrix in Jordan form is a lower triangular matrix with the eigenvalues of T displayed on the diagonal. Since Z is a Euclidean domain. Now suppose F is a field and V is a finitely generated F -module.e. These are independent sections added on at the end. is given in this chapter. A proof of this. This classical and very powerful technique allows an easy proof of the canonical forms. it is stated without proof that the determinant of the product is the product of the determinants. it is a free R-module. The organization is mostly a linearly ordered sequence except for the last two sections on determinants and dual spaces. then V becomes an F [x]-module by defining vx = T (v). If T : V → V is a linear transformation. which depends upon the classification of certain types of alternating multilinear forms. Two of the main goals are to characterize finitely generated abelian groups and to prove the Jordan canonical form.
then a or b ∈ I. Note that ∼ is an equivalence relation. Later on it will be shown that every Euclidean domain is a principle ideal domain. I is maximal means I = R and there are no ideals properly between I and R. then ∃ a maximal ideal I of R with a ∈ I. the student should work the exercise in group theory at the end of this section (see page 114). If a ∼ b. b ∈ R. Theorem 0 is a prime ideal of R iff R is ¯ 0 is a maximal ideal of R iff R is ¯ Suppose J ⊂ R is an ideal. The ideal V = Vt does not contain 1 and ¯ t∈T thus is not equal to R. This collection contains a maximal monotonic collection {Vt }t∈T . Note To properly appreciate this proof.
Definition Suppose R is a domain and a. Thus every Euclidean domain is a unique factorization domain. then a
. Then we say a ∼ b iff there exists a unit u with au = b. J = R. Definition Suppose R is a commutative ring and I ⊂ R is an ideal. If a ∈ R is not a unit.Chapter 6
Appendix Prime and Maximal Ideals and UFDs
109
In the first chapter on background material.
I is prime means I = R and if a. Here it will be shown that this property holds for any principle ideal domain. J is a prime ideal iff R/J is J is a maximal ideal iff R/J is Maximal ideals are prime. it was shown that Z is a unique factorization domain. Consider {J : J is an ideal of R containing a with J = R}.
Theorem
Corollary Proof Theorem
Every field is a domain.
Proof This is a classical application of the Hausdorff Maximality Principle. b ∈ R have ab ∈ I. Therefore V is equal to some Vt and is a maximal ideal containing a.
Note also ∃ a unit u and primes p1 . Note a is prime ⇒ a is irreducible. while the only ¯ associate of 0 is 0 itself. a is prime if it generates a prime ideal. Theorem equivalent. it is called a prime element. ¯ ¯ If F is a field and g ∈ F [x] is a non-zero polynomial.110 and b are said to be associates. . Definition 1) 2) Suppose R is a domain and a ∈ R is a non-zero non-unit. If an element of a domain generates a non-zero prime ideal. 1) 2) 3) Suppose R is a domain and a. then au is irreducible (prime). b ∈ (R − 0). Then the following are ¯
a ∼ b. An element a divides b (a|b) if ∃! c ∈ R with ac = b. .
Note If a is a prime and a|c1 c2 · · · cn . then a|ci for some i. then a is irreducible (prime) iff b is irreducible (prime). then its associates are n and −n. aR = bR. Note If a ∼ b. then a ∼ ci for some i.
Parts 1) and 3) above show there is a bijection from the associate classes of R to the principal ideals of R. Thus if R is a PID.. if d = b1 b2 · · · bn = c1 c2 · · · cm with each bi and each ci prime. This is immediate from the definitions. then n = m and for some permutation σ of the indices. .
a is irreducible if it does not factor. . This follows from the definition and induction on n. i. p2 .e. i. bi and cσ(i) are associates for every i. a = bc ⇒ b or c is a unit.
Theorem Factorization into primes is unique up to order and associates. The following theorem is elementary. a|b and b|a. In other words. then the associates of g are all cg where c is a non-zero constant. the associates of 1 are the units of R. if a is irreducible (prime) and u is a unit.e.
Appendix
Chapter 6
Examples If R is a domain..e. pt where no two are associates and du = ps1 ps2 · · · pst . i.. If n ∈ Z is not zero. there is a bijection from the associate classes of R to the ideals of R. t 1 2
. but it shows how associates fit into the scheme of things. If each cj is irreducible. a|bc ⇒ a|b or a|c.
then a factors into a finite product of irreducibles. Then the following are equivalent. Suppose R is a UFD.
Proof We already know 1) ⇒ 2).e. Theorem Suppose R is a PID and a ∈ R is non-zero non-unit.
R is a UFD. a is a prime element. Part 2) ⇒ 1) because factorization into primes is always unique. i. that one of these irreducibles is an associate of a. Since b divides a. If either Proof b or c is a unit or is zero. a is irreducible.
Proof Every maximal ideal is a prime ideal.. aR is a prime ideal. 1) 2) 3) aR is a maximal ideal. It follows from the uniqueness of the factorization of ad = bc.
Definition R is a factorization domain (FD) means that R is a domain and if a is a non-zero non-unit element of R. the element b is a unit or an associate of a. ∃ b ∈ R with I = bR. Fortunately.
. Then the following are equivalent. This means I = R or I = aR.Chapter 6 Proof
Appendix
111
This follows from the notes above. then R is a UFD iff each irreducible element generates a prime ideal. If I is an ideal containing aR. principal ideal domains have this property. so 2) ⇒ 3). Now suppose a is irreducible and show aR is a maximal ideal.. Therefore the element a is a prime. Each of b and c factors as the product of irreducibles and the product of these products is the factorization of bc. elements factor as the product of primes. then a divides one of them. and thus a|b or a|c. then a is irreducible ⇔ a is prime. Theorem If R is a UFD and a is a non-zero non-unit of R. Every prime element is an irreducible element. Definition R is a unique factorization domain (UFD) means R is a FD in which factorization is unique (up to order and associates). as seen in the next theorem. so suppose each of b and c is a non-zero non-unit element of R. There exists an element d with ad = bc. i. Thus in a UFD. so 1) ⇒ 2).e. Every irreducible element of R is prime. Theorem 1) 2) Suppose R is a FD. and a|bc. a is an irreducible element of R. If R is a FD. This is a revealing and useful theorem. a irreducible ⇔ a is prime.
Since R is Noetherian.. i. ∃ t0 ≥ 1 such that It = It0 for all t ≥ t0 . The element c must be reducible. Each of aR and bR properly contains cR. Let I be an ideal of R.112
Appendix
Chapter 6
Our goal is to prove that a PID is a UFD. nor need it be a maximal ideal of the ring R. Theorem A Noetherian domain is a FD. We shall see below that this is a useful concept which fits naturally into the study of unique factorization domains. Using the two theorems above. The proof will not require that ideals be principally generated. Having three definitions makes this property useful and easy to use. an } ⊂ R such that I = a1 R + a2 R + · · · + an R. It need not contain all the ideals of the collection. then J = I. and thus 1) is true.
Proof Suppose 1) is true and show 3). a2 . . each ideal of R is finitely generated.) Definition If R satisfies these properties. and it must be I itself. see the next theorem. and consider the collection of all finitely generated ideals contained in I..
Proof Suppose there is a non-zero non-unit element that does not factor as the finite product of irreducibles. but we slide over that.. ∃ a maximal one cR. is a monotonic sequence of ideals. a PID is a FD.. ... Any non-void collection of ideals of R contains an ideal I which is maximal in the collection.) If I1 ⊂ I2 ⊂ I3 ⊂ . with each properly contained in the next.. Thus 3) is true. i. so suppose 2) is false and show 3) is false. it only remains to show that a PID is a FD. . Theorem 1) 2) Suppose R is a commutative ring. We now have 2)⇒1)⇒3). and therefore 3) is false. By 2) there is a maximal one. (Actually this construction requires the Hausdorff Maximality Principle or some form of the Axiom of Choice. (The ideal I is maximal only in the sense described. . and so each
. Thus it is possible to construct a sequence of ideals I1 ⊂ I2 ⊂ I3 . . For example. This turns out to be equivalent to the property that any collection of ideals has a "maximal" element. Consider all ideals dR where d does not factor.e. is finitely generated and ∃ t0 ≥ 1 such that It0 contains those generators. Then the following are equivalent.
3)
If I ⊂ R is an ideal. This property is satisfied by many of the classical rings in mathematics. In particular. Now suppose 2) is true and show 1). The ideal I = I1 ∪ I2 ∪ . R is said to be Noetherian. c = ab where neither a nor b is a unit. This means if J is an ideal in the collection with J ⊃ I. but only that they be finitely generated. ∃ a finite set {a1 .e. So there is a collection of ideals of R such that any ideal in the collection is properly contained in another ideal of the collection. or it is said to satisfy the ascending chain condition.
. Bourbaki. Corollary A PID is a UFD.2 of An Introduction to Complex Analysis in Several Variables by L. m ∈√ Z}. In particular 2 · 2 = (1 − 5) · (−1 − 5) are two distinct irreducible factorizations of 4. xn ]] is a UFD.6. which is not a domain. ...)
Domains With Non-unique Factorizations Next are presented two of the standard examples of Noetherian domains that are not unique factorization domains.. (This follows immediately from the definition. Theorem If R is a UFD then R[x1 . Show R is isomorphic to Z[x]/(x2 − 5). which is a contradiction. o Theorem Suppose R is a commutative ring. Thus if F is a field. This gives a finite factorization of c into irreducibles. xn ]] is a UFD.... xn ]] are Noetherian.. √ √ Exercise Let R = Z( 5) = {n + m 5 : n.. So Z is a UFD and if F is a field. (There is a UFD R where R[[x]] is not a UFD.) Theorem Germs of analytic functions on Cn form a UFD. then the formal power series R[[x1 .. F [[x1 .. H¨rmander. See page 566 of Commutative Algebra by N. Thus if F is a field. Show that R is a subring of √ R which is not a UFD. . xn ] and R[[x1 . . then R/I is Noetherian.
Proof See Theorem 6. F [x] is a UFD.Chapter 6
Appendix
113
of a and b factors as a finite product of irreducibles.. where (x2 − 5) represents the ideal (x2 − 5)Z[x]. and R/(2) is isomorphic to Z2 [x]/(x2 − [5]) = Z2 [x]/(x2 + [1]).) If R is a PID. This and the previous theorem show that Noetherian is a ubiquitous property in ring theory. .) Theorem If R is Noetherian and I ⊂ R is a proper ideal.. (This is the famous Hilbert Basis Theorem. Then R is Noetherian ⇒ R[x1 .
You see the basic structure of UFDs is quite easy... F [x1 ..
.. xn ] is a UFD. which are stated here only for reference. . xn ] is a UFD. . (This theorem goes all the way back to Gauss.. It takes more work to prove the following theorems.
In particular. and g : B → C is a surjective homomorphism with kernel K. z] → R[x. and f (z) = y 2 . Show x2 − yz is irreducible and thus prime in R[x. An easier ¯ x approach is to let f : R[x. Show R ¯ is not a UFD. Show that H is maximal iff G/H ≈ Zp for some prime p. K must be a summand of B. y. a subgroup H of G is said to be maximal if H = G and there are no subgroups properly between H and G. In particular x · x = y · z are two distinct irreducible factorizations ¯ ¯ ¯ ¯ of x2 . g has a right inverse. i. Splitting Short Exact Sequences Suppose B is an R-module and K is a submodule of B. Q contains no maximal subgroups. (h is called a splitting map. Then the following are equivalent. Now suppose 2) is true and h : C → B is a right inverse of g. f (y) = x2 .. then there is a maximal subgroup H of Q which contains a. If u ∈ R[x. Then h is injective.)
Proof Suppose 1) is true. 1) 2) K is a summand of B. then h defined by h = i ◦ (g|L)−1 is a right inverse of g. let u ∈ R be the coset containing u. y. z]/(yz). This is used below to show that if R is a PID. and y 2 are irreducible in S and (xy)(xy) = (x2 )(y 2 ) are two distinct irreducible factorizations of (xy)2 in S. y] be the ring homomorphism defined by f (x) = xy. If i : L → B is inclusion. x2 .e. consider the case G = Q.e. i.
. Note that xy. ∃ a homomorphism h : C → B with g ◦ h = I : C → C. When is K a summand ¯ of B? It turns out that K is a summand of B iff there is a splitting map from B/K to B. z]. ¯ Thus K ⊕ h(C) = B. then every submodule of Rn is free. Which one of the following is true? 1) 2) If a ∈ Q. Then (g|L) : L → C is an isomorphism. Exercise In Group Theory If G is an additive abelian group. Theorem 1 Suppose R is a ring. y. x2 . In this case we write K ⊕ L = B. B and C are R-modules. For simplicity. y. As defined in the chapter on linear algebra. Show R/(¯) is isomorphic to R[y.. suppose ∃ a submodule L of B with K ⊕ L = B.114
Appendix
Chapter 6
Exercise Let R = R[x. Then S = R[xy. K is a summand of B provided ∃ a submodule L of B with K + L = B and K ∩ L = 0. K + h(C) = B and K ∩ h(C) = 0. z]. which is not a domain. if B/K is free. y 2 ] is the image of f and S is isomorphic to R. z]/(x2 − yz).
Chapter 6
Appendix
115
Definition Suppose f : A → B and g : B → C are R-module homomorphisms.e. If C is a free R-module. Although this theorem is transparent. iff B → C has a splitting map.s) means f is injective. is A → A ⊕ C → C where f = i1 and g = π2 . A short exact sequence is said to split if ∃ ≈ an isomorphism B → A ⊕ C such that the following diagram commutes. Showing these properties are equivalent to the splitting of the sequence is a good exercise in the art of diagram chasing. f g The statement that 0 → A → B → C → 0 is a short exact sequence (s.1 A short exact sequence 0 → A → B → C → 0 splits iff f (A) is a summand of B. Proof From the previous theorem we know this is true for n = 1. Suppose n > 1 and the theorem is true for submodules of Rn−1 . Proof We know from the previous theorem f (A) is a summand of B iff B → C has a splitting map. g is surjective and f (A) = ker(g).
. Thus subgroups of Zn are free Z-modules of dimension ≤ n. Now suppose C has a free basis T ⊂ C. 0→ A f
E
B
c
g
&2 π
E & b & &
C →0
i 1
~
≈
A⊕C
&
&
We now restate the previous theorem in this terminology. Suppose A ⊂ Rn is a submodule. The canonical split s.e. then A is a free R-module of dimension ≤ n. and g : B → C is surjective.
This theorem restates the ring property of PID as a module property. there is a splitting map and thus the sequence splits. There exists a function h : T → B such that g ◦ h(c) = c for each c ∈ T . Theorem 2 1) 2) If R is a domain.s. then the following are equivalent. Theorem 1. The function h extends to a homomorphism from C to B which is a right inverse of g. Theorem 3 If R is a PID and A ⊂ Rn is a submodule.
R is a PID. 1)⇒2) is a precursor to the following classical result. Every submodule of RR is a free R-module of dimension ≤ 1.
and is sometimes omitted from the definition. it is free of dimension 1 and thus the sequence splits by Theorem 1. r ∈ R such that a = bq + r with r = 0 or φ(r) < φ(b). is used only in Theorem 2. ¯ ¯ F [x] where F is a field with φ(f = a0 + a1 x + · · · + an xn ) = deg(f ). If π(A) = 0.116
Appendix
Chapter 6
Consider the following short exact sequences. This is one of the great classical theorems of abstract algebra. ¯ In either case. ¯ Examples of Euclidean Domains Z with φ(n) = |n|. Furthermore the first axiom. where f : R n−1 → Rn−1 ⊕ R is inclusion and g = π : Rn−1 ⊕ R → R is the projection. φ(a) ≤ φ(ab).s. b ∈ Z} = Gaussian integers with φ(a + bi) = a2 + b2 .1. Here N will denote the set of all non-negative integers. and the efficiency of this abstraction is displayed in this section. Show A ×3 is a free Z-module of dimension 1. ¯ If π(A) = 0. then ¯ 1) φ(a) ≤ φ(ab). and you don't have to worry about it becoming obsolete. (16. 2) ∃ q. 24). Definition A domain R is a Euclidean domain provided ∃ φ : (R − 0) −→ N such ¯ that if a. Z[i] = {a + bi : a. Exercise Let A ⊂ Z2 be the subgroup generated by {(6. b ∈ (R − 0). Euclidean Domains The ring Z possesses the Euclidean algorithm and the polynomial ring F [x] has the division algorithm (pages 14 and 45). then M is the sum of cyclic modules. Also show the s. Anyway it is possible to just play around with matrices and get some deep results. A is a free submodule of dimension ≤ n. not just the set of positive integers. 0 −→ Rn−1 −→ Rn−1 ⊕ R −→ R −→ 0 0 −→ A ∩ Rn−1 −→ A −→ π(A) −→ 0 By induction. 64)}. Z4 −→ Z12 −→ Z3 splits ×2 ×2 but Z −→ Z −→ Z2 and Z2 −→ Z4 −→ Z2 do not (see top of page 78). A ∩ Rn−1 is free of dimension ≤ n − 1. then A ⊂ Rn−1 . The concept of Euclidean domain is an abstraction of these properties.
f π
. If R is a Euclidean domain and M is a finitely generated R-module. A field F with φ(a) = 1 ∀ a = 0 or with φ(a) = 0 ∀ a = 0.e.
If E ∈ Rn . ¯ ¯ Theorem 2 If R is a Euclidean domain and a. If E is invertible. This means that row and column operations on A do not change the ideal I. This is a good exercise. and this will turn out to be the d1 displayed in the theorem.j ). . Since R is a PID.
Proof
The following remarkable theorem is the foundation for the results of this section. . However. if E ∈ Rt is invertible and J is the ideal generated by the elements of AE. . then ∃ b ∈ I − 0 satisfying φ(b) ≤ φ(a) ∀ a ∈ I − 0. then by elementary row and column operations (ai. Now r ∈ I and ¯ r = 0 ⇒ φ(r) < φ(b) which is impossible.
117
Proof If I is a non-zero ideal. r with a = bq + r. . row and column operations on (ai. dm 0 0
0
0
where each di = 0. However it is unnecessary for Theorem 3 below.j ) ∈ Rn.t is a non-zero matrix.j ). Theorem 3 If R is a Euclidean domain and (ai. b ∈ R − 0. then R is a PID and thus a UFD.Chapter 6 Theorem 1
Appendix If R is a Euclidean domain. ¯ ¯ Then b generates I because if a ∈ I − 0.j ) may produce elements with smaller
. In the same manner. Proof Let I ⊂ R be the ideal generated by the elements of the matrix A = (ai.j ) can be transformed to
d1 0 · · · 0 d2 . then J = I.. and di |di+1 for 1 ≤ i < m. ¯ a and b are associates ⇒ φ(a) = φ(b). there is an element d1 with I = d1 R. ∃ q. then J = I. ¯ a is a unit in R iff φ(a) = φ(1). The matrix (ai. then ¯ φ(1) is the smallest integer in the image of φ. then the ideal J generated by the elements of EA has J ⊂ I.j ) has at least one non-zero element d with φ(d) a miminum. Also d1 generates the ideal of R ¯ generated by the entries of (ai. Thus r = 0 and a ∈ bR so I = bR.
¯ and di |di+1 for 1 ≤ i < t. where each di = 0.i wn . Theorem 4 Suppose R is a Euclidean domain.. and thus I = d1 R. ∃ invertible matrixes U ∈ Rn and V ∈ Rt such that
. a2 .... Then ∃ free bases {a1 .t where vi = a1. By the previous theorem. This is an example of a theorem that is easy to prove playing around at the blackboard. Yet it must be a deep theorem because the next two theorems are easy consequences.. vt }. and such that each ai = di bi . Thus B/A ≈ R/d1 ⊕ R/d2 ⊕ · · · ⊕ R/dt ⊕ Rn−t . the matrix may be changed to the following form.j ) by a finite number of row and column operations.. .j ) be one which has an entry d1 = 0 with φ(d1 ) a minimum. Among these. . Thus by column operations of type 3. Proof By Theorem 3 in the section Splitting Short Exact Sequences.j . By elementary operations of type 2.i w2 + · · · + an. the other entries of the first row may be made zero.. w2 . The composition Rt −→ A −→ B −→ Rn ei −→ vi wi −→ ei
≈ ⊂ ≈
is represented by a matrix (ai. . Let {w1 . In a similar manner. . .118
Appendix
Chapter 6
φ values. at } for A and {b1 . 0 cij
Note that d1 divides each ci..j ) ∈ Rn. where n ≥ t. 1) place in the matrix... else we could obtain an entry with a smaller φ value. Then d1 will divide the other entries in the first row. with t ≤ n. wn } be a free basis for B.. the entry d1 may be moved to the (1. To consolidate this approach. v2 .
d1 0 · · · 0 0 .i w1 + a2. bn } for B. consider matrices obtained from (ai. b2 . A has a free basis {v1 . B is a finitely generated free Rmodule and A ⊂ B is a non-zero submodule. . by row operations of type 3.. The proof now follows by induction on the size of the matrix. let (bi.
.s. there is no prime common to their prime factorizations. d2 . dt are ¯ called invariant factors.) Proof There exists an a ∈ R with aR = bR + cR.) Proof
s If i = j.j )V =
0
···
≈
dt 0
with di |di+1 . . Then t ≈ the natural map R/d −→R/ps1 ⊕ · · · ⊕ R/pst is an isomorphism of R-modules. . If R = Z and we select the di to be positive.. The result now follows from the previous theorem. then the elements d1 . If R = F [x] and we select the di to be monic. the theorem follows. then they are unique... R/di = 0. Theorem 6 Suppose R is a PID and d is a non-zero non-unit element of R. 0 d2 0 ··· 0 0 . and B/A ≈ M. t 1 s (The elements pi i are called elementary divisors of R/d. then M ≈ R/d1 ⊕ R/d2 ⊕ · · ·⊕ R/dt ⊕ Rm where each di = 0. . i.Chapter 6
Appendix
119 d1 0 . v2 .e. they are s
. The way Theorem 5 is stated. They are unique up to associates. Since changing the isomorphisms Rt −→ A and B −→ Rn corresponds to changing the bases {v1 . they are unique. The splitting in Theorem 5 is not the ultimate because the modules R/di may split into the sum of other cyclic modules. Let A be the kernel.. s1 s2 Assume d = p1 p2 · · · pst is the prime factorization of d (see bottom of p 110). and for such di . Suppose b and c are relatively prime. some or all of the elements di may be units.. Then bR and cR are comaximal ideals.. so R = bR + cR. wn }. By the Lemma above. (See p 108 for comaximal. .e..
≈
U (ai. To prove this we need the following Lemma. a is a unit. w2 . but we do not bother with that here. and di |di+1 for 1 ≤ i < t... pi i and pj j are relatively prime. Lemma Suppose R is a PID and b and c are non-zero non-unit elements of R. Theorem 5 If R is a Euclidean domain and M is a finitely generated R-module. vt } and {w1 . . If we assume that no di is a unit. Since a|b and a|c. ¯ Proof By hypothesis ∃ a finitely generated free module B and a surjective homo⊂ morphism B −→ M −→ 0.. . so 0 −→ A −→ B −→ M −→ 0 is a s.
splits. Then the following s..
.e. M ≈ Rm . An element m ∈ M is said to be a torsion element if ∃ r ∈ R with r = 0 and mr = 0.
Theorem 9 Suppose R is a Euclidean domain and M is a finitely generated R-module. Suppose R is a PID. Then M is a free R-module. the natural map is a ring isomorphism (page 108). Proof This is a simple exercise.e.120
Appendix
Chapter 6
comaximal and thus by the Chinese Remainder Theorem. If R = Z. and thus there is a splitting map. Since the natural map is also an R-module homomorphism. Theorem 7 Suppose M is a module over a domain R. M/T (M ) is torsion free. Denote by T (M ) the set of all torsion elements of M . it is the same as saying m has finite order. as seen by the next exercise. i.
Definition Suppose M is a module over a domain R. If T (M ) = 0. because Theorem 5 gives a splitting of M into a torsion part and a free part. it is an R-module isomorphism. M/T (M ) is a free R-module. Proof This follows immediately from Theorem 5. Then T (M ) is a submodule of M and M/T (M ) is torsion free. This theorem carries the splitting as far as it can go. and s ≥ 1.
Theorem 8 Suppose R is a Euclidean domain and M is a finitely generated R-module which is torsion free. 0 −→ T (M ) −→ M −→ M/T (M ) −→ 0 Proof By Theorem 7. Then the Exercise R-module R/ps has no proper submodule which is a summand.s. we say that M is torsion ¯ free. By Theorem 8. p ∈ R is a prime element. This is the same as ¯ ¯ saying m is dependent. Of course this theorem is transparent anyway.
Torsion Submodules
This will give a little more perspective to this section.
Thus t = 1 and so G is cyclic. The multiplicative group G is isomorphic to an additive group Z/d1 ⊕ Z/d2 ⊕ · · · ⊕ Z/dt where each di > 1 and di |di+1 for 1 ≤ i < t. V depends upon the splitting map and is unique only up to isomorphism.
To complete this section. but the complementary summand V is not unique..
0
dn
where each di = 0 and di |di+1 for 1 ≤ i < n. Every u in the additive group has the property that udt = 0. Theorem 11 Suppose R is a Euclidean domain and A ∈ Rn is a matrix with non-zero determinant. Thus if A is invertible. Thus if p is a prime. the equation will have degree ¯ ¯ less than the number of roots. Theorem 10 Suppose T is a domain and T ∗ is the multiplicative group of units of T . A may be transformed to a diagonal matrix
d1 d2 0 . If t > 1. Also d1 generates the ideal generated ¯ by the entries of A. Thus if F is a finite field. (Zp )∗ is cyclic. A is the product of elementary matrices. Exercise For which primes p and q is the group of units (Zp ×Zq )∗ a cyclic group?
We know from Exercise 2) on page 59 that an invertible matrix over a field is the product of elementary matrices. which is impossible.Chapter 6
Appendix
121
Note It follows from Theorem 9 that ∃ a free submodule V of M such that T (M )⊕ V = M . If G is a finite subgroup of T ∗ . ¯ So every g ∈ G is a solution to xdt − 1 = 0. then G is a cyclic group. The first summand T (M ) is unique. Proof This is a corollary to Theorem 5 with R = Z. Then by elementary row and column operations. the multiplicative group F ∗ is cyclic. This result also holds for any invertible matrix over a Euclidean domain. here are two more theorems that follow from the work we have done. Furthermore A is invertible iff each di is a unit.
. .
and C(q) will be the matrix of this endomorphism with respect to this basis. is zero iff h(x) ∈ qR[x]. The companion matrix
. V is a torsion module over the ring R[x]. 3 11 3 11 and D = . Write D as the product of elementary matrices. Find the characteristic polynomials of A and D. (See the last part of the last theorem on page 46. the matrix A is invertible iff the diagonal matrix is invertible. Perform elementary 0 4 1 4 operations on A and D to obtain diagonal matrices where the first diagonal element divides the second diagonal element. . xn−1 }. Let T : V → V be defined by T (v) = vx. while the Jordan canonical form will exist iff the characteristic polynomial factors as the product of linear polynomials. Exercise Let R = Z. If each di is a unit. it is the product of elementary matrices. q = a0 + a1 x + · · · + an−1 xn−1 + xn ∈ R[x] is a monic polynomial of degree n ≥ 1. . V has a free basis {1. and h(T ) is the zero homomorphism iff h(x) ∈ qR[x]. Find an elementary matrix B over Z such that B −1 AB is diagonal.) Multiplication by x defines an R-module endomorphism on V . A Jordan block displays its eigenvalue on the diagonal. x2 . Theorem Let V have the free basis {1. If h(x) ∈ R[x]. Note that the Jordan block B(q) is the sum of a scalar matrix and a nilpotent matrix. The homomorphism from R[x]/q to R[x]/q given by multiplication by h(x). ¯ Furthermore. it follows that each di = 0. but as an R-module. That is to say q(T ) = a0 I + a1 T + · · · + T n is the zero homomorphism. All of this is supposed to make the next theorem transparent. and is more interesting than the companion matrix C(q). which is true iff each di is a unit. and V is the R[x]-module V = R[x]/q. Therefore if A is invertible.122
Appendix
Chapter 6
Proof It follows from Theorem 3 that A may be transformed to a diagonal matrix with di |di+1 . then the diagonal matrix is the product of elementary matrices of type 1. . Since the determinant of A is not zero.. But as we shall see later. x2 . Show C cannot be selected in Q2 .. h(T ) is the R-module homomorphism given by multiplication by h(x). Find an invertible matrix C in R2 such that C −1 DC is diagonal. . Suppose R is a commutative ring. we define the two special types of square matrices used in the Rational and Jordan canonical forms. the Rational canonical form will always exist.. x. . x. A = Jordan Blocks In this section. xn−1 }.
A submodule of VF [x] is a submodule of VF which is invariant under T . Now we say all this again with a little more detail. In the section on Jordan Blocks. x2 . and T : V → V is an R-module homomorphism. If all the eigenvalues for T are in F . R[x] is a Euclidean domain and so we know almost everything about V as an R[x]-module. . xm−1 } as the F -basis for F [x]/di where m is the degree of the polynomial di . the matrix representing T is
VF [x] ≈ F [x]/d1 ⊕ F [x]/d2 ⊕ · · · ⊕ F [x]/dt
C(d1 ) C(d2 )
. Theorem 2 With respect to this basis. Questions about the transformation T are transferred to questions about the module V over the ring R[x]. Since V is finitely generated as an F -module. we suppose R is a field F . V is an R[x]-module. x. From Theorem 5 in the section on Euclidean Domains. V is a finitely generated F -module. Theorem 1 Under this scalar multiplication. . Pick {1. T : V → V is a linear transformation and V is an F [x]-module with vx = T (v).. . C(dt )
. . it follows that where each di is a monic polynomial of degree ≥ 1. we pick another basis for each of the cyclic modules (see the second theorem in the section on Jordan Blocks). We know VF [x] is the sum of cyclic modules from Theorems 5 and 6 in the section on Euclidean Domains. a basis is selected for these cyclic modules and the matrix representing T is described. Now in this section.
This is just an observation. And in the case R is a field. and di |di+1 . V is an R-module.124
Appendix
Chapter 6
Suppose R is a commutative ring. Then the matrix representing T is called the Jordan Canonical Form. the free part of this decomposition will be zero. but it is one of the great tricks in mathematics. Define a scalar multiplication V × R[x] → V by v(a0 + a1 x + · · · + ar xr ) = va0 + T (v)a1 + · · · + T r (v)ar . This gives the Rational Canonical Form and that is all there is to it.
. Our goal is to select a basis for V such that the matrix representing T is in some simple form.
this section is loosely written. and to show the diagonal elements are all zero. but it is assumed for this exercise. Note that the λi need not be distinct. λ2 . Suppose V is an n-dimensional vector space over a field F and T : V → V is a linear transformation. Then λi = 0 for 1 ≤ i ≤ n. its trace is the sum of the eigenvalues and its determinant is the product of the eigenvalues. So suppose T is in Jordan form and trace (T i ) = 0 for 1 ≤ i ≤ n. and the result follows by induction on the size of T . As before we make V a module over F [x] with T (v) = vx. Of course a diagonal matrix is about as canonical as you can get. Thus trace ¯ (p(T )) = a0 n where a0 is the constant term of p(x). Show T is nilpotent. n 2 ¯ ¯
Minimal polynomials To conclude this section here are a few comments on the minimal polynomial of a linear transformation. This is based on the fact that there exists a field F containing F ¯ as a subfield. a0 = 0 ¯ ¯ ¯ and so 0 is an eigenvalue of T . However this exercise can still be worked ¯ using Jordan form. Corollary Suppose F is a field of characteristic 0. Note A diagonal matrix is in Rational canonical form and in Jordan canonical form.126
Appendix
Chapter 6
The characteristic polynomial of T is p = (x − λ1 )s1 · · · (x − λr )sr and p(T ) = 0. λn ) ∈ F n i satisfies λ1 + λi + · · +λi = 0 for each 1 ≤ i ≤ n.. We know p(T ) = 0 and thus ¯ trace (p(T )) = 0. This fact is not proved in this ¯ book. n ≥ 1. T is nilpotent iff U T U is nilpotent. and (λ1 . Since the field has characteristic 0. and thus T may have no conjugate in Fn which is in Jordan form. . Finally. So ∃ an invertible matrix U ∈ Fn so that −1 −1 U T U is in Jordan form. This means that one block of T is a strictly lower ¯ triangular matrix. such that p factors into linears in F [x]. The polynomial p may not factor into linears in F [x]. This ¯ is called the Jordan canonical form for T. Removing this block leaves a smaller matrix which still satisfies the hypothesis. This part should be studied only if you need it. It also has a cute corollary. This exercise illustrates the power and facility of Jordan form.
. This is the case where each block is one by one. so it is important to use the transpose principle to write three other versions of the last two theorems. The point is that it sufficies to consider the case where T is in Jordan form..
Exercise Suppose F is a field of characteristic 0 and T ∈ Fn has trace(T i ) = 0 ¯ for 0 < i ≤ n. Note also that if a matrix is in Jordan form. Let p ∈ F [x] be the characteristic polynomial of T . and of course. and thus a0 n = 0.
If A is given to start with. If p(x) ∈ F [x] is the characteristic polynomial of T . The polynomial u is called the minimal polynomial of T. Recall that q is the characteristic polynomial and the minimal polynomial of each of these matrices. This is a non-zero ideal of F [x] and is thus generated by a unique ¯ monic polynomial u(x) ∈ F (x). The polynomial u is also called the minimal polynomial of A. ¯ Now we state this again in terms of matrices. h(A) = 0 iff h is a multiple of ¯ ¯ u in F [x].
.
. Find the characteristic and minimal polynomials of A. then p(A) = 0 and ¯ thus p is a multiple of u. Suppose A is the matrix displayed in Theorem 4 above. Find the characteristic and minimal polynomials of A. let B(q) ∈ Fn be the Jordan block matrix also defined in that section. Suppose A ∈ Fn is a matrix representing T . Exercise 1) 2) 3) Suppose A ∈ Fn . Now suppose q ∈ F [x] is a monic polynomial and C(q) ∈ Fn is the companion matrix defined in the section Jordan Blocks. Note that these properties hold for any matrix representing T ... . If p(x) ∈ F [x] is the characteristic polynomial of A. and thus similar matrices have the same minimal polynomial. Suppose A is the matrix displayed in Theorem 3 above.
0
polynomial. which satisfy V h = 0. use the linear transformation T : F n → F n determined by A to define the polynomial u.
Find the characteristic polynomial
Suppose A is the matrix displayed in Theorem 2 above. Find the characteristic and minimal polynomials of A. This together with the rational form and the Jordan form will allow us to understand the relation of the minimal polynomial to the characteristic polynomial.e. Ann(VF [x]) = uF [x]. p(T ) = 0 and thus p is a multiple of u. h(T ) = 0 iff ¯ ¯ h is a multiple of u in F [x]. Whenever q(x) = (x − λ)n . Note that u(T ) = 0 and if h(x) ∈ F [x]. i. Exercise Suppose Ai ∈ Fni has qi as its characteristic polynomial and its minimal
A1 A2 0 .Chapter 6
Appendix
127
Definition Ann(VF [x]) is the set of all h ∈ F [x] which annihilate V . and A =
Ar
and the minimal polynomial of A. Then u(A) = 0 and if h(x) ∈ F [x].
bn ) = f (bτ (1) . . .128
Appendix
Chapter 6
4) Suppose λ ∈ F . 2. . B2 . . Find the characteristic and minimal 6) Let F = R and A = 0 −3 1 −1 polynomials of A. . so is f1 + f2 . n ≥ 2.
Determinants In the chapter on matrices. C is an R-module. . The purpose of this section is to give a proof of this. . it is stated without proof that the determinant of the product is the product of the determinants (see page 63). . . then (f r) is R-multilinear. Also if f is R-multilinear and r ∈ R. We suppose R is a commutative ring. b2 . (This funny looking exercise is a little delicate. . . .) 5 −1 3 2 0 . bn ) defines an R-linear map from Bi to C. . . . From here on. Show λ is a root of the characteristic polynomial of A iff λ is a root of the minimal polynomial of A. . . . . . Definition A map f : B1 ⊕ B2 ⊕ · · · ⊕ Bn → C is R-multilinear means that if 1 ≤ i ≤ n. Definition 1) 2) f is symmetric means f (b1 . the set of all functions from B1 ⊕ B2 ⊕ · · · ⊕ Bn to C is an R-module (see page 69). Bi . . ¯ 5) Suppose F is a field containing F as a subfield. . bτ (n) ) for all τ .
Proof From the first exercise in Chapter 5. Bn is a sequence of R-modules. . Show that the minimal polynomial of A ∈ Fn is the same as the minimal polynomial of A considered as a ¯ matrix in Fn . It is easy to see that if f1 and f2 are R-multilinear. bτ (n) ) for all permutations τ on {1. Theorem The set of all R-multilinear maps is an R-module. bn ) = sign(τ )f (bτ (1) . Show that if λ is a root. and B1 . . its order in the characteristic polynomial is at least as large as its order in the minimal polynomial.
. . It must be seen that the R-multilinear maps form a submodule. . . and bj ∈ Bj for j = i. then f |(b1 . suppose B1 = B2 = · · · = Bn = B. . . . f is skew-symmetric if f (b1 . . n}. . | 677.169 | 1 |
Trigonometry: A Complete Introduction is the most comprehensive yet easy-to-use introduction to Trigonometry. Written by a leading expert, this book will help you if you are studying for an important exam or essay, or if you simply want to improve your knowledge. The book covers all areas of trigonometry including the theory and equations of tangent, sine and cosine, using trigonometry in three dimensions and for angles of any magnitude, and applications of trigonometry including radians, ratio, compound angles and circles related to triangles areas of trigonometry, including: the theory and equations of tangent, sine and cosine, dimensions and angles, radians, ratios, compound angles and circles related to trianglesigonometry | 677.169 | 1 |
ADULTS CAN DO ALGEBRA
You probably already know more than you think you do - you just don't know the mathematical terms in order to put a name to the math you are already doing.
Now there is an EASY TO UNDERSTAND SELF-PACED ONLINE COURSE just for you!
I'M TERRIBLE AT MATH – HOW CAN I POSSIBLY LEARN ALGEBRA?! AND WHY WOULD I WANT TO ANYWAY? This online course will start with working with basic numbers. These are things you already do in your life – think checkbook balancing, tip figuring, and so on. But we will start to look at numbers in different ways - with mathematical terms. Each lesson only looks only at one specific math skill at a time, and so it's not overwhelming. Each lesson starts with an easy to understand explanation, then gives you some examples, which you then apply to practice problems. Once you have a different way of thinking about doing math, and the language to put a name to the math you are doing, the last step is to apply that to algebra. "Algebra" is only a name for using numbers in certain ways, and you've possibly been doing this too, without knowing you were doing algebra. The course not only culminates with lessons in how algebra works, but also shows you how you use it in your every day life. As a bonus at the end, we will also use algebra to create graphs. You may never need to create a graph in your life, but you probably often come across them – in news stories, advertisements, sports stats, science stories, and so on – and you will have a better understanding of how to interpret what you see. This course is also useful for homeschooling parents! HOW DOES IT WORK This is a self-paced online course. The course is made up of 4 units, which have 3 to 6 lessons each, for a total of 16 lessons. You progress at your own speed, and you aren't given the next lesson until you've completed – and understood - the one you are working on. There are no tests. This course is entirely online, but no specialized computer knowledge is needed to participate. Each lesson is e-mailed to you as a regular attachment. You print out the attachments (easy to understand instructions, examples, and worksheet) and complete the worksheet. Then when you're ready, scan the worksheet and e-mail it back. A scanned copy of the corrected worksheet is e-mailed back to you. The next lesson is also e-mailed to you at that time. But, if you still need some time on a previous lesson, because you don't understand a concept (which after all is why you are here), support is available by direct e-mail with the instructor. You don't have to start the next lesson until you are ready. There is no pressure to keep up with any classmates. There is no pressure to finish up something you don't fully understand because of a deadline. There is no pressure to complete any lesson in any specified time frame. You work on the course at your own pace. Each lesson is made up of:
An easy to understand explanation.
Examples of how to apply the explanation to the problems.
A worksheet of practice problems.
The completed practice problems returned to you. If you understand that lesson's concept you receive the next lesson. If you didn't get it yet (after all that's why we're here, because we don't get it yet) then the instructor will personally work with you further.
The instructor is always available for questions at any time (allow 24 hours for e-mailed responses).
HOW MUCH DOES IT COST? The total cost of the course is just $240 (because it shouldn't cost more to exercise your mind than it does to exercise your body):
Non-refundable registration fee of $40.
Balance due of $200 before you start your course, which is completely refundable through the completion of the second lesson, if you find this course really isn't for you for any reason.
(Checks and Visa-M/C are accepted.) There is no time limit for each lesson, however if you haven't completed the 16 lessons within a year of your registration your account will be closed. Contact Us to request a registration form. WHO IS THE INSTRUCTOR? Hi! I'm Beth Rand. I'm bit of a math geek, and also a tech theatre geek (as you can see by the rest of this website). I have teaching certifications in math and theatre, and I have taught both in schools in the past. I have many people in my life who say they 'hate' math, or they were 'never any good at' math, yet I see them using - and understanding - math all the time. The difference between what they see and what I see is often simply nomenclature - in other words, I know a term for what they are doing (often times in their heads!), and because they don't know the term they think they don't know math. For example, I see people using "Distributive and Associative Properties" quite often, yet they also tell me they've never been any good at math. "Algebra" is also simply a term for a way of manipulating numbers to solve a question you have about something, but it's often times when a person reaches algebra that they start to tune out of mathematics. For these reasons I decided to create a course, which is self-paced (so no pressure, no comparing yourself to other classmates), which teaches the math needed for Pre-Alegbra up to basic Algebraic equations and graphing, and which then allows you to apply this knowledge in your everyday life. I hope you will enjoy (yes, enjoy!) algebra as much as I do!
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Math application
i was wandering if there is like a math program/application that can help me with math problems like homework and all. i need something that can solve a problem for me or help me step by step?
any ideas
thanks a bunch!
That's a double edged sword! A competent math program like MuPAD will certainly let you cover a lot more territory in a short time. However, if you haven't mastered the basics, which you learn by DOING, then simply seeing more answers won't really help you master the basics in the long run. Used properly, such programs can be of great benefit.
As said before, nothing is quite as good as trying to solve the problem yourself. There are a lot of programs which solve things and that is good for checking answers, but that's all computer programs are really good for (until you get a lot more advanced).
If you're in the U.K (and you probably look up the needed stuff if you live in any other country) then has a lot of good examples for almost any problems from early school to A Level with step by step on how to solve. But really it's no substitute for a teacher. | 677.169 | 1 |
University of Dallas Mathematics
Much of mathematics has its roots in science but the spirit of mathematical inquiry
is not bound to any specific area. Mathematics is an important discipline for every
educated person.
All students at the university are therefore required to study some mathematics. The
goal of the requirement is to strengthen the student's imaginative and deductive powers
through the discipline imposed by rigorous mathematical thinking. The precise use
of language and logic characteristic of mathematics is developed in the courses which
the student may select to meet the core requirement.
There are many areas of mathematics from which the student may choose. Each of these
areas deals with profound ideas that play an important part in our culture. The courses
in Euclidean and Non-Euclidean Geometry and Linear Point Set Theory are designed explicitly
for this purpose. In certain circumstances Calculus I, II, and Linear Algebra also
serve the purpose of the core requirement, as do the other more advanced courses in
the Department | 677.169 | 1 |
Calculus Help and Problems
This section contains in depth discussions and explanations on key topics that appear throughout Calculus 1 and 2 up through Vector Calculus. The topics are arranged in a natural progression catering typically to late highschool and early college students, covering the foundations of calculus, limits, derivatives, integrals, and vectors.
Calculus is an advanced branch of mathematics, incorporating algebra, geometry, and
trigonometry. Known as the study of change and motion, core calculus concepts include limits,
derivatives, and integrals of functions. In this section, learn about calculus' history, how it's the
similar to (and different from!) other branches of math, and how people use calculus in real life.
One of the first topics introduced in any calculus class, limits introduce the component of
infinity to math problems. How do you find the limit of a function? This lesson defines limits and
provides a variety of examples to understand the concept.
The process of finding the derivative of a function at any point is called differentiation, and
differential calculus is the field that studies this process. This overview of differential calculus
introduces different concepts of the derivative and walks you through example problems.
Integral calculus involves the concept of integration. Alongside differentiation, integration is one
of the main operations in calculus. Integration is the process of finding the integral of a function
at any point on a graph. This lesson defines integration and also covers Riemann integration
and the general power rule.
Multivariable calculus, also called vector calculus, deals with functions of two variables in three-
dimensional space. Multivariable calculus extends concepts found in differential and integral
calculus. This group of lessons introduces important concepts such as vectors in two and three-
dimensional space and vector functions.
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Note - you only need to make one application for A-levels. You can specify the additional A-Levels you wish to study when you fill out the application form
Level: 3 Location: Joseph Wright Centre Years: 2 Interview: Y
Course Summary
This exciting new course will help you to use maths in a range of 'real-world' contexts to prepare for success in your studies, future career and life.
Half the size of an A-level, this two-year course will develop your mathematical skills and thinking appropriate for a Level 3 qualification. You will become more confident and competent in using maths and the qualification will be considered for UCAS points if you are planning to study at university in future.
The course is particularly useful for students taking subjects such as A-level Biology, Business Studies, Economics, Computing, Geography, Psychology, BTEC Applied Science, Health and Social Care, and IT – as well as technical and vocational qualifications.
Entry requirements
You need a grade 5 or higher tier GCSE Maths, alongside the entry requirements for your other A-level courses.
Course Content
You will learn and apply real-world maths throughout your studies. This will equip you with skills for scenarios and contexts you are likely to encounter in vocational and academic study, future employment and life.
As part of the programme, you will study data analysis, maths for personal finance, estimation, critical path and critical analysis.
How will I be assessed?
There are two examinations at the end of the second year.
Are there any additional costs or specialist equipment required?
No
What can I do after this course?
Further Study
This course will provide additional UCAS points to support your university application. You will gain 24 UCAS points if you achieve a grade A, 20 for grade B, 16 for grade C, 12 for grade D and 8 for grade E.
Careers
Your mathematical skills will be enhanced and improved, providing you with an important asset to impress potential employers across many sectors.
This course is run at the Joseph Wright Centre
Yikai Shen
Aiming for a career in financial services
Aiming for a career in financial services, Yikai Shen chose to study A-levels in Maths, Further Maths, Physics and Business at Derby College before continuing his education at Imperial College, London.
A British education is recognised throughout the world as the best. My biggest challenge at College was obviously the language but I worked hard and gained a lot of support from tutors at the College.
Derby is a very nice city and I settled in well. I made a lot of good friends and everyone was very welcoming | 677.169 | 1 |
Calculators
Graphing calculators offer numerous tools for students to use that promote deeper conceptual understanding of mathematics by exposing students to multiple representations of concepts. Furthermore, graphing calculators increase students' higher order thinking skills, improve student motivation, and provide students and teachers comprehensive methods to investigate, explore and discover mathematical concepts (Ozel, Yetkiner, & Capraro, 2008). Research has shown that spending more time using graphing calculators for algebra courses leads to higher end-of-course exam scores. However, the research also recommends having occasional periods of not using the calculator to ensure students do not become too dependent upon them (Heller, Curtis, Jaffe, & Verboncoeur, 2005). It is important for the teacher to become experienced with calculators in order to understand their capabilities and limitations. Programs and applications can be transferred from a computer to the calculator, as well as from one calculator to another. | 677.169 | 1 |
This up-to-date book will prepare students for the new Geometry (Common Core) Regents exam. It features the first two actual Regents exams administered for the updated Geometry Regents, all answers thoroughly explained, study tips, test-taking strategies, score analysis charts, and more.
This Algebra Essentials Practice Workbook with Answers provides ample practice for developing fluency in very fundamental algebra skills - in particular, how to solve standard equations for one or more unknowns. These algebra 1 practice exercises are relevant for students of all levels - from grade 7 thru college algebra. With no pictures, this workbook is geared strictly toward learning the material and developing fluency through practice. This workbook is conveniently divided up into seven chapters so that students can focus on one algebraic method at a time. Skills include solving linear equations with a single unknown (with a separate chapter dedicated toward fractional coefficients), factoring quadratic equations, using the quadratic formula, cross multiplying, and solving systems of linear equations. Not intended to serve as a comprehensive review of algebra, this workbook is instead geared toward the most essential algebra skills. Each section begins with a few pages of instructions for how to solve the equations followed by a few examples. These examples should serve as a useful guide until students are able to solve the problems independently. Answers to exercises are tabulated at the back of the book. This helps students develop confidence and ensures that students practice correct techniques, rather than practice making mistakes. The copyright notice permits parents/teachers who purchase one copy or borrow one copy from a library to make photocopies for their own children/students only. This is very convenient for parents/teachers who have multiple children/students or if a child/student needs additional practice. An introduction describes how parents and teachers can help students make the most of this workbook. Students are encouraged to time and score each page. In this way, they can try to have fun improving on their records, which can help lend them confidence in their math skills. | 677.169 | 1 |
Algebra I Reference (formula) Chart -- the formula chart students are allowed to use on EOC STAAR test Algebra I Performance Level Descriptors -- the criteria to reach the three levels set by the state based on Bloom's Taxonomy.
Homework Help
There is a math teacher available 30 minutes before school and 30 minutes after school every school day. Even if it is not your teacher, any math teacher can help you with any course.
itutoring -- This website has a course designed for SAT preparation and two practice math SAT's username: mwisd password: rams
Khan Academy -- a free website that covers SAT preparation for math, reading, and writing. It has a complete test with worked out examples and extra practice for each section.
Purple Math -- a website dedicated to showing step by step solutions for any given topic.
Educreations -- a website where teachers create step by step video solutions for selected problems. See your teacher for their course code | 677.169 | 1 |
Many worksheets are here for arithmetic, pre-algebra, algebra, and basic geometry. Useful for elementary through...
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Many worksheets are here for arithmetic, pre-algebra, algebra, and basic geometry. Useful for elementary through developmental college level.
Material Type:
Drill and Practice
Author:
Unknown
Date Added:
Aug 07, 2009
Date Modified:
Sep Math Coordinator, Susan McClory (San Jose State University), discusses her experience working with here...
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Developmental Math Coordinator, Susan McClory (San Jose State University), discusses her experience working with here department to transform the remedial math course on her campus.
Material Type:
Case Study
Author:
Susan McClory step | 677.169 | 1 |
Jessica Flores, FL
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How is doing operations (adding, subtracting, multiplying, and dividing) with rational expressions similar to or different from doing operations with fractions? Can understanding how to work with one kind of problem help understand how to work another type? W | 677.169 | 1 |
The Sharp EL-8106 ELSIMATE is an arithmetic calculator with 8 digits precision and algebraic logic. It has 6 functions, 21 keys and a VFD (vacuum fluorescent) display. The power source is 3xAA batteries. | 677.169 | 1 |
Abstract
Tese de mestrado, Educação (Didática da Matemática), Universidade de Lisboa, Instituto de Educação, 2011Esta investigação tem por objetivo caracterizar o sentido de símbolo de alunos na fase final do ensino secundário e a sua relação com a aprendizagem da Álgebra. Duas questões orientam este trabalho: (i) Que sentido de símbolo revelam esses alunos no modo como resolvem questões envolvendo expressões algébricas, equações, problemas e funções? (ii) Qual a relação entre o desenvolvimento do sentido de símbolo dos alunos e a sua capacidade de realização de questões sobre conteúdos específicos do 12.º ano?
A investigação incide num grupo de vinte e um alunos, e aborda com mais profundidade dois alunos do 12.º ano. A recolha de dados é feita através de um teste diagnóstico, duas entrevistas e documentos escritos pelos alunos. Um quadro de referência é a base da caracterização do sentido de símbolo dos alunos, a partir da forma como estes resolvem questões envolvendo expressões algébricas, equações, problemas e funções. A metodologia é essencialmente qualitativa (estudos de caso), complementada por alguns aspetos quantitativos, e insere-se no paradigma interpretativo.
O quadro de referência mostrou ser uma ferramenta útil e adequada à caracterização do sentido de símbolo dos alunos. Os resultados indicam uma heterogeneidade quanto ao sentido de símbolo dos diversos alunos, mas apontam para uma maior facilidade no trabalho com as expressões algébricas e maior dificuldade no trabalho com problemas. A manipulação simbólica surge como o aspeto do sentido de símbolo mais desenvolvido e a utilização dos símbolos para conjeturar e generalizar, o aspecto menos desenvolvido. O sentido de símbolo de cada aluno é visível e influencia o seu trabalho com conteúdos específicos de Matemática A: num deles a abordagem tem caráter dominantemente instrumental e no outro a abordagem revela compreensão na forma como desenvolve processos de matematização e abstração.This research aims to characterize the symbol sense of students in their final stage of secondary education and its relation with the learning of algebra. Two questions conduct this research: (i) What symbol sense do those students show, when working in questions involving algebraic expressions, equations, problems and functions? (ii) What is the relation between the students' symbol sense and their ability to solve questions regarding specific topics of the year 13 syllabus?
The research focuses on a group of twenty one students and involves a deeper study of two students of year 13. Data collection is carried out through a diagnostic test, two interviews and documents written by the students. A reference framework is used to characterize the students' symbol sense, based on the way they solve questions involving algebraic expressions, equations, problems and functions. The methodology is essentially qualitative (case study), complemented by a small quantitative element, and is based on the interpretative paradigm.
The reference framework proved to be a useful tool, adequate to characterize the students' symbol sense. The results indicate heterogeneity regarding the students' symbol sense, but also show that the work with algebraic expressions is easier for the students while they show difficulty when working with problems. Symbolic manipulation appears to be the more developed aspect of symbol sense, while the use of symbols to conjecture and generalize is the less developed one. Regarding the specific topics of Mathematics A, each student symbol sense is visible and influences their work: the approach of one of the students is essentially instrumental, while the other reveals understanding as he develops mathematization and abstraction processes | 677.169 | 1 |
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Various levels of calculus/diff-eq are probably the only 'essential' uni-level maths (the quotes meaning that you can get by without, but you're going to limit yourself significantly or otherwise have a hard time of it). [!IMO!]
Depending on your interests, statistics or a math-heavy economics course might be of great use. (but then again certain psychology/sociology courses might be of better benefit)
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I would say you need to have strong math skills in problem solving, algebra, geometry. However some people take this math thing a bit overboard. Not to brag but I had a 96% average in my Math Pure courses however when I program I only use algebra and problem solving nothing too heavy. I think you need good math skills when it comes to physics in games, 3D games.
Don't let your lack of math hold you down from being a programmer you can always learn what you had trouble with or missed online. ^_^0Well I don't really have any bounds in my head right now in terms of what kinds of games i'd like to make, but as for my intended area of expertise, I'd like to be skilled in Gameplay/User Interface programming.
BTW, its nice that you guys mentioned Stats as thats in my planned curriculum.
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Original post by K_I_L_E_R The skills I found necessary to develop games and programming in general:
Linear algebra is absolutely essential as it is also used in statistics and calculus.
Single variable and multivariate calculus.
Elementary level statistics (1 year of statistics at my uni).
Statistics will be the most helpful. A good example is planning a table of data and trying to fit that data to some model.
Yup I pretty much 2nd everything in your list especially if you are going to work with D3D or Opengl anything 3D really. Linear algebra will probably the most helpful since right away you will have to work with matrices as soon as you do anything 3D. If you are going to use any sort of realistic physics in your game calculus will be necessary to understand the equations you work with. Numerical Analysis would be helpful so you can distinguish between garbage in vs garbage out.
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Statistics comes with probability calculations which is very useful for AI among other things. Without beeing a math guru I can easily see how Im gonna use it to make an "intelligent", self learning AI =). Another useful tool when writing AI is fuzzy logic. Something your school probably wont teach you unless it is geared towards game programming or something similar.
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I agree with the last poster. _Basic_ math knowledge about the topic mentioned by last poster is helpful, _although_ not necessary in my opinion. After all, graphics programming is very visual and if i don't get something, i see it on screen and then know what to fix or alter. I don't need to calculate or do any advanced trigonometry. Ofcourse, once you are are an advanced programmer and you want to pull of cool visual effects, then everything you have learned in physics or math will ofcourse give you a very real edge over those who have not. It could also give you some advantage (trig) to make some faster algorithms for example.
Don't forget, most things new programmers learn are tricks from others anyway. Pretty much all things like BSP trees, or graphic tricks they learn from reading articles or other programmers showing them or copying code. I actually think that if you want to show off nice graphic tricks you first implement them (sometimes by accident) and then, once you understand what you have done you can shortcut it into an algorithm by using your math knowledge.
But seriously, it also all depends on what you want to do. If you do not want to create a visual spectacle, but just use some basic 3d objects and space, you don't need any math. That goes for any basic game, but also a basic space sim.
But that's just my opinion and i'm _not_ an experienced opengl programmer. Although i can say that i'm not an expert in advanced math, i can say that i'm pretty far in creating opengl software that i visualized from the beginning. And this is after 2-3 months or something.
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Original post by serratemplar How about Differential Equations? What is the value of that in comp sci and should aspiring graphics programmers take that? =)
They have their uses in game development. Basic differential equation knowledge can be useful when understanding integrators as part of a physics engine, for example. In general computer science, they also have their uses in computer graphics and geometry (level set representations) and systems modelling (hybrid temporal automata), and some others I might be forgetting.
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Original post by ToohrVyk Disc levelI have taken some basic Discrete Math and I can definitely see myself applying that in Game Programming, especially when it comes to A.I.
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Original post by Ezbez Seeingwell said, im in algebra 2 and the total for these two are so far: matrices: 5 class periods (45 min ea with a horrible teacher) vectors: 0
then again we live in a tiny Ohio town where the average IQ is about five and a half... lol one day i was programming in C++ on the school comp and the teacher thought i was hacking the school's network :D
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Here's the most "classical" example of floating point math breaking the rules.
10^60 - 10^60 + 1 = 1
You agree - I hope =) - and most compilers will agree too, because they'll group the first two together as identical symbols, zero them about due to the subtraction, and give us one. However...
10^60 + 1 - 10^60 = 0 according to machine.
Because 10^60 + 1 = 10^60 (1 just gets discarded as it's beyond the scope of the float's precision) and 10^60 - 10^60 = 0. To some degree, addition is not associative in floating point math. This isn't consistent though, because if you substitute a number "big enough" for 1, the math will start working as you expect it to.
I didn't learn this all in a math course though; I learned it in architecture and assembly class =) | 677.169 | 1 |
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A streamlined course designed to prepare students for the calculus sequence (MATH 047–048). Properties and graphs of elementary functions. Emphasis on developing conceptual understanding and problem-solving skills.
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Concurrently, students must enroll in a pre-calculus lab, MATH 003L, designed to strengthen their algebraic skills. | 677.169 | 1 |
Introducing our revised OCR Level 3 FSMQ: Additional Maths (6993)
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18/01/2018
Following the recent reforms in both GCSE and A level maths, we are pleased to now bring you our revised Level 3 FSMQ: Additional Maths qualification. Developed in conversation with teachers and education stakeholders, you'll discover the revised qualification continues to enrich students' learning and understanding of GCSE (9-1) Mathematics content.
The revised Level 3 FSMQ: Additional Maths will have its first assessment in summer 2019.
Rationale for the re-launch
The legacy Additional Maths was launched back in 2002. Since then it has grown to be an incredibly valued qualification, but recent educational reforms have brought in a need for revisions.
Most students now follow a parallel programme, rather than studying Additional Maths after sitting their GCSE in year 10.
Reforms have increased the level of demand of GCSE Mathematics and expanded its content to now include content previously introduced to students through Additional Maths.
Scientific calculators have greater functionality, with numerical calculus, statistical distributions, and equation solving functions now common. More candidates are also using graphical calculators in examinations.
Care has been taken to ensure that Additional Maths is not an acceleration course covering the first part of the AS Maths course. Instead Additional Maths provides a Level 3 enrichment of GCSE (9–1) Mathematics, whilst at the same time providing a taster of A level Maths and Further Maths.
Summary of the key content changes
'Enumeration' content, which expands the binomial distribution with permutations and combinations. The systematic listing content of the GCSE is built upon.
'Numerical methods' content, which expands upon the informal graphical approximations in GCSE and encourages more formal links graphs (with the algebra and calculus of Additional Maths).
'Exponentials and Logarithms' content, which develops the growth and decay content and the graphs section of GCSE (this content was identified as one of the main gaps in knowledge that students not doing A level Maths have when following other post-16 courses with a numerical aspect).
Sequences using subscript notation have been included to support the iterative work on numerical methods and building upon new content in the reformed GCSE.
Summary of other key qualification changes
An integrated question and answer paper, to reflect the majority of GCSE examinations.
Removal of the separate sections A and B, to a more traditional building of demand through the paper (as at GCSE).
The Assessment Objectives have been changed to meet the reformed GCSE and A level Maths qualifications.
A list of Assessment command words has been included in the Specification. This was a popular feature of OCR's reformed A levels and makes clear the level of written response needed.
About the author
Steven Walker - Subject Advisor - Mathematics
Steven joined OCR in July 2014 as a Subject Advisor, working on the GCSE and GCE reform and leading the redevelopment of R449 Entry Level Certificate Mathematics and 6993 Level 3 FSMQ: Additional Mathematics.
Steven graduated in Material Engineering, and after an extensive period working and travelling around the world, including four years as a volunteer maths and science teacher with VSO, he gained a PGCE in secondary mathematics from Homerton College, Cambridge and has taught maths in a number of schools in UK and overseas. | 677.169 | 1 |
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Your Guide to a Better Grade in Math.
To be competitive for college admissions, you need to maintain an excellent G.P.A. in high school, especially in core subjects such as math. "Essential Review: High School Mathematics I" uses targeted math review, in-depth practice drills, and Kaplan's proven strategies for performing well on tests, to help you develop the skills you need to get the math grade you want. | 677.169 | 1 |
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Practical Algebra If you studied algebra years ago and now need a refresher course in order to use algebraic principles on the job, or if you're a student who needs an introduction to the subject, here's the perfect book for you. Practical Algebra is an easy and fun-to-use workout program that quickly puts you in command of all the basic concepts and tools of algebra. With the aid of practical, real-life examples and applications, you'll learn: * The basic approach and application of algebra to problem solving* The number system (in a much broader way than you have known it from arithmetic)* Monomials and polynomials; factoring algebraic expressions; how to handle algebraic fractions; exponents, roots, and radicals; linear and fractional equations* Functions and graphs; quadratic equations; inequalities; ratio, proportion, and variation; how to solve word problems, and more Authors Peter Selby and Steve Slavin emphasize practical algebra throughout by providing you with techniques for solving problems in a wide range of disciplines - from engineering, biology, chemistry, and the physical sciences, to psychology and even sociology and business administration. Step by step, Practical Algebra shows you how to solve algebraic problems in each of these areas, then allows you to tackle similar problems on your own, at your own pace. Self-tests are provided at the end of each chapter so you can measure your mastery.
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Written by an experienced maths teacher, this book aims to help you master number and data handling. Whether you are just starting at college or university, or a mature student, it explains all you need to know.Read more... | 677.169 | 1 |
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IX. Units
The TI-89 will convert from one unit to another.
This is a very useful menu. There are also many built-in constants available
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Saturday, September 7, 2013
Linear programming
A pictorial representation of a simple linear program with 2 variables & six inequalities. The set of feasible solutions is depicted in light red & forms a 2-dimensional polytope. The linear cost function is represented by the red line & the arrow: The red line is a level set of the cost function, & the arrow indicates the direction in which we are optimizing.
Linear programming is a mathematical method for determining a way to achieve the best outcome (such as maximum profit /or lowest cost) in a given mathematical model for some list of requirements represented as linear relationships. Linear programming(LP) is a specific case of mathematical programming (mathematical optimization).
More formally, linear programming(LP) is a technique for the optimization of a linear objective function, subject to linear equality & linear inequality constraints. Its feasible region is a convex polyhedron, which is a set defined as the intersection of finitely many half spaces, each of which is defined by a linear inequality. Its objectives function is a real-valued affine function defined on this polyhedron. A linear programming algorithm finds a point in the polyhedron where this function has the smallest (/or/ largest) value if such point exists.
Linear programs are problems; that can be expressed in canonical form:
T
Maximize c x
Subject to Ax < b
--
and x > 0
--
where x represents the vector of variables (to be determined), c & b are vectors of (known) coefficients, A is a (known) matrix of coefficients, & (.)T is the matrix transpose. The expression to be maximized /or/ minimized is called the objective function (cTx in this case). The inequalities Ax ≤ b are the constraints, which specify a convex polytope over which the objective function is to be optimized. In this context, 2 vectors are comparable when they have the same dimensions. If every entry in the first is < or equal-to the corresponding entry in the second then we can say the first vector is less-than(<) or equal-to(=) the second vector.
Linear programming(LP) can be applied to various fields of study. It is used in business & economics, but can also be utilized for some engineering problems. Industries that use linear programming models include telecommunications, energy, transportation, & manufacturing. It has proved useful in modeling diverse types of problems in planning, routing, scheduling, assignment, & design. | 677.169 | 1 |
In a few courses, all it requires to move an examination is observe taking, memorization, and recall. Nonetheless, exceeding inside a math course takes a unique sort of exertion. You can't just demonstrate up to get a lecture and watch your instructor "talk" about geometry and . You discover it by undertaking: being attentive in school, actively learning, and resolving math difficulties – even when your instructor hasn't assigned you any. When you end up struggling to perform very well within your math class, then go to finest internet site for fixing math complications to find out how you may become a better math pupil.
Affordable math professionals on the web
Math classes adhere to a normal progression – each builds on the awareness you have obtained and mastered with the prior program. In case you are locating it challenging to stick to new principles in class, pull out your outdated math notes and critique prior content to refresh oneself. Be sure that you satisfy the stipulations prior to signing up for your class.
Evaluate Notes The Night time In advance of Course
Hate any time a trainer calls on you and you have forgotten how you can address a certain trouble? Avoid this instant by examining your math notes. This may allow you to establish which principles or concerns you'd choose to go more than in school the next working day.
The thought of undertaking research every single evening may seem bothersome, however, if you'd like to succeed in , it is actually important that you consistently apply and master the problem-solving strategies. Make use of your textbook or on line guides to operate through top math difficulties on a weekly foundation – even when you've no research assigned.
Utilize the Health supplements That come with Your Textbook
Textbook publishers have enriched modern-day publications with additional material (which include CD-ROMs or on the net modules) that could be used to enable students attain additional apply in . Some of these resources may contain an answer or clarification guidebook, which may assist you with doing the job by math complications by yourself.
Browse Ahead To remain In advance
If you prefer to minimize your in-class workload or even the time you shell out on research, use your spare time after college or on the weekends to read through forward to the chapters and ideas that could be included the subsequent time you happen to be in class.
Overview Outdated Exams and Classroom Examples
The work you do in class, for homework, and on quizzes can offer you clues to what your midterm or remaining test will glance like. Use your aged exams and classwork to make a private review guideline for the forthcoming test. Glance within the way your trainer frames thoughts – this is probably how they may look on your examination.
Learn to Operate With the Clock
This is the well-known examine tip for men and women taking timed examinations; primarily standardized tests. Should you have only forty minutes for a 100-point test, then you can optimally expend four minutes on each and every 10-point issue. Get data about how prolonged the take a look at is going to be and which sorts of inquiries might be on it. Then program to attack the better concerns to start with, leaving oneself plenty of time and energy to invest on the a lot more challenging ones.
Maximize your Sources to obtain math homework enable
If you are obtaining a tough time understanding principles in school, then make sure to get aid outside of class. Ask your pals to produce a analyze group and pay a visit to your instructor's office environment hrs to go about difficult troubles one-on-one. Attend analyze and evaluation periods when your teacher announces them, or employ a non-public tutor if you want 1.
Converse To Yourself
Any time you are reviewing problems for an exam, try to elucidate out loud what strategy and strategies you accustomed to obtain your remedies. These verbal declarations will arrive in helpful for the duration of a exam any time you should remember the ways you ought to get to locate a solution. Get more apply by striving this tactic having a close friend.
Use Review Guides For More Apply
Are your textbook or course notes not supporting you comprehend whatever you ought to be learning in class? Use review guides for standardized tests, such as the ACT, SAT, or DSST, to brush up on old content, or . Study guides normally occur equipped with extensive explanations of how to clear up a sample difficulty, , therefore you can generally find where by will be the superior purchase mathtroubles.
December 25 | 677.169 | 1 |
Algebra I
Algebra I
Algebra I is a full year, high school credit course that is intended for the student who has successfully mastered the core algebraic concepts covered in the prerequisite course, Pre-Algebra. Within the Algebra I course, the student will explore basic algebraic fundamentals such as evaluating, creating, solving and graphing linear, quadratic, and polynomial functions.
Upon successfully completing the course, the student should have mastered the following concepts:
Solve and graph single variable, absolute value, and linear equations and inequalities.
Solve linear, quadratic and exponential systems of equations using graphing, substitution or elimination.
Evaluate and solve quadratic equations and inequalities using graphing, factoring, quadratic formula, and completing the square.
Interpret and apply the relationship between the independent and dependent variable in a linear, exponential, and quadratic function through algebraic modeling and applications.
Understand and know how to apply the distance, midpoint, and slope formulas as well as the Pythagorean Theorem.
Form an equation of a line using the slope-intercept, point-slope and standard forms of a line. Apply basic fundamental rules of exponents.
Be able to construct a formula or equation necessary to solve algebraic word problems involving area, perimeter, and linear systems of equations, basic probability and statistical reasoning, distance, and compounding interest. | 677.169 | 1 |
Integrals
This Integrals worksheet also includes:
In this integrals learning exercise, students solve and complete 22 various types of problems. First, they evaluate each of the listed integrals on the sheet. Then, students solve each of the given differential equations by substituting the given values for each variable. | 677.169 | 1 |
This resource enables high school students to explore mathematical principles through independent, hands-on investigations, and also allows teachers to enhance their customary classroom approach. The 54 projects cover a range of levels and expertise in number theory, algebra, geometry, trigonometry | 677.169 | 1 |
Calculus is the greatest mathematical breakthrough since the pioneering discoveries of the ancient Greeks. Without it, we wouldn't have spaceflight, skyscrapers, jet planes, economic modeling, accurate weather forecasting, modern medical technologies, or any of the countless other achievements we take for granted in today's world. Accomplish Mathematical Wonders
Ideally, every photo we take would be perfect: perfect exposure, perfect white balance, no backlighting, no harsh shadows. Of course the reality is that some images need to be fixed, and in this course we will look at ways to deal with common problems. In each lesson Dave will fix a problem image, real-time,... | 677.169 | 1 |
Category: "mathematics" requires | 677.169 | 1 |
Review of the Art of Problem Solving
I was recently introduced to The Art of Problem Solving, which is a series of math textbooks aimed at gifted middle and high school students, especially those who are interested in math competitions. I only looked through one of the books, Introduction to Algebra, by Richard Rusczyk , and this review will be about that text only. Other books in the series delve into topics such as geometry and probability.
The Art of Problem Solving bills itself as a book for 6th to 10th grades. This evaluation is quite ambitious! It is, however, in character with the rest of the book, as ambitious is the best word I can come up with to describe the overall tone. (According to the author's biography, he was a high level math competition champion as a child, and I think it would be fair to suggest that he wrote this book with his younger self in mind.) I would warn parents and teachers to take the pre-test provided on the website very seriously. If your student(s) can not get a perfect score on the pre-test without your help, they are not ready for this book, regardless of their age.
As an adult who is comfortable with math, I loved this book. Both the text and the problems are thoughtfully written and very interesting. The explanations provided are lucid. If time was not a constraint, I would joyfully devote an hour or two a day to methodically working through this book- it would probably take me a year or so to finish, and I have no doubt that I would learn a great deal. However, while my endorsement of this book is strong, it is also very limited and specific. So that you can understand, let me tell you a little bit about myself.
As a child, I was an insanely conscientious student. Not surprisingly, I did well in school and was placed in an accelerated math program in middle school. Nevertheless, I found no joy in math (and always had the nagging feeling that my success on tests and report cards was due to some sort of cosmic mistake rather than real achievement on my part.) In high school, I stopped pursuing math as soon as I decently was able. I never took pre-calculus, never mind calculus. I chose my college partly based on where I would be able to major in biology without taking higher math classes. Fortunately, I experienced an epiphany at the age of 21.
My epiphany was the result of a research project that I was perusing- I was researching certain aspects of ancient salt marshes, and my advisor told me that if I could successfully do a statistical analysis of my data, it could most likely be published. With that enormous inducement, I began studying elementary statistics, and with almost no instruction except from a textbook, I soon understood statistics well enough to analyze my data. My paper was published and, much more importantly, my fear of math was conquered.
Years past, and I became a tutor. I teach test preparation and science as well as math, but I spend the largest portion of my time teaching math to 8-14 year olds. (I've hired other tutors to teach more advanced math.) I'm very good at what I do, and I think it is in large part because I have a very thorough understanding of math through high school algebra, a genuine affection for the subject and, simultaneously, a clear memory of a time when math was not my friend.
All of this history is a roundabout way of explaining why I feel like I have a lot to learn from this book- although it starts out with basic algebra, it ends up covering topics normally reserved for pre-calculus. Furthermore, when I look at The Art of Problem Solving, Introduction to Algebra through the lens of my remembered childhood feelings about math, I see a terrifying tome. It does not gently lead the student forward, first with easy problems and then with gradually more challenging ones. Instead, it dashes ahead and dives straight into hard problems. This approach is great for a motivated, interested person with a solid background in the pre-requisites, but it could easily prove miserable, frustrating, and ultimately counter productive for students who do not meet that description.
I intend to begin using The Art of Problem Solving, Introduction to Algebra, but only with a select group of students who are already robustly successful in math and who are coming to me for enrichment. For example, I will incorporate Art of Problem Solving problems into my work preparing students for the Hunter College High School and Anderson School entrance exams.
I wish to offer one further warning about The Art of Problem Solving, Introduction to Algebra, specifically to homeschool families. If your child is ready for this textbook and eager for the challenge it presents, then that is a wonderful thing. However, if you plan on integrating your child into a school environment, you should be aware that the book does not touch on topics that are important in both middle and high school curriculums (primarily geometry and probability) and you might therefore want to provide supplementation in these topics. | 677.169 | 1 |
Mathematics is the language of science. It is the art of problem-solving. And if we care to look closely enough at the world around us, we will be amazed at the mathematics at play. It is my hope as your professor that you will take from this class a newfound appreciation of how mathematics fits into your world. Whether you are interested in the microscopic cells and chemical reactions occuring in your body, the local geography of the beautiful Upper Peninsula of Michigan, the global changes in our climate, the large-scale behavior of our universe, or maybe you are just interested in playing video games and listening to music - all of these things can be better understood, better explained, better appreciated with mathematics.
The catch is, however, that it takes a lot of work to understand the math behind the world. But the rewards are exponentially greater than the pain required to learn something new. This semester you will learn a lot of new mathematics and you will perhaps be surprised to see how far it will take you toward understanding your world a bit better.
Learning Objectives
In Calculus, the concepts are as important as are the computations. Perhaps more importantly, you will gain an appreciation of the power and ability of mathematics in modeling the real world.
Resources
Math is not a spectator sport. You will have to get your hands dirty and it will not always be easy. There are many resources available to help you via the links on right.
At the links to the right you will find updated information, such as the syllabus for the course which contains information on prerequisites, grading policy, homework, study resources and a tentative course calender. See the box in the upper right for more links and information for the course. The required textbook is pictured above, and can be found in the library online. | 677.169 | 1 |
Build a solid understanding of logarithmic functions and equations. Five lessons in the module begin by developing the concept of a logarithm. The next lessons address graphing logarithmic functions, logarithmic properties, and solving logarithmic equations. | 677.169 | 1 |
Math Physics Reference Advice
I'm a physics graduate student (going for my masters) currently taking math physics looking for useful books on math physics. I really want to find books/websites that shows the step by step process of how to do for example PDE's, ODE's, cauchy-riemann equations, tips, tricks, involving math in physics. I also would like the books/websites to explain the purpose of the mathematics. I went into graduate school for physics with knowledge of Calc 1 through 3, discrete math, and linear algebra. Sadly I wish I had time for other math classes but now I need to gain this knowledge on my own. So any advice will be helpful. I'm not afraid of hard work to understand the concepts and methods. I'm working very hard to understand everything I can so I can make my future classes more enjoyable.
Key notes: I need books/websites with great in depth examples. Books/websites without heavy math terminology like a book for dummies. The books/websites should also explain the purpose of the math used. <-- I know this is like looking for the fountain of youth but maybe there are great guides out there.
Any advice on what worked for you when you didn't have knowledge on a specific math class will be helpful too. | 677.169 | 1 |
Linear Systems
Abstract
The solution of systems of linear equations arises in various parts of mathematics and is of central importance in numerical analysis. To illustrate the significance of linear systems, we will start this chapter by providing some examples of their occurrence as part of the numerical solution of differential and integral equations. After seeing the examples, we will proceed with the solution of systems of linear equations. In principle, we have to distinguish between two groups of methods for the solution of linear systems:
1
In the so-calleddirect methodsorelimination methodsthe exact solution, in principle, is determined through a finite number of arithmetic operations (in real arithmetic leaving aside the influence of roundoff errors).
2
In contrast to thisiterative methodsgenerate a sequence of approximations to the solution by repeating the application of the same computational procedure at each step of the iteration. Usually, they are applied for large systems with special structures that ensure convergence of the successive approximations. | 677.169 | 1 |
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