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Systems of equations project
The National Center and State Collaborative (NCSC) is a project led by five centers and 24 states to build an alternate assessment based on alternate achievement. Disabilities, Opportunities, Internetworking, and Technology. Compendium of all course descriptions for courses available at Reynolds Community College Bored with Algebra? Confused by Algebra? Hate Algebra? We can fix that. Coolmath Algebra has hundreds of really easy to follow lessons and examples. Algebra 1.
Designed to "illuminate" the new NCTM Principles and Standards for School Mathematics. [inside math] inspiration. A professional resource for educators passionate about improving students' mathematics learning and performance [ watch our trailer . Progressions Documents for the Common Core Math Standards Funded by the Brookhill Foundation Progressions. Draft Front Matter; Draft K–6 Progression on Geometry. - This video provides an overview of how to perform Critical Path Method (CPM) to find the Critical Path and Float using. A non-profit research and education organization dedicated to the advancement of science and math education, specifically through the use of modeling and simulation. System dynamics (SD) is an approach to understanding the nonlinear behaviour of complex systems over time using stocks, flows, internal feedback loops, and time. Compendium of all course descriptions for courses available at Reynolds Community College.
Systems of equations project
Global Positioning System; Country of origin: United States: Operator(s) AFSPC: Type: Military, civilian: Status: Operational: Coverage: Global: Precision: 5 meters. Electrical Engineering and Technology Welcome to this open and free electrical engineering study site. A strong team of well experienced electrical engineers in. Bored with Algebra? Confused by Algebra? Hate Algebra? We can fix that. Coolmath Algebra has hundreds of really easy to follow lessons and examples. Algebra 1. Nucor Building Systems is a leading manufacturer of custom pre-engineered metal building systems with over 1,000 Authorized Builders across North America. Magnetic flux (Φ B) is the number of magnetic field lines (also called "magnetic flux density") passing through a surface (such as a loop of wire). PlanetMath is a virtual community which aims to help make mathematical knowledge more accessible. PlanetMath's content is created collaboratively: the main feature is.
Engage students with immersive content, tools, and experiences. Part of the world's leading collection of online homework, tutorial, and assessment products, Pearson. Welcome! InterAct Math is designed to help you succeed in your math course! The tutorial exercises accompany the end-of-section exercises in your Pearson textbooks. Systems (ISSN 2079-8954) is an international open access journal on systems engineering and systems management, published quarterly online by MDPI. Our homepage is configured to use MathJax's CommonHTML mode with web fonts to display the equations, which produces uniform layout and typesetting across. 19 Online homework and grading tools for instructors and students that reinforce student learning through practice and instant feedback.
Global Positioning System; Country of origin: United States: Operator(s) AFSPC: Type: Military, civilian: Status: Operational: Coverage: Global: Precision: 5 meters. Solve systems of equations with linear algebra operations on vectors and matrices. System dynamics (SD) is an approach to understanding the nonlinear behaviour of complex systems over time using stocks, flows, internal feedback loops, and time. Disabilities, Opportunities, Internetworking, and Technology. Maxima, a Computer Algebra System. Maxima is a system for the manipulation of symbolic and numerical expressions, including differentiation, integration, Taylor. Math Worksheets Dynamically Created Math Worksheets. Math-Aids.Com provides free math worksheets for teachers, parents, students, and home schoolers. The Bachelor of Science in Information Systems at CityU prepares students for technology jobs working with operating systems, networks, databases and more.
Welcome! InterAct Math is designed to help you succeed in your math course! The tutorial exercises accompany the end-of-section exercises in your Pearson textbooks.
The National Center and State Collaborative (NCSC) is a project led by five centers and 24 states to build an alternate assessment based on alternate achievement.
Online homework and grading tools for instructors and students that reinforce student learning through practice and instant feedback.
- This video provides an overview of how to perform Critical Path Method (CPM) to find the Critical Path and Float using.
19 Systems (ISSN 2079-8954) is an international open access journal on systems engineering and systems management, published quarterly online by MDPI. Magnetic flux (Φ B) is the number of magnetic field lines (also called "magnetic flux density") passing through a surface (such as a loop of wire). Electrical Engineering and Technology Welcome to this open and free electrical engineering study site. A strong team of well experienced electrical engineers in. Engage students with immersive content, tools, and experiences. Part of the world's leading collection of online homework, tutorial, and assessment products, Pearson. | 677.169 | 1 |
Introduction to Linear & Nonlinear Programming
内容简介
· · · · · ·... of this type is the connection between the purely analytical character of an optimization problem, expressed perhaps by properties of the necessary conditions, and the behavior of algorithms used to solve a problem. This was a major theme of the first edition of this book and the second edition expands and further illustrates this relationship.</P>
"Linear and Nonlinear Programming" covers the central concepts of practical optimization techniques. It is designed for either self-study by professionals or classroom work at the undergraduate or graduate level for technical students. Like the field of optimization itself, which involves many classical disciplines, the book should be useful to system analysts, operations researchers, numerical analysts, management scientists, and other specialists from the host of disciplines from which practical optimization applications are drawn. </P> | 677.169 | 1 |
No matter what grade you teach, you can be sure your students will enjoy this project, which is one of my favorite abstract art lessons. Much early work in algebra, as the Arabic origin of its name suggests, was done in the Middle East, by Persian [4] [5] mathematicians such as al-Khwārizmī (780–850) [6] and Omar Khayyam (1048–1131). [7] Elementary algebra differs from arithmetic in the use of abstractions, such as using letters to stand for numbers that are either unknown or allowed to take on many values. [8] For example, in The word algebra comes from the Arabic الجبر (al-jabr "restoration") from the title of the book Ilm al-jabr wa'l-muḳābala by al-Khwarizmi.
Standard grade revision maths questions, card games for teaching exponents, multiply by one or adding zero. Claims of anything new and profound are general pompous bullstuff. This is a call to all catalysts of change to Join the Education in the Philippines Forum and make a statement AND an action. Over the years, it has further broadened and diversified its scope to include a.o. numerical analysis, integral transforms, signal processing, group representation theory, Lie (super)algebras, ….
Description of use: The algebraic expression, "4x = 8," can be represented with four plates ("4x"). Schemes for data mining and combination in some higher abstract structure level. We're sorry, but there's no news about "Abstract algebra" right now. Solving systems of equation worksheet, practice for the 7th grade NY state exam, free math algebra worsheet software. The text is intended as a first introduction to the ideas of proof and abstraction in mathematics, as well as to the concepts of abstract algebra (groups and rings).
To start viewing messages, select the forum that you want to visit from the selection below. Normally, we'd track this in a spreadsheet. Base Blocks – Illustrate addition and subtraction in a variety of bases. Detect the relationships between Euclidean, principal ideal, unique factorization, and integral domains. Algebraic structures, with their associated homomorphisms, form mathematical categories. Each chapter rests upon a central theme, usually a specific application or use.
This conference is a continuation of the earlier conferences and workshops on Operator theory and Operator algebras held in Indian Statistical Institute, Bangalore. You've already violated my trust once, so you're going to have to prove to me that you know the word by telling me the word. We welcome your feedback, comments and questions about this site or page. Printing tips - For best printed results. It is a large text with enough material for a senior level sequence in mathematical statistics, or a more advanced graduate sequence in mathematical statistics.
It was not realized until the second half of the 19th century that several different areas of study all fit under the same umbrella of abstract group theory. Digplanet also receives support from Searchlight Group. I believe this book makes a strong case for open-source textbooks, and is in the vanguard of a revolution which will completely change the way future textbooks are developed, adapted, and utilized.
Class lectures may cover the material from different points of views of the textbook. At a more complex level, suppose you are doing one of those problems where it takes one guy 3 days to do a job by himself, another guy 2 days, and a third guy a day and a half, and they want to know how long it will take them all working together to do the job. Report any bugs in the programming to me please. (12/1/97) Uploaded a_qbn1.pdf, Algeboard, the Algebra Quiz Board Game. Economies of scale are more easily realized the bigger the redesign project is.
I think part of my confusion stemmed from the examples. Proof The integers are cyclic, hence any PowerShow.com is a leading presentation/slideshow sharing website. Ks3 trigonometry questions, pre-algebra with pizzazz, Book AA, TI-83 programs for help with ACT, Graphing Pictures worksheet.. I think that it takes quite a bit of mathematical maturity to get through D&F. In America, one generally then looks left before stepping off the curb, and right once one gets to the middle of the street.
A lot of math concepts can be applied to other fields. Answer Dynamic references are constructed on the fly as a result of calling various methods, such as System. He represented mathematical symbols using characters from the Arabic alphabet. 1535: Nicolo Fontana Tartaglia and others mathematicians in Italy independently solved the general cubic equation. 1545: Girolamo Cardano publishes Ars magna -The great art which gives Fontana's solution to the general quartic equation. 1572: Rafael Bombelli recognizes the complex roots of the cubic and improves current notation. 1591: Francois Viete develops improved symbolic notation for various powers of an unknown and uses vowels for unknowns and consonants for constants in In artem analyticam isagoge. 1682: Gottfried Wilhelm Leibniz develops his notion of symbolic manipulation with formal rules which he calls characteristica generalis. 1680s: Japanese mathematician Kowa Seki, in his Method of solving the dissimulated problems, discovers the determinant, and Bernoulli numbers. 1750: Gabriel Cramer, in his treatise Introduction to the analysis of algebraic curves, states Cramer's rule and studies algebraic curves, matrices and determinants. 1824: Niels Henrik Abel proved that the general quintic equation is insoluble by radicals. 1832: Galois theory is developed by Évariste Galois in his work on abstract algebra.
Here are some notes on the commutator subgroup of a group. The total cost savings was $51,418 or 19%. This is a significant cost savings at a small college like Cleveland State with a math department of eight faculty members and one staff member. Algebra helps us to answer questions where there is an unknown value (not necessarily the bit after '=') or for when there is a range of answers which will be correct for …a mathematic equation. For example, if you had an equation: 2A + 16 = 18, it is easy to see that A=1 in this case. You can also use algebra in cases where two equations use the same letter within, corresponding to the same value in each, and this can be used to solve when there are 2 unknown values. | 677.169 | 1 |
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A Practical Approach to Dynamical Systems for Engineers takes the abstract mathematical concepts behind dynamical systems and applies them to real-world systems, such as a car traveling down the road, the ripples caused by throwing a pebble into a pond, and a clock pendulum swinging back and forth. Many relevant topics are covered, including modeling systems using differential equations, transfer functions, state-space representation, Hamiltonian systems, stability and equilibrium, and nonlinear system characteristics with examples including chaos, bifurcation, and limit cycles. In addition, MATLAB is used extensively to show how the analysis methods are applied to the examples. It is assumed readers will have an understanding of calculus, differential equations, linear algebra, and an interest in mechanical and electrical dynamical systems.Presents applications in engineering to show the adoption of dynamical system analytical methodsProvides examples on the dynamics of automobiles, aircraft, and human balance, among others, with an emphasis on physical engineering systemsMATLAB and Simulink are used throughout to apply the analysis methods and illustrate the ideasOffers in-depth discussions of every abstract concept, described in an intuitive manner, and illustrated using practical examples, bridging the gap between theory and practiceIdeal resource for practicing engineers who need to understand background theory and how to apply it | 677.169 | 1 |
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Calculus is a collection of tools, such as differentiation and integration, for solving problems in mathematics which involve "rates of change" and "areas". In the second of two articles aimed specially at students meeting calculus for the first time, Chris Sangwin tells us how to move on from first principles to differentiation as we know and love it!
Calculus is a collection of tools, such as differentiation and integration, for solving problems in mathematics which involve "rates of change" and "areas". In the first of two articles aimed specially at students meeting calculus for the first time, Chris Sangwin tells us about these tools - without doubt, the some of the most important in all | 677.169 | 1 |
Synopsis
The subject matter loosely called "Riemann surface theory" has been the starting point for the development of topology, functional analysis, modern algebra, and any one of a dozen recent branches of mathematics; it is one of the most valuable bodies of knowledge within mathematics for a student to learn. Professor Cohn's lucid and insightful book presents an ideal coverage of the subject in five pans. Part I is a review of complex analysis analytic behavior, the Riemann sphere, geometric constructions, and presents (as a review) a microcosm of the course. The Riemann manifold is introduced in Part II and is examined in terms of intuitive physical and topological technique in Part III. In Part IV the author shows how to define real functions on manifolds analogously with the algebraic and analytic points of view outlined here. The exposition returns in Part V to the use of a single complex variable z. As the text is richly endowed with problem material — 344 exercises — the book is perfect for self-study as well as classroom use. Harvey Cohn is well-known in the mathematics profession for his pedagogically superior texts, and the present book will be of great interest not only to pure and applied mathematicians, but also engineers and physicists. Dr. Cohn is currently Distinguished Professor of Mathematics at the City University of New York Graduate Center.
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calculusCliffsQuickReview course guides cover the essentials of your toughest classes. You're sure to get a firm grip on core concepts and key material and be ready for the test with this guide at your side.
Whether you're new to functions, analytic geometry, and matrices or just brushing up on those topics, "CliffsQuickReview Precalculus" can help. This guide introduces each topic, defines key terms, and walks you through each sample problem step-by-step. In no time, you'll be ready to tackle other concepts in this book such as Arithmetic and algebraic skills Functions and their graphs Polynomials, including binomial expansion Right and oblique angle trigonometry Equations and graphs of conic sections Matrices and their application to systems of equations
"CliffsQuickReview Precalculus" acts as a supplement to your textbook and to classroom lectures. Use this reference in any way that fits your personal style for study and review -- you decide what works best with your needs. You can either read the book from cover to cover or just look for the information you want and put it back on the shelf for later. What's more, you can Use the free Pocket Guide full of essential information Get a glimpse of what you'll gain from a chapter by reading through the Chapter Check-In at the beginning of each chapter Use the Chapter Checkout at the end of each chapter to gauge your grasp of the important information you need to know Test your knowledge more completely in the CQR Review and look for additional sources of information in the CQR Resource Center Use the glossary to find key terms fast.
With titles available for all the most popular high school and college courses, CliffsQuickReview guides are a comprehensive resource that can help you get the best possible grades | 677.169 | 1 |
Description
Course content of this class is the third and the last of the series started from class sixth. It is a continuation of the processes initiated to help the students in abstraction of ideas and principles of mathematics. Our students, to be able to deal with mathematical ideas and use them, need to have the logical foundations to abstract and use postulates and construct new formulations.
Learning to abstract helps formulate and understand arguments. The capacity to see interrelations among concepts helps us deal with ideas in other subjects as well. It also helps us understand and make better patterns, maps, appreciate area and volume and see similarities between shapes and sizes. Once students do that they need to be able to find the way to use the knowledge they have and reach where the problem requires them to go. They need to identify and define a problem, select or design possible solutions and revise or redesign the steps, if required.
Class VIII course has attempted to bring together the different aspects of mathematics and emphasise the commonality. Unitary method, Ratio and proportion, Interest and dividends are all part of one common logical framework. The idea of variable and equations is needed wherever we need to find an unknown quantity in any branch of mathematics. | 677.169 | 1 |
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Michaelmas Term Week 4
Talk: How to prepare a talk
The purpose of the Part III year is to transform undergraduates into working mathematicians, capable of reading papers, doing research, writing papers, giving talks and generally being part of a mathematical community. In the course of the Part III year, you will have two chances to give a talk, and be part of a seminar group, a chance to read papers, and a chance to write what may be your first mathematical paper.
The Michaelmas Term Part III Seminar series offers all Part III students the chance to talk for about half an hour on a subject of their choice. "How to prepare a talk" goes through the stages of planning and rehearsing a talk. | 677.169 | 1 |
Mathematics
courses provide students with the basic concepts and relationships
fundamental to existence in a highly technical society. Understanding
the nature of mathematics encourages critical thinking, precise
expression, and analytical approaches to problem solving. Students need
two years of Math to be promoted to high school. The following link
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The Tao of Mathematics
Terence Tao is amazing. Of course he is extremely gifted- he was awarded the Fields Medal at the last ICM in Madrid in 2006- but in addition, he is articulate, prolific, and intellectually very generous. He can not only write at the most advanced level- we have his Analysis I and II listed in our Mathematics section (both a part of the superb TRIM series of books in mathematics from the HBA, New Delhi), but also for a less specialist audience.
This two-volume introduction to real analysis by Fields medalist Terence Tao, is intended for honours undergraduates, who have already been exposed to calculus. The emphasis is on rigour and on foundations. The material starts at the very beginning - the construction of the number systems and set theory, then goes onto the basics of analysis (limits, series, continuity, differentiation, Riemann integration), through to power series, several variable calculus and Fourier analysis, and finally to the Lebesgue integral. These are almost entirely set in the concrete setting of the real line and Euclidean spaces, although there is some material on abstract metric and topological spaces. There are also appendices on mathematical logic and the decimal system. The course material is deeply intertwined with the exercises, as it is intended for the student to actively learn the material and to practice thinking and writing rigorously. Vol I: Rs 550, 365 pages, ISBN: 9788185931944 Vol II: Rs. 350, 236 pages, ISBN: 9789380250656 | 677.169 | 1 |
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Translated from the popular French edition, this book offers a detailed introduction to various basic concepts, methods, principles, and results of commutative algebra. It takes a constructive viewpoint in commutative algebra and studies algorithmic approaches alongside several abstract classical theories. Indeed, it revisits these traditional topics with a new and simplifying manner, making the subject both accessible and innovative.The algorithmic aspects of such naturally abstract topics as Galois theory, Dedekind rings, Prufer rings, finitely generated projective modules, dimension theory of commutative rings, and others in the current treatise, are all analysed in the spirit of the great developers of constructive algebra in the nineteenth century. This updated and revised edition contains over 350 well-arranged exercises, together with their helpful hints for solution. A basic knowledge of linear algebra, group theory, elementary number theory as well as the fundamentals of ring and module theory is required. Commutative Algebra: Constructive Methods will be useful for graduate students, and also researchers, instructors and theoretical computer scientists | 677.169 | 1 |
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All high school math courses are planned out and sequentially paced. Courses include: Pre-Algebra, Algebra 1, Geometry, Algebra 2, Pre-Calculus, AP Calculus AB, and AP Statistics. In a nice 1-page format for coherent access. You will know what to teach each day and know what topics are coming up. Non-teaching days are taken into account, such as testing, staff development days, etc. I hope that this helps you in your planning! Send me a message if you'd like more information. | 677.169 | 1 |
Calculus
Calculus has two main parts: differential calculus and integral calculus. Differential calculus studies the derivative and integral calculus studies (surprise!) the integral. The derivative and integral are linked in that they are both defined via the concept of the limit: they are inverse operations of each other (a fact sometimes known as the fundamental theorem of calculus): and they are both fundamental to much of modern science as we know it.
Linear algebra is the branch of mathematics concerning vector spaces and linear mappings between such spaces. It includes the study of lines, planes, and subspaces, but is also concerned with properties common to all vector spaces. | 677.169 | 1 |
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103 Trigonometry Problems ; From the Training of the USA IMO
Then you see a lot of these in circuit analysis and heat transfer or Fourier series. Real-life examples include parabolic trajectories and the use of cubic splines in designing sailboats and computer-generated teapots. Note that the triangle needs to "hug" the x axis, not the y axis: We find the values of the composite trig functions (inside) by drawing triangles, using SOH-CAH-TOA, or the trig definitions found here in the Right Triangle Trigonometry Section, and then using the Pythagorean Theorem to determine the unknown sides.
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Publisher: BIRKHAUSER (2005)
ISBN: 8181283392
Trigonometric Travelogue
PRACTICAL MATHEMATICS BEING THE ESSENTIALS OF ARTIHMETIC, GEOMETRY ALGEBRA AND TRIGONOMETRY PART IV TRIGONOMETRY AND LOGARITHMS
Attacking Trigonometry Problems (Dover Books on Mathematics)
Now let the rotating arm continue to rotate, in the same direction as before, until it arrives back at its original position on OA Elements of geometry and read online read online. Those with the proper high school background but a lower placement score should talk to a math faculty member or take a lower level algebra course first. It is usually better to take Math 141 College Algebra first, but students have been successful taking this course immediately following Math 118 Intermediate Algebra or its equivalent , source: Precalculus: Concepts Through download online subtractionrecords.com. Pursuant to our rights under the Digital Millennium Copyright Act ("DMCA"), 17 U. C. §512, we have designated a copyright agent to receive copyright infringement notices for claims of infringement related to materials found on this Site epub. Math exercises for grade 7 for free, how to solve a third order polynomial, free maths work booklets, algebraic substitution ks2, 4th grade factor worksheets, algebra swf , e.g. Plane and Spherical read online It has many practical applications in engineering, physics, and other branches of science. Trigonometry Algebra Geometry read online Trigonometry Algebra Geometry. I suppose I can't blame CliffsNotes, since that is given in the title of the book. I bought this book when I was beginning to learn Trigonometry, along with another textbook. I found that a lot of the things the textbook went over was not even mentioned in this Quick Review book. If you have taken Trigonometry in the past and want to refresh some of the basic trigonometry concepts, then this book is for you , cited: Teach yourself trigonometry download for free And I've replaced this trig integral with something that doesn't involve trig functions at all. This is u^(n+1) / (n+1) plus a constant, and I've done the integral. But I'm not quite done with the problem yet. Because to be nice to your reader and to yourself, you should go back at this point, probably, go back and get rid of this new variable that you introduced ref.: Plane trigonometry with tables, (Addison-Wesley mathematics series) read here.
On the other hand, if you are too ill to take the exam, contact the instructor as soon as possible to arrange for a makeup (and you should be prepared to send the instructor a note from your doctor in this case) online. Maths Wizard se Kuber Klaskamer bied Di.. An analytical treatise on plane and spherical trigonometry, and the analysis of angular sections; designed for the use of students in the University of London Convert units of Acceleration, Angle, Area, Density, Energy, Force, Length, Mass/Weight, Power, Pressure, Speed, Temperature, Time, and Volume. Complex calculations using factorial, gamma, combinations, permutations, logarithms, standard and hyperbolic TRIG functions, BASE-N conversions, logical operations. 26 separate memory registers download. Not any three line segments may serve as the sides of a triangle. They do iff they satisfy the triangle inequality, or rather three triangle inequalities , e.g. Plane and spherical trigonometry: With stereographic projections, Plane and spherical trigonometry: With.
Logarithms, Trigonometry, Statistics
Students will generate and solve linear systems with two equations and two variables and will create new functions through transformations. (4) Statements that contain the word "including" reference content that must be mastered, while those containing the phrase "such as" are intended as possible illustrative examples. (1) Mathematical process standards The Elements Of Algerbra And Trigonometry read here. A significant historical contribution of Jain mathematicians lay in their freeing Indian mathematics from its religious and ritualistic constraints , source: Modern Algebra and Trigonometry: Structure and Method, Book 2 subtractionrecords.com. Prerequisite(s): MATH 13300 or MATH 15200 or MATH 16200. Note(s): Recommended sequence for ECON majors: MATH 19620, STAT 23400, ECON 21000 in consecutive quarters. Mathematical Methods for Physical Sciences I-II. This sequence is intended for students who are majoring in a department in the Physical Sciences Collegiate Division other than mathematics. Mathematical Methods for Physical Sciences I. 100 Units , e.g. An Elementary Treatise on download online An Elementary Treatise on Plane and Problems and Solutions in download online download online. Please contact your local customs office for further information before placing your order, and please refer to the Frequently Asked Questions Page of our Site for further information. Please also note that it is your responsibility to understand and comply with all applicable laws and regulations of the country for which the products are destined ref.: ISM * College Algebra, 5th download for free subtractionrecords.com. I highly recommend ratios or gcf without fuss, please drop me a line Thanks Have you ever sat in your math class asking yourself, "Why do I need to learn trigonometry and calculus? Who uses trigonometry and calculus in the real world?" Civil engineering professors and students from the University of Hawaii College of Engineering visited McKinley High School and Kaimuki High School to teach upper division math students how civil engineers use trigonometry and calculus to design and construct buildings, bridges, and other structures Benson's Microbiological Applications: Laboratory Manual in General Microbiology, Short Version
College Algebra and Trigonometry Updated edition by Dugopolski, Mark published by Addison-Wesley Hardcover
Addison-Wesley Algebra Problem Bank
By Charles P. McKeague - Trigonometry: 6th (sixth) Edition
We will now use an example to show how these rules are applied to solve a triangle when two sides and a non-included angle are given. Example: A triangle has sides a=5 and b=7, and a non-included angle A=30�. Solve for the unknown side and the two unknown angles. Often this type of triangle problem has two solutions Geometry: Plane Trigonometry. Chain Surveying. Compass Surveying. Transit Surveying subtractionrecords.com. Start a 30-day free trial of our Premium eReader, and gain access to our huge Business eBook library Try for free As the teacher I will be monitoring these discussion, listening for misconceptions and understanding of the material. One quiz and one unit test will be given as formal assessments, each of these are included at the end of the document. Another formal assessment for this unit is the Geometry of Baseball project, also included below. �The process of evaluation begins with the objectives of the educational program,� (Tyler, 1949, p. 110) , e.g. High (Secondary School) Grades download pdf High (Secondary School) Grades 11 & 12 - online. Take a right angle triangle with two 45 degree angles, and with sides of 1 unit length. By the Theorem of Pythagoras, the hypotenuse of this triangle is of length √2. This is what this triangle looks like: So then, from these values and SOHCAHTOA, you can obtain the trig values for this special angle of 45 degrees ref.: NY Algebra 2 & Trigonometry, read pdf Any unauthorized use of the Streaming Service or its contents will terminate the limited license granted by us and will result in the cancellation of your membership. Some restrictions are placed on the extent to which we accept orders for Products or Services from specific countries , source: Technical trigonometry download epub download epub. Consider a right-angled triangle ABC as shown in the figure below. Hypotenuse: The side opposite to the right angle in a triangle is called the hypotenuse ref.: Algebra and Trigonometry download for free A diameter is a chord which passes through the centre of the circle. It divides the circle into two equal parts called semi-circles. A segment is a part of a circle bounded by a chord and the arc which it cuts off online. If you still have doubts, here are some reasons why you should purchase your college essays with EssayOnlineStore: You Will Get a 100% Original Paper Our promise to each of our customers is to deliver absolutely unique work Practical Mathematics Part Iv: Trigonometry And Logarithms... Practical Mathematics Part Iv:. May not be taken in addition to 0050, 0060, or 0070; 0100 may not be taken in addition to MA 0170 Constructive Text-Book of read online Luckily, people such as Phil Plait put the media back on the proper astronomical track. This is a top-notch site authored by a well-informed, well-credentialled astronomer. Please pay a visit to: Here's a good website that teaches basic math resources and has calculators: Consider this right angled triangle: In right angled triangles, the terms 'opposite' and �adjacent� is never used to refer to the longest side opposite to the right angle pdf. The student is expected to: (A) graph a set of parametric equations; (B) convert parametric equations into rectangular relations and convert rectangular relations into parametric equations; (C) use parametric equations to model and solve mathematical and real-world problems; (D) graph points in the polar coordinate system and convert between rectangular coordinates and polar coordinates; (E) graph polar equations by plotting points and using technology; (F) determine the conic section formed when a plane intersects a double-napped cone; (G) make connections between the locus definition of conic sections and their equations in rectangular coordinates; (H) use the characteristics of an ellipse to write the equation of an ellipse with center (h, k); and (I) use the characteristics of a hyperbola to write the equation of a hyperbola with center (h, k). (4) Number and measure , e.g. Plane Trigonometry (Second Edition) Plane Trigonometry (Second Edition). | 677.169 | 1 |
LAST 10 YEARS CBSE BOARD PAPERS CLASS 12 MATHS
Last 10 years cbse board papers class 12 maths
Arrows represent the paths that relate to dependencies. FREEMyMaths for KS3 Penguin Origami PDF An origami excercise for KS3 to explore shapes and anglesFREEKey Stage 3 Spreadsheets Homework SheetA homework activity sheet mainly aimed at year 7 or 8 students. By insisting on constant informal contact and by preserving that contact year after year, The New American Academy has the potential to create richer, mentorlike or even familylike relationships for students who are not rich in those things. Read more Read less See all buying options Algebra Connections Volume 1 CPM Available algebra 2 help online these sellers. A major theme of Polya's writing was to do mathematics, to reflect on problems solved or attempted, and to think 27,28. Nikhil was very easy to follow along with and understand. The graphing calculator as an aid to teaching algebra. All of IXL's skills are fully aligned area of the semicircle formula the Common Core State Standards. 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HOW DO I GET MY SON TO DO HIS HOMEWORK
How do i get my son to do his homework
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What if I was wrong and you had never questioned it. Brown and Walter provide a wide variety of situations implementing this strategy including a discussion of the development of non-Euclidean geometry. This opportunity is offered from 2:45 to 3:45. He's committed to providing the highest quality resources to help you succeed. Click the button below to post your assignment and get math answers right now. Wondering if there is a way to help you out conduct a simple regression analysis. In the above example, the user is on question 5. On Thursday Mabel handled 90 transactions. GlencoeMcGraw Hill Holliday, etAlgebra 2 California Edition 2005 By Glencoe Mcgraw Hill Algebra 2 help answers for algebra 2 homework problems, Algebra 2 helpStudent Edition, Volume 2, a book by McGraw-Hill Education. I have not claimed that problems are useless. Also, be sure to go back and re-check the word problem for what it actually wants. 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Some solution methods I was not familiar with, but it was easy to follow to the end. Make our publisher: click here. Fill in the correct unit of measure for each answer you choose. Your child can learn to: skip-count by whole numbers, compare and order numbers, identify ordinal position to the twentieth, identify and complete patterns, add and subtract multidigit numbers, divide by single-digit divisors, add positive and negative numbers, picture, name, and order fractions, add and subtract fractions with common denominators, understand and calculate measurements, compare and measure mass, identify function rules, graph ordered pairs on a coordinate graph, identify angles, identify lines of symmetry. Tutoring sessions focus on learning general topics rather than completing specific homework assignments. How many holt geometry workbook answers does it rotate in one second. CD Set Video Lectures on CD-ROM This optional CD-ROM set delivers the exact same video lectures delivered online, but without an internet connection. Study Island Students may use Study Island to review important skills and to gain additional practice aligned with the Colorado Academic Standards. Many of these are for more advanced users. Introduction to gradients and intercepts. Multiplying and dividing, bar graphs, pronouns, possessives, weather and climate, geography, and more. They criticize teachers who use direct instruction everyday but they want the discovery approach everyday. LCM using ladder method, Factoring trinomials Cheat Sheet, trigonometry factoring calculator. Send your problems, which you can't manage to solve without homework help via e-mail or fax to us and our math tutors will help your to understand, that studing math with tutor is not so dreadfull as it seems from the first look: Email: an10 homeworktutoring. Here's the most smileys you got in a row. You cannot get credit for it since we are not a school. The radius of the face of a circular clock. Algebra 1 Chapter 7 Resource Book 16. But the next time you see a how do i get my son to do his homework like this, you already know the trick - and you know exactly what to do. Asked by framirez3433 on June 17, 2016 at 1:35 AM via web 1 educator answer. SEATTLE AP - A Seattle hospital says about 650 dialysis patients since 2011 might have been exposed to hepatitis B because of a lapse in screening procedures. Khan Academy's video fun problem solving questions are justifiably famous and cover various subjects including mathematics. How do i get my son to do his homework Universal WolframAlpha is a much more math answers for geometry a math app, but the computational and mathematical abilities it possesses are phenomenal. Chapter 7 How do i get my son to do his homework Book. Graphic calculator in teaching product rule, frustrated with algebra, multiplying rational expression calculator, what are the pros and cons of using the quadratic formula?. Sign up for FREE Math Counts how do i get my son to do his homework for students in 6th, 7th and 8th grade. Arithmetic Sequences Problems with Solutions. If you're seeing this message, it means we're having trouble loading external resources for Khan Academy. Login Contact Us Pre-Calculus Not a Pre-Calculus Question. One approach is to try working a simpler version of the problem, and use the solution to the problem to get insights that are useful in solving the original problem. The gas tank contains 9 gallons. Topics include:Linear Equations and Their Graphs introduces you to the concept of a line and how how do i get my son to do his homework graph and interpret lines in various applications using the different forms of a linear equation, including slope-intercept and standard. Sometimes they just have to do a few problems to remember how to do it on the test. Learning it with context made all the pieces fall into place in a satisfying and calming way. Buy the one and practice all the questions if you can. Simplifying rational expressions solver, basic algebra by brown, answers for prentice hall algebra 2, how to cheat with a ti-83, algebra function worksheets, algebra using the 5 step method illinois cdl written test. Near the end of most pages is a "Your Turn" section. Very helpful for my students. Brooks notes that: The Solving for x with fractions American Academy has two big advantages as a reform model. Thanks to Marco2013 for this resource. This trio of horrors are: math, physics and chemistry. Algebra is a big part of the curriculum at the high school level but even college level algebra students can benefit from seeing complete solutions and solved equations which are produced by a math genius. You can unlock the pro version with an in-app purchase.
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Be sure to SCROLL DOWN to find YOUR Class' Homework Calendar. I must say that there is no alternative for paying attention in class, writing down notes, studying at home, and doing a lot of independent homeworks south bend using pencil and paper. Like it's the case with the Solutions Manual, you will only need one DIVE CD even if you have more than one student. It is actually more comprehensive than Mathway. I maths skills test thought this was the standard view. Y, YY, Z, ZZ, A, AA, B, BB, C, and CC tell us nothing about what XX is equal to. You may also view your score by logging into the Math Placement Test and selecting the Test Scores button in the left menu. Whitefish bay family access probability practice online free, examples of intermediate algebra equations, grade 10 algebra worksheets, mathimatical trivia about algebra.
I am no more dependent on anyone except on this little piece of software. My example is personal. He thought no one had noticed his paper, but in fact it had attracted the attention of just the right people to solve it. I think some of the CPM activities are great as supplements but really the books are mostly a collection of activities. We have worked, to the best of our ability, to ensure accurate and correct information on each page and solutions to practice problems and exams. The problem is: Verify that this implicit equation: 1. It leads to substantial gains in reading comprehension. I love looking through them, too. Department of Education released a report designating 10 math programs as "exemplary" or "promising. Question 2: Find the area of square whose one side is 12 cm. A similar problem for a parabola was solved by Archimedes, t 8 sec. Become a Shmoopscriber and get access to Shmoop Mcgraw hill connect online book Prep and Online Courses. Pre algebra usually requires a book, calculator, paper, algebra 2 equation solver, and a writing utensil. To get the free app, enter your email address or mobile phone number. Shannon's Corrections: 6- I'm just confused on this one. Find the zeros of a polynomial function by factoring, radicals with negative powers, multiplying radicals, algerbra math. Click the button to go on to the next activity.
Not always real math. AXYZ - AHPV AOGW - ANMU AABC s AEDF ZEPPE Answers 6 - 11: LAPLE B D ASKM r ANGJ Answers 12 17: AEFG - - AHIG ATLC - AOKC " ARHX s AWL LG D-41 TOPIC 3. Don't such simple problems asvap practice test algebra books just encourage students to forget common sense and simple arithmetic. His riddle involving Achilles, the solving equations involving fractions worksheet from Homer's Iliad and a tortoise went something like: The tortoise challenged Achilles to a race and Achilles, full of typical hubris, accepted and even gave the Tortoise a 10 foot head start. Linear equations calculator, algebra books online, adding and subtracting unlike mixed numbers worksheets. I have never seen anything quite like it before. This means that you don't even need to launch the app when you need to perform calculations - just swipe down from the top of your screen no matter what you're doing and written driving test questions and answers started. Please write this information down so you can sign in anywhere. Don't forget to also join as a fan on my Facebook page. Here are two things I did to address curriculum and scheduling problems at Berkeley High. How many 11 year olds can weld a beautiful bead. Figure out the general form of the kind of thing you don't understand. This is an amazing resource. Will the ball have enough tance. An extensive list of other algebra how do i get my son to do his homework is located below. Geometry Homework Help Online Free Free math lessons and math homework help from basic math to algebra, geometry and beyond. | 677.169 | 1 |
language isn't English Feel good about yourself knowing that you accomplished something amazing. Get GED Test For Dummies and put yourself on the road to greater successLearning Analytics Explained draws extensively from case studies and interviews with experts in order to discuss emerging applications of the new field of learningThis book can form the basis of a second course in algebraic geometry. As motivation, it takes concrete questions from enumerative geometry and intersection theory, and provides intuition and technique, so that the student develops the ability to solve geometric problems. The authors explain key ideas, including rational equivalence, Chow rings, Schubert calculus and Chern classes, and readers will appreciate the abundant examples, many provided as exercises with solutions available online. Intersection is concerned with the enumeration of solutions of systems of polynomial equations in several variables. It has been an active area of mathematics since the work of Leibniz. Chasles' nineteenth-century calculation that there are 3264 smooth conic plane curves tangent to five given general conics was an important landmark, and was the inspiration behind the title of this book. Such computations were motivation for Poincaré's development of topology, and for many subsequent theories, so that intersection theory is now a central topic of modern mathematics access | 677.169 | 1 |
Sunday, November 30, 2008
Solve your math problem in the internet.
stuck on a math problem? figuring out how to get the answer ? Then head over to and let it solve and explain it for you.
Let's here what webmath says about it's service and I believe\trust on it.
"
Webmath is not a database of questions and answers, or an online math testing site. Webmath is a math-help web site that generates answers to specific math questions, as entered by a user of the site, at any particular moment.......... .......
In addition to the answers, Webmath also attempts to show to the student how to arrive at the answer as well. For example, if the user wants to know how to square the quantity (x+2), Webmath does not just display the answer x2+4x+4, but a step-by-step solution as well. ..........
It contains a sophisticated computer math "engine" that is actually able to recognize and "do the math" on a particular problem it is presented with. For this reason, Webmath is a very dynamic website, in that most of the replies a user will encounter are created the instant they are sent to the user. .....
We acknowledge though, that programming a computer to solve math problems "on the fly" is not an easy task, and Webmath is not a a complete math solution system. Many computer methods and algorithms related to having a computer solve a math problem remain research-level topics at computer science labs around the world. This difficulty in computer math programming remains at the heart of the limited offerings of this site, and shortcomings therein. Nevertheless, the developmental goal of this site is to slowly and steadily increase and improve upon its math solving capabilities, so frustrated students have a place to at least try to find the answer to their math problem, at the time they need it. | 677.169 | 1 |
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Winner of the 1983 National Book Award!"e;...a perfectly marvelous book about the Queen of Sciences, from which one will get a real feeling for what mathematicians do and who they are. The exposition is clear and full of wit and humor..."e; - The New Yorker (1983 National Book Award edition)Mathematics has been a human activity for thousands of years. Yet only a few people from the vast population of users are professional mathematicians, who create, teach, foster, and apply it in a variety of situations. The authors of this book believe that it should be possible for these professional mathematicians to explain to non-professionals what they do, what they say they are doing, and why the world should support them at it. They also believe that mathematics should be taught to non-mathematics majors in such a way as to instill an appreciation of the power and beauty of mathematics. Many people from around the world have told the authors that they have done precisely that with the first edition and they have encouraged publication of this revised edition complete with exercises for helping students to demonstrate their understanding. This edition of the book should find a new generation of general readers and students who would like to know what mathematics is all about. It will prove invaluable as a course text for a general mathematics appreciation course, one in which the student can combine an appreciation for the esthetics with some satisfying and revealing applications.The text is ideal for 1) a GE course for Liberal Arts students 2) a Capstone course for perspective teachers 3) a writing course for mathematics teachers. A wealth of customizable online course materials for the book can be obtained from Elena Anne Marchisotto (elena.marchisotto@csun.edu) upon request delivery | 677.169 | 1 |
Introductory Algebra
This course is being developed as part of a screen design project at Arizona State University, for EDT 503 with Dr. Atkinson. Please refrain from contributing directly until the project submission date of 12/08/2010 has passed.
Bastetswarrior respectfully asks that people use the discussion page, their talk page or email them, rather than contribute to this page at this time. This page might be under construction, controversial, used currently as part of a brick and mortar class by a teacher, or document an ongoing research project. Please RESPECT their wishes by not editing this page.
This course is intended for adult learners returning to college or university studies. Some common classes covering roughly these topics are the College Algebra class sequence at University of Phoenix and Axia College, and the Basic Algebra class sequence at Devry. You might not have used academic algebra for many years, but don't worry! Introductory algebra can be both fun and useful! | 677.169 | 1 |
13
E. T. S. I. Caminos, Canales y Puertos13 LU decomposition LU decomposition - The LU decomposition is a method that uses the elimination techniques to transform the matrix A in a product of triangular matrices. This is specially useful to solve systems with different vectors b, because the same decomposition of matrix A can be used to evaluate in an efficient form, by forward and backward sustitution, all cases.
15
E. T. S. I. Caminos, Canales y Puertos15 LU decomposition LU decomposition is very much related to Gauss method, because the upper triangular matrix is also looked for in the LU decomposition. Thus, only the lower triangular matrix is needed. Surprisingly, during the Gauss elimination procedure, this matrix L is obtained, but one is not aware of this fact. The factors we use to get zeroes below the main diagonal are the elements of this matrix L. Substract | 677.169 | 1 |
General Mathematics
Course Description
General Mathematics focuses on mathematical skills and techniques which have direct application to everyday activity. The course content is written in five areas of study, with an emphasis on application of specific skills and on tasks that involve integrating mathematical skills and techniques across a range of familiar and unfamiliar situations. These tasks may draw from more than one area of study, and encourage transfer of knowledge across the entire course, as well as linking with study in other Stage 6 subjects.
The course is fully prescribed, and is designed to support TAFE and other vocational courses. It provides an appropriate mathematical background for students who do not wish to pursue the formal study of mathematics at tertiary level, while giving a strong foundation for university study in the areas of business, humanities, nursing and paramedical sciences. | 677.169 | 1 |
17 Februari 2010
Understanding algebra in more detail
Most people think "Learning algebra is stupid". Have you ever noticed that the math and science is far more important for the foundation of a modern technological society than a simple social rules others. So, why do we only hear complaints about the math and science? Maybe because they are hard to? Because they require work and discipline?! Because they are not always "easy "?!
Educators of modern philosophy in the United States seems to say that education should be "fun" and "entertaining" can be justified. Today, students often absorb ethics, but a simple thing, they do not have to bother. But the most precious things in life will require some effort. If you want a big job, career interest, that the future is open, you almost certainly will need math skills. And algebra 2 is the basis, foundation, tool-boxes, for their skills.
"Is Algebra 1 Answers even be 'relevant' in the future?" When work and their specific skills-set can change from time to time, math is not. Twenty years from now, two plus two will still be four. In order to get a job in the first place, you will need some background knowledge and skills. Foundation that includes mathematics.
And mathematics patterns are also important. If all you take from the comfort Algebra Problems with variables and formulas, the ability to interpret graphs and logical thinking, and willingness to use abstraction when you try to solve the problem, then you have gained some life skills are very useful.
And there is only one of the best to help you or your children to easily answer the Algebra 1 Problems. He is Tutorvista. Get answers to all Algebra problems online with TutorVista. Their online Algebra tutoring program is designed to help you get all the answers to your Algebra word problems giving you the desired edge in excelling in the subject. Solve Algebra 1 problems, work on basic concepts and get help with your Algebra 1 homework too with TutorVista. | 677.169 | 1 |
Math
The Mathematics major provides students with the knowledge and skills necessary to enter a career in teaching, insurance, finance and industry, or to continue into graduate school. With instruction in discrete math, calculus, algebra, geometry, math modeling, computer programming, and physics, students gain a wide range of mathematical understanding. A math history focus also allows students to expand their knowledge of leading mathematicians and their revolutionary discoveries.
The Mathematics for Teachers (Mathematics Single Subject) degree equips students with the mathematical knowledge required to teach grades 7 through 12. Students learn to communicate and teach mathematics in formal and informal settings. Through this course of study, students develop the tools and skills necessary for careers in industry, education, ministry, and even graduate studies.
Assistant Professor of Mathematics Math Department Chair
Profile:
Dr. Kelly has been teaching mathematics at Simpson University since 2013. He is the faculty advisor for the Simpson University Math club and runs the KME honors society. His research is in stochastic processes that model evolutionary dynamics. Before coming to Simpson he was a postdoc for one year at the University of Minnesota. Dr. Kelly writes computer programs as a hobby.
Educational Background:
Ph.D., University of California - San Diego
M.S. University of California - San Diego
B.S., Oklahoma State University
Courses Taught:
Discrete Math
Math History
Math Modeling
Computer Programming
Mathematical Statistics and Probability
Real Analysis
Complex Analysis
Putnam Problem Seminar
Modern Algebra
Senior Project Reading
Senior Project Writing
Links:
Assistant Professor of Mathematics
Profile:
Dr. Wenjing Li is an Assistant Professor of Mathematics in the Mathematics Department at the Simpson University. Wenjing Li received her Ph.D. in Mathematics from the University of Georgia, under the direction of William Graham. Dr. Li's research interest is in Lie theory.
Educational Background:
Ph.D., University of Georgia
M.A., University of Georgia
M.S. University of South Alabama
Courses Taught:
Modern Algebra
Calculus
Statistics
Links:
Construct and develop rigorous mathematical reasoning, including the use of standard mathematical concepts;
Effectively communicate rigorous mathematical reasoning, including the use of appropriate visual aids; and
Expand his or her own mathematical understanding through a process of self-directed learning. | 677.169 | 1 |
Showing 1 to 14 of 14
Beginning Algebra
MTH 095
FALL 2011
Moraine Valley Community College
Chapter 1.2: Pac-Math
A quick note as we begin our journey, everyone: some of this stuff is going to
come very easily to you in the beginning. A few of you might even say, Wow,
Im in Col
Beginning Algebra
MTH 095
FALL 2011
Moraine Valley Community College
1.8: Properties of Real Numbers
Were not going to go over everything in this section because some stuff is
more important than others. Id like to try and stick to the really important
st
Beginning Algebra
MTH 095
FALL 2011
Moraine Valley Community College
Chapter 1.7: Multiplying and Dividing Real Numbers
Do you remember how I mentioned a while ago that, for most things,
multiplying and dividing are easier than adding and subtracting? Its
Beginning Algebra
MTH 095
FALL 2011
Moraine Valley Community College
Chapter 1.6: Subtracting Real Numbers
If you feel solid with adding real numbers Im not going to do too much
to add (pun intended) to that to help you understand how to subtract real
num
Beginning Algebra
MTH 095
FALL 2011
Moraine Valley Community College
Chapter 1.5: Adding (and Subtracting) Real Numbers
What Im about to show you isnt difficult but it might require a little bit
of practice. Its how to add and subtract numbers with basica
Beginning Algebra
MTH 095
FALL 2011
Moraine Valley Community College
1.4: Intro to Variables and some Order of Operations Stuff
Hows everybody doing? Ready to rock? Excellent!
Order of Operations
The order of operations tells you what to do first in a pro
Beginning Algebra
MTH 095
FALL 2011
Moraine Valley Community College
Chapter 1.3: Fractions
Go ahead and get it out of your system. AAAAAHHHH! Fractions! Feel
any better? Listen I know that you hate fractions. Its okay; you dont have
to like them to get t | 677.169 | 1 |
Sorting Expressions and Polynomial Vocabulary
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this file type before downloading and/or purchasing.
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Product Description
Steps:
1. Sort expressions into two columns of polynomials and non polynomials.
2. Glue the expressions in the correct columns.
3. For the polynomials, students need to identify each: name by number of terms, the leading terms, the degree and lead coefficient of the polynomials
This activity is a good review of what makes an expression a polynomial and give students a chance to practice using polynomial related vocabulary. | 677.169 | 1 |
Overview
Calculus II For Dummies by Mark Zegarelli
An easy-to-understand primer on advanced calculus topics
Calculus II is a prerequisite for many popular college majors, including pre-med, engineering, and physics. Calculus II For Dummies offers expert instruction, advice, and tips to help second semester calculus students get a handle on the subject and ace their exams.
It covers intermediate calculus topics in plain English, featuring in-depth coverage of integration, including substitution, integration techniques and when to use them, approximate integration, and improper integrals. This hands-on guide also covers sequences and series, with introductions to multivariable calculus, differential equations, and numerical analysis. Best of all, it includes practical exercises designed to simplify and enhance understanding of this complex subject.
Introduction to integration
Indefinite integrals
Intermediate Integration topics
Infinite series
Advanced topics
Practice exercises
Confounded by curves? Perplexed by polynomials? This plain-English guide to Calculus II will set you straight!
Product Details
About the Author
Mark Zegarelli, a math tutor and writer with 25 years of professional experience, delights in making technical information crystal clear — and fun — for average readers. He is the author of Logic For Dummies and Basic Math & Pre-Algebra For Dummies.
Most Helpful Customer Reviews
This book is a lifesaver! After taking calculus 2 and failing, I decided to look for books that dealt with integration (the concept of calculus 2). I was thrilled when I purchased this book. It is full of excellent explainations and examples that really helped me get a better understanding of calculus 2. I would recommend this book to all current and future calculus students!
Kourtlandt
More than 1 year ago
About 5/8ths of the book is a review of Calculus I for Dummies... however, they are different authors so I will not complain too much.
Included are tear out cheat sheets on front and back.... however, I did notice one small error on page 1. It states that " d/dx arcsin(x) = -1/sqrt(1-x^2) "
and that is wrooooooongggg!!!! hopefully alot of students will notice that and not memorize the wrong thing.... that little "-" sign can be the difference in an A and a B. kind of a minor but big screw up Wiley Publishing. The error was fixed everywhere else in the book but the problem is that the error IS on the tear out cheat sheet that is designed to be studied and memorized.
anywho... not a bad book, definitely resourceful. 4 out of 5 stars! | 677.169 | 1 |
Description
This book was written by a licensed Pharmacist who specializes in the training of Pharmacy Technicians and has done so for the past eleven years.In this book, she talks about the basic things that stand in a person's way when it comes to learning pharmacy math. This book lays the foundational framework necessary in order to advance to more challenging calculations.It is part analysis, part inspiration, part calculation, and all of it is necessary for successful learning!Vanessa uses the same conversational tone in this book that has had success with in her classroom. Hundreds of students have taken and passed the PTCB Exam because of her ability to bring lessons to life. Many technicians have come to her classes as math inept and left as being very skilled.She has proven that pharmacy math does not have to be scary!show more | 677.169 | 1 |
Saturday, January 12, 2008
First, the price tag on this book can be scary. Most bookshops list it for $55.00 on average. But you are getting enough ideas to cover mathematics from first grade through eighth. And I mean you are getting all you need in this book.
When the book first arrived, I was disappointed. "Another small Waldorf book," I thought, "I hope it has some practical information in it."
Then I paged through it and thought: "This looks way over my head. At least I can resell it for what I paid for it."
But then I decided to give it a try and sat down with a notebook and pen and began to read. Wow! That is all I can say. This book is excellent. It has some philosophy but not too much. And lots of great practical ideas on what to do and what order to do it in.
Best of all he gives you examples of the kind of problems to do and questions to ask but helps you and encourages you to create your own.
So basically, you can gear the math curriculum completely to the needs and interests of the child. Now, I have only read through the information for first through third grade but from looking forward I expect the rest of the book to be as good.
This book is worth the price tag and worth taking the time to read and take notes. | 677.169 | 1 |
This unit introduces sequences and series and gives some simple examples of each. It also explores particular types of sequence known as arithmetic progressions (APs) and geometric progressions (GPs) and the corresponding series. | 677.169 | 1 |
Mathematics Department
The mathematics curriculum inspires and equips students to be persistent and creative problem solvers. Though our curricula purposefully align with standardized tests (ACT, SAT, SAT Subject Tests), our instruction also challenges students to see through a variety of mathematical lenses and to appreciate not only the subject's practical applications but also its inherent elegance.
Every mathematics student is encouraged, challenged, and supported to go as far in the discipline as his skills and goals will take him. Many students advance earlier still to and through the AP level. All students take calculus and/or statistics before they graduate.
In addition to cultivating a mathematics culture that encourages curiosity and promotes mastery, the Mathematics Department offers an array of robust computer science courses – including post-AP electives in Data Structures and Networking.
A dedicated faculty with deep and varied experience in both industry and educational arenas, including collegiate teaching, forges relationships with students within intimate classes, through one-on-one tutorials, and in co-curricular opportunities that allow the most accomplished and motivated young mathematicians to explore their passion and to be competitive in the discipline outside the walls of the school | 677.169 | 1 |
Category: Emerging Trends years How To Teach Mathematics
This third edition is a lively and provocative tract on how to teach mathematics in today's new world of online learning tools and innovative teaching devices. The author guides the reader through the joys and pitfalls of interacting with modern undergraduates–telling you very explicitly what to do and what not to do. This third edition has been streamlined from the second edition, but still includes the nuts and bolts of good teaching, discussing material related to new developments in teaching methodology and technique, as well as adding an entire new chapter on online teaching methods.
Title: Explaining and Exploring Mathematics
Author: Christian Puritz
English | 2017 | ISBN: 1138680192, 1138680214 | 226 pages
Explaining and Exploring Mathematics is designed to help you teach key mathematical concepts in a fun and engaging way by developing the confidence that is vital for teachers. This practical guide focuses on improving students mathematical understanding, rather than just training them for exams. Covering many aspects of the secondary mathematics curriculum for ages 11-18, it explains how to build on students current knowledge to help them make sense of new concepts and avoid common misconceptions.
Focusing on two main principles to improve students understanding: spotting patterns and extending them to something new, and relating the topic being taught to something that the pupils already understand, this book helps you to explore mathematics with your class and establish a successful teacher-student relationship.
Structured into a series of lessons, Explaining and Exploring Mathematics is packed full of practical advice and examples of the best way to answer frequently asked questions such as:
Do two minuses really make a plus?
Why doesnt 3a + 4b equal 7ab?
How do you get the area of a circle?
Why do the angles of a triangle add up to 180°?
How can you integrate 1/x and calculate the value of e?
This book will be essential reading for all trainee and practising teachers who want to make mathematics relevant and engaging for their students.
Title: Algebra Teacher's Activity Kit
Authors: Judit A. Muschla, Gary Rober Muschla and Erin Muschla-Berry
English | 2015 | ISBN: 1119045746 | 336 pages
Help your students succeed with classroom-ready, standards-based activities. The Algebra Teacher's Activities Kit: 150 Activities That Support Algebra in the Common Core Math Standards helps you bring the standards into your algebra classroom with a range of engaging activities that reinforce fundamental algebra skills. This newly updated second edition is formatted for easy implementation, with teaching notes and answers followed by reproducibles for activities covering the algebra standards for grades 6 through 12. Coverage includes whole numbers, variables, equations, inequalities, graphing, polynomials, factoring, logarithmic functions, statistics, and more, and gives you the material you need to reach students of various abilities and learning styles. Many of these activities are self-correcting, adding interest for students and saving you time.
This book provides dozens of activities that
Engage students and get them excited about math
Directly address each Common Core algebra standard
Are tailored to a diverse range of levels and abilities
Reinforce fundamental skills and demonstrate everyday relevance
Algebra lays the groundwork for every math class that comes after it, so it's crucial that students master the material and gain confidence in their abilities. The Algebra Teacher's Activities Kit helps you face the challenge, well-armed with effective activities that help students become successful in algebra class and beyond.
Title: Learning Spaces
Space, whether physical or virtual, can have a significant impact on learning. Learning Spaces focuses on how learner expectations influence such spaces, the principles and activities that facilitate learning, and the role of technology from the perspective of those who create learning environments: faculty, learning technologists, librarians, and administrators.
Information technology has brought unique capabilities to learning spaces, whether stimulating greater interaction through the use of collaborative tools, videoconferencing with international experts, or opening virtual worlds for exploration. This e-book represents an ongoing exploration as we bring together space, technology, and pedagogy to ensure learner success.
Title: Learning Spaces: Interdisciplinary Applied Mathematics
Authors: Jean-Claude Falmagne, Jean-Paul Doignon
2011 | pages: 434 | ISBN: 3642010385
Learning spaces offer a rigorous mathematical foundation for practical systems of educational technology. Learning spaces generalize partially ordered sets and are special cases of knowledge spaces. The various structures are investigated from the standpoints of combinatorial properties and stochastic processes.
Leaning spaces have become the essential structures to be used in assessing students' competence of various topics. A practical example is offered by ALEKS, a Web-based, artificially intelligent assessment and learning system in mathematics and other scholarly fields. At the heart of ALEKS is an artificial intelligence engine that assesses each student individually and continously.
The book is of interest to mathematically oriented readers in education, computer science, engineering, and combinatorics at research and graduate levels. Numerous examples and exercises are included, together with an extensive bibliography.
The ability to have or to find space in academic life seems to be increasingly difficult since we seem to be consumed by teaching and bidding, overwhelmed by emails and underwhelmed by long arduous meetings. This book explores the concept of learning spaces, the idea that there are diverse forms of spaces within the life and life world of the academic where opportunities to reflect and critique their own unique learning position occur.
Title: The Design of Learning Spaces
Learning can take place anywhere. So does the detail of the physical surroundings provided by schools matter? After many years of minimal investment in school premises, schools in the UK are in the midst of a wave of planning, building and using new schools. This includes all English secondary schools, being renewed through Building Schools for the Future (BSF), as well as schemes for English primaries and programmes of school construction in Scotland and Wales.
Immersive learning has grown in popularity with the development of open-source immersive 3D learning spaces. Those in e-learning have been working to find ways to capitalize on immersive learning through simulations, digital kiosks, live virtual events, live interactivity, instructor-facilitated learning, AI-driven robots, and hyper-realistic experiences. Virtual Immersive and 3D Learning Spaces: Emerging Technologies helps push the conceptual and applied boundaries of virtual immersive learning. Virtual immersive spaces bring with them plenty of promise, of sensory information-rich learning experiences that will enable a much wider range of experiential learning and trainingdelivered to computer desktops, augmented reality spaces, digital installations, and mobile projective devices. This work explains how these spaces may be exploited for effective learning in terms of the technologies, pedagogical strategies, and directions.
Over the last decade, the practices by which scholarly knowledge is produced – both within and across disciplines – have been substantially influenced by the appearance of digital information resources, communication networks and technology enhanced research tools.
Title: Children and their Environments: Learning, Using and Designing Spaces
Examining theories of children's perceptions of space and place, this book explores how these theories are applied to the world of children. Its focus is on children in large real world spaces; places that children live in, explore and learn from. These include classrooms, playgrounds, homes and yards, towns, communities, countryside, natural environments, and the wider world. An international team of authors compares the experiences of children from different cultures and backgrounds by linking research on children's comprehension and daily lives to recommendations for practice. | 677.169 | 1 |
Product Description
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Accelerated Christian Education (ACE) curriculum has Scripture as its foundation, fully integrating biblical principles, wisdom, and character-building concepts into education. Students move at their own speed through the self-instructional "PACE" workbooks. Following the mastery approach, PACEs are formatted for students to complete the exercises found throughout the workbook, take a practice "self test," and conclude with a "final test" (torn out from the center) to measure understanding.
Students will develop foundational math skills needed for higher education and practical life skills with ACE's Math curriculum. This set includes Math PACEs 1133-1138 which covers:
Unknown measurements involving triangles.
Values of the trigonometric functions for any angle.
Use of trigonometric identities.
Graphing trigonometric functions with and without a graphing calculator.
Using the inverse trigonometry functions to find angle measures.
Solving equations involving the trigonometric functions.
Graphing functions in the polar plane.
Performing operations with complex numbers.
Recognizing how trigonometry models our world.
6 Paperback booklets. Grade 12. PACEs 1133-1138.
Answers are not included, but are available in the sold-separately SCORE key | 677.169 | 1 |
Plane and spherical trigonometry : with tables
She assumes some previous acquaintance with geometry and simple algebra and analytic geometry. That's the red horizontal line in the drawing. In fact, in real life when something is pushed towards a central point in proportion to its distance from that point (as in pendulum s, weights on spring s, molecule s trapped in solid s, and musical instruments - we call this ' simple harmonic motion ') it will indeed move in a sine curve, which is why trigonometry is the mathematics of oscillation s as well as triangles and circles.
By the way, you can remember one of the nice things about doing an integral is it's fairly easy to check your answer. You can always differentiate the thing you get, and see whether you get the right thing when you go back. It's not too hard to use the power rules and the differentiation rule for the cosine to get back to this if you want to check the work , cited:. In the eighteenth century, the definitions of trigonometric functions were broadened by being defined as points on a unit circle ref.: Determining Functions Using Regression - This activity allows students to look for functions within a given set of data. After analyzing the data, students should be able to determine what type of function best represents the data online. The User can fully customize their dice and use any combination of up to 10 dice. Includes , cited: Factorize fractions, simplifying expressions with rational exponents fractions, simplifying radicals expressions answers, free help on how to change a fraction into a decimal Though I find theoretical math exciting and captivating, I realize not all fourteen year olds feel the same way download. By the beginning of the seventeenth century, the science of trigonometry had become a sophisticated technique used in calculating more and more accurate tables for use in astronomy and navigation, and had been instrumental in fundamentally changing man's concept of his world. See the notes to this article to read some thoughts on the value of teachig the history of mathematics. Y and Chichester Wiley Skywatching in three ancient cultures: Megalithic Astronomy, the Maya and the Inca , e.g.
Unless Detailed solutions for all the tests help students compare their solutions with model answers provided. Highlights of important concepts in a chapter, written in easy language serve as valuable notes for use during the revision sessions before exams. These crisp revision notes along with key diagrams & illustrations help save valuable time and reinforce concepts already studied in the chapter Topics include: matrices, linear equations, determinants, characteristic polynomials, and eigenvalues; vector spaces and linear transformations; inner products; Hermitian, orthogonal, and unitary matrices; bilinear forms; elementary divisors and Jordan normal forms Choose from a pre-made algebra review game, download a blank template and create your own. A unique approach to math education that integrates popular movies, music, sports and more. Enjoy free math help with fractions, algebra, and geometry. Buy Math Training Aids include Math Foldables, Math Board and Card Games, Powerpoint Bingo Games, Math Proficiency Exam Study Guide, Math Scanvenger Hunt Games , cited: Berlin, Heidelberg, and New York: Springer-Verlag. 46. Keller, Agathe (2006), Expounding the Mathematical Seed epub ref.: Accordingly, jaib was translated into the Latin sinus, which can mean "fold" (in a garment), "bosom," "bay," or even "curve." source: Spherical coordinates, (rho, theta, pi), are three-dimensional coordinates which define a point in three-dimensional space. The radius of the sphere is rho, also known as the radial coordinate. The angle in the xy-plane (around the z-axis) is theta, also known as the azimuthal coordinate , source: Basic. 2004. 0465043550 I really haven't gotten around to this area yet. Secondly, I prefer to learn most physics from specialized sources (for example to study mechanics, how about using a book just on mechanics?) | 677.169 | 1 |
About this title:
Synopsis: This text has been written in clear and accurate language that students can read and comprehend. The author has minimized the number of explicitly state theorems and definitions, in favor of dealing with concepts in a more conversational manner. This is illustrated by over 250 worked out examples. The problems are extremely high quality and are regarded as one of the text's many strengths. This book also allows the instructor to select the level of technology desired. Trench has simplified this by using the symbols C and L. C exercises call for computation and/or graphics, and L exercises are laboratory exercises that require extensive use of technology. Several sections include informal advice on the use of technology. The instructor who prefers not to emphasize technology can ignore these exercises.
Book Description Brooks Cole 1999-10-28, 1999. Hardcover. Book Condition: Good. 1. 0534368417 Your purchase benefits those with developmental disabilities to live a better quality of life. wear on edges and cornerc over wear peeling/bubbling of cover peeling of corners sun damage stains/markings on edges may have highlighting/writing on pages. Bookseller Inventory # 1464-060617-DR-018 763906 | 677.169 | 1 |
Full Disclosure: Like most reviewers here, I purchased this book for a calculus class. I chose Amazon because the price on the book was much better than in my local bookstore.
The Good: - Most problems in the book are not terribly difficult to understand. - It appears that this book is useful for Calculus I and II (AB and BC) at most colleges/high schools where it is used. - Formatting in the physical edition is entirely acceptable, and for the most part easy to follow
The Bad: - Practice problems introducing the section are entirely useless. - Understanding the textbook is mostly teacher-dependent. - The vast majority of what a teacher might like to refer to as "challenge problems" are even-numbered. - As some reviewers have mentioned, this book focuses on proofs more than it should, essentially asking for explanations on how to solve a problem when it should be explaining the problems. - Kindle formatting is absolutely terrible. I was compelled to buy the hardcover the second I previewed the Kindle edition. - Several sections make you want to take the Google way out. - The back of the book has the occasional wrong answer.
THE BOTTOM LINE: - Mileage will vary. If your instructor is helpful and refers to problems from the book in class, you'll find this to be fairly easy. If you aren't so lucky, this book will not teach you what you need to know. I am a bit biased, belonging to the former category; having this in mind, I dropped my review to two stars after a lengthy discussion with a few of my classmates on just how useful the book is.
If you have any questions about this product or this review that I did not cover here, feel free to comment and I will respond shortly. If you found that this review was unsatisfactory and warrants an unhelpful rating, let me know why below, and I will do my best to address your concerns in my review.Read more ›
There are two things to review about this book: the book itself, and the kindle format of the book. Both are bad.
The itself is fairly useless in terms of in-chapter examples and explanations, often completely neglecting to cover common scenarios altogether. The examples are cryptic, short on explanation and extremely hard to follow unless you already know what's going on. I found the WikiUniversity book to be much more helpful in many areas. I would rate the book itself at BEST a 2.5 out of 5.
But then there's the kindle format as well. If you're on this page, you're probably thinking (as I did) to save a bundle of money by buying or renting the book for Kindle. It is NOT worth it, ESPECIALLY if you have a difficult time already with math. The whole book seems like it was scanned through an algorithm to convert print books to kindle books, but with no library of mathematical symbols, and no human input. Sometimes characters will render as actual parts of the kindle text, while most of the time they will render as small, lo-res images of the symbol or equation from the book. As there's no way to zoom in on it, it's extremely hard to read many problems. Even worse are the problems where the text DID scan into the kindle format, which makes it IMPOSSIBLE to tell whether something is a power or coefficient. This confusion led to many things the book said being flat-out wrong. INCREDIBLY frustrating when trying to learn integration. Also, the non-scanned text (at least a third of the book) is unsearchable.
Oh, and if your professor is assigning things by page number, you're screwed. The pages are not numbered, and the "page location" feature of the kindle is completely useless. what's even worse is that the text would randomly switch pages. I'd scroll through some problems, and problem 10 would be at the bottom of one page, but then if I came back to it, it'd be at the top of a different page. I'm actually getting aggravated just thinking about this whole mess.
Oh, and the table of integrals at the back of the book is just completely missing most of the pages.
TL;DR: This book is NOT worth the savings. If you want to do yourself a real favor, go get the free wikipedia calculus textbook. It's not perfect, but it's better than this tripe in almost every way.Read more ›
Pros: I pre-ordered this book for about $70, which I thought was a good deal especially for a hard cover textbook. The book seems pretty sturdy and would be good if you needed an expensive large paper weight, a heavy object to throw, or to fill up a shelf with random textbooks to delude your friends into thinking you're smart and actually read and understood the book.
Cons: The book does an awful job of explaining everything and anything, I'm pretty sure I learned more from checking the answer of a problem and working backwards than actually reading, my gosh it's horrible. If it weren't for my professor who gave us worksheets to do and showed how to do it I'm sure I wouldn't have done well in my class.
Overall I would give two stars because of the great price, but this textbook material is just awful and "I HATE IT". Of course I'm being bias and this textbook could work for others, but I found this textbook to be the worst of all textbooks I've had. If this textbook is required for your class I would just recommend renting it and sending it back for another poor soul to go through its torment.
For some reason this was the book that was used in my university classes. Calculus is a very interesting subject and this book manages to make students hate it. I always thought this book was terrible, and when I saw its ratings and reviews here on Amazon I learned that this is a common opinion. This book will frustrate you if you are trying to teach yourself calculus. It gives you problems in the exercise sections that must be solved using methods that aren't even taught in the book. The presentation is boring and left me with myriad questions unanswered. Returning to calculus outside of college I've found that without a professor to clarify things I am completely helpless with this book. If you want a good calculus book, get Calculus: An Intuitive and Physical Approach by Morris Kline. Only when I started reading Kline's excellent book did I realize just how poorly written is the above. University teachers, PLEASE stop making your students use this book and give them Kline instead. They will love calculus, and they'll actually understand it, even without your help. | 677.169 | 1 |
The courses offered by the Department of Mathematics serve several purposes; they supply the mathematical preparation necessary for students specializing in the physical, life or social sciences, in business administration, in engineering, and in education; they provide a route by which students may achieve a level of competence to do research in any of several special mathematical areas; they allow students to prepare themselves for work as mathematicians and statisticians in industry and government; and they give an opportunity to all inquisitive students to learn something about modern mathematical ideas.
Exponents and radicals, solving polynomial and other types of equations and inequalities, graphs and systems of linear equations, introduction to functions, elementary geometry. Offered Every Term.
Prerequisites: ([MAT 0993 with a minimum grade of D-Applications of mathematics to issues of current interest including patterns, paradoxes, limitations, and possibilities in voting, apportionment and division processes, using sampling methods, and developing information to support decisions. Offered Every Term.
Algebra: properties of the real number system, equations and inequalities, lines, graphs, introduction to functions, exponents, logarithms. Geometry and trigonometry: basic concepts, introduction to trigonometric functions, solving right triangles. Mathematics, mathematics education, science, and engineering majors should elect the 7-credit version of this course. If elected for 5 credits, only 2 credits apply toward degree; if elected for 7 credits, only 3 credits apply toward degree. Offered Every Term.
Prerequisites: ([MAT 0993 with a minimum grade of CNC 1800 Elementary Functions Cr. 4
Basic definition and concept of function. Definitions, properties and graphs of polynomial, rational, exponential, logarithmic, trigonometric, and inverse trigonometric functions. Only two degree credits after MAT 1500. Offered Every Term.
Prerequisites: ([MAT 1050 with a minimum grade of C- 1990 Precalculus Workshop Cr. 2
Students work cooperatively in groups to solve challenging problems related to precalculus. Learning is through discovery rather than by lecture. Offered Every Term.
MAT 2010 Calculus I Cr. 4
Calculus as the study of change. Definitions, concepts, and interpretations of the derivative and the definite and indefinite integrals; differentiation, integration, applications. No credit after former MAT 1510. Offered Every Term.
Prerequisites: ([MAT 1800 with a minimum grade of C- and MAT 1800 with a minimum grade of C-] OR [Math Permit to Reg - (L00-L03) with a test score minimum of 30000-39999 and Math Permit to Reg - (L00-L03) with a test score minimum of 31701-39999] OR [MAT Permit to Reg ACT/SAT with a test score minimum of 30000-39999 and MAT Permit to Reg ACT/SAT with a test score minimum of 31701-39999] OR [MQE Math Permit to Reg-(L0-L3) with a test score minimum of 30000-39999 and MQE Math Permit to Reg-(L0-L3) with a test score minimum of 31701-39999] OR [MAT 2010 with a minimum grade of C-] OR [MAT 2020 with a minimum grade of C-])
Continuation of MAT 3430, including logarithmic and exponential functions, first and second order ordinary differential equations, vectors, polar coordinates, Laplace transforms, Taylor series, and Fourier series. No degree credit in College of Liberal Arts and Sciences. Offered Every Term.
Development of Euclidean geometry as a mathematical system; related historical topics; introduction to other geometries; selected topics such as transformations and tessellations. No credit toward a major or minor for secondary mathematics teaching. Offered for undergraduate credit only. Offered Fall, Winter.
Topics from elementary theory of numbers which underlie middle school mathematics; historical connections; role of abstraction and proof in mathematics. No credit toward a major or minor for secondary mathematics teaching. Offered for undergraduate credit only. Offered Fall, Winter.
Prerequisites: ([MAT 5600 with a minimum grade of C-] OR [PHI 1850 with a minimum grade of C-] OR [PHI 1860 with a minimum grade of C-] OR [PHI 5050 with a minimum grade of C-] OR [MAT 5420 with a minimum grade of C-])
Prerequisites: ([MAT 2030 with a minimum grade of C- and MAT 2250 with a minimum grade of C-] OR [MAT 2030 with a minimum grade of C- and MAT 2350 with a minimum grade of C-] OR [MAT 2030 with a minimum grade of C- and MAT 2150 with a minimum grade of C-])
MAT 5710 Introduction to Stochastic Processes Cr. 3
Non-measure-theoretic introduction to the theory of stochastic processes and its applications, with emphasis on Markov processes in both discrete and continuous time, the Poisson process, and Brownian motion. Offered Biannually.
Concrete problems used to explore concepts in the theory of interest, including measurement of interest, annuities, yield rates, amortization, bonds, and stochastic approaches. Students prepare for certain professional actuarial examinations. Offered Yearly.
Time series models, moving average models, autoregressive models, non-stationary models, and more general models; point estimators, confidence intervals, and forecast in the time domain. Statistical analysis in the frequency domain; spectral density and periodogram. Offered Biannually.
Undergraduates who elect this course must be mathematics majors of honors caliber. Content will vary to satisfy needs of individual student. Offered Every Term.
Repeatable for 8 Credits
MAT 5992 Teaching Mathematics in College Cr. 1
Preparation for first semester of teaching in developmental-level mathematics course. Content presentation, test-writing, grading, classroom management, use of technology. Students are videotaped and critiqued. Offered Fall.
MAT 5993 (WI) Writing Intensive Course in Mathematics Cr. 0
Disciplinary writing assignments under the direction of a faculty member. Must be selected in conjunction with a course designated as a corequisite. See section listing in Schedule of Classes for corequisites available each term. Satisfies the University General Education Writing-Intensive Course in the Major requirement. Required for all majors. Offered Every Term.
Rings: basic definitions; properties; examples including the integers, rationals, reals, and complex numbers; ideals; homomorphisms; and divisibility. Connections to high school algebra. Students will be involved in the mathematical processes of exploration, conjecture, and proof. Only two credits after MAT 5420; no credit after MAT 5430. Offered Irregularly.
Statistics
Prerequisites: ([MAT 0993 with a minimum grade of CNC] OR [MAT 0900 with a minimum grade of CNC] OR [MAT Permit to Reg ACT/SAT with a test score minimum of 11601-19999] OR [MAT Permit to Reg ACT/SAT with a test score minimum of 01601-09999] OR [MAT Permit to Reg ACT/SAT with a test score minimum of 21601-29999] OR [MAT Permit to Reg ACT/SAT with a test score minimum of 31601-39999] OR [Math Permit to Reg - (L00-L03) with a test score minimum of 11601-19999] OR [Math Permit to Reg - (L00-L03) with a test score minimum of 21601-29999] OR [Math Permit to Reg - (L00-L03) with a test score minimum of 31601-39999] OR [MQE Math Permit to Reg-(L0-L3) with a test score minimum of 11601-19999] OR [MQE Math Permit to Reg-(L0-L3) with a test score minimum of 21601-29999] OR [MQE Math Permit to Reg-(L0-L3) with a test score minimum of 31601-39999] OR [MAT 2010 with a minimum grade of C-] OR [MAT 1050 with a minimum grade of C-] OR [MPR2/MPE is outdated with a test score minimum of 31509-39999] OR [MPR2/MPE is outdated with a test score minimum of 21509-29999] OR [MPR2/MPE is outdated with a test score minimum of 11509-19999] OR [MPR1/ACTMath is outdated with a test score minimum of 31509-39999] OR [MPR1/ACTMath is outdated with a test score minimum of 21509-29999] OR [MPR1/ACTMath is outdated with a test score minimum of 01509-09999] OR [MPR1/ACTMath is outdated with a test score minimum of 11509-19999]) | 677.169 | 1 |
Linear algebra
Linear algebra is the branch of mathematics concerning vector spaces and linear mappings between such spaces. It includes the study of lines, planes, and subspaces, but is also concerned with properties common to all vector spaces.
The set of points with coordinates that satisfy a linear equation forms a hyperplane in an n-dimensional space. The conditions under which a set of n hyperplanes intersect in a single point is an important focus of study in linear algebra. Such an investigation is initially motivated by a system of linear equations containing several unknowns. Such equations are naturally represented using the formalism of matrices and vectors.
Linear algebra is central to both pure and applied mathematics. For instance, abstract algebra arises by relaxing the axioms of a vector space, leading to a number of generalizations. Functional analysis studies the infinite-dimensional version of the theory of vector spaces. Combined with calculus, linear algebra facilitates the solution of linear systems of differential equations. effort EliteMost Extreme Elimination Challenge MXC 102 Donors vs AddictsVisit our website:
Accent Reduction: An Introduction
Are you a faculty member, instructor, or graduate student teaching at USC? Would accent reduction help you communicate more effectively with your students and c...30:28
MY FIRST GOLDEN GUN! - OVERWATCH 3v3 ELIMINATION with Josh & Simon
We play some Overwatch. Leave a like rating if you enjoy!
Overwatch Playlist:
MATHEMATICS SHORTCUT PART - I (OPTION ELIMINATION ...
MY FIRST GOLDEN GUN! - OVERWATCH 3v3 ELIMINATION central government and the Jakarta administration plan to jointly construct loop lines in an effort to eliminate railway crossings along the tracks of commuter trains in the capital city ... "Currently, there are too many railway crossings ... Read also. Train hits Transjakarta bus, car at rail crossing ... (bbn). Topics .. .... | 677.169 | 1 |
Navigating through Reasoning and Proof
in Grades 9-12
Expanding from 'reasoning to obtain solutions' to 'reasoning to justify ideas and validate results' is a significant development for the student of mathematics. Activities highlight the cycle of exploration, conjecture and justification. Supported by blackline masters, the activities build on and extend problem solving and reasoning to include proof; through Number, Algebra, Geometry, Data Analysis and Probability and Measurement. The book is accompanied by a CD-ROM which includes further reading for teachers, files for the blackline masters, and computer applications for students to use in their investigations. | 677.169 | 1 |
Computerized Mathematics
Illustrations of important ideas
In the graph above, we see a section of a "parabola" (y = x2); the "derivative" y' = 2x means that the slope of the curve, at any point on the x-axis, is twice the x value at that point -- thus, the green line above, at x = .5, has a slope of 1 (45o).
This "differentiation" can be used in "curve analysis", as shown below, where the red point represents the "peak", and the green point is the "inflection point" (at which the curvature "changes direction"):
y = xe-x
Many "graphing calculators" do this "numerically" (which can take time); or, a language like Mathematica can do it automatically, with code like this:
"Integrating" y = x2, between the limits of x = .2 and .8,
yields the result .168, meaning that the shaded area between the x-axis and the curve, between x = .2 and .8, is .168 units of the graph's area.
"Differentiating" either y = x2, or y = x2 + 3,
yields "2x", because the slope of parabolas, when they are raised or lowered in the y direction, is still the same at all x values; so, integrating y' = 2x
yields "x2", but, more completely, the answer is "x2 + c", with "c" being the "constant of integration" -- thus, y = x2 + c is the solution of the "differential equation" y' = 2x, so that this one equation, instead of describing only one curve, describes a field of curves, as shown below.
This is only a very rudimentary illustration of a huge subject: this may involve, for example, a 3-dimensional description of interacting magnetic fields in a motor, with a fourth dimension, in the form of time, allowing a computer to make a moving simulation of what happens as the motor runs; or, it might show stress fields in a bridge, varying in time as a heavy truck crosses it; or, the explosion of a "supernova" star. | 677.169 | 1 |
What is Math 8? Math 8 is a rigorous 8th grade level study of mathematics. It is designed to help students build a strong foundation for further study of algebra and geometry in high school. Students will study linear equations and functions, slope, the real number system, Pythagorean theorem, preservation of size and shape, congruence of triangles, similarity, lines and angle relationships, systems of linear equations and functions, properties of exponents, volumes and 3D figures, and data in two variables.
Extra Credit Assignments on TenMarks Due Monday May 16th
Five Extra Credit assignments on TenMarks- Effects of Dilation,Effects of Rotation,Effects of Translation,Effects of Reflection,Effects of all types of Transformations. Each assignments completed to 70% or higher is worth 2 points of extra credit.
Extra Credit Assignments on TenMarks due Monday May 2.
Extra Credit for Week 8
There are five assignments posted on TenMarks for extra credit this weekend. These topics are new content and you will need to watch the videos and use the hints to understand and complete these tasks. Each assignment completed with a score of 70% or higher will earn 2 points extra credit.
Extra Credit for Week 7
I posted 5 assignments on TenMarks for extra credit. Each assignments completed to 70% or higher by Monday February 22 will earn 2 points extra credit
Extra Credit For Week 6
There are 10 assignments posted on TenMarks for extra credit this weekend. Each assignment completed with a score of 70% or higher is worth 2 points extra credit. Do as many or as few as you want. Have a great weekend!
Extra Credit Week 4 Second Semester
I have posted 3 extra credit assignments on TenMarks. You must complete all 3 assignments with a score of 70% or higher by Monday February 1 to receive extra credit.
Extra Credit Week 3 Second Semester
I have posted 3 assignments on Tenmarks for extra credit. You must complete all 3 assignments to a score of 70% or higher by Monday Jan. 25 to receive credit.
Extra Credit for Week 1 2nd Semester
There are 3 assignments on TenMarks for extra credit. All 3 must be complete by Tuesday 1/19/16 for extra credit.
Growth Mindset
Watch this video to learn the basic ideas of the growth mindset.
Extra Credit for Week 18
Three extra credit assignments are posted on Ten Marks. All three assignments must be completed to a score of 70% or higher by Monday December 14 to receive credit.
Extra Credit for Week 17
Three extra credit assignments are posted on Ten Marks. Students must complete all 3 assignments to a score higher than 70% by Monday Dec. 7 in order to receive extra credit points.
Extra Credit Week 15
I have six extra credit assignments posted in Ten Marks for over the break. You must complete all six by Sunday Nov. 29 to receive credit.
Extra credit for Week 14
3 extra credit assignments on 10 marks for this weekend. Each assignment must be completed by Monday Nov. 16 to a score higher than 70% to receive credit. | 677.169 | 1 |
Problem 1
Correctly find force on
a cross-section with
justification (for the
hemisphere)
Correctly find force on
a cross-section with
justification (for the
cylinder)
Correctly find distance
traveled by water at a
cro
MATH 112
WORKSHEET 4 (6.7)
NAME:
SECTION:
In the problems below, make sure to show all work to get full credit. In particular, you should clearly communicate your solution by providing arguments and statements that are complete, logically ordered, and use
Calc For Sci & Eng II Ma1 Advice
Showing 1 to 2 of 2
Christian was a great professor who actually helped me make sense of Calculus. Calc 2 is a difficult course, but with the right professor it can make sense.
Course highlights:
I learned how to be speedy with my calculus, how to do problems that apply to the real world, and, best of all, how to approach a difficult Calculus course.
Hours per week:
6-8 hours
Advice for students:
Practice your problems. Before tests, run through each type of problem that might be on the test. Don't skimp out on your review sessions!
Course Term:Spring 2017
Professor:ChristianWesthof
Course Required?Yes
Course Tags:Math-heavyBackground Knowledge Expected
Feb 24, 2016
| Would recommend.
Not too easy. Not too difficult.
Course Overview:
This course is very necessary for all majors here at Mines, as it is a core math class.
Course highlights:
I was able to learn many different ways to solve many different problems.
Hours per week:
3-5 hours
Advice for students:
I would advise to do all of the homework assigned, and if you are falling behind to always go into office hours. The professors are extremely helpful and are always willing to be flexible around your schedule. | 677.169 | 1 |
Core Maths
Core Maths
Level 3 Certificate
Core Maths is a new course for those who want to develop their valuable maths skills, but do not want to study the same level of algebra that is in A Level Mathematics. It is different to GCSE Maths because it is often in the context of real-life situations and requires you consider what approach you are going to take to solve the problems.
This is a one-year course which will enable students to achieve a Level 3 qualification in mathematics worth up to 20 UCAS points in the new 2017 tariff (equivalent to an AS Level).
Entry Requirements
You need to have a Grade 4 or above in GCSE Mathematics. Please note:
-Core Maths cannot be taken alongside A Level Mathematics.
-Core Maths is a one-year course, so there will be no opportunity to continue this subject into Year 13. | 677.169 | 1 |
Competition and efficient pro forma Mathematics Learning and
I used to have an advisor who once explained that its contributions as a researcher involved solving problems that appeared in engineering & physics literatures but that the authors didn't had the tools and/or time and/or interest to work them out. Dashboard and detailed reports show each student's progress and comprehension level on curriculum units based on the Common Core and other standards. This type engagement is a prerequisite to learning skills with meaning and being able to apply those skills to solve problems.
Pages: 0
Publisher: China Press (January 1, 2000)
ISBN: 7560269370
Problem-Solving Research in Competitive Mathematics(Chinese Edition)
Advances in Mathematical Research
Come to Boise State and learn Mathematics from scholarly Mathematicians, Statisticians, and Mathematics Educators who care about teaching. If you have any questions about the Department of Mathematics, please send email to office@math.boisestate.edu. If you have any comments or questions about this website, please send email to webmaster@math.boisestate.edu download. Teachers can easily scan through the skills and concepts presented in prior years to identify background knowledge necessary for student success. The concrete–pictorial–abstract approach is well-suited for English language learners since it is less text-heavy and has more illustrations and visuals. The Teacher's Edition provides lesson-specific suggestions for facilitating instruction for English language learners , source: Current and Future Directions read online Our theoretical framework Vygotskian notion of semiotic mediation. The article describes in detail the actions, gestures, graph tures served as means of semiotic mediation. Specifically, they supported the interpretation and the ematical signs and culturally accepted mathematical meaning; and they enable linking bodily actions movement and semiotics Mathematical Results In download here But for others, the form is seen as a resource enabling such a move. These contradictory responses are warranted in competing but complementary notions, grounded on the corpus of teacher responses, that teachers hold about the nature of authentic mathematical activity when proving online. A typology of responsesAuthors: Matthew Inglis; Juan Pablo Mejia-Ramos • [4]To be or to become: how dynamic geometry changes discourseAuthors: Nathalie Sinclair; Violeta Yurita • [5]A diagrammatic view of the equals sign: arithmetical equivalence as a means, not an endAuthor: Ian • [6]Paradoxes as a window to infinityAuthors: Ami Mamolo; Rina Zazkis • [8]The mathematical competence of adults returning to learning on a university foundation programme: a selective comparison of performance with the CSMS studyAuthor: Mary Dodd[9] • [10]Mathematics and dyslexics: classroommanagement skills and children's response to noiseAuthor: Mari • [11]A synthesis of existing frameworks used to analyse mathematics curriculaAuthor: Nusrat Fatima Rizvi • [12]Beginning elementary teachers' use of representations in mathematics teachingAuthor: Fay Turner • [13]Observing students' use of images through their gestures and gazesAuthor: Tracy Wylie of primary identity in a mathematics classroom, which was recently published in [2]ZDM download.
To contrast with the Wittgensteinian position, a Platonist position is presented and the two [1]ZDM recently published another (online first) article, called: [2]Comparing theoretical frameworks enacted in experimental research: TELMA experience. The article is written developed by six European research teams. The methodology is: A new (online first) article has been published by [1]International Journal of Science and Mathematics Education, called [2]Number Sense Strategies Used by Pre-Service Teachers in Taiwan , cited: Primary Mathematics Teaching read epub Primary Mathematics Teaching and. Currently, areas of focus include fluid dynamics, plasma physics, statistical mechanics, molecular dynamics and dynamical systems. The tradition at the Institute is to investigate fundamental questions as well as to solve problems with direct, real-world applications ref.: Create Your Own Mathematics read here read here. The Supporting Standards encompass 60-70% of the eligible TEKS and will comprise 35-40% of the STAAR® assessment. The Product Development Team for Motivation Math developed this resource to help students in Levels 3-8 achieve mastery of identified Readiness and Supporting Standards epub. Founded in 1989 as part of the SPIC Science Foundation, it has been an autonomous institute since 1996. The research groups in Mathematics and Computer Science at CMI are among the best known in the country. The Institute has nurtured an impressive collection of PhD students. In 1998, CMI took the initiative to bridge the gap between teaching and research in India by starting BSc and MSc programmes in Mathematics and allied subjects Test research (Series 7) read pdf arsenaultdesign.com. If you feel that it is important to present a proof, remember that you need to keep things easy to understand. Rather than going through every step, discuss the main points and the conclusion. If you like, you can write out the entire proof and include it in a handout so that folks who are interested in the details can look at them later Computational Mechanics: Proceedings of the 2007 International Symposium on Computational Mechanics in Beijing
Mathematical Proficiency for All Students: Toward a Strategic Research and Development Program in Mathematics Education
Twenty-Four Vincent van Gogh's Paintings (Collection) for Kids
Fibonacci's Liber Abaci: A Translation into Modern English of Leonardo Pisano's Book of Calculation (Sources and Studies in the History of Mathematics and Physical Sciences)
Error in Digital Computation Vol 2
The Spectrum of a Module Category (Memoirs of the American Mathematical Society)
Lectures on plasma effects of charged particles in space (Tata Institute of Fundamental Research lectures on mathematics and physics : Physics ; 48)
Mathematical Aspects of Finite Elements in Partial Differential Equations (Publication of the Mathematics Research Center, University of Wisconsin--Madison ; no. 33)
Bibliography and research manual of the history of mathematics
To help difficult students learn proper behavior it is sometimes necessary to develop contracts. These contracts, which are designed to be used by administrators, place student behavior in a closely watched, productive environment This is not available 029968 read pdf projectbaseline.org. Scientific computing with primary emphasis to date on numerical modeling for geoscience problems (seismic imaging and wave propagation, reservoir simulation, and mechanical deformation modeling) The Motive Force of Time and read for free The Motive Force of Time and its. The Geometry Garden features curiosities found in nature from crystals to seashells to sculptures Pacific Journal of Applied read epub projectbaseline.org Twenty-Four Henri Matisse's Paintings (Collection) for Kids The group is augmented by faculty in the Computer Science Department, including Endre Szemeredi (Theoretical computer science, graph theory, combinatorial number theory and geometry), and Mario Szegedy (Quantum computing, computational complexity and combinatorics), and by faculty in RUTCOR download. Locally free groups are in a way the opposite of quotient divisible groups. In particular, rings are never locally free, except for a few extremely simple and uninteresting cases. So this entryway was not available in my study of locally free groups. In fact, I had a hard time thinking of any examples of locally free groups on which a multiplication was naturally defined pdf. And so I needed to see what goes wrong if one applies this recipe to an almost completely decomposable group. What one sees in this proof is that completely decomposable groups exist in layers, corresponding to types (which more or less correspond to rank-one groups) On the waiting time in the download for free xn--traverserlanage-fjb.com. The price you pay for your order depends on several factors and is defined individually for every order. It means you can manipulate the price by adjusting some pricing parameters. Speaking of the price, our professional writing services don't cost much! If you have ever used essay writing services, you might be under the impression that it costs a lot The Nonlinear Limit-Point/Limit-Circle Problem Research articles also include innovative and mathematical theories of inventory, manufacturing, and distribution; organization, finance, and marketing; routing, queuing, and scheduling; data and storage management; location, reliability, search, measurement, and service; and artificial intelligence Symbolic Dynamics and Its Applications: American Mathematical Society, Short Course, January 4-5, 2002, San Diego, California (Proceedings of Symposia in Applied Mathematics) S. and lodging expenses are paid for, and every participant receives a $3,000 stipend (Brown students typically receive an UTRA grant in lieu of the S@I stipend) , cited: Free Boundary Problems: v. 3: Theory and Applications (Chapman & Hall/CRC Research Notes in Mathematics Series) read for free. Singapore Math® is a trademark owned by Singapore Math Inc. and Marshall Cavendish Education Pts ref.: Theory of Named Sets (Mathematics Research Developments) Writing essay here is always affordable due to our elaborated pricing policy. The price you pay for your order depends on several factors and is defined individually for every order. It means you can manipulate the price by adjusting some pricing parameters. Speaking of the price, our professional writing services don't cost much ref.: Turing knowledge How to Solve: research Collection Mathematics Competition (2) download online! Teachers and other educational leaders should consistently help students and parents to understand that an increased emphasis on the importance of effort is related to improved mathematics performance (The National Mathematics Advisory Panel, 2008). Research indicates that the more parents are excited and involved in the learning of their children, the more successful a child can be academically Handbook of Solitons: download online | 677.169 | 1 |
Elementary and Intermediate Algebra
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Read More analogy that "the price of a hamburger plus a Coke is the same as a Coke plus a hamburger". Given the importance of examples within a math book, the author has paid close attention to the most important details for solving the given topic."Dugopolski" includes a double cross-referencing system between the examples and exercise sets, so no matter which one the students start with, they will see the connection to the other. Finally, the author finds it important to not only provide quality, but also a good quantity of exercises and applications. "The Dugopolski" series is known for providing students and faculty with the most quantity and quality of exercises as compared to any other developmental math series on the market. In completing this revision, Dugopolski feels he has developed the clearest and most concise developmental math series on the market, and he has done so without comprising the essential information every student needs to become successful in future mathematics courses. The book is accompanied by numerous useful supplements, including McGraw-Hill's online homework management system, MathZone 0073384356 | 677.169 | 1 |
What are some tips for studying college algebra?
A:
Quick Answer
Tips for studying college algebra include spending significant time learning the basics of college algebra such as inequalities and exponents and mastering simple algebraic concepts before moving on to more complex topics. Additional tips for studying college algebra are to appreciate the sequential nature of a college algebra course and to spend time reviewing homework in order to reinforce missed concepts.
Keep Learning
When studying college algebra, it is important to understand that two sides of an equation are not always equal. Sometimes, one number or phrase is greater than another number or phrase. In college algebra, a mathematician expresses this inequality with the appropriate mathematical symbol.
Another tip for studying college algebra is understanding exponents. An exponent is a small number written above and to the right of a regular number. This is a mathematical function, and it means to multiply the large number by itself, the number of times indicated by the little number. Understanding this basic function helps a student study college algebra because the student can comfortably work with fractional exponents or the exponents of negative numbers once the student masters the concept of exponential numbers.
When studying college algebra, it is important to remember that the subject is sequential. Unlike other subjects, what a person learns on any particular day in college algebra builds on what she learned earlier in the course. It is very important to attend class each day and master simple concepts before moving on to complex ones. Keeping up with homework is also critical to college algebra mastery, because a student can't succeed in a sequential course without attending class prepared to undertake new concepts. In addition, reviewing returned homework is critical, because reviewing missed problems reinforces concepts that appear on exams. | 677.169 | 1 |
Calculus I
Grading
Each student is expected to master the material in a variety of ways including: multi-step problems, applications, and conceptual understanding. If any concerns arise regarding grading, please contact the instructor. Students earn "math points" (MP) for demonstration of mathematical thinking in their solutions.
Please enable javascript to compute the grading scale.
Warm-up activities
Warm-up activities are sometimes assigned to prepare students for in-class activities. These activities are more effective when everyone attends class fully prepared. "Eyeglasses" in the calendar will indicate warm-up activities.
Online practice (+10 MP bonus)
WeBWorK problem sets are used to practice skills from class. Feedback is limited to correct/incorrect so if problems are difficult please see the instructor for better feedback.
Each student earns bonus points based on their percent of correct answers. For example, say a given student completes 70% of the problems. This student receives 70% of 10, or 7 bonus points.
If you feel that WeBWorK has not given credit for a correct answer, please contact the instructor immediately.
Group quizzes (10 × 10 MP)
During most weeks a group quiz will be given. Only one quiz per group will be scored with each group member receiving the same score. It is expected that all group members contribute to quiz responses and that all answers are fully explained. Notes are not allowed for quizzes, but approved calculators are allowed.
Exams (3 × 25 MP)
There will be three in-class exams. A note card (5×7, handwritten, both sides) is allowed for each exam as are approved calculators.
Assessments are a part of the learning experience. As such, assessments will not only require mastery of class material, but the ability to apply class material in new situations (Rittle-Johnson, Seigler, Alibali, 2001).
Final exam (25 MP, +5 MP bonus)
This course uses a comprehensive, common final with the other sections. The final exam is typically 25 multiple choice questions. Students gain +1 MP for every 5 correct answers.
Remember that university policy does not allow students to take an examination prior to its scheduled time, so plan accordingly | 677.169 | 1 |
Product Description
The Calculus Workbook NCEA Level 3 offers students the tools they need for success in Calculus.In the first part there are exercises at Achievement, Merit and Excellence levels. The author walks the student through each new concept, provides several worked examples and then a set of problems for students to solve. The fully worked answers follow immediately, encouraging students to carefully compare their work with the worked examples and worked answers, and improve their results. With a 'practice makes perfect' attitude, the Calculus Workbook NCEA Level 3 also contains six full-length practice exams, written for the 2010 assessment specifications, with full answers, as well as Unit Standards practice. For those students who have mastered their Level 3 work, there follows a scholarship training programme. Students can work their own way through this material with minimum supervision from the teacher. | 677.169 | 1 |
28 Apr 2017
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This video shows how calculus is both interesting and useful. Its history, practical uses, place in mathematics and wide use are all covered. If you are wondering why you might want to learn calculus, start here!AP Calculus
Advanced Placement Calculus (also known as AP Calculus, AP Calc AB / AP Calc BC, or AP Calc) is used to indicate one of two distinct Advanced Placement courses and examinations offered by College Board in calculus: AP Calculus AB and AP Calculus BC.
AP Calculus AB
AP Calculus AB is an Advanced Placement calculus course taken by high school students. The course is traditionally taken after precalculus and is the first calculus course offered at most schools except for the regular calculus class. The Pre-Advanced Placement pathway for math will help prepare students for further Advanced Placement classes and exams.
Purpose
The AP Program includes specifications for two calculus courses and the exam for each course. The two courses and the two corresponding exams are designated as Calculus AB and Calculus BC. Calculus AB can be offered as an AP course by any school that can organize a curriculum for students with advanced mathematical ability. Calculus? (Mathematics)Calculus - Introduction to CalculusCalculus: What Is It? p...
published: 28 Apr 2017 studyW...
published: 29 Sep 2016 19 Nov 2015 geom... Calculus? (Mathematics)
What is calculus? In this video, we give you a quick overview of calculus and introduce the limit, derivative and integral.
We begin with the question "Who inv...Calculus - Introduction to Calculus
This video will give you a brief introduction to calculus. It does this by explaining that calculus is the mathematics of change. A couple of examples are pre...Calculus: What Is It?
This video shows how calculus is both interesting and useful. Its history, practical uses, place in mathematics and wide use are all covered. If you are wonderi...
This video shows how calculus is both interesting and useful. Its history, practical uses, place in mathematics and wide use are all covered. If you are wondering why you might want to learn calculus, start here!... calculus? In this video, we give you a quick overview of calculus and introduce th...This video will give you a brief introduction to calculus. It does this by explaining tha...This video shows how calculus is both interesting and useful. Its history, practical uses,...0:18
Calculus in 20 Seconds
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Burden Of Grief
The war is over The last battles are gone Swords laying broken My bloodwork is all done I sit down for calming My breath is lessening I�m starting to tremble My sight is clearing My head is weary A dreadful awakening What has driven me Into insanity Awaking from this dreadful tragedy I return to myself Beginning to dwell in this elegy Put my anger on the shelf Awaking from this dreadful tragedy I return to myself Beginning to dwell in this elegy Put my anger back on the shelf I look around As I raise from my rest Discover what I�ve done No life I have left My heart is in pieces My soul is laying bare Awaking from this dreadful tragedy I return to myself Beginning to dwell in this elegy Put my anger on the shelf Awaking from this dreadful tragedy I return to myself Beginning to dwell in this elegy weekend in Charlottesville, a driver mowed down peaceful protesters and killed 32-year-old Heather HeyerCatherine Rampell Washington Post... But he also didn't need to turn to either of these factions for inspiration CoachJay Gruden's commands ... "We're trying to win here! We're trying to win!" ... Throughout the NFL, it's a calculus that all coaches are working out ... Among them ... ... ....
The discovery of Thomas Parran's papers confirms that he was much more involved in Tuskegee than anyone ever knew ... It was while Gregory J ... He would go on to describe scientific research on humans as "an immoral business." How else, de Kruif would argue, does one explain Reed's belief that to successfully fight tropical disease one had "to risk human lives," and accept the cruel calculus that "you must kill men to save them." ... Dr ... Dr ... Dr ... ....
Over the last several decades, the proportion of Americans who get married has greatly diminished—a development known as well to those who lament marriage's decline as those who take issue with it as an institution ... The divide in the timing of childbirth is even starker ... * * * ... They may also make the relationships in that household less transactional—that is, less dominated by a calculus that tallies what one partner does for another ... ....
INDIANAPOLIS (AP) — A study of Indiana schools found that size may have an impact on academic performance. More than half of Indiana districts can't operate efficiently because they're too small, Ball State University's Center for Business and Economic Research said ... The study also found that smaller districts are less likely to offer calculus, physics and Advanced Placement courses ... ....
But seven months into the Trump administration, we can see that this comparison was unfair. For one thing, Caligula did not, as far as we know, foment ethnic violence within the empire ... It found a way to get rid of him ... system ... If so, we have to hope that our country somehow stumbles through the next year and a half without catastrophe and that the midterm elections transform the political calculus and make the Constitution great again The plain truth is that India's post-Cold War foreign policy calculus will be severely put to test for the first time if a conflict with China ensues....
(CNN)In the 1976TV movie "The Boy in the PlasticBubble," the main character, played by John Travolta, is a kid who lacks an immune system to protect him from infection. He suffers from the isolation of life inside a sealed environment ... It was one of private schools and great wealth, and a lavish gilded lifestyle with armed guards who kept the world at bay ... In the Trump calculus, this close circle only makes sense. As Donald Trump Jr ... ....
In fact, Cruz graduated from Agricultural Engineering at Araneta University in 1982... That was not that, though ... "It was never easy for me because imagine—I had just graduated from college but I had to take lessons again to study science-based subjects and higher math subjects like Integral and DifferentialCalculus? And while I trained, I took on sideline jobs because like I said, my family needed me to work." ... On to Australia ... Rivalry ... ....
Limits aims to make learning maths, specifically calculus, fun ... To solve puzzles, you need calculus ... In order to connect the beam of energy with the thing you need to unlock or power, you must solve calculus problems ... If you're as unfamiliar with calculus as I am, don't worry ... But by the time I reached the third level, I realised that I wasn't going to progress without a basic understanding of calculus principles....
So the odds are that we're stuck with a malevolent, incompetent president whom nobody knowledgeable respects, and many consider illegitimate.If so, we have to hope that our country somehow stumbles through the next year and a half without catastrophe, and that the midterm elections transform the political calculus......
https.// But seven months into the Trump administration, we can see that this comparison was unfair ... It found a way to get rid of him. -snip- ... -snip- ... If so, we have to hope that our country somehow stumbles through the next year and a half without catastrophe, and that the midterm elections transform the political calculus and make the Constitution great again ... ....
Last year, a young Chinese citizen and I were trying to connect for a conversation. Skype seemed to be the most convenient medium, although once our chat commenced, she started apologising. That was not just for the quality of the connection, which kept dropping as during a cellphone conversation in Delhi...Freedom of expression, in their corporate calculus, is an optional extra, one, that at times is unaffordable....
When President Trump assigned blame to "both sides" for the deadly violence in Charlottesville in remarks this week, he was met with condemnation from both sides of the aisle—and then from the other side of the Atlantic... But the nature of Trump's remarks—and his seeming willingness to defend white nationalist and neo-Nazi protesters as including "some very fine people"—seemed to change the traditional calculus ... .... | 677.169 | 1 |
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Eliminate the frustration and fear of multiple step problem solving in your students by providing them with Core Rigor's Mathematical Procedure Manuals.This product is designed to be used as a procedure manual in the application of algebraic thinking skills, which in turn will prepare students for higher education, employment upon high school graduation, and problem solving in life. It may also be printed as individual handouts for math notebooks or enlarged to be classroom anchor charts. It includes anchor charts on order of operations, square roots, powers and exponents, and coordinate graphing. Of course each skill will have to be introduced and taught in the beginning. The manual has room for additional note taking, and is conducive to the use of highlighters and colored writing utensils. Students will use the manual through out the school year in multiple step problem solving. Designed with the busy teacher in mind with printing being the only preparation required. | 677.169 | 1 |
Do I have to study things in the order shown on the Study Path page?
No. You may study this material in whatever order you like. However, the order shown is recommended because math concepts tend to build on each other, so it may be harder for you to master concepts if you skip over earlier material. | 677.169 | 1 |
Geometry is probably the most accessible branch of mathematics, and can provide an easy route to understanding more complex mathematical ideas. This book introduces readers to the major geometrical topics taught at the undergraduate level, in a manner that is both accessible and rigorous. The author uses world measurement as a synonym for geometry -- hence the importance of numbers, coordinates and their manipulation -- and has included over 300 exercises, with answers to most of them. Bookseller Inventory #
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Book Description Springer, 2003. Book Condition: Very Good. *Price HAS BEEN temporarily REDUCED by 10% until Monday, Aug. 21. Order now for best savings* this is the third printing (2003), 313 pp., Paperback, previous owner's name to front free endpaper else very good. Bookseller Inventory # ZB1057938
Book Description Springer32609
Book Description Springer London LtdInt, 2000. Paperback. Book Condition: NEW. 9781852330583 This listing is a new book, a title currently in-print which we order directly and immediately from the publisher. Bookseller Inventory # HTANDREE0311433
Book Description Springer London Ltd, United Kingdom, 2007. Paperback. Book Condition: New. 1st ed. 2001. Uncorr. 4th printing 2007. Language: English . This book usually ship within 10-15 business days and we will endeavor to dispatch orders quicker than this where possible. Brand New Book. LIE9781852330583 | 677.169 | 1 |
6
A lab manual! You can get this one at Johnny Print, which looks like this: Actually, I'm not certain these are ready yet, but when they are, Johnny Print is on S Limestone
7
The tests, quizzes, and homeworks for this class assume you have access to a calculator The labs require a TI-83 or TI-84 series calculator (eg. TI-84 plus, TI-83 plus pink shiny edition, etc)
8
Register your clicker: This is possible at Steve Ellis (your lecturer!)'s website: (Hint: Now would be a good time to start writing things down if you haven't been.) Read through the syllabi for this course (including lecture/recitation and lab— two separate syllabi!) You can find them at Steve's site or on Blackboard.
9
We'll be using an online homework system for homework in this course This is located at (or.com, either works) You need the following CLASS KEY to register: uky 6402 7919
10
Physics is not like most classes. Memorization is not that important Formula sheets Physical constants Problem solving is very important Learn how to solve problems—not the solution to individual problems Figuring out what to do is often more difficult than doing it—unless you struggle with the mathematics
11
The more math you can use, the better off you'll be Algebra Solid foundation absolutely required If you can't readily: Solve two equations with two unknowns Solve quadratic equations Deal with exponents, cancel factors appropriately, work with negative numbers then you are behind already. Use the next few days to brush up on math. Trigonometry. You can (re)learn as you go, but you'll need to know it by the end
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Geometry We'll be working with physical objects—or at least idealized approximations of physical objects These take the form of shapes—knowing about shapes is good. Know your Euclidean Geometry
13
If you feel confident that, at a moment when you least expect it, somebody could hand you a pen and a math exam (say, college algebra, trig, or geometry) and you could pull a decent grade on it without too much trouble? Math probably won't be your Achilles Heel in this course.
14
Like I said before—figuring out what to do is often more difficult in this class than doing it. If you like solving puzzles, you'll probably like physics Seeing a problem worked out shows you the sort of math you'll need to be capable of—but it doesn't show you the logical leaps necessary to get to the math!
15
You can read and memorize every problem on the homework and in the solutions manual and still fail If you do not put pen(cil) to paper, you won't get the problem solving experience you need. It's not about the final answer, it's about the journey to get there.
19
The following exercise is for completion only—don't stress if you don't get it just yet. Vectors are one of the mathematical tools we don't expect every student to have experience with—the first thing you'll learn about in this physics course is the math with vectors! We just want to see where you stand right now with vectors. | 677.169 | 1 |
Grade 8 » Introduction
In Grade 8, instructional time should focus on three critical areas: (1) formulating and reasoning about expressions and equations, including modeling an association in bivariate data with a linear equation, and solving linear equations and systems of linear equations; (2) grasping the concept of a function and using functions to describe quantitative relationships; (3) analyzing two- and three-dimensional space and figures using distance, angle, similarity, and congruence, and understanding and applying the Pythagorean Theorem.
Students use linear equations and systems of linear equations to represent, analyze, and solve a variety of problems. Students recognize equations for proportions (y/x = m or y = mx) as special linear equations (y = mx + b), understanding that the constant of proportionality (m) is the slope, and the graphs are lines through the origin. They understand that the slope (m) of a line is a constant rate of change, so that if the input or x-coordinate changes by an amount A, the output or y-coordinate changes by the amount m·A. Students also use a linear equation to describe the association between two quantities in bivariate data (such as arm span vs. height for students in a classroom). At this grade, fitting the model, and assessing its fit to the data are done informally. Interpreting the model in the context of the data requires students to express a relationship between the two quantities in question and to interpret components of the relationship (such as slope and y-intercept) in terms of the situation.
Students strategically choose and efficiently implement procedures to solve linear equations in one variable, understanding that when they use the properties of equality and the concept of logical equivalence, they maintain the solutions of the original equation. Students solve systems of two linear equations in two variables and relate the systems to pairs of lines in the plane; these intersect, are parallel, or are the same line. Students use linear equations, systems of linear equations, linear functions, and their understanding of slope of a line to analyze situations and solve problems.
Students grasp the concept of a function as a rule that assigns to each input exactly one output. They understand that functions describe situations where one quantity determines another. They can translate among representations and partial representations of functions (noting that tabular and graphical representations may be partial representations), and they describe how aspects of the function are reflected in the different representations.
Students use ideas about distance and angles, how they behave under translations, rotations, reflections, and dilations, and ideas about congruence and similarity to describe and analyze two-dimensional figures and to solve problems. Students show that the sum of the angles in a triangle is the angle formed by a straight line, and that various configurations of lines give rise to similar triangles because of the angles created when a transversal cuts parallel lines. Students understand the statement of the Pythagorean Theorem and its converse, and can explain why the Pythagorean Theorem holds, for example, by decomposing a square in two different ways. They apply the Pythagorean Theorem to find distances between points on the coordinate plane, to find lengths, and to analyze polygons. Students complete their work on volume by solving problems involving cones, cylinders, and spheres.
Grade 8 Overview
The Number System
Know that there are numbers that are not rational, and approximate them by rational numbers.
Expressions and Equations
Work with radicals and integer exponents.
Understand the connections between proportional relationships, lines, and linear equations. | 677.169 | 1 |
Building on what you have learned about limits and continuity, investigate derivatives, which are the foundation of differential calculus. Develop the formula for defining a derivative, and survey the history of the concept and its different forms of notation. | 677.169 | 1 |
Mathematics
Mathematics is a successful and popular choice in the Sixth Form. It is a challenging subject which offers a
great deal of enjoyment and satisfaction, whilst offering useful support for a range of other A Levels. All further courses and careers welcome Mathematics A Level in combination with other subjects. Employers actively look for A Level Mathematics as it is such a desirable qualification. It is essential to have Mathematics A Level when progressing on to study many of the sciences, Engineering, Computing and Mathematics at university.
Course content
The Mathematics specifications are still under review for first teaching in September 2017.
Whatever board is chosen, students will study pure, statistics and mechanics mathematics. There are two papers that are taken after the end of Year 1 covering the pure, statistics and mechanics topics. At the end of Year 2, students will take 3 exams, testing all of the topics in more depth. | 677.169 | 1 |
Mathematics (Secondary Level) Natural, rational and irrational numbers, real numbers and their decimal expansions, operations on real numbers, laws of exponents for real numbers, rational numbers and their decimal expansions. Zeroes of a polynomial. Relationship between zeroes and coefficients of a polynomial. Division algorithm for polynomials. Algebraic methods of solution of pair of linear equations in two variables.
Mensuration : Surface area of a cuboid and a cube, right circular cylinder, right circular cone, sphere. Volume of a cuboid, cylinder, right circular cone and sphere, Surface area and volume of a combination of solids conversion of solid from shape to another. | 677.169 | 1 |
3051895
ISBN: 0073051896
Edition: 6
Publication Date: 2005
Publisher: McGraw-Hill College
AUTHOR
Burton, David M.
SUMMARY
David Burton covers the history behind the topics typically covered in an undergraduate maths curriculum or in elementary or high schools. He illuminates the people, stories, and social context behind mathematics' greatest historical advances, while maintaining appropriate focus on the mathematical concepts themselves.Burton, David M. is the author of 'History of Mathematics An Introduction', published 2005 under ISBN 9780073051895 and ISBN 00730518 | 677.169 | 1 |
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Essential MATLAB for Engineers and Scientists, Third Edition, is an essential guide to MATLAB as a problem-solving tool. It presents MATLAB both as a mathematical tool and a programming language, giving a concise and easy-to-master introduction to its potential and power. Stressing the importance of a structured approach to problem solving, the text provides a step-by-step method for program design and algorithm development. It includes numerous simple exercises for hands-on learning, a chapter on algorithm development and program design, and a concise introduction to useful topics for solving problems in later engineering and science courses: vectors as arrays, arrays of characters, GUIs, advanced graphics, and simulation and numerical methods. The text is ideal for undergraduates in engineering and science taking a course on Matlab.Numerous simple exercises give hands-on learningA chapter on algorithm development and program design Common errors and pitfalls highlightedConcise introduction to useful topics for solving problems in later engineering and science courses: vectors as arrays, arrays of characters, GUIs, advanced graphics, simulation and numerical methodsA new chapter on dynamical systems shows how a structured approach is used to solve more complex problems.Text and graphics in four colour | 677.169 | 1 |
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Description
In the first sentence of the prologue of the book "Portfolio Management Formulas" history of the book, I wrote that it is devoted to the mathematical tools. This book - about the mechanisms.
We take the tools and build a more advanced, more powerful tools - mechanisms, where the whole is greater than the sum of its parts, and try to understand the structure of these mechanisms, otherwise they would have been just a "black box". At the same time, we will refrain from a detailed review of all somehow related topics (which would make this book impossible). For example, the argument about how to build a jet engine, can be quite detailed, without the need to teach you chemistry, to know how the jet fuel. The same can be said about this book, which relies on many areas, particularly the statistics, and covers computational methods. I'm not trying to teach mathematics but necessary to understand the text. However, I tried to write a book so that if you know the computational methods (or statistics), then for you it will be fully understood, and if you do not know these disciplines, there will be little, if any, loss of meaning, and you still You can use and understand (at most) the material disclosed in the book without appreciable | 677.169 | 1 |
Algebra 1/Algebra 2 Transition- Please remember that a student's grade in Accelerated Algebra 1 or Algebra 1 Part B is used to determine eligibility for Geometry as well as for Algebra 2. Their performance in Geometry will also be used in the recommendation process, assessing readiness based on~the Characteristics of an Honors Student on the Westford Academy Math Department website.
Outside Courses and Opt-Out Testing - Courses taken outside of Westford Public Schools do not replace the courses offered at Westford Academy. Grades from outside courses will not be used to determine placement or eligibility for courses offered at WA. Additionally, beginning in the 2017-2018 school year, students will no longer have the option of testing out of courses at WA, including Geometry Honors, Intro to Programming, and Java Honors.
Geometry Honors - Starting in the 2018-2019 school year,all students must take a Geometry course at WA either their Freshmen or Sophomore year. Students must earn a 95 in Geometry CP1 or an 83 in Geometry Honors at Westford Academy to be recommended for Trig & Pre-Calc Honors.
Doubling Up - Students have the option to double up their math courses in either Algebra 1B with Geometry their Freshmen year or Geometry with Algebra 2 their Sophomore year. Doubling up is not recommended by the Math Department, particularly in a student's Freshmen year, as students are encouraged to explore their passions and interests by taking electives. When possible, students are encouraged to wait to double up their Sophomore year instead. Students are expected to take a math course each year, even if they choose to take two math classes in one year.
Algebra 2 Enrollment - Starting in the 2018-2019 school year, Algebra 2 enrollment will be limited to students in grades 10-12. We believe the material presented in this course is more developmentally appropriate for students in these levels. Freshmen who are particularly interested in furthering their Math education are encouraged to enroll in Exploring Computer Science. The collaboration and problem solving opportunities in this course are designed to help prepare students for success in both college and career in the STEM field and beyond. The application of mathematical logical reasoning in this course will be extremely beneficial in the growth and development of all students.
This department offers a wide variety of high quality, engaging classes to our students. We have different levels of courses with ambitious expectations to meet everyone's needs. It is crucial that both students and parents consider our recommendations for course placement seriously when it is time to make choices for the following year. Our recommendations are based upon the knowledge that an essential understanding of prerequisite skills and academic benchmarks are needed to ensure proper placement in a course that will appropriately challenge the student, while allowing them to find success. Please note: The courses taken elsewhere are for remedial or enrichment purposes only. They will not be recognized as a replacement for a course offered during the academic year at Westford
Academy. Important exchanges of ideas occur between classmates and teachers throughout the year, which are very important to the growth of our students. For this reason, students are not allowed to "place out" of courses.
Our courses are based on the strands outlined in the Massachusetts Curriculum Framework and we prepare our students to do well on the MCAS by the end of their sophomore year.~ We pride ourselves on the top AP scores and SAT scores that our students have been able to achieve over the years. The course sequences, flow charts, and grade prerequisites can be obtained through the links on the left.
The members of this department at Westford Academy are dedicated professionals willing to help their students succeed. They will work around their schedules to make themselves available to give extra help to a student who requests the extra time needed to ensure they understand the concepts being taught.
Test out for Geometry Honors on April 5, 2017 from 11:30-1:30 pm in the Lecture Hall
Incoming Freshmen who have previously taken a formal Geometry Honors course out of district may want to take a test to opt-out of this course at WA and take Algebra 2 Honors in 9th grade. It is required that students earn an 87 on the test to opt out of the course. Some of the students may decide to double up in both of these courses if they don't reach the required score.
Please know, opting out of Geometry Honors is not recommended as many of the skills and methods taught at WA are carried through the other courses in the sequence. Taking Geometry Honors as freshmen allows students to take AP Calculus BC in their senior year without doubling up. ~If students finish their math requirements early, they may need to take math at night at a college since they do need to have 4 years of a math for graduation. ~Neither the previous Geometry Honors course nor the out-of-district college course will be counted towards their GPA.
We will be administering the test on April 5, 2017, as the student may need to choose another course if they are successful on the test. ~Some electives fill up fast so we need to know if they need Geometry Honors or not. ~Parents will need to provide the transportation.
The test is based on our Honors Geometry Curriculum which covers all 12 chapters of the Prentice Hall Geometry Book with ISBN # 0-13-062560-4 for your reference. ~The students will be allowed to use a non-graphing calculator; the test will be mostly open response/short answer with a couple of proofs.
Please note:
There is only 1 test out given, and the minimum grade is 87 or better in order to skip Geometry Honors and go directly into Algebra 2 Honors Java Honors on April 5 from 11:30-1:30 pm in the Lecture Hall
Students need to be familiar and comfortable with Java programming and complete the pre-AP, full year Java programming Hons course as a prerequisite in order to qualify for the Computer Science AP class. However, if the student has extensive Java knowledge and programming experience outside of WA, we do allow a test out.
The format for the test out is a combination of multiple choice, short answers, writing Java methods and programs (on paper, not on a computer). The student is expected to know Java syntax and write programs using packages TerminalIO for input / output and BreezySwing for GUI. They may not use any textbooks or online resources for help.
There is only 1 test out given, and the minimum grade is 85 or better in order to skip Intro to Programming on April 11 from 2-2:50 pm in Room 248
Students need to be familiar and comfortable with programming and complete an introductory programming course as a prerequisite (either the semester Intro to Programming or full year Intro to Programming Hons course) prior to qualifying for Java Programming Hons which is a rigorous pre-AP class. However, if the student has learned programming outside of WA, we do allow a test out.
For the test out, students will be asked to write a basic program within 50 minutes using variables, input / output, if and looping statements in any language of their choice (they need to make sure their program is able to run on computers in Westford Academy room 248 since it will be tested there). For further information on what is covered in the introductory courses, you can refer to Mrs. Trehan's website at
Please note:
There is only 1 test out given, and the minimum grade is 85 or better in order to skip | 677.169 | 1 |
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A PPT showing three different methods, one familiar (the scientific method), and two novel methods, both used to solve a variety of math problems. The PPT features Polya's heuristics featured in his famous work "How to Solve It". Examples from an Algebra2 curriculum are given in the PPT. | 677.169 | 1 |
books.google.co.uk - This... of Hyperstructure Theory
Applications of Hyperstructure Theory
This 7) cryptography; 8) median algebras, relation algebras; 9) combinatorics; 10) codes; 11) artificial intelligence; 12) probabilities. Audience: Graduate students and researchers. | 677.169 | 1 |
Vectors, Tensors, and the Basic Equations of Fluid. understanding of the underlying principles of fluid mechanics. EGM 6812 - Fluid Mechanics 1.Tensors and the Basic Equations of Fluid Mechanics. It applies the mathematics of Cartesian and general tensors to physical.
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Vectors, Tensors and the Basic Equations of Fluid Mechanics. By:.The specific topics covered will include vector and tensor analysis,.Vectors, Tensors and the Basic Equations of Fluid Mechanics (Dover Books on Mathematics) by Rutherford Aris, Mathematics,.Review of Vector Calculus. and a vector is a tensor of rank one. Vectors, tensors, and the basic equations of fluid mechanics.FREE TEXTBOOK (Represents about 80% of finished project.) Introduction to Tensor Calculus and Continuum Mechanics.
Tensors and the Basic Equations of Fluid Mechanics. It applies the mathematics of Cartesian and general tensors.Modern mathematics also relies upon. as Fluid Mechanics,...
Download Vectors, Tensors and the Basic Equations of Fluid Mechanics by Rutherford Aris or any other file from Books category.Simply click the Sign Up Now button below to complete your registration for the Dover.Vectors, Tensors and the Basic Equations of Fluid Mechanics. Introduction to Continuum Mechanics for.
Vectors, Tensors and the Basic Equations of Fluid Mechanics by Rutherford Aris,. | 677.169 | 1 |
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KEY MESSAGE: Presented in a clear and concise style, theAkst/Braggseries teaches by example while expanding understanding with applications that are fully integrated throughout the text and exercise sets. Akst/Bragg's user-friendly design offers a distinctive side-by-side format that pairs each example and its solution with a corresponding practice exercise. The concise writing style keeps readers'interest and attention by presenting the mathematics with minimal distractions, and the motivating real-world applications demonstrate how integral mathematical understanding is to a variety of disciplines, careers, and everyday situations. KEY TOPICS: Whole Numbers, Fractions, Decimals, Basic Algebra: Solving Simple Equations, Ratio and Proportion, Percents, Signed Numbers, Basic Statistics, More on Algebra, Measurement and Units, Basic Geometry MARKET: For all readers interested in Basic Mathematics
Author Biography | 677.169 | 1 |
Elements of Plane and Spherical Trigonometry: With Practical
In addition to being very useful in probability calculations, factorials are frequently seen in important mathematical series, including those used to calculate e, sin, and cos. Do you know each of the different ways sine can be described? Robert of Chester's translation of al-Khwarzmi's treatise on algebra opens with the words dixit Algorithmi, "Algorithmi says." So this exercise is restricted to how the Mandelbrot picture is constructed on the complex plane. FREE online interactive quizzes on Math, Trigonometry problems, Trigonometric Functions, Trigonometric Identities & Trigonometry Word Problems
Pages: 186
Publisher: Cornell University Library (October 21, 2009)
ISBN: B002WYJET0
Unit conversions are quick and easy using the scrolling picker control. Equation Genius will help you do you math in seconds. It supports the following: Calculator XL is a great free calculator that takes advantage of the iPad's large screen and support for all orientations pdf. Update 2/12/2004: -The program can now solve unknown angles. ----------End Update---------- Update 2/11/2004: -HUGE revisions done to the program.. it now takes up way less space (due to previous superfluous code), and it fixes a few bugs from the older version. ----------End Update---------- Update 2/10/2004: -Created blocks for more than one unknown (which previously crashed the program) Mathematical Methods for Physical Sciences I-II. This sequence is intended for students who are majoring in a department in the Physical Sciences Collegiate Division other than mathematics Your placement result will be a percentage score. A minimum placement score is required to enroll in any entry level Mathematics course above College Algebra (Math 0031), as shown below. There are several reasons the MathsWatch resource works as the best Maths revision aid on the market today: It revises every single topic just like a teacher would, at the board ref.: Three points should be noted. (1) sin-I X stands for an angle: thus sin - I ~ = 30°. (2) The '- l ' is not an index, but merely a sign to denote inverse notation. (3) (sin x)" I is not used, because by section 31 it would mean the reciprocal of sin x and this is cosec x , cited: Must I work in my ALEKS Prep and Learning Module between placement assessments? Yes, after completing the assessment, you must spend a minimum of 5 hours in your Prep and Learning Module before repeating the assessment. Yes, you have 24 hours to complete a placement assessment once it has begun. How long will a placement assessment take to complete? Placement assessments require approximately 90 minutes to complete, but the amount of time will vary by student online. | 677.169 | 1 |
Core 3 coursework help
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Coursework help can be found on our service! This is a finest resource for all students where one can order coursework online ! Call us any time you need. Coursework Core 300+ About. Browse books; Site directory; About Scribd; Meet the team;. Help; FAQ; Accessibility; Press; Purchase help; AdChoices; Legal. Terms. Core 3 coursework help: how to write coursework and exam essays: argumentative research paper on homelessness: how to start a definition essay on love. No Coursework Phd Sl.Nodeliver you expert coursework assistance. We guarantee quality performance and 100% unique resultCoursework Help. When compared to all. A description of a good coursework example that students can follow and apply to minimize the potential of errors while writing their own paper.
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Core 3 coursework help. Of education are required.Coursework help can be found on our service! This is a finest resource for all students where one can order. Are you scared of your English coursework writing? Then you can get the best English coursework help online from PHD experts and enjoy reading literary classics. Cpc Coursework College Admission Essay Guidelines Discuss The Importance Of Literature. ocr mei core 3 coursework. Thank you for your commendable help. Core 3 Coursework Help Help with narrative essay This resource begins with a general description of essay writing and moves to help with narrative essay a discussion. Coursework In-progress or Planned. Enter college terms in progress or planned through the completion of the summer 2017. Enter all of your in-progress or planned. SUBMIT BY EMAIL ONLY Coursework Numerical Methods 3 7 7 7 7 1 2 3 4 5 6 7 7 7 7 1 2 3 4 x y. Be able to explain how the methods work with the help of graphs. Core 3 MEI OCR Coursework Be warned. I have previously entered this coursework and it got 16/18 so they will know if you copy. | 677.169 | 1 |
Department of Mathematics and Philosophy
Undergraduate Research-Math 444
Areas of Interest: Mathematics education; the concept of function; technology in the classroom, guided discovery approaches to learning and teaching mathematics; interdisciplinary math and science courses.
Description: I am especially interested in how activities (science experiments, interactive games, technology simulations, open-ended problems) can be used in fostering and developing mathematical understanding. In particular, I am researching the role of ADAGE (activity, data, analysis, generalizations, and extensions) based lessons in the mathematics curriculum. I would invite interested undergraduate students to devise, research, and assess activity-based instructional materials and then comment on their effectiveness in the math classroom.
I am also interested in studying how students learn the concept of function and the importance of this concept in all branches of mathematics. This is a broad topic which could include research involving the development of the function concept and definition, the multiple representations or uses of functions, graphing techniques, rates of change, or concept maps. | 677.169 | 1 |
BC Calculus Audit Syllabus4-b students to ... studentsto class is discussed using analytical, graphical, and numerical approaches. The students learn to use a tabular representation. They also have to verbalize what does the solution mean. III Technology component Technology is woven into the calculus curriculum. We use a TI-83/84 calculator. Students know how to graph functions, their derivatives, find the intersection of two curves, evaluate derivatives and integrals as a result of our every day calculator activities. Graphing Calculator software developed by Pacific Tech (2002) is used to graph polar and parametric equations, as well functions in 3 variables. The latter we do after the AP Exam. Calculus in Motion by Audrey Weeks is used to demonstrate volumes of revolution, area problems and other concepts. Technology removes the necessity for tedious calculation and table manipulation, allowing more time to develop concepts and interpret results. IV Assessments (a) Quizzes/ pop-Quizzes; (b) Tests (calculator and non-calculator parts); (c) Projects after AP Exam (usually in pairs); (d) Partner quizzes or tests with a Take-Home Part; (e) Working at a white board (I assess it too). (f) Take-Home Tests
Quizzes and tests are all cumulative. Students expect that all topics covered before a test could be given on the test. On each give test, there are two relatively equal parts – a calculator part and a non-calculator part. Starting at the end of October, I give Take-Home Tests every 10 days. There are 5-6 old AP problems in each Take-Home Test. On all oral or written assignments the students are asked to use precise language in describing their answer, and always justify their answer. V Topics after AP Exam 1. Hyperbolic Functions; 2.Parametrics in Physics World: Three-Dimensional Approach; 3. Vectors in 3-D; 4. Second-Order Linear Homogeneous Differential Equations; 5. As a culmination of my students' learning of Calculus, they do the final project usingtheir knowledge of Calculus. They choose a topic that should include Calculus. Among the topics were History of Calculus, Models that demonstrate the Volumes of revolutions, Volumes by Discs and Washers, TI-84 Calculator programs designed to improve Numerical evaluations of Integrals (in particular, using left, right, midpoint, and trapezoidal and Simpson's approximations). VI Course texts: Main Textbook: Calculus by Anton, Bivens, Davis; Seventh edition. Other sources: 1. Instructor's Resourse CD-ROM, Single and Multivariable Calculus by Hughes-Hallett, Gleason, McCallum et al.,Third Edition; 2. Mathematics 4C , Math Department, Phillips Exeter Academy; 3. Multiple-Choice & Free-Response Questions in Preparation for the AP Calculus BC Examination , David Lederman; 4. Solutions, AP Calculus Problems Part II AB and BC 1987 – 2001 J. Broadwin, G. Lenchner, and M. Rudolph; 5. All Free-Response Questions from the recent years (2002 – 2006); 6. Calculus in Motion by Audrey Weeks (software); 7. Graphing Calculator Version 2 by Pacific Tech. VII AP Calculus Course Outline Unit 1 Review of Functions (Week 1) The students receive the comprehensive Summer Assignment before the course starts and turn in the Take –Home Test on the first day of school. They also take the in-school test to show the mastery of the functions described below. 1. Linear, Quadratic, Logarithmic, Exponential, Polynomial, Rational, Six Trigonometric,
Absolute Value, Radical Functions and their graphs are reviewed; 2. Inverse Functions and Their Graphs; 3. Inverse Trigonometric Functions and Their Graphs; 4. Transformations, Composition of Functions; 5. Parametric Equations and Graphs. 6. Vectors in plane. Students study vectors in Pre-Calculus class and review the properties of vectors at the beginning of the year. We make connection between parametric equations and vectors, such as: if the position of a particle in the plane is x(t) = 2+3t and y(t) = -1 + t, then the displacement vector can be represented as [2 + 3t, -1 + t], or r = (2 + 3t) i (-1 + t) j with + the velocity vector to be [3,1]. Later in the course (Unit 5 - parametric functions revisited and Unit 10 - tangent lines and arc length for parametric curves), we discuss more deeply how we represent a position, velocity and acceleration vectors (in plane) in general. Unit 2 Limits and Continuity (Week 2 - Week 4) 1. Intuitive Approach to Limits; 2. One-sided Limits; 3. Infinite Limits and Vertical Asymptotes; 4. Limits at Infinity and Horizontal Asymptote; End Behavior of a Function; 5. Techniques of Computing Limits: (a)Properties of Limits; Some Basic Limits; (b) Limits of Polynomial and rational Functions as ; (c) Indeterminate forms: , , ; (d) Limits Involving Radicals; (e) Limits of Piecewise-Defined Functions; (f) Limits of Polynomials as ; (g) Limits of Rational Functions as . 6. Limits discussed rigorously with the formal definition and "epsilon and delta" notation; 7. Continuity: (a) Definition of Continuity; (b) Continuity on an interval; (c) Continuity of Polynomials; (d) Properties of Continuous Functions; (e) Continuity of Rational Functions; (f) Continuity of Compositions; (g) Continuity on a closed interval [a,b]; (h) The Intermediate Value Theorem. (i) Continuity of Trigonometric Functions;
8. The Squeezing Theorem.
Unit 3 The derivative (Week 5 – Week 9) 1. Secant Lines, Tangent Lines, Velocity, Slopes, Rate of Change; 2. Average Rate of Change and Instantaneous Rate of Change; 3. The Derivative Function: (a) Difference Quotient; (b) Slope of a Curve and Tangent Lines; (c) Definition of the derivative of f at x= ; (d) Tangent Line to the graph of f at the point ( , P( )); (e) Interpretation of the Derivative; (f) Differentiability (corner, vertical tangent line, and discontinuity); (g) Differentiability and Continuity; (h) Different Notations for the Derivative. 4. Techniques of Differentiation: we prove most of the rules of differentiation (a) Derivative of a constant; (b) Derivative of a constant times a function; (c) The Power rule; (d) Derivative of Sums and Difference; (e) The Derivatives of a Product; (f) The Derivatives of a Quotient; (f) Higher Derivatives. 6. Derivatives of Trigonometric Functions; 7. The Chain Rule; 8. Related Rates; 9. Local Linear Approximation, Differentials; 10. Implicit Differentiation; 11. Derivatives of Logarithmic and Exponential Functions; 12. Derivatives of Inverse Trigonometric Functions. The following is a an example of class notes with additional exercises for better understanding a concept of the derivatives Class Notes 1. Average Velocity (geometric interpretation) 3.1.1 2. Instantaneous Velocity see 3.1.2 3. Average rate of change (over an interval) vs. instantaneous rate of change at a point
this is a slope of secant line between two points on a graph (x, f(x)) and ( )
this is the slope of the tangent line at the point . We will do ##2,3,6,8,10 12 on page 176 in class. 4. Difference quotient, symmetric difference quotient Difference Quotient
Symmetric Difference Quotient 5. Derivative of f with respect to x, notation: 6. In class ##6,8,26,28,30,44 of section 3.1. 7. Tangent line to the graph of f at a point x=
Using the definition of the derivative
, we write
the equation of the tangent line to the graph of f at the point as . Usea calculator to 'linearize' the curve of at the point x=2 Zoom in several times until you have a locally linear graph, then Evaluate the slope. For instance, we have two points (2.0345745,4.1394933) and (1.9780585, 3.9127155), and the slope is m=4.01263 8. Differentiability a) has to exist; b) has to exist. If f is differentiable at every point , then we say that f is differentiable on (a, b). 9. A function has to be locally linear in the neighborhood of to be differentiable at . 10. A function is not differentiable at a) a point of vertical tangency; See
Ex.: b) corner; See Ex: y=|x| c) at a point of discontinuity (obvious?!) 11. Derivative<0 corresponds to a decreasing function; Derivative>0 corresponds to an increasing function; Derivative=0 max or min of a function, or an inflection point. 12. Ex.:
Unit 5 Integration (Week 12 – Week 15) 1. Overview of the Area problem 2. The Indefinite Integral 3. Integration by Substitution 4 Sigma Notation. The Definition of the Area as a Limit; 5. Numerical approximations of areas (a) Left endpoint approximation; (b) Right endpoint approximation; (c) Midpoint approximation; (d) Trapezoidal approximation; 6. The Definite Integral 7. The Fundamental Theorem of Calculus and its Applications: (a) Distance as the definite integral of speed; (b)Rectilinear Motion using Integration (see examples 1,3 below); (c) Problems using the integral of a rate of change to determine total or accumulated change (see example 1 & 2 below ) (d) Average Value of a Function (see example 3 below). 8. Parametric Equations and Vector functions revisited (see the examples below). At this point in time, we discuss the motion of a particle in plane (extending it to space, too). A position function is introduced parametrically as x(t) = f(t), and y(t) = g(t) and by using a vector notation r (t) = [f(t), g(t)]. The instantaneous velocity and acceleration of the particle at time t are given as r (t) = v (t) = [f (t), g (t)] and r (t) = a (t) = [f (t), g (t)], respectively. 9. Length of the Plane Curve: (a) Arc length L of the curve y = f(x) over the interval [a,b]; (b) Arc length L of the curve x = f(y) over the interval [c,d]; (c) Arc length of parametric curves. (d) Finding Arc length by numerical methods. As a good example of Vector Functions , I discuss Question 1 of AP Calculus BC (2004, Form B). Another example that I always use is as follows: The position of a particle at any time t ≥ 0 is given by a vector function r = ( ) i + ( ) j (a) Find the magnitude of the velocity vector at t = 4. (b) Find the acceleration vector for t = 4. (c) Find the total distance traveled by the particle from t = 0 to t = 4. (d) Write an equation for the line tangent to the curve at t = 4.
Below are the examples of the class work that the students have to do in groups for 10-15 minutes. Ex 1 BC Calculus
Class Work (working an groups)
Names:
Ex 2 BC Calculus
2.
Class work (in pairs)
De inite Inte ral o a Rate o Chan
e .
Names:
Ex 3 BC Calculus 1.
2.
Short Quiz Average Value of a Function
Name:
3. Suppose that the position function of a particle moving along a coordinate line is . Find the average velocity over the time interval 1 ≤ t ≤ 4. Unit 6 More Applications of the Definite Integral (Week 16 – Week 17) 1. Area between two curves: (a)Area between y = f(x) and y = g(x) (b) Area between x = v(y) and x = w(y) 2. Volumes by slicing: (a) Solids of Revolution; (b) Volumes by disks perpendicular to the x-axis; (c) Volumes by washers perpendicular to the x-axis; (d) Volumes by disks and washers perpendicular to the y-axis; 3. Volumes by cylindrical Shells (a) Volume by Cylindrical Shells about the y-axis; (b) Volume by Cylindrical Shells about the x-axis; 5. Area of a Surface of Revolution; 6. Work. The Role of Work in Physics and Engineering. | 677.169 | 1 |
Table of Contents
SYLLABUS FOR MATH 304-FALL 2015
The course covers basics of elementary linear algebra. It is not just computation but the intent is that some concepts and understanding give and gain significance from the computation. It is meant to be at the level of a Second year mathematically competent student.
The text is book written by Matthew Brin and Gerald Marchesi ;Linear Algebra 16-th edition. We will cover the first 6 chapters fairly thoroughly and if time provides delve into Chapter 7. A weekly schedule plan is laid out below.
WEEK 1
Aug 31-Sept4
Chapter 1
Sections 1.1-1.3.1
WEEK 2
Sept 8-Sept 11
..
.. 1.3.2-1.5
WEEK 3
Sept 16-Sept 18
..
.. 1.6.1-1.6.9
WEEK 4
Sept 21;Sept 24-25
Chapter 2
Sections 2.1-2.3
WEEK 5
Sept 28-Oct 2
..
.. 2.4-2.7
WEEK 6
Oct 5-Oct 9
..
.. 2.8-2.10; Test 1
Chapter 3
.. 3.1-3.2
WEEK 7
Oct 12-Oct 16
..
.. 3.3-3.6
WEEK 8
Oct 19-Oct 23
..
.. 3.7-3.8
Chapter 4
.. 4.1-4.2
MIDTERM EXAM
WED. OCT 21, 7:00 PM- 9:00 PM
Location To Be Announced
WEEK 9
Oct 26-Oct 30
..
.. 4.2-4.7
WEEK 10
Nov 2- Nov 6
..
.. 4.8- 4.9
WEEK 11
Nov 9- Nov 13
Chapter 5
.. 5.1-5.3.2
WEEK 12
Nov 16-Nov 20
..
.. 5.3.3-5.5
WEEK 13
Nov 23-Nov 24
..
.. Review Test 2; THANKSGIVING
WEEK 14
Nov 30-Dec 4
Chapter 6
.. 6.1-6.2.3
WEEK 15
Dec 7- Dec 11
..
.. 6.2.4; 6.2.6 OR 7.1 ; QUIZ
WEEK 16
Dec 14-15
REVIEW
Final exam date and location to be announced.
Exams
Tests 1 and 2, and the Quiz at the end will be prepared for the individual sections and graded by the instructor(some sharing of writing and grading may occur). The Midterm is scheduled Oct. 21 in the evening. Please plan accordingly for class conflicts and extreme hardship we will schedule a make-up date. The time and location of the Final exam will be determined by the University during the scheduled times for final exams.
The exams will be pretty detailed and will require your mastery of the material to do well. Typically the percentage score expectation for the various grades are: A 85-100 ; B 70-85 ; C 50-70 ; D 40-50 with + and - at the extremes.
Weighting of the contributions to final grade
There will be 1000 possible for the course comprised as follows;
Final exam
400 pts.
Midterm
250 pts.
Hourly Exams
200 pts. (100 each)
End Quiz
50 pts.
Additionally, there will be 100pts. contributed by the instructor to be based on work in the class. Each instructor will choose how to use attendance, in class quizzes, homework or special assignments to determine this contribution. | 677.169 | 1 |
Read Analytic Geometry PDF, azw (Kindle)
Maths Wizard's Virtual Classroom offers Direct One... All of them could make computations more accurate near certain angles, but I don't know which ones were commonly used and which ones were named* analogously to other functions but rarely actually used. In this, all the faces are shown, as though the body were made of transparent material, those edges which could not otherwise be seen being indicated by dotted lines.
Pages: 496
Publisher: Cengage Learning; 6 edition (October 25, 1995)
ISBN: 0534948545
A Treatise on Plane and Spherical Trigonometry, and Its Applications to Astronomy and Geodesy
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Studyguide for Algebra and Trigonometry by Axler, Sheldon
An elementary treatise on plane trigonometry: with numerous examples and applications: designed for the use of high schools and colleges
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A Treatise of Trigonometry, Plane and Spherical, Theoretical and Practical ...: As Likewise a Treatise of Stereographick and Orthographick Projection ... of the Several Cases in Right and Oblique
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Math is Everywhere: Applications of Finite Math
Equations of lines can allow computers to create fonts, store them quite compactly, and render them at essentially any desired resolution.
Plotting the graph of a polynomial can affect how you play Angry Birds as you strive to dislodge the pesky pigs.
Linear systems model the performance of sports teams and influence which college football teams play in the new year bowl games.
You can create your own linear equations to help you create a bracket for March Madness.
Finally, probability and simulation lies at the core of the mathematical algorithm that catapulted Google as a leader in search engines.
You benefit from applications of finite math every day. Through this course, you can better understand how you benefit from applications of finite math in your every day life. Along the way, you will likely learn new mathematical ideas, too | 677.169 | 1 |
Note: This is a standalone book, if you want the book/access card please order the ISBN listed below: 0321837533 / 9780321837530 A Survey of Mathematics with Applications plus MyMathLab Student Access Kit59664 / 9780321759665 Survey of Mathematics with Applications, A
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SURVEYING: PRINCIPLES & APPLICATIONS, 9/e is the clearest, easiest to understand, and most useful introduction to surveying as it is practiced today. It brings together expert coverage of surveying principles, remote sensing and other new advances in technological instrumentation, and modern applications for everything from mapping to engineering. Designed for maximum simplicity, it also covers sophisticated topics typically discussed in advanced surveying courses. This edition has been reorganized and streamlined to align tightly with current surveying practice, and to teach more rapidly and efficiently. It adds broader and more valuable coverage of aerial, space and ground imaging, GIS, land surveying, and other key topics. An extensive set of appendices makes it a useful reference for students entering the workplace Mathematics 0321935446 / 9780321935441 Mathematics with Applications In the Management, Natural, and Social Sciences Plus NEW MyMathLab with Pearson eText1931076 / 9780321931078 Mathematics with Applications In the Management, Natural and Social Sciences
A Transition to Advanced Mathematics promotes the goals of a ``bridge'' course in mathematics, helping to lead students from courses in the calculus sequence to theoretical upper-level mathematics courses. The text simultaneously promotes the goals of a ``survey'' course, describing the intriguing questions and insights fundamental to many diverse areas of mathematics. | 677.169 | 1 |
* A branch of mathematics concerning the study of finite or countable discrete structures. Aspects of combinatorics include counting the structures of a given kind and size (enumerative combinatorics), deciding when certain criteria can be met, and constructing and analyzing objects meeting the criteria (as in combinatorial designs andmatroid theory), finding "largest", "smallest", or "optimal" objects (extremal combinatorics and combinatorial optimization), and studying combinatorial structures arising in analgebraic context, or applying algebraic techniques to combinatorial problems (algebraic combinatorics).
* The branch of mathematics studying the enumeration, combination, and permutation of sets of elements and the mathematical relations that characterize their properties.
PROBABILITY OR LIKELIHOOD
* A measure or estimation of how likely it is that something will happen or that a statement is true. Probabilities are given a value between 0 (0% chance or will not happen) and 1 (100% chance or will happen). The higher the degree of probability, the more likely the event is to happen, or, in a longer series of samples, the greater the number of times such event is expected to happen.
GRAPH THEORY
* The study of graphs, which are mathematical structures used to model pair wise relations between objects from a certain collection. A "graph" in this context is a collection of "vertices" or "nodes" and a collection of edges that connect pairs of vertices. A graph may be undirected, meaning that there is no distinction between the two vertices associated with each edge, or its edges may be directed from one vertex to another; see graph (mathematics) for more detailed definitions and for other variations in the types of graph that are commonly considered. Graphs are one of the prime objects of study in discrete mathematics.
* A branch of mathematics concerned about how networks can be encoded and their properties...
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In the current system of communication between people and computer, screen and keyboard are the primary channels of information exchange. Advances in technology and the art of computer programming have given rise to a new generation of methods for providing computer access to teachers and students.
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Discrete mathematics is the study of mathematical structures that are fundamentally discrete rather than continuous. In contrast to real numbers that have the property of varying "smoothly", the objects studied in discrete mathematics – such as integers, graphs, and statements in logic – do not vary smoothly in this way, but have distinct, separated values. Discrete mathematics therefore excludes topics in "continuous mathematics" such as calculus and analysis. Discrete objects can often be enumerated by integers. More formally, discrete mathematics has been characterized as the branch of mathematics dealing with countable sets (sets that have the same cardinality as subsets of the integers, including rational numbers but not real numbers). However, there is no exact, universally agreed, definition of the term "discrete mathematics." Indeed, discrete mathematics is described less by what is included than by what is excluded: continuously varying quantities and related notions.
The set of objects studied in discrete mathematics can be finite or infinite. The term finite mathematics is sometimes applied to parts of the field of discrete mathematics that deals with finite sets, particularly those areas relevant to business.
Research in discrete mathematics increased in the latter half of...
...SIAM REVIEW Vol. 41, No. 1, pp. 135–147
c 1999 Society for Industrial and Applied Mathematics
The Discrete Cosine Transform∗
Gilbert Strang†
Abstract. Each discrete cosine transform (DCT) uses N real basis vectors whose components are π cosines. In the DCT-4, for example, the jth component of vk is cos(j + 1 )(k + 1 ) N . These 2 2 basis vectors are orthogonal and the transform is extremely useful in image processing. If the vector x gives the intensities along a row of pixels, its cosine series ck vk has the coefficients ck = (x, vk )/N . They are quickly computed from a Fast Fourier Transform. But a direct proof of orthogonality, by calculating inner products, does not reveal how natural these cosine vectors are. We prove orthogonality in a different way. Each DCT basis contains the eigenvectors of a symmetric "second difference" matrix. By varying the boundary conditions we get the established transforms DCT-1 through DCT-4. Other combinations lead to four additional cosine transforms. The type of boundary condition (Dirichlet or Neumann, centered at a meshpoint or a midpoint) determines the applications that are appropriate for each transform. The centering also determines the period: N − 1 or N in the established transforms, N − 1 or N + 1 in the other four. The key point is that all these "eigenvectors 2 2 of cosines" come from simple and familiar matrices. Key words. cosine transform, orthogonality, signal processing AMS subject...
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Math Curse By: Jon Scieszka and Lane Smith
Math Curse, written by Jon Scieszka and Lane Smith, takes us on a journey with a small child who is cursed by math. His teacher's name is Mrs. Fibonacci, who was a well know mathematician who connected a mathematical sequence found in nature. Of course Mrs. Fibonacci told her class and this child how easily math can be seen in the outside world. Our main character goes on amath rampage that drives him crazy. Scieszka and Smith do a great job a combining mathematical concepts as well as rhymes and brain games. The book is continuously rhyming accompanied by humorous art work that gives the story a kind of flow. Want a little bit of a challenge? Try answering a number of math questions asked throughout the book. The math used consisted mainly of patterns if not basic math of a 3rd grader. Fractions were mentioned but as any 3rd grader would be our main character was terrified of them. So much so that he may have considered answering the question in French instead of math.
Overall the book seemed good for the target audience. There was appealing art work on each page, as well as rhymes. The Rhyming scheme made a big difference because it made the story have a sense of flow. Our authors also made the story interesting for an older more sophisticated audience with the introduction of Ms. Fibonacci who...
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DeGeorge, B., Santoro, A. (2004). "Manipulatives: A Hands-On Approach to Math." Principal, 84 (2), (28-28).
This article speaks about the importance and significance of the use of manipulatives in the classroom, specifically in the subject of math. Manipulatives have proven to be valuable when used in a math class and are even more valuable to the children when they are young, and are learning new math concepts. Students are able to physically visualize the math concepts and gain knowledge because they understand what they're learning a whole lot better and they also are able to gain insights on those concepts. Different examples of manipulatives may include counting with beans or M&M's, using pattern blocks, puzzles, tangrams, and flash cards, just to name a few.
Using manipulatives in a math class are beneficial to both the student and the teacher because the teacher is able to explain the concepts to the students in a much easier manner using the hands-on technique, rather than explaining it verbally. It's especially beneficial to the student because by incorporating these manipulatives into their learning process, they are able to pick up the concepts much quicker and in a way that they better understand, yet are having fun while doing it. When they have the concepts down, the students' self-esteem goes up and they feel encouraged to keep on going.
After...
...The Discrete Cosine Transform
(DCT):
Theory and Application
1
Syed Ali Khayam
Department of Electrical & Computer Engineering
Michigan State University
March 10th 2003
1
This document is intended to be tutorial in nature. No prior knowledge of image processing concepts is
assumed. Interested readers should follow the references for advanced material on DCT.
ECE 802 – 602: Information Theory and Coding
Seminar 1 – The Discrete Cosine Transform: Theory and Application
1. Introduction
Transform coding constitutes an integral component of contemporary image/video processing
applications. Transform coding relies on the premise that pixels in an image exhibit a certain
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adjacent pixels in consecutive frames2 show very high correlation. Consequently, these
correlations can be exploited to predict the value of a pixel from its respective neighbors. A
transformation is, therefore, defined to map this spatial (correlated) data into transformed
(uncorrelated) coefficients. Clearly, the transformation should utilize the fact that the information
content of an individual pixel is relatively small i.e., to a large extent visual contribution of a
pixel can be predicted using its neighbors.
A typical image/video transmission system is outlined in Figure 1. The objective of the source
encoder is to exploit the redundancies in image...
...Classification of simulation models
1. Monte carlo simulation – describe system w/c are both stochastic and static
2. Continuous simulation – the system modeled are dynamic but may be deterministic and stochastic
3. Discrete event simulation – used to model systems which are assumed to change only at discrete set point of time
4. Combined/Discrete/Continuous simulation – combination of discrete and continuous
Steps in building a model and simulation
1. Define an achievable goal
2. Put together a complete mix of skills on the team
3. Involve the end-user
4. Choose the appropriate simulation tools
5. Model the appropriate level of detail
6. Start early to collect the necessary input data
7. Provide adequate and on-going documentation
8. Develop a plan for adequate model verification
9. Develop a plan for model validation
10. Develop a plan for statistical output analysis
Model levels
1. Conceptual – very high level
* What are the state variable w/c are dynamic
* How concept should the model be
2. Specification – on paper
* May involve pseudocode or equation
* How will the model receive input
3. Computational – a computer program
Discrete event system – it is one way of building up models to observe the time based behavior of the system
Queuing characteristic
1....
...Discrete wavelet transform 2
Others
Other forms of discrete wavelet transform include the non- or undecimated wavelet transform (where downsampling
is omitted), the Newland transform (where an orthonormal basis of wavelets is formed from appropriately
constructed top-hat filters in frequency space). Wavelet packet transforms are also related to the discrete wavelet
transform. Complex wavelet transform is another form.
Properties
The Haar DWT illustrates the desirable properties of wavelets in general. First, it can be performed in
operations; second, it captures not only a notion of the frequency content of the input, by examining it at different
scales, but also temporal content, i.e. the times at which these frequencies occur. Combined, these two properties
make the Fast wavelet transform (FWT) an alternative to the conventional Fast Fourier Transform (FFT).
Time Issues
Due to the rate-change operators in the filter bank, the discrete WT is not time-invariant but actually very sensitive to
the alignment of the signal in time. To address the time-varying problem of wavelet transforms, Mallat and Zhong
proposed a new algorithm for wavelet representation of a signal, which is invariant to time shifts.[3] According to this
algorithm, which is called a TI-DWT, only the scale parameter is sampled along the dyadic sequence 2^j (j∈Z) and
the wavelet transform is calculated for each point in time.[4][5]... | 677.169 | 1 |
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Math And Owning A Restaraunt
This essay Math And Owning A Restaraunt has a total of 741 words and 3 pages.
Math And Owning A Restaraunt
Math is an
essential asset in the business world. Without mathematics businesses
wouldnít be able to operate effectively. In order to run a restaurant math plays
an important role in a lot of different areas. For instance the items on the menu
may change due to the way it sells. Bookkeeping and math allow you to both
figures out what items are profitable and what items are selling. The business
world revolves around math, from profit and loss statements, to graphs, to
taxes. Everything in business requires mathematics. Owning a restaurant is no
different then any other field of business when it comes to math. The simplest
things in a restaurant could not happen without math such as paying for your
meal. Math is used to add up the total cost of a personís bill as well as adding
in the sales tax. More advanced math is used in the restaurant business as
well. Using equations to determine what your business can afford to buy as
well as the difference in the cost of the product and the profit it turns over is all
determined by math. Jobs you might not even think require math do, such as
portioning products or prepping food. When you are preparing food you need
to measure amounts of ingredients and measurements are a form of
mathematics. Wheatley-2 Keeping your books up to date requires math as
well. When keeping records of your restaurantís sales you can keep track of
your busy periods to know when you are required to order more food or
alcohol. Equations are used then to determine what and how much you need
of a product. Using math to determine a product markup is an important
process to a successful business as well. Determining the overall cost of a
dinner special and factoring in each amount of the ingredients used all goes
towards the final cost of a dinner special. The main objective in owning a
restaurant or any business for that matter is to turn profit. Math allows you to
take a look at just how productive your restaurant is and what steps you need
to take in order to keep it successful. Profit and Loss statements are a big part
of owning a restaurant. Profit and Loss statements revolve around math too.
These statements using your bookkeeping records show what times you
turned profit and what times you took a loss. Using this mathematical step an
owner can get ideas about what was working for them and what wasnít.
Doing your business taxes requires a lot of math. Showing your business
expenses and your sales are required. A lot of restaurant owners hire
bookkeepers and accountants to do the math work for them because it is
such a vital part of the restaurantís success. Math is used to figure out all your
restaurantís expenses such as,advertising, cost of goods and employees
wages. A restaurant owner needs to be aware of the exchanges in money
both coming in and going out. Without math it would be impossible to
determine what wages the owner can afford to pay his/her employees. Wages
are Wheatley-3 determined by the bookkeeping and the knowledge of how
much profit is being brought into the business. Restaurant owners often use
graphs for visuals of what is working for their company. Graphs are also used
to look at how your business is doing next to your competition. Math allows
an owner to determine the best time to do markup or markdown prices on
their menu. For restaurant owners the markup period is usually in the
summertime. A lot of marketing research comes into play in determining how
your business should be run. Determining location and types of food is one
aspect of the decision process. Another is taking a look at records of average
wages and spending of the particular area you want to have your restaurant in.
By using math these steps help determine the price range of your menu and
what products you as an owner can afford to use to turn profit. If you are
located in a geographical area that is not known for itís money then lobster
and caviar are not going to be items on your menu. Algebraic equations,
graphs, accounting, and simple math such as adding subtracting multiplying
and dividing are used everyday when owning a restaurant. Looking over this
paper I hope it shows just how important math is in the restaurant and
business world. Without math business wouldnít exist, as we know it and it
would be very
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Plane CrashPlane Crash Instructor: Greg Alston Abstract This paper examines the in-flight separation of the number two pylon and engine from a Boeing 747-121 shortly after takeoff from the Anchorage International Airport on March 31, 1993. The safety issues discussed focus on the inspection of Boeing 747 engine pylons, meteorological hazards to aircraft, the lateral load-carrying capability of engine pylon structures, and aircraft departure routes at Anchorage International Airport during turbulent weather | 677.169 | 1 |
Maths for Caribbean Schools 3(Paperback)
Synopsis
This new, third edition of Mathematics for Caribbean Schools (book 3) retains its popular structure and syllabus coverage, but provides extra examination practice for students to consolidate their learning at the end of each chapter. The series contains the following features: * many exercises in every chapter * worked examples for each topic * full cyllabus coverage * relevant group exercises to encourage investigation and discussion * practice exams in Books 1 and 2 *CSEC Mathematics exam papers in Books 3 and 4 * a free Teacher's Guide for each level with extra exercises, answers and guidance * a colourful and lively design to engage students | 677.169 | 1 |
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This review was created by one of my tutoring partners. He gave me express permission to post it here. We have found great success in using this with students that are taking Algebra in College, or that need a review of skills before starting their College Algebra course.
There are also skills in the review that align with Common Core Math I, Math II, and Algebra II
This is an all encompassing review of Algebra and useful as a pretest, practice, or review before a final.
There are graphing problems on this review, so a graphing calculator will be needed for those. | 677.169 | 1 |
Summation Notation Sigma Notation Task Cards plus Flip Book
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Activity Based Learning with Task Cards really does work to help reinforce your lessons. Task and station cards get your students engaged and keep them motivated. Use all at once or as many as you like. Directions for using Task Cards included in this resource.
Proficiency with Sigma Notation aka Summation Notation is a very important skill usually taught in PreCalculus and also in Calculus 1 and 2 (AB and BC). This activity can be used in all three courses.
One of the things I have noticed in higher level courses, is that even though the students do very well with task and station cards, they leave the class empty handed. The cards and response sheets go back to the teacher. I have started giving them organizers and flip books to do along with the task cards so they can go back and study. This reinforces the lesson even more. Great for interactive notebooks.
Included in the Resource:
✓ 18 Task cards plus two blanks for you to customize.
# 1 - 10 students write sums without sigma notation and find the sum. Some of these include trig functions. All problems are Multiple Choice.
# 11 - 18 students write a sum in sigma notation
✓ Answer key
✓ Student response sheet
✓ Foldable Flip Book with 6 problems from the Task Cards in free response format. One sheet of paper makes one book printed two sided (full directions included). No glue or tape, just one cut or careful tear, and a fold | 677.169 | 1 |
Insights Into Algebra 1: Teaching for Learning is a production of Thirteen/WNET New York. Copyright 2004, The Annenberg Foundation. All rights reserved.
Thirteen/WNET
A major American cultural and educational institution for nearly four decades, Thirteen/WNET supplies more than one-third of all primetime programs aired on PBS, including acclaimed cultural, science, and public affairs series and specials. The award-winning Children's and Educational Programming group is a leading and innovative provider of programming for a variety of projects, from teacher professional development to instructional television and interactive multimedia. Broadcast series that further the station's educational mission include the daily animated PBS Kids math program Cyberchase, the history series for families Freedom: A History of US, ZOOM Local/National, What's Up in the Environment/Technology/Factories?, and In the Mix specials. Many projects promote implementation of national and state education standards. These include The Expanding Canon: Teaching Multicultural Literature in High School, Science ... Simply Amazing, Learning Science Through Inquiry for Annenberg Media, and PBS TeacherLine, Mathline, and Scienceline.
Thirteen is also a trusted source of Web-based content for educators. Thirteen Ed Online, an award-winning Web site that features classroom materials, resources and support for teachers, and professional development workshops, reaches thousands of teachers, teachers-in-training, administrators, and others involved in pre-K-12 instruction. Ed Online workshops are currently used by professors and students at Harvard University, Teachers College at Columbia University, and in school districts throughout the country. Ed Online also produces education companion pieces to national series' Web sites and original online content to complement Thirteen's educational initiatives. Projects include Human Rights 101, Teaching Heritage, Afterschool Exchange, Cyberchase Parents and Teachers, Innovation, Religion & Ethics NewsWeekly, The Rise and Fall of Jim Crow, and Wide Angle. | 677.169 | 1 |
Monday, August 11, 2014 - Thursday, August 14, 2014
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About This Activity
This one-week course is for students entering high school as freshmen. We have found that the curriculum and areas covered in eighth grade vary school to school and district to district. The goal of this course is to make a smoother transition to high school math, to strengthen a rising freshman's math foundation and to gain a head start. This course will sharpen pre-algebra skills taught in eighth grade and try to bring them to the level they need to be in entering Algebra I or Honors Algebra I. This course is strongly recommended for students taking Algebra I | 677.169 | 1 |
Introduction
This is an introduction to linear algebra. The main part of the book features row operations
and everything is done in terms of the row reduced echelon form and specific algorithms.
At the end, the more abstract notions of vector spaces and linear transformations on vector
spaces are presented. However, this is intended to be a first course in linear algebra for
students who are sophomores or juniors who have had a course in one variable calculus
and a reasonable background in college algebra. I have given complete proofs of all the
fundamental ideas but some topics such as Markov matrices are not complete in this book but
receive a plausible introduction. The book contains a complete treatment of determinants
and a simple proof of the Cayley Hamilton theorem although these are optional topics.
The Jordan form is presented as an appendix. I see this theorem as the beginning of more
advanced topics in linear algebra and not really part of a beginning linear algebra course.
There are extensions of many of the topics of this book in my on line book [9]. I have also
not emphasized that linear algebra can be carried out with any field although I have done
everything in terms of either the real numbers or the complex numbers. It seems to me this
is a reasonable specialization for a first course in linear algebra.
7
8
INTRODUCTION
Fn
2.0.1
Outcomes
A. Understand the symbol, Fn in the case where F equals the real numbers, R or the
complex numbers, C.
B. Know how to do algebra with vectors in Fn , including vector addition and scalar
multiplication.
C. Understand the geometric significance of an element of Fn when possible.
The notation, Cn refers to the collection of ordered lists of n complex numbers. Since
every real number is also a complex number, this simply generalizes the usual notion of
Rn , the collection of all ordered lists of n real numbers. In order to avoid worrying about
whether it is real or complex numbers which are being referred to, the symbol F will be
used. If it is not clear, always pick C.
Definition 2.0.1 Define Fn ≡ {(x1 , · · · , xn ) : xj ∈ F for j = 1, · · · , n} .
(x1 , · · · , xn ) = (y1 , · · · , yn )
if and only if for all j = 1, · · · , n, xj = yj . When (x1 , · · · , xn ) ∈ Fn , it is conventional
to denote (x1 , · · · , xn ) by the single bold face letter, x. The numbers, xj are called the
coordinates. The set
{(0, · · · , 0, t, 0, · · · , 0) : t ∈ F}
for t in the ith slot is called the ith coordinate axis. The point 0 ≡ (0, · · · , 0) is called the
origin. Elements in Fn are called vectors.
Thus (1, 2, 4i) ∈ F3 and (2, 1, 4i) ∈ F3 but (1, 2, 4i) 6= (2, 1, 4i) because, even though the
same numbers are involved, they don't match up. In particular, the first entries are not
equal.
The geometric significance of Rn for n ≤ 3 has been encountered already in calculus or
in pre-calculus. Here is a short review. First consider the case when n = 1. Then from the
definition, R1 = R. Recall that R is identified with the points of a line. Look at the number
line again. Observe that this amounts to identifying a point on this line with a real number.
In other words a real number determines where you are on this line. Now suppose n = 2
and consider two lines which intersect each other at right angles as shown in the following
picture.
9
FN
10
· (2, 6)
6
(−8, 3) ·
3
2
−8
Notice how you can identify a point shown in the plane with the ordered pair, (2, 6) .
You go to the right a distance of 2 and then up a distance of 6. Similarly, you can identify
another point in the plane with the ordered pair (−8, 3) . Go to the left a distance of 8 and
then up a distance of 3. The reason you go to the left is that there is a − sign on the eight.
From this reasoning, every ordered pair determines a unique point in the plane. Conversely,
taking a point in the plane, you could draw two lines through the point, one vertical and the
other horizontal and determine unique points, x1 on the horizontal line in the above picture
and x2 on the vertical line in the above picture, such that the point of interest is identified
with the ordered pair, (x1 , x2 ) . In short, points in the plane can be identified with ordered
pairs similar to the way that points on the real line are identified with real numbers. Now
suppose n = 3. As just explained, the first two coordinates determine a point in a plane.
Letting the third component determine how far up or down you go, depending on whether
this number is positive or negative, this determines a point in space. Thus, (1, 4, −5) would
mean to determine the point in the plane that goes with (1, 4) and then to go below this
plane a distance of 5 to obtain a unique point in space. You see that the ordered triples
correspond to points in space just as the ordered pairs correspond to points in a plane and
single real numbers correspond to points on a line.
You can't stop here and say that you are only interested in n ≤ 3. What if you were
interested in the motion of two objects? You would need three coordinates to describe
where the first object is and you would need another three coordinates to describe where
the other object is located. Therefore, you would need to be considering R6 . If the two
objects moved around, you would need a time coordinate as well. As another example,
consider a hot object which is cooling and suppose you want the temperature of this object.
How many coordinates would be needed? You would need one for the temperature, three
for the position of the point in the object and one more for the time. Thus you would need
to be considering R5 . Many other examples can be given. Sometimes n is very large. This
is often the case in applications to business when they are trying to maximize profit subject
to constraints. It also occurs in numerical analysis when people try to solve hard problems
on a computer.
There are other ways to identify points in space with three numbers but the one presented
is the most basic. In this case, the coordinates are known as Cartesian coordinates after
Descartes1 who invented this idea in the first half of the seventeenth century. I will often
not bother to draw a distinction between the point in space and its Cartesian coordinates.
The geometric significance of Cn for n > 1 is not available because each copy of C
corresponds to the plane or R2 .
1 Ren´
e Descartes 1596-1650 is often credited with inventing analytic geometry although it seems the ideas
were actually known much earlier. He was interested in many different subjects, physiology, chemistry, and
physics being some of them. He also wrote a large book in which he tried to explain the book of Genesis
scientifically. Descartes ended up dying in Sweden.
2.1. ALGEBRA IN FN
2.1
11
Algebra in Fn
There are two algebraic operations done with elements of Fn . One is addition and the other
is multiplication by numbers, called scalars. In the case of Cn the scalars are complex
numbers while in the case of Rn the only allowed scalars are real numbers. Thus, the scalars
always come from F in either case.
Definition 2.1.1 If x ∈ Fn and a ∈ F, also called a scalar, then ax ∈ Fn is defined by
ax = a (x1 , · · · , xn ) ≡ (ax1 , · · · , axn ) .
Fn is often called n dimensional space. With this definition, the algebraic properties
satisfy the conclusions of the following theorem.
Theorem 2.1.2 For v, w ∈ Fn and α, β scalars, (real numbers), the following hold.
v + w = w + v,
(2.3)
(v + w) + z = v+ (w + z) ,
(2.4)
v + 0 = v,
(2.5)
v+ (−v) = 0,
(2.6)
the commutative law of addition,
the associative law for addition,
the existence of an additive identity,
The geometric meaning is especially significant in the case of Rn for n = 2, 3. Here is a
short discussion of this topic.
Definition 2.2.1 Let x = (x1 , · · · , xn ) be the coordinates of a point in Rn . Imagine an
arrow with its tail at 0 = (0, · · · , 0) and its point at x as shown in the following picture in
the case of R3 .
´
(x1 , x2 , x3 ) = x
r
3
´´
´
¡
¡
¡
Then this arrow is called the position vector of the point, x. Given two points, P, Q
whose coordinates are (p1 , · · · , pn ) and (q1 , · · · , qn ) respectively, one can also determine the
position vector from P to Q defined as follows.
−−→
P Q ≡ (q1 − p1 , · · · , qn − pn )
Thus every point determines a vector and conversely, every such vector (arrow) which
has its tail at 0 determines a point of Rn , namely the point of Rn which coincides with the
point of the vector. Also two different points determine a position vector going from one to
the other as just explained.
Imagine taking the above position vector and moving it around, always keeping it pointing in the same direction as shown in the following picture.
(x1 , x2 , x3 ) = x
r
3
´
´´
´
3 ´´
3
´
´´
´
´
´
¡
´
´
¡
3
´´
¡
After moving it around, it is regarded as the same vector because it points in the same
direction and has the same length.2 Thus each of the arrows in the above picture is regarded
as the same vector. The components of this vector are the numbers, x1 , · · · , xn . You
should think of these numbers as directions for obtainng an arrow. Starting at some point,
(a1 , a2 , · · · , an ) in Rn , you move to the point (a1 + x1 , · · · , an ) and from there to the point
(a1 + x1 , a2 + x2 , a3 · · · , an ) and then to (a1 + x1 , a2 + x2 , a3 + x3 , · · · , an ) and continue
this way until you obtain the point (a1 + x1 , a2 + x2 , · · · , an + xn ) . The arrow having its
tail at (a1 , a2 , · · · , an ) and its point at (a1 + x1 , a2 + x2 , · · · , an + xn ) looks just like the
arrow which has its tail at 0 and its point at (x1 , · · · , xn ) so it is regarded as the same
vector.
2.3
Geometric Meaning Of Vector Addition
It was explained earlier that an element of Rn is an n tuple of numbers and it was also
shown that this can be used to determine a point in three dimensional space in the case
2 I will discuss how to define length later. For now, it is only necessary to observe that the length should
be defined in such a way that it does not change when such motion takes place.
2.3. GEOMETRIC MEANING OF VECTOR ADDITION
13
where n = 3 and in two dimensional space, in the case where n = 2. This point was specified
relative to some coordinate axes.
Consider the case where n = 3 for now. If you draw an arrow from the point in three
dimensional space determined by (0, 0, 0) to the point (a, b, c) with its tail sitting at the
point (0, 0, 0) and its point at the point (a, b, c) , this arrow is called the position vector
of the point determined by u ≡ (a, b, c) . One way to get to this point is to start at (0, 0, 0)
and move in the direction of the x1 axis to (a, 0, 0) and then in the direction of the x2 axis
to (a, b, 0) and finally in the direction of the x3 axis to (a, b, c) . It is evident that the same
arrow (vector) would result if you began at the point, v ≡ (d, e, f ) , moved in the direction
of the x1 axis to (d + a, e, f ) , then in the direction of the x2 axis to (d + a, e + b, f ) , and
finally in the x3 direction to (d + a, e + b, f + c) only this time, the arrow would have its
tail sitting at the point determined by v ≡ (d, e, f ) and its point at (d + a, e + b, f + c) . It
is said to be the same arrow (vector) because it will point in the same direction and have
the same length. It is like you took an actual arrow, the sort of thing you shoot with a bow,
and moved it from one location to another keeping it pointing the same direction. This
is illustrated in the following picture in which v + u is illustrated. Note the parallelogram
determined in the picture by the vectors u and v.
Thus the geometric significance of (d, e, f ) + (a, b, c) = (d + a, e + b, f + c) is this. You
start with the position vector of the point (d, e, f ) and at its point, you place the vector
determined by (a, b, c) with its tail at (d, e, f ) . Then the point of this last vector will be
(d + a, e + b, f + c) . This is the geometric significance of vector addition. Also, as shown
in the picture, u + v is the directed diagonal of the parallelogram determined by the two
vectors u and v. A similar interpretation holds in Rn , n > 3 but I can't draw a picture in
this case.
Since the convention is that identical arrows pointing in the same direction represent
the same vector, the geometric significance of vector addition is as follows in any number of
dimensions.
Procedure 2.3.1 Let u and v be two vectors. Slide v so that the tail of v is on the point
of u. Then draw the arrow which goes from the tail of u to the point of the slid vector, v.
This arrow represents the vector u + v.
This is called the distance formula. Thus |x| ≡ |x − 0| . The symbol, B (a, r) is defined
by
B (a, r) ≡ {x ∈ Rn : |x − a| < r} .
This is called an open ball of radius r centered at a. It means all points in Rn which are
closer to a than r. The length of a vector x is the distance between x and 0.
First of all note this is a generalization of the notion of distance in R. There the distance
between two points, x and y was given by the absolute value of their difference. Thus |x − y|
³
´1/2
2
is equal to the distance between these two points on R. Now |x − y| = (x − y)
where
the square root is always the positive square root. Thus it is the same formula as the above
definition except there is only one term in the sum. Geometrically, this is the right way to
define distance which is seen from the Pythagorean theorem. Often people use two lines
to denote this distance, ||x − y||. However, I want to emphasize this is really just like the
absolute value. Also, the notation I am using is fairly standard.
Consider the following picture in the case that n = 2.
(y1 , y2 )
(x1 , x2 )
(y1 , x2 )
There are two points in the plane whose Cartesian coordinates are (x1 , x2 ) and (y1 , y2 )
respectively. Then the solid line joining these two points is the hypotenuse of a right triangle
2.4. DISTANCE BETWEEN POINTS IN RN LENGTH OF A VECTOR
15
which is half of the rectangle shown in dotted lines. What is its length? Note the lengths
of the sides of this triangle are |y1 − x1 | and |y2 − x2 | . Therefore, the Pythagorean theorem
implies the length of the hypotenuse equals
³
2
|y1 − x1 | + |y2 − x2 |
2
´1/2
³
´1/2
2
2
= (y1 − x1 ) + (y2 − x2 )
which is just the formula for the distance given above. In other words, this distance defined
above is the same as the distance of plane geometry in which the Pythagorean theorem
holds.
Now suppose n = 3 and let (x1 , x2 , x3 ) and (y1 , y2 , y3 ) be two points in R3 . Consider the
following picture in which one of the solid lines joins the two points and a dotted line joins
the points (x1 , x2 , x3 ) and (y1 , y2 , x3 ) .
(y1 , y2 , y3 )
³
´1/2
2
2
2
= (y1 − x1 ) + (y2 − x2 ) + (y3 − x3 )
,
which is again just the distance formula above.
This completes the argument that the above definition is reasonable. Of course you
cannot continue drawing pictures in ever higher dimensions but there is no problem with
the formula for distance in any number of dimensions. Here is an example.
Example 2.4.2 Find the distance between the points in R4 , a = (1, 2, −4, 6) and b = (2, 3, −1, 0)
Use the distance formula and write
2
2
2
2
2
|a − b| = (1 − 2) + (2 − 3) + (−4 − (−1)) + (6 − 0) = 47
FN
16
√
Therefore, |a − b| = 47.
All this amounts to defining the distance between two points as the length of a straight
line joining these two points. However, there is nothing sacred about using straight lines.
One could define the distance to be the length of some other sort of line joining these points.
It won't be done in this book but sometimes this sort of thing is done.
Another convention which is usually followed, especially in R2 and R3 is to denote the
first component of a point in R2 by x and the second component by y. In R3 it is customary
to denote the first and second components as just described while the third component is
called z.
Example 2.4.3 Describe the points which are at the same distance between (1, 2, 3) and
(0, 1, 2) .
Let (x, y, z) be such a point. Then
q
Since these steps are reversible, the set of points which is at the same distance from the two
given points consists of the points, (x, y, z) such that 2.11 holds.
There are certain properties of the distance which are obvious. Two of them which follow
directly from the definition are
|x − y| = |y − x| ,
|x − y| ≥ 0 and equals 0 only if y = x.
The third fundamental property of distance is known as the triangle inequality. Recall that
in any triangle the sum of the lengths of two sides is always at least as large as the third
side. I will show you a proof of this later. This is usually stated as
|x + y| ≤ |x| + |y| .
Here is a picture which illustrates the statement of this inequality in terms of geometry.
3
´
x + y ´´¢¢¸
´ ¢y
´
´
-¢
x
2.5. GEOMETRIC MEANING OF SCALAR MULTIPLICATION
2.5
17
Geometric Meaning Of Scalar Multiplication
As discussed earlier, x = (x1 , x2 , x3 ) determines a vector. You draw the line from 0 to
x placing the point of the vector on x. What is the length of this vector? The length
of
p this vector is defined to equal |x| as in Definition 2.4.1. Thus the length of x equals
x21 + x22 + x23 . When you multiply
x by a scalar, α, you get (αx1 , αx2 , αx3 ) and the length
r³
´
p
2
2
2
of this vector is defined as
(αx1 ) + (αx2 ) + (αx3 ) = |α| x21 + x22 + x23 . Thus the
following holds.
|αx| = |α| |x| .
In other words, multiplication by a scalar magnifies the length of the vector. What about
the direction? You should convince yourself by drawing a picture that if α is negative, it
causes the resulting vector to point in the opposite direction while if α > 0 it preserves the
direction the vector points.
You can think of vectors as quantities which have direction and magnitude, little arrows.
Thus any two little arrows which have the same length and point in the same direction are
considered to be the same vector even if their tails are at different points.
£±
£
£
£ £±
£±
£
£
£
£
£
£±
You can always slide such an arrow and place its tail at the origin. If√the resulting
point of the vector is (a, b, c) , it is clear the length of the little arrow is a2 + b2 + c2 .
Geometrically, the way you add two geometric vectors is to place the tail of one on the
point of the other and then to form the vector which results by starting with the tail of the
first and ending with this point as illustrated in the following picture. Also when (a, b, c)
is referred to as a vector, you mean any of the arrows which have the same direction and
magnitude as the position vector of this point. Geometrically, for u = (u1 , u2 , u3 ) , αu is any
of the little arrows which have the same direction and magnitude as (αu1 , αu2 , αu3 ) .
1
³³
µ
³³ ¡
³
¡
³
¡
£±
u £
¡u + v
¡
£
£ ¡
1
³
£ ¡³³³
³
³
£¡ v
The following example is art which illustrates these definitions and conventions.
Exercise 2.5.1 Here is a picture of two vectors, u and v.
FN
18
µ
¡
¡
¡
¡u
¡
HH
HH
v H
j
Sketch a picture of u + v, u − v, and u+2v.
First here is a picture of u + v. You first draw u and then at the point of u you place the
tail of v as shown. Then u + v is the vector which results which is drawn in the following
pretty picture.
H
µ
¡
HH
vHH
:
»
¡
»j
»
u
»
»
¡ »» u + v
¡
»»»
¡
Next consider u − v. This means u+ (−v) . From the above geometric description of
vector addition, −v is the vector which has the same length but which points in the opposite
direction to v. Here is a picture.
Suppose you push on something. What is important? There are really two things which are
important, how hard you push and the direction you push. This illustrates the concept of
force.
Definition 2.6.1 Force is a vector. The magnitude of this vector is a measure of how hard
it is pushing. It is measured in units such as Newtons or pounds or tons. Its direction is
the direction in which the push is taking place.
Vectors are used to model force and other physical vectors like velocity. What was just
described would be called a force vector. It has two essential ingredients, its magnitude and
its direction. Geometrically think of vectors as directed line segments or arrows as shown in
the following picture in which all the directed line segments are considered to be the same
vector because they have the same direction, the direction in which the arrows point, and
the same magnitude (length).
£
£±
£±
£
£
£ £±
£±
£
£
£
£
Because of this fact that only direction and magnitude are important, it is always possible
→ be a directed line segment or
to put a vector in a certain particularly simple form. Let −
pq
−
→
vector. Then it follows that pq consists of the points of the form
p + t (q − p)
where t ∈ [0, 1] . Subtract p from all these points to obtain the directed line segment consisting of the points
0 + t (q − p) , t ∈ [0, 1] .
The point in Rn , q − p, will represent the vector.
→ was slid so it points in the same direction and the base is
Geometrically, the arrow, −
pq,
at the origin, 0. For example, see the following picture.
£
£
£±
£
£
£±
£
£
£±
In this way vectors can be identified with points of Rn .
Definition 2.6.2 Let x = (x1 , · · · , xn ) ∈ Rn . The position vector of this point is the
vector whose point is at x and whose tail is at the origin, (0, · · · , 0). If x = (x1 , · · · , xn )
is called a vector, the vector which is meant is this position vector just described. Another
term associated with this is standard position. A vector is in standard position if the tail
is placed at the origin.
FN
20
It is customary to identify the point in Rn with its position vector.
→ is just the distance
The magnitude of a vector determined by a directed line segment −
pq
between the point p and the point q. By the distance formula this equals
Ã
n
X
!1/2
(qk − pk )
2
= |p − q|
k=1
and for v any vector in Rn the magnitude of v equals
¡Pn
vk2
k=1
¢1/2
= |v|.
Example 2.6.3 Consider the vector, v ≡ (1, 2, 3) in Rn . Find |v| .
First, the vector is the directed line segment (arrow) which has its base at 0 ≡ (0, 0, 0)
and its point at (1, 2, 3) . Therefore,
p
√
|v| = 12 + 22 + 32 = 14.
What is the geometric significance of scalar multiplication? If a represents the vector, v
in the sense that when it is slid to place its tail at the origin, the element of Rn at its point
is a, what is rv?
Ã
|rv| =
n
X
!1/2
2
Ã
(rai )
=
Ã
!1/2
k=1
¡ ¢1/2
= r2
n
X
a2i
n
X
!1/2
2
r2 (ai )
k=1
= |r| |v| .
k=1
Thus the magnitude of rv equals |r| times the magnitude of v. If r is positive, then the
vector represented by rv has the same direction as the vector, v because multiplying by the
scalar, r, only has the effect of scaling all the distances. Thus the unit distance along any
coordinate axis now has length r and in this rescaled system the vector is represented by a.
If r < 0 similar considerations apply except in this case all the ai also change sign. From
now on, a will be referred to as a vector instead of an element of Rn representing a vector
as just described. The following picture illustrates the effect of scalar multiplication.
£
£±
£± £ £
v£ 2v£ £ −2v
£
£°
Note there are n special vectors which point along the coordinate axes. These are
ei ≡ (0, · · · , 0, 1, 0, · · · , 0)
where the 1 is in the ith slot and there are zeros in all the other spaces. See the picture in
the case of R3 .
z
¡
x¡
e3 6 ¡
¡
¡
¡
e2
e1¡
ª
¡
y
2.6. VECTORS AND PHYSICS
21
The direction of ei is referred to as the ith direction. Given a vector, v = (a1 , · · · , an ) ,
ai ei is the ith component of the vector. Thus ai ei = (0, · · · , 0, ai , 0, · · · , 0) and so this
vector gives something possibly nonzero only in the ith direction. Also, knowledge of the ith
component of the vector is equivalent to knowledge of the vector because it gives the entry
in the ith slot and for v = (a1 , · · · , an ) ,
v=
n
X
a i ei .
k=1
What does addition of vectors mean physically? Suppose two forces are applied to some
object. Each of these would be represented by a force vector and the two forces acting
together would yield an overall force acting on the object which
Pnwould also be a force
Pn vector
known as the resultant. Suppose the two vectors are a = k=1 ai ei and b = k=1 bi ei .
Then the vector, a involves a component in the ith direction, ai ei while the component in
the ith direction of b is bi ei . Then it seems physically reasonable that the resultant vector
should have a component in the ith direction equal to (ai + bi ) ei . This is exactly what is
obtained when the vectors, a and b are added.
a + b = (a1 + b1 , · · · , an + bn ) .
n
X
=
(ai + bi ) ei .
i=1
Thus the addition of vectors according to the rules of addition in Rn which were presented
earlier, yields the appropriate vector which duplicates the cumulative effect of all the vectors
in the sum.
What is the geometric significance of vector addition? Suppose u, v are vectors,
u = (u1 , · · · , un ) , v = (v1 , · · · , vn )
Then u + v = (u1 + v1 , · · · , un + vn ) . How can one obtain this geometrically? Consider the
−
→
directed line segment, 0u and then, starting at the end of this directed line segment, follow
−−−−−−→
the directed line segment u (u + v) to its end, u + v. In other words, place the vector u in
standard position with its base at the origin and then slide the vector v till its base coincides
with the point of u. The point of this slid vector, determines u + v. To illustrate, see the
following picture
1
³³
µ
³³ ¡
³
³
¡
£±
¡
u £
¡u + v
£
¡
£ ¡
1
³
£ ¡³³³
³
³
£¡ v
Note the vector u + v is the diagonal of a parallelogram determined from the two vectors u and v and that identifying u + v with the directed diagonal of the parallelogram
determined by the vectors u and v amounts to the same thing as the above procedure.
An item of notation should be mentioned here. In the case of Rn where n ≤ 3, it is
standard notation to use i for e1 , j for e2 , and k for e3 . Now here are some applications of
vector addition to some problems.
Example 2.6.4 There are three ropes attached to a car and three people pull on these ropes.
The first exerts a force of 2i+3j−2k Newtons, the second exerts a force of 3i+5j + k Newtons
FN
22
and the third exerts a force of 5i − j+2k. Newtons. Find the total force in the direction of
i.
To find the total force add the vectors as described above. This gives 10i+7j + k
Newtons. Therefore, the force in the i direction is 10 Newtons.
As mentioned earlier, the Newton is a unit of force like pounds.
Example 2.6.5 An airplane flies North East at 100 miles per hour. Write this as a vector.
A picture of this situation follows.
¡
µ
¡
¡
¡
¡
The vector has length 100. Now using that vector as the
√ hypotenuse of a right triangle
having equal
sides,
the
sides
should
be
each
of
length
100/
2. Therefore, the vector would
√
√
be 100/ 2i + 100/ 2j.
This example also motivates the concept of velocity.
Definition 2.6.6 The speed of an object is a measure of how fast it is going. It is measured
in units of length per unit time. For example, miles per hour, kilometers per minute, feet
per second. The velocity is a vector having the speed as the magnitude but also specifying
the direction.
√
√
Thus the velocity vector in the above example is 100/ 2i + 100/ 2j.
Example 2.6.7 The velocity of an airplane is 100i + j + k measured in kilometers per hour
and at a certain instant of time its position is (1, 2, 1) . Here imagine a Cartesian coordinate
system in which the third component is altitude and the first and second components are
measured on a line from West to East and a line from South to North. Find the position of
this airplane one minute later.
Consider the vector (1, 2, 1) , is the initial position vector of the airplane. As it moves,
the position vector changes. After one minute the airplane has moved in the i direction a
1
1
distance of 100 × 60
= 53 kilometer. In the j direction it has moved 60
kilometer during this
1
same time, while it moves 60 kilometer in the k direction. Therefore, the new displacement
vector for the airplane is
µ
¶ µ
¶
5 1 1
8 121 121
(1, 2, 1) +
, ,
=
,
,
3 60 60
3 60 60
Example 2.6.8 A certain river is one half mile wide with a current flowing at 4 miles per
hour from East to West. A man swims directly toward the opposite shore from the South
bank of the river at a speed of 3 miles per hour. How far down the river does he find himself
when he has swam across? How far does he end up swimming?
Consider the following picture.
2.7. EXERCISES WITH ANSWERS
23
6
3
¾
4
You should write these vectors in terms of components. The velocity of the swimmer in
still water would be 3j while the velocity of the river would be −4i. Therefore, the velocity
of the swimmer is −4i + 3j. Since the component of velocity in the direction across the river
is
The speed at which he travels is
√ 3, it follows the trip takes 1/6 hour or 10 minutes.
42 + 32 = 5 miles per hour and so he travels 5 × 16 = 65 miles. Now to find the distance
downstream he finds himself, note that if x is this distance, x and 1/2 are two legs of a
right triangle whose hypotenuse equals 5/6 miles. Therefore, by the Pythagorean theorem
the distance downstream is
q
2
2
2
(5/6) − (1/2) = miles.
3
2.7
Exercises With Answers
1. The wind blows from West to East at a speed of 30 kilometers per hour and an airplane
which travels at 300 Kilometers per hour in still air is heading North West. What is
the velocity of the airplane relative to the ground? What is the component of this
velocity in the direction North?
Let the positive y axis point in the direction North and let the positive x axis point in
the direction East. The velocity of the wind is 30i. The plane moves in the direction
i + j. A unit vector in this direction is √12 (i + j) . Therefore, the velocity of the plane
√
√ ¢
¡
√ (i + j) = 150 2j + 30 + 150 2 i. The component
relative to the ground is 30i+ 300
2
√
of velocity in the direction North is 150 2.
2. In the situation of Problem 1 how many degrees to the West of North should the
airplane head in order to fly exactly North. What will be the speed of the airplane
relative to the ground?
In this case the unit vector will be − sin (θ) i + cos (θ) j. Therefore, the velocity of the
plane will be
300 (− sin (θ) i + cos (θ) j)
and this is supposed to satisfy
300 (− sin (θ) i + cos (θ) j) + 30i = 0i+?j.
Therefore, you need to have sin θ = 1/10, which means θ = . 100 17 radians. Therefore,
the degrees should be .1×180
= 5. 729
In this case the velocity vector of the
π
´
³ 6 degrees.
plane relative to the ground is 300
√
99
10
j.
3. In the situation of 2 suppose the airplane uses 34 gallons of fuel every hour at that air
speed and that it needs to fly North a distance of 600 miles. Will the airplane have
enough fuel to arrive at its destination given that it has 63 gallons of fuel?
³√ ´
The airplane needs to fly 600 miles at a speed of 300 1099 . Therefore, it takes
600
³
√
³
300
99
10
´´
= 2. 010 1 hours to get there. Therefore, the plane will need to use about
68 gallons of gas. It won't make it.
24
FN
4. A certain river is one half mile wide with a current flowing at 3 miles per hour from
East to West. A man swims directly toward the opposite shore from the South bank
of the river at a speed of 2 miles per hour. How far down the river does he find himself
when he has swam across? How far does he end up swimming?
The velocity of the man relative to the earth is then −3i + 2j. Since the component
of j equals 2 it follows he takes 1/8 of an hour to get across. During this time he is
swept downstream at the rate
3 miles per hour and so he ends up 3/8 of a mile
q¡ of
¢
¡ ¢2
2
3
down stream. He has gone
+ 12 = . 625 miles in all.
8
5. Three forces are applied to a point which does not move. Two of the forces are
2i − j + 3k Newtons and i − 3j − 2k Newtons. Find the third force.
Call it ai + bj + ck Then you need a + 2 + 1 = 0, b − 1 − 3 = 0, and c + 3 − 2 = 0.
Therefore, the force is −3i + 4j − k.
Systems Of Equations
3.0.1
Outcomes
A. Relate the types of solution sets of a system of two or three variables to the intersections
of lines in a plane or the intersection of planes in three space.
B. Determine whether a system of linear equations has no solution, a unique solution or
an infinite number of solutions from its echelon form.
C. Solve a system of equations using Gauss elimination.
D. Model a physical system with linear equations and then solve.
3.1
Systems Of Equations, Geometric Interpretations
As you know, equations like 2x + 3y = 6 can be graphed as straight lines in R2 . To find
the solution to two such equations, you could graph the two straight lines and the ordered
pairs identifying the point (or points) of intersection would give the x and y values of the
solution to the two equations because such an ordered pair satisfies both equations. The
following picture illustrates what can occur with two equations involving two variables.
y
In the first example of the above picture, there is a unique point of intersection. In the
second, there are no points of intersection. The other thing which can occur is that the
two lines are really the same line. For example, x + y = 1 and 2x + 2y = 2 are relations
which when graphed yield the same line. In this case there are infinitely many points in the
simultaneous solution of these two equations, every ordered pair which is on the graph of
the line. It is always this way when considering linear systems of equations. There is either
no solution, exactly one or infinitely many although the reasons for this are not completely
comprehended by considering a simple picture in two dimensions, R2 .
Example 3.1.1 Find the solution to the system x + y = 3, y − x = 5.
25
26
SYSTEMS OF EQUATIONS
You can verify the solution is (x, y) = (−1, 4) . You can see this geometrically by graphing
the equations of the two lines. If you do so correctly, you should obtain a graph which looks
something like the following in which the point of intersection represents the solution of the
two equations.
@
r
(x, y) = (−1, 4) -@¡
¡@
¡
¡
x
Example 3.1.2 You can also imagine other situations such as the case of three intersecting
lines having no common point of intersection or three intersecting lines which do intersect
at a single point as illustrated in the following picture.
y
y
@
@¡
¡@
¡
@
@
¡
¡
@
¡»
»
@»
»»¡
»»» ¡ @@
x
x
In the case of the first picture above, there would be no solution to the three equations
whose graphs are the given lines. In the case of the second picture there is a solution to the
three equations whose graphs are the given lines.
The points, (x, y, z) satisfying an equation in three variables like 2x + 4y − 5z = 8 form
a plane 1 and geometrically, when you solve systems of equations involving three variables,
you are taking intersections of planes. Consider the following picture involving two planes.
¡¡@@
¡
@
¡
@
¡
¡
¡
@
¡@
¡
@
¡
¡
@
¡¡
¡
¡
¡
¡
¡@
¡ ¡
¡
¡¡ ¡
¡
¡
¡
¡
@
¡
¡
¡
@¡
@
@
@
¡
¡
@ ¡
@¡
1 Don't worry about why this is at this time. It is not important. The following discussion is intended
to show you that geometric considerations like this don't take you anywhere. It is the algebraic procedures
which are important and lead to important applications.
3.1. SYSTEMS OF EQUATIONS, GEOMETRIC INTERPRETATIONS
27
Notice how these two planes intersect in a line. It could also happen the two planes
could fail to intersect.
Now imagine a third plane. One thing that could happen is this third plane could have
an intersection with one of the first planes which results in a line which fails to intersect the
first line as illustrated in the following picture.
¡@
New Plane
¡ @
¡
@
¡
ª
¡
¡
¡ ¡
¡¡
¡
¡
¡¡
¡ ¡¡@
@
¡¡
¡
¡
¡¡
¡
¡¡
¡
¡
¡ @
¡ ¡¡ ¡¡
¡
¡
¡
¡
@ ¡ ¡
¡
¡
¡
¡@
@
¡
¡
¡
¡
¡
@
¡
¡
¡
¡
@¡
¡ ¡
@
¡
@
¡
@
¡
@ ¡
@¡
Thus there is no point which lies in all three planes. The picture illustrates the situation
in which the line of intersection of the new plane with one of the original planes forms a line
parallel to the line of intersection of the first two planes. However, in three dimensions, it
is possible for two lines to fail to intersect even though they are not parallel. Such lines are
called skew lines. You might consider whether there exist two skew lines, each of which
is the intersection of a pair of planes selected from a set of exactly three planes such that
there is no point of intersection between the three planes. You can also see that if you tilt
one of the planes you could obtain every pair of planes having a nonempty intersection in a
line and yet there may be no point in the intersection of all three.
It could happen also that the three planes could intersect in a single point as shown in
the following picture.
¡
¡
¡
¡
¡
ª
¡
¡
¡
¡
¡
@
¡@
¡ @
¡
¡
New Plane
¡
¡
¡
¡
¡
¡@
¡ ¡
¡
¡
¡¡
r
@
¡
¡
@¡
@
@
@
¡
@ ¡
@¡
In this case, the three planes have a single point of intersection. The three planes could
Thus in the case of three equations having three variables, the planes determined by
these equations could intersect in a single point, a line, or even fail to intersect at all. You
see that in three dimensions there are many possibilities. If you want to waste some time,
you can try to imagine all the things which could happen but this will not help for more
variables than 3 which is where many of the important applications lie.
Relations like x + y − 2z + 4w = 8 are often called hyper-planes.2 However, it is
impossible to draw pictures of such things.The only rational and useful way to deal with
this subject is through the use of algebra not art. Mathematics exists partly to free us from
having to always draw pictures in order to draw conclusions.
The set of ordered pairs, (x, y) which solve both equations is called the solution set.
You can verify that (x, y) = (5, 2) is a solution to the above system. The interesting
question is this: If you were not given this information to verify, how could you determine
the solution? You can do this by using the following basic operations on the equations, none
of which change the set of solutions of the system of equations.
Definition 3.2.2 Elementary operations are those operations consisting of the following.
1. Interchange the order in which the equations are listed.
2 The evocative semi word, "hyper" conveys absolutely no meaning but is traditional usage which makes
the terminology sound more impressive than something like long wide flat thing.Later we will discuss some
terms which are not just evocative but yield real understanding.
3.2. SYSTEMS OF EQUATIONS, ALGEBRAIC PROCEDURES
29
2. Multiply any equation by a nonzero number.
3. Replace any equation with itself added to a multiple of another equation.
Example 3.2.3 To illustrate the third of these operations on this particular system, consider the following.
x+y =7
2x − y = 8
The system has the same solution set as the system
x+y =7
.
−3y = −6
To obtain the second system, take the second equation of the first system and add -2 times
the first equation to obtain
−3y = −6.
Now, this clearly shows that y = 2 and so it follows from the other equation that x + 2 = 7
and so x = 5.
Of course a linear system may involve many equations and many variables. The solution
set is still the collection of solutions to the equations. In every case, the above operations
of Definition 3.2.2 do not change the set of solutions to the system of linear equations.
Theorem 3.2.4 Suppose you have two equations, involving the variables, (x1 , · · · , xn )
E1 = f1 , E2 = f2
(3.2)
where E1 and E2 are expressions involving the variables and f1 and f2 are constants. (In
the above example there are only two variables, x and y and E1 = x + y while E2 = 2x − y.)
Then the system E1 = f1 , E2 = f2 has the same solution set as
E1 = f1 , E2 + aE1 = f2 + af1 .
(3.3)
Also the system E1 = f1 , E2 = f2 has the same solutions as the system, E2 = f2 , E1 = f1 .
The system E1 = f1 , E2 = f2 has the same solution as the system E1 = f1 , aE2 = af2
provided a 6= 0.
Proof: If (x1 , · · · , xn ) solves E1 = f1 , E2 = f2 then it solves the first equation in
E1 = f1 , E2 +aE1 = f2 +af1 . Also, it satisfies aE1 = af1 and so, since it also solves E2 = f2
it must solve E2 + aE1 = f2 + af1 . Therefore, if (x1 , · · · , xn ) solves E1 = f1 , E2 = f2 it
must also solve E2 + aE1 = f2 + af1 . On the other hand, if it solves the system E1 = f1 and
E2 + aE1 = f2 + af1 , then aE1 = af1 and so you can subtract these equal quantities from
both sides of E2 +aE1 = f2 +af1 to obtain E2 = f2 showing that it satisfies E1 = f1 , E2 = f2 .
The second assertion of the theorem which says that the system E1 = f1 , E2 = f2 has the
same solution as the system, E2 = f2 , E1 = f1 is seen to be true because it involves nothing
more than listing the two equations in a different order. They are the same equations.
The third assertion of the theorem which says E1 = f1 , E2 = f2 has the same solution
as the system E1 = f1 , aE2 = af2 provided a 6= 0 is verified as follows: If (x1 , · · · , xn ) is a
solution of E1 = f1 , E2 = f2 , then it is a solution to E1 = f1 , aE2 = af2 because the second
system only involves multiplying the equation, E2 = f2 by a. If (x1 , · · · , xn ) is a solution
of E1 = f1 , aE2 = af2 , then upon multiplying aE2 = af2 by the number, 1/a, you find that
E2 = f2 .
Stated simply, the above theorem shows that the elementary operations do not change
the solution set of a system of equations.
30
SYSTEMS OF EQUATIONS
Here is an example in which there are three equations and three variables. You want to
find values for x, y, z such that each of the given equations are satisfied when these values
are plugged in to the equations.
Example 3.2.5 Find the solutions to the system,
x + 3y + 6z = 25
2x + 7y + 14z = 58
2y + 5z = 19
(3.4)
To solve this system replace the second equation by (−2) times the first equation added
to the second. This yields the system
x + 3y + 6z = 25
y + 2z = 8
2y + 5z = 19
(3.5)
Now take (−2) times the second and add to the third. More precisely, replace the third
equation with (−2) times the second added to the third. This yields the system
x + 3y + 6z = 25
y + 2z = 8
z=3
(3.6)
At this point, you can tell what the solution is. This system has the same solution as the
original system and in the above, z = 3. Then using this in the second equation, it follows
y + 6 = 8 and so y = 2. Now using this in the top equation yields x + 6 + 18 = 25 and so
x = 1. This process is called back substitution.
Alternatively, in 3.6 you could have continued as follows. Add (−2) times the bottom
equation to the middle and then add (−6) times the bottom to the top. This yields
x + 3y = 7
y=2
z=3
Now add (−3) times the second to the top. This yields
x=1
y=2 ,
z=3
a system which has the same solution set as the original system. This avoided back substitution and led to the same solution set.
3.2.2
Gauss Elimination
A less cumbersome way to represent a linear system is to write it as an augmented matrix.
For example the linear system, 3.4 can be written as
1 3 6 | 25
2 7 14 | 58 .
0 2 5 | 19
It has exactly
the same
it is understood there is
informationas the
original system but
here
1
3
6
an x column, 2 , a y column, 7 and a z column, 14 . The rows correspond
0
2
5
3.2. SYSTEMS OF EQUATIONS, ALGEBRAIC PROCEDURES
31
to the equations in the system. Thus the top row in the augmented matrix corresponds to
the equation,
x + 3y + 6z = 25.
Now when you replace an equation with a multiple of another equation added to itself, you
are just taking a row of this augmented matrix and replacing it with a multiple of another
row added to it. Thus the first step in solving 3.4 would be to take (−2) times the first row
of the augmented matrix above and add it to the second row,
1 3 6 | 25
0 1 2 | 8 .
0 2 5 | 19
Note how this corresponds to 3.5. Next
third,
1
0
0
take (−2) times the second row and add to the
3
1
0
6
2
1
| 25
|8
|3
This augmented matrix corresponds to the system
x + 3y + 6z = 25
y + 2z = 8
z=3
which is the same as 3.6. By back substitution you obtain the solution x = 1, y = 6, and
z = 3.
In general a linear system is of the form
a11 x1 + · · · + a1n xn = b1
..
.
,
(3.7)
am1 x1 + · · · + amn xn = bm
where the xi are variables and the aij and bi are constants. This system can be represented
by the augmented matrix,
a11 · · · a1n | b1
..
..
.
(3.8)
.
.
| .. .
am1 · · · amn | bm
Changes to the system of equations in 3.7 as a result of an elementary operations translate
into changes of the augmented matrix resulting from a row operation. Note that Theorem
3.2.4 implies that the row operations deliver an augmented matrix for a system of equations
which has the same solution set as the original system.
Definition 3.2.6 The row operations consist of the following
1. Switch two rows.
2. Multiply a row by a nonzero number.
3. Replace a row by a multiple of another row added to it.
Gauss elimination is a systematic procedure to simplify an augmented matrix to a
reduced form. In the following definition, the term "leading entry" refers to the first
nonzero entry of a row when scanning the row from left to right.
32
SYSTEMS OF EQUATIONS
Definition 3.2.7 An augmented matrix is in echelon form if
1. All nonzero rows are above any rows of zeros.
2. Each leading entry of a row is in a column to the right of the leading entries of any
rows above it.
Definition 3.2.8 An augmented matrix is in row reduced echelon form if
1. All nonzero rows are above any rows of zeros.
2. Each leading entry of a row is in a column to the right of the leading entries of any
rows above it.
3. All entries in a column above and below a leading entry are zero.
4. Each leading entry is a 1, the only nonzero entry in its column.
Example 3.2.9 Here are some augmented matrices which are in row reduced echelon form.
Definition 3.2.12 A pivot position in a matrix is the location of a leading entry in an
echelon form resulting from the application of row operations to the matrix. A pivot column
is a column that contains a pivot position.
For example consider the following.
Example 3.2.13 Suppose
This is still not in reduced echelon form. Replace the bottom row by −1 times the middle
row added to the bottom. This yields
1 2
3 | 4
0 −4 −8 | −6
0 0
0 | 0
which is in echelon form, although not in reduced echelon form. Therefore, the pivot positions in the original matrix are the locations corresponding to the first row and first column
and the second row and second columns as shown in the following:
1
3
4
2
2
4
3 |
1 |
4 |
4
6
10
Thus the pivot columns in the matrix are the first two columns.
The following is the algorithm for obtaining a matrix which is in row reduced echelon
form.
Algorithm 3.2.14
This algorithm tells how to start with a matrix and do row operations on it in such a
way as to end up with a matrix in row reduced echelon form.
1. Find the first nonzero column from the left. This is the first pivot column. The
position at the top of the first pivot column is the first pivot position. Switch rows if
necessary to place a nonzero number in the first pivot position.
2. Use row operations to zero out the entries below the first pivot position.
3. Ignore the row containing the most recent pivot position identified and the rows above
it. Repeat steps 1 and 2 to the remaining sub-matrix, the rectangular array of numbers
obtained from the original matrix by deleting the rows you just ignored. Repeat the
process until there are no more rows to modify. The matrix will then be in echelon
form.
4. Moving from right to left, use the nonzero elements in the pivot positions to zero out
the elements in the pivot columns which are above the pivots.
5. Divide each nonzero row by the value of the leading entry. The result will be a matrix
in row reduced echelon form.
34
SYSTEMS OF EQUATIONS
This row reduction procedure applies to both augmented matrices and non augmented
matrices. There is nothing special about the augmented column with respect to the row
reduction procedure.
Example 3.2.15 Here is a matrix.
0
0
0
0
0
0
1
0
0
0
2
1
1
0
0
3
4
2
0
2
2
3
2
0
1
Do row reductions till you obtain a matrix in echelon form. Then complete the process by
producing one in reduced echelon form.
The pivot column is the second. Hence the pivot position is the one in the first row and
second column. Switch the first two rows to obtain a nonzero entry in this pivot position.
0 1 1 4 3
0 0 2 3 2
0 0 1 2 2
0 0 0 0 0
0 0 0 2 1
Step two is not necessary because all the entries below the first pivot position in the resulting
matrix are zero. Now ignore the top row and the columns to the left of this first pivot
position. Thus you apply the same operations to the smaller matrix,
2 3 2
1 2 2
0 0 0 .
0 2 1
The next pivot column is the third corresponding to the first in this smaller matrix and the
second pivot position is therefore, the one which is in the second row and third column. In
this case it is not necessary to switch any rows to place a nonzero entry in this position
because there is already a nonzero entry there. Multiply the third row of the original matrix
by −2 and then add the second row to it. This yields
0 1 1 4
3
0 0 2 3
2
0 0 0 −1 −2 .
0 0 0 0
0
0 0 0 2
1
The next matrix the steps in the algorithm are applied to is
−1 −2
0
0 .
2
1
The first pivot column is the first column in this case and no switching of rows is necessary
because there is a nonzero entry in the first pivot position. Therefore, the algorithm yields
The above algorithm is the way a computer would obtain a reduced echelon form for a
given matrix. It is not necessary for you to pretend you are a computer but if you like to do
so, the algorithm described above will work. The main idea is to do row operations in such
a way as to end up with a matrix in echelon form or row reduced echelon form because when
this has been done, the resulting augmented matrix will allow you to describe the solutions
to the linear system of equations in a meaningful way.
Example 3.2.16 Give the complete solution to the system of equations, 5x+10y−7z = −2,
2x + 4y − 3z = −1, and 3x + 6y + 5z = 9.
The augmented matrix for this system is
2 4 −3
5 10 −7
3 6
5
| −1
| −2
| 9
Multiply the second row by 2, the first row by 5, and then take (−1) times the first row and
add to the second. Then multiply the first row by 1/5. This yields
2 4 −3 | −1
0 0 1 | 1
3 6 5 | 9
Now, combining some row operations, take (−3) times the first row and add this to 2 times
the last row and replace the last row with this. This yields.
2 4 −3 | −1
0 0 1 | 1 .
0 0 1 | 21
One more row operation, taking (−1) times the second row and adding to the bottom yields.
2 4 −3 | −1
0 0 1 | 1 .
0 0 0 | 20
This is impossible because the last row indicates the need for a solution to the equation
0x + 0y + 0z = 20
and there is no such thing because 0 6= 20. This shows there is no solution to the three given
equations. When this happens, the system is called inconsistent. In this case it is very
easy to describe the solution set. The system has no solution.
Here is another example based on the use of row operations.
Example 3.2.17 Give the complete solution to the system of equations, 3x − y − 5z = 9,
y − 10z = 0, and −2x + y = −6.
The entry, 3 in this sequence of row operations is called the pivot. It is used to create
zeros in the other places of the column. Next take −1 times the middle row and add to the
bottom. Here the 1 in the second row is the pivot.
3 −1 −5 | 9
0 1 −10 | 0
0 0
0
| 0
Take the middle row and add to the top and then divide the top row which results by 3.
1 0 −5 | 3
0 1 −10 | 0 .
0 0
0
| 0
This is in reduced echelon form. The equations corresponding to this reduced echelon form
are y = 10z and x = 3 + 5z. Apparently z can equal any number. Lets call this number,
t. 3 Therefore, the solution set of this system is x = 3 + 5t, y = 10t, and z = t where t
is completely arbitrary. The system has an infinite set of solutions which are given in the
above simple way. This is what it is all about, finding the solutions to the system.
There is some terminology connected to this which is useful. Recall how each column
corresponds to a variable in the original system of equations. The variables corresponding to
a pivot column are called basic variables . The other variables are called free variables.
In Example 3.2.17 there was one free variable, z, and two basic variables, x and y. In describing the solution to the system of equations, the free variables are assigned a parameter.
In Example 3.2.17 this parameter was t. Sometimes there are many free variables and in
these cases, you need to use many parameters. Here is another example.
Example 3.2.18 Find the solution to the system
x + 2y − z + w = 3
x+y−z+w =1
x + 3y − z + w = 5
The augmented matrix is
1 2
1 1
1 3
−1
−1
−1
1 |
1 |
1 |
3
1 .
5
Take −1 times the first row and add to the second. Then take −1 times the first row and
add to the third. This yields
1 2 −1 1 | 3
0 −1 0 0 | −2
0 1
0 0 | 2
3 In
This matrix is in echelon form and you see the basic variables are x and y while the free
variables are z and w. Assign s to z and t to w. Then the second row yields the equation,
y = 2 while the top equation yields the equation, x + 2y − s + t = 3 and so since y = 2, this
gives x + 4 − s + t = 3 showing that x = −1 + s − t, y = 2, z = s, and w = t. It is customary
to write this in the form
−1 + s − t
x
y
2
.
(3.10)
z =
s
w
t
This is another example of a system which has an infinite solution set but this time
the solution set depends on two parameters, not one. Most people find it less confusing
in the case of an infinite solution set to first place the augmented matrix in row reduced
echelon form rather than just echelon form before seeking to write down the description of
the solution. In the above, this means we don't stop with the echelon form 3.9. Instead we
first place it in reduced echelon form as follows.
1 0 −1 1 | −1
0 1 0 0 | 2 .
0 0 0 0 | 0
Then the solution is y = 2 from the second row and x = −1 + z − w from the first. Thus
letting z = s and w = t, the solution is given in 3.10.
The number of free variables is always equal to the number of different parameters
used to describe the solution. If there are no free variables, then either there is no solution
as in the case where row operations yield an echelon form like
1 2 | 3
0 4 | −2
0 0 | 1
or there is a unique solution as in the case
1 2
0 4
0 0
Also, sometimes there are free variables and no solution as in the following:
1 2 2 | 3
0 4 3 | −2 .
0 0 0 | 1
There are a lot of cases to consider but it is not necessary to make a major production of
this. Do row operations till you obtain a matrix in echelon form or reduced echelon form
and determine whether there is a solution. If there is, see if there are free variables. In this
case, there will be infinitely many solutions. Find them by assigning different parameters
to the free variables and obtain the solution. If there are no free variables, then there will
be a unique solution which is easily determined once the augmented matrix is in echelon
3.2. SYSTEMS OF EQUATIONS, ALGEBRAIC PROCEDURES
39
or row reduced echelon form. In every case, the process yields a straightforward way to
describe the solutions to the linear system. As indicated above, you are probably less likely
to become confused if you place the augmented matrix in row reduced echelon form rather
than just echelon form.
In summary,
Definition 3.2.19 A system of linear equations is a list of equations,
a11 x1 + a12 x2 + · · · + a1n xn = b1
a21 x1 + a22 x2 + · · · + a2n xn = b2
..
.
am1 x1 + am2 x2 + · · · + amn xn = bm
where aij are numbers, and bj is a number. The above is a system of m equations in the n
variables, x1 , x2 · · · , xn . Nothing is said about the relative size of m and n. Written more
simply in terms of summation notation, the above can be written in the form
n
X
aij xj = fj , i = 1, 2, 3, · · · , m
j=1
It is desired to find (x1 , · · · , xn ) solving each of the equations listed.
As illustrated above, such a system of linear equations may have a unique solution, no
solution, or infinitely many solutions and these are the only three cases which can occur for
any linear system. Furthermore, you do exactly the same things to solve any linear system.
You write the augmented matrix and do row operations until you get a simpler system in
which it is possible to see the solution, usually obtaining a matrix in echelon or reduced
echelon form. All is based on the observation that the row operations do not change the
solution set. You can have more equations than variables, fewer equations than variables,
etc. It doesn't matter. You always set up the augmented matrix and go to work on it.
Definition 3.2.20 A system of linear equations is called consistent if there exists a solution. It is called inconsistent if there is no solution.
These are reasonable words to describe the situations of having or not having a solution. If you think of each equation as a condition which must be satisfied by the variables,
consistent would mean there is some choice of variables which can satisfy all the conditions.
Inconsistent would mean there is no choice of the variables which can satisfy each of the
conditions.
You have now solved systems of equations by writing them in terms of an augmented matrix
and then doing row operations on this augmented matrix. It turns out such rectangular
arrays of numbers are important from many other different points of view. Numbers are
also called scalars. In these notes numbers will always be either real or complex numbers.
I will refer to the set of numbers as F sometimes when it is not important to worry about
whether the number is real or complex. Thus F can be either the real numbers, R or the
complex numbers, C.
A matrix is a rectangular array of numbers. Several of them are referred to as matrices.
For example, here is a matrix.
1 2 3 4
5 2 8 7
6 −9 1 2
The size or dimension of a matrix is defined as m × n where m is the number of rows and n
is the number of columns. The above matrix is a 3 × 4 matrix because there are three rows
and four columns. The first row is (1 2 3 4) , the second row is (5 2 8 7) and so forth. The
41
42
MATRICES
1
first column is 5 . When specifying the size of a matrix, you always list the number of
6
rows before the number of columns. Also, you can remember the columns are like columns
in a Greek temple. They stand upright while the rows just lay there like rows made by
a tractor in a plowed field. Elements of the matrix are identified according to position in
the matrix. For example, 8 is in position 2, 3 because it is in the second row and the third
column. You might remember that you always list the rows before the columns by using
the phrase Rowman Catholic. The symbol, (aij ) refers to a matrix. The entry in the ith
row and the j th column of this matrix is denoted by aij . Using this notation on the above
matrix, a23 = 8, a32 = −9, a12 = 2, etc.
There are various operations which are done on matrices. Matrices can be added multiplied by a scalar, and multiplied by other matrices. To illustrate scalar multiplication,
consider the following example in which a matrix is being multiplied by the scalar, 3.
3
6
9 12
1 2 3 4
6
24 21 .
3 5 2 8 7 = 15
6 −9 1 2
18 −27 3
6
The new matrix is obtained by multiplying every entry of the original matrix by the given
scalar. If A is an m × n matrix, −A is defined to equal (−1) A.
Two matrices must be the same size to be added. The sum of two matrices is a matrix
which is obtained by adding the corresponding entries. Thus
1 2
−1 4
0
6
3 4 + 2
8 = 5 12 .
5 2
6 −4
11 −2
Two matrices are equal exactly when they are the same size and the corresponding entries
are identical. Thus
µ
¶
0 0
0 0
0 0 6=
0 0
0 0
because they are different sizes. As noted above, you write (cij ) for the matrix C whose
ij th entry is cij . In doing arithmetic with matrices you must define what happens in terms
of the cij sometimes called the entries of the matrix or the components of the matrix.
The above discussion stated for general matrices is given in the following definition.
Definition 4.1.1 (Scalar Multiplication) If A = (aij ) and k is a scalar, then kA = (kaij ) .
µ
¶ µ
¶
2 0
14
0
Example 4.1.2 7
=
.
1 −4
7 −28
Definition 4.1.3 (Addition) If A = (aij ) and B = (bij ) are two m × n matrices. Then
A + B = C where
C = (cij )
for cij = aij + bij .
Example 4.1.4
µ
1 2
1 0
3
4
¶
µ
+
5 2 3
−6 2 1
¶
µ
=
6 4 6
−5 2 5
¶
4.1. MATRIX ARITHMETIC
43
To save on notation, we will often use Aij to refer to the ij th entry of the matrix, A.
Definition 4.1.5 (The zero matrix) The m × n zero matrix is the m × n matrix having
every entry equal to zero. It is denoted by 0.
µ
¶
0 0 0
Example 4.1.6 The 2 × 3 zero matrix is
.
0 0 0
Note there are 2 × 3 zero matrices, 3 × 4 zero matrices, etc. In fact there is a zero matrix
for every size.
Definition 4.1.7 (Equality of matrices) Let A and B be two matrices. Then A = B means
that the two matrices are of the same size and for A = (aij ) and B = (bij ) , aij = bij for all
1 ≤ i ≤ m and 1 ≤ j ≤ n.
The following properties of matrices can be easily verified. You should do so.
• Commutative Law Of Addition.
A + B = B + A,
• Rule for Multiplication by 1.
As an example, consider the Commutative Law of Addition. Let A + B = C and
B + A = D. Why is D = C?
Cij = Aij + Bij = Bij + Aij = Dij .
Therefore, C = D because the ij th entries are the same. Note that the conclusion follows
from the commutative law of addition of numbers.
44
4.1.2
MATRICES
Multiplication Of Matrices
Definition 4.1.8 Matrices which are n×1 or 1×n are called vectors and are often denoted
by a bold letter. Thus the n × 1 matrix
x1
x = ...
xn
is also called a column vector. The 1 × n matrix
(x1 · · · xn )
is called a row vector.
Although the following description of matrix multiplication may seem strange, it is in
fact the most important and useful of the matrix operations. To begin with consider the
case where a matrix is multiplied by a column vector. First consider a special case.
µ
¶
7
1 2 3
8 =?
4 5 6
9
One way to remember this is as follows. Slide the vector, placing it on top the two rows as
shown and then do the indicated operation.
7 8 9
µ
¶ µ
¶
1
2
3
7×1+8×2+9×3
50
→
=
.
7 8 9
7×4+8×5+9×6
122
4 5 6
multiply the numbers on the top by the numbers on the bottom and add them up to get a
single number for each row of the matrix as shown above.
In more general terms,
µ
¶
µ
¶
x1
a11 x1 + a12 x2 + a13 x3
a11 a12 a13
x2 =
.
a21 x1 + a22 x2 + a23 x3
a21 a22 a23
x3
Another way to think of this is
µ
¶
µ
¶
µ
¶
a11
a12
a13
x1
+ x2
+ x3
a21
a22
a23
Thus you take x1 times the first column, add to x2 times the second column, and finally
x3 times the third column. In general, here is the definition of how to multiply an (m × n)
matrix times a (n × 1) matrix.
Definition 4.1.9 Let A = Aij be an m × n matrix and let v be an n × 1 matrix,
v1
v = ...
vn
Then Av is an m × 1 matrix and the ith component of this matrix is
(Av)i = Ai1 v1 + Ai2 v2 + · · · + Ain vn =
n
X
j=1
Aij vj .
4.1. MATRIX ARITHMETIC
45
Thus
Pn
A1j vj
..
Av =
.
Pn .
j=1 Amj vj
j=1
(4.9)
In other words, if
A = (a1 , · · · , an )
where the ak are the columns,
Av =
The next task is to multiply an m × n matrix times an n × p matrix. Before doing so,
the following may be helpful.
For A and B matrices, in order to form the product, AB the number of columns of A
must equal the number of rows of B.
these must match!
[
n)
(n × p
(m ×
)=m×p
Note the two outside numbers give the size of the product. Remember:
If the two middle numbers don't match, you can't multiply the matrices!
Definition 4.1.11 When the number of columns of A equals the number of rows of B the
two matrices are said to be conformable and the product, AB is obtained as follows. Let
A be an m × n matrix and let B be an n × p matrix. Then B is of the form
B = (b1 , · · · , bp )
where bk is an n × 1 matrix or column vector. Then the m × p matrix, AB is defined as
follows:
AB ≡ (Ab1 , · · · , Abp )
(4.10)
where Abk is an m × 1 matrix or column vector which gives the k th column of AB.
Example 4.1.12 Multiply the following.
µ
First check if it is possible. This is of the form (3 × 3) (2 × 3) . The inside numbers do not
match and so you can't do this multiplication. This means that anything you write will be
absolute nonsense because it is impossible to multiply these matrices in this order. Aren't
they the same two matrices considered in the previous example? Yes they are. It is just
that here they are in a different order. This shows something you must always remember
about matrix multiplication.
Order Matters!
Matrix Multiplication Is Not Commutative!
This is very different than multiplication of numbers!
4.1.3
The ij th Entry Of A Product
It is important to describe matrix multiplication in terms of entries of the matrices. What
is the ij th entry of AB? It would be the ith entry of the j th column of AB. Thus it would
be the ith entry of Abj . Now
B1j
bj = ...
Bnj
and from the above definition, the ith entry is
n
X
This shows the following definition for matrix multiplication in terms of the ij th entries of
the product coincides with Definition 4.1.11.
Definition 4.1.14 Let A = (Aij ) be an m × n matrix and let B = (Bij ) be an n × p matrix.
Then AB is an m × p matrix and
(AB)ij =
n
X
Aik Bkj .
(4.12)
k=1
Another way to write this is
(AB)ij =
¡
Ai1
Ai2
···
Ain
¢
B1j
B2j
..
.
Bnj
Note that to get (AB)ij you involve the ith
1
Example 4.1.15 Multiply if possible 3
2
row of A and the j th column of B.
µ
¶
2
2 3 1
1
.
7 6 2
6
First check to see if this is possible. It is of the form (3 × 2) (2 × 3) and since the inside
numbers match, the two matrices are conformable and it is possible to do the multiplication.
The result should be a 3 × 3 matrix. The answer is of the form
µ ¶
µ ¶
µ ¶
1 2
1 2
1 2
3 1 2 , 3 1 3 , 3 1 1
7
6
2
2 6
2 6
2 6
where the commas separate the columns in the
equals
16 15
13 15
46 42
resulting product. Thus the above product
5
5 ,
14
a 3 × 3 matrix as desired. In terms of the ij th entries and the above definition, the entry in
the third row and second column of the product should equal
X
a3k bk2 = a31 b12 + a32 b22
j
=
2 × 3 + 6 × 6 = 42.
You should try a few more such examples to verify the
entries works for other entries.
1 2
2
Example 4.1.16 Multiply if possible 3 1 7
2 6
0
above definition in terms of the ij th
3
6
0
1
2 .
0
4.1. MATRIX ARITHMETIC
49
This is not possible because it is of the form (3 × 2) (3 × 3) and the middle numbers
don't match. In other words the two matrices are not conformable in the indicated order.
2 3 1
1 2
Example 4.1.17 Multiply if possible 7 6 2 3 1 .
0 0 0
2 6
This is possible because in this case it is of the form (3 × 3) (3 × 2) and the middle
numbers do match so the matrices are conformable. When the multiplication is done it
equals
13 13
29 32 .
0 0
Check this and be sure you come up with the same answer.
1
¡
¢
Example 4.1.18 Multiply if possible 2 1 2 1 0 .
1
In this case you are trying to do (3 × 1) (1 × 4) .
do it. Verify
1
¡
¢
2 1 2 1 0 =
1
4.1.4
The inside numbers match so you can
1
2
1
2
4
2
0
0
0
1
2
1
Properties Of Matrix Multiplication
As pointed out above, sometimes it is possible to multiply matrices in one order but not
in the other order. What if it makes sense to multiply them in either order? Will the two
products be equal then?
µ
¶µ
¶
µ
¶µ
¶
1 2
0 1
0 1
1 2
Example 4.1.19 Compare
and
.
3 4
1 0
1 0
3 4
The first product is
The second product is
µ
µ
1 2
3 4
0 1
1 0
¶µ
¶µ
0
1
1
0
1
3
2
4
¶
µ
=
¶
µ
=
2
4
1
3
3
1
4
2
¶
.
¶
.
You see these are not equal. Again you cannot conclude that AB = BA for matrix multiplication even when multiplication is defined in both orders. However, there are some
properties which do hold.
Proposition 4.1.20 If all multiplications and additions make sense, the following hold for
matrices, A, B, C and a, b scalars.
A (aB + bC) = a (AB) + b (AC)
Another important operation on matrices is that of taking the transpose. The following
example shows what is meant by this operation, denoted by placing a T as an exponent on
the matrix.
T
µ
¶
1 4
1 3 2
3 1 =
4 1 6
2 6
What happened? The first column became the first row and the second column became the
second row. Thus the 3 × 2 matrix became a 2 × 3 matrix. The number 3 was in the second
row and the first column and it ended up in the first row and second column. Here is the
definition.
Definition 4.1.21 Let A be an m × n matrix. Then AT denotes the n × m matrix which
is defined as follows.
¡ T¢
A ij = Aji
Example 4.1.22
µ
1
3
2
5
−6
4
¶T
1 3
= 2 5 .
−6 4
The transpose of a matrix has the following important properties.
Lemma 4.1.23 Let A be an m × n matrix and let B be a n × p matrix. Then
T
(AB) = B T AT
(4.16)
4.1. MATRIX ARITHMETIC
51
and if α and β are scalars,
T
(αA + βB) = αAT + βB T
Proof: From the definition,
³
´
T
(AB)
=
ij
=
(AB)ji
X
Ajk Bki
k
=
(4.17)
X¡
BT
¢ ¡
ik
AT
¢
kj
k
=
¡ T T¢
B A ij
The proof of Formula 4.17 is left as an exercise and this proves the lemma.
Definition 4.1.24 An n × n matrix, A is said to be symmetric if A = AT . It is said to
be skew symmetric if A = −AT .
Example 4.1.25 Let
2
A= 1
3
1
5
−3
3
−3 .
7
Then A is symmetric.
Example 4.1.26 Let
0
A = −1
−3
1 3
0 2
−2 0
Then A is skew symmetric.
4.1.6
The Identity And Inverses
There is a special matrix called I and referred to as the identity matrix. It is always a
square matrix, meaning the number of rows equals the number of columns and it has the
property that there are ones down the main diagonal and zeroes elsewhere. Here are some
identity matrices of various sizes.
1 0 0 0
µ
¶
1 0 0
0 1 0 0
1 0
(1) ,
, 0 1 0 ,
0 0 1 0 .
0 1
0 0 1
0 0 0 1
The first is the 1 × 1 identity matrix, the second is the 2 × 2 identity matrix, the third is
the 3 × 3 identity matrix, and the fourth is the 4 × 4 identity matrix. By extension, you can
likely see what the n × n identity matrix would be. It is so important that there is a special
symbol to denote the ij th entry of the identity matrix
Iij = δ ij
where δ ij is the Kroneker symbol defined by
½
1 if i = j
δ ij =
0 if i =
6 j
It is called the identity matrix because it is a multiplicative identity in the following
sense.
52
MATRICES
Lemma 4.1.27 Suppose A is an m × n matrix and In is the n × n identity matrix. Then
AIn = A. If Im is the m × m identity matrix, it also follows that Im A = A.
Proof:
(AIn )ij
X
=
Aik δ kj
k
=
Aij
and so AIn = A. The other case is left as an exercise for you.
Definition 4.1.28 An n×n matrix, A has an inverse, A−1 if and only if AA−1 = A−1 A =
I. Such a matrix is called invertible.
It is very important to observe that the inverse of a matrix, if it exists, is unique. Another
way to think of this is that if it acts like the inverse, then it is the inverse.
Theorem 4.1.29 Suppose A−1 exists and AB = BA = I. Then B = A−1 .
Proof:
¡
¢
A−1 = A−1 I = A−1 (AB) = A−1 A B = IB = B.
Unlike ordinary multiplication of numbers, it can happen that A 6= 0 but A may fail to
have an inverse. This is illustrated in the following example.
µ
¶
1 1
Example 4.1.30 Let A =
. Does A have an inverse?
1 1
One might think A would have an inverse because it does not equal zero. However,
¶ µ ¶
¶µ
µ
0
−1
1 1
=
0
1
1 1
and if A−1 existed, this could not happen because you could write
µ ¶
µµ ¶¶
µ µ
¶¶
0
0
−1
= A−1
= A−1 A
=
0
0
1
µ
¶
µ
¶ µ
¶
¡ −1 ¢ −1
−1
−1
= A A
=I
=
,
1
1
1
a contradiction. Thus the answer is that A does not have an inverse.
µ
Example 4.1.31 Let A =
for the second. Lets solve the first system. Take (−1) times the first row and add to the
second to get
µ
¶
1 1 | 1
0 1 | −1
Now take (−1) times the second row and add to the first to get
µ
¶
1 0 | 2
.
0 1 | −1
Putting in the variables, this says x = 2 and y = −1.
Now solve the second system, 4.19 to find z and w. Take (−1) times the first row and
add to the second to get
¶
µ
1 1 | 0
.
0 1 | 1
Now take (−1) times the second row and add to the first to get
µ
¶
1 0 | −1
.
0 1 | 1
Putting in the variables, this says z = −1 and w = 1. Therefore, the inverse is
µ
¶
2 −1
.
−1 1
Didn't the above seem rather repetitive? Note that exactly the same row operations
were used in both systems. In each case, the end result was something ofµ the ¶form (I|v)
x
where I is the identity and v gave a column of the inverse. In the above,
, the first
y
µ
¶
z
column of the inverse was obtained first and then the second column
.
w
and read off the inverse as the 2 × 2 matrix on the right side.
This is the reason for the following simple procedure for finding the inverse of a matrix.
This procedure is called the Gauss-Jordan procedure.
Procedure 4.1.32 Suppose A is an n × n matrix. To find A−1 if it exists, form the
augmented n × 2n matrix,
(A|I)
and then, if possible do row operations until you obtain an n × 2n matrix of the form
(I|B) .
(4.20)
When this has been done, B = A−1 . If it is impossible to row reduce to a matrix of the form
(I|B) , then A has no inverse.
1 2 2
Example 4.1.33 Let A = 1 0 2 . Find A−1 if it exists.
3 1 −1
Set up the augmented matrix, (A|I)
1 2 2
1 0 2
3 1 −1
Next take (−1) times the first row and
row added to the last. This yields
1 2
0 −2
0 −5
| 1 0
| 0 1
| 0 0
0
0
1
add to the second followed by (−3) times the first
2
0
−7
| 1
| −1
| −3
0 0
1 0 .
0 1
Then take 5 times the second row and add to -2 times the last row.
1
2
2 | 1 0 0
0 −10 0 | −5 5 0
0
0
14 | 1 5 −2
Next take the last row and add to
−7
0
0
¡
¢
and proceed to do row operations attempting to obtain I|A−1 . Take (−1) times the top
row and add to the second. Then take (−2) times the top row and add to the bottom.
1
0
0
2 2
−2 0
−2 0
| 1
| −1
| −2
0 0
1 0
0 1
Next add (−1) times the second row to the bottom row.
1
0
0
2 2
−2 0
0 0
| 1
| −1
| −1
0 0
1 0
−1 1
At this point, you can see there will be no inverse because you have obtained a row of zeros
in the left half of the augmented matrix, (A|I) . Thus there will be no way to obtain I on
the left.
1 0
1
Example 4.1.35 Let A = 1 −1 1 . Find A−1 if it exists.
1 1 −1
Form the augmented matrix,
Always check your answer because if you are like some of us, you will usually have made a
mistake.
Example 4.1.36 In this example, it is shown how to use the inverse of a matrix to find
the solution to a system of equations. Consider the following system of equations. Use the
inverse of a suitable matrix to give the solutions to this system.
x+z =1
x − y + z = 3 .
x+y−z =2
The system of equations can be written in terms of matrices as
1 0
1
x
1
1 −1 1 y = 3 .
1 1 −1
z
2
(4.21)
More simply, this is of the form Ax = b. Suppose you find the inverse of the matrix, A−1 .
Then you could multiply both sides of this equation by A−1 to obtain
¡
¢
x = A−1 A x = A−1 (Ax) = A−1 b.
This gives the solution as x = A−1 b. Note that once you have found the inverse, you can
easily get the solution for different right hand sides without any effort. It is always just
A−1 b. In the given example, the inverse of the matrix is
1
1
0
2
2
1 −1
0
1 − 21 − 12
This was shown in Example 4.1.35. Therefore, from what was just explained the solution
to the given system is
5
1
1
x
0
1
2
2
2
y = 1 −1
0 3 = −2 .
z
2
1 − 12 − 12
− 32
4.1. MATRIX ARITHMETIC
57
What if the right side of 4.21 had been
0
1 ?
3
What would be the solution to
1
1
1
0
−1
1
By the above discussion, it is just
x
0
y = 1
z
1
x
0
1
1 y = 1 ?
3
z
−1
1
2
−1
− 12
0
2
0 1 = −1 .
3
−2
− 12
1
2
This illustrates why once you have found the inverse of a given matrix, you can use it to
solve many different systems easily.
58
MATRICES
Vector Products
5.0.8
Outcomes
1. Evaluate a dot product from the angle formula or the coordinate formula.
2. Interpret the dot product geometrically.
3. Evaluate the following using the dot product:
(a) the angle between two vectors
(b) the magnitude of a vector
(c) the work done by a constant force on an object
4. Evaluate a cross product from the angle formula or the coordinate formula.
5. Interpret the cross product geometrically.
6. Evaluate the following using the cross product:
(a) the area of a parallelogram
(b) the area of a triangle
(c) physical quantities such as the torque and angular velocity.
7. Find the volume of a parallelepiped using the box product.
8. Recall, apply and derive the algebraic properties of the dot and cross products.
5.1
The Dot Product
There are two ways of multiplying vectors which are of great importance in applications.
The first of these is called the dot product, also called the scalar product and sometimes
the inner product.
Definition 5.1.1 Let a, b be two vectors in Rn define a · b as
a·b≡
n
X
ak bk .
k=1
With this definition, there are several important properties satisfied by the dot product.
In the statement of these properties, α and β will denote scalars and a, b, c will denote
vectors.
59
60
VECTOR PRODUCTS
Proposition 5.1.2 The dot product satisfies the following properties.
a·b=b·a
Furthermore equality is obtained if and only if one of a or b is a scalar multiple of the other.
Proof: First note that if b = 0 both sides of 5.6 equal zero and so the inequality holds
in this case. Therefore, it will be assumed in what follows that b 6= 0.
Define a function of t ∈ R
f (t) = (a + tb) · (a + tb) .
Then by 5.2, f (t) ≥ 0 for all t ∈ R. Also from 5.3,5.4,5.1, and 5.5
f (t) = a · (a + tb) + tb · (a + tb)
= a · a + t (a · b) + tb · a + t2 b · b
2
2
= |a| + 2t (a · b) + |b| t2 .
Now this means the graph, y = f (t) is a polynomial which opens up and either its vertex
touches the t axis or else the entire graph is above the x axis. In the first case, there exists
some t where f (t) = 0 and this requires a + tb = 0 so one vector is a multiple of the other.
Then clearly equality holds in 5.6. In the case where b is not a multiple of a, it follows
f (t) > 0 for all t which says f (t) has no real zeros and so from the quadratic formula,
2
2
2
(2 (a · b)) − 4 |a| |b| < 0
which is equivalent to |(a · b)| < |a| |b|. This proves the theorem.
You should note that the entire argument was based only on the properties of the dot
product listed in 5.1 - 5.5. This means that whenever something satisfies these properties,
the Cauchy Schwartz inequality holds. There are many other instances of these properties
besides vectors in Rn .
The Cauchy Schwartz inequality allows a proof of the triangle inequality for distances
in Rn in much the same way as the triangle inequality for the absolute value.
5.2. THE GEOMETRIC SIGNIFICANCE OF THE DOT PRODUCT
61
Theorem 5.1.6 (Triangle inequality) For a, b ∈ Rn
|a + b| ≤ |a| + |b|
(5.7)
and equality holds if and only if one of the vectors is a nonnegative scalar multiple of the
other. Also
||a| − |b|| ≤ |a − b|
(5.8)
Proof : By properties of the dot product and the Cauchy Schwartz inequality,
2
|a + b| = (a + b) · (a + b)
= (a · a) + (a · b) + (b · a) + (b · b)
2
= |a| + 2 (a · b) + |b|
2
2
2
2
2
≤ |a| + 2 |a · b| + |b|
≤ |a| + 2 |a| |b| + |b|
2
= (|a| + |b|) .
Taking square roots of both sides you obtain 5.7.
It remains to consider when equality occurs. If either vector equals zero, then that
vector equals zero times the other vector and the claim about when equality occurs is
verified. Therefore, it can be assumed both vectors are nonzero. To get equality in the
second inequality above, Theorem 5.1.5 implies one of the vectors must be a multiple of the
other. Say b = αa. If α < 0 then equality cannot occur in the first inequality because in
this case
2
2
(a · b) = α |a| < 0 < |α| |a| = |a · b|
Therefore, α ≥ 0.
To get the other form of the triangle inequality,
a=a−b+b
so
|a| = |a − b + b|
≤ |a − b| + |b| .
Therefore,
|a| − |b| ≤ |a − b|
(5.9)
|b| − |a| ≤ |b − a| = |a − b| .
(5.10)
Similarly,
It follows from 5.9 and 5.10 that 5.8 holds. This is because ||a| − |b|| equals the left side of
either 5.9 or 5.10 and either way, ||a| − |b|| ≤ |a − b| . This proves the theorem.
5.2
5.2.1
The Geometric Significance Of The Dot Product
The Angle Between Two Vectors
Given two vectors, a and b, the included angle is the angle between these two vectors which
is less than or equal to 180 degrees. The dot product can be used to determine the included
In words, the dot product of two vectors equals the product of the magnitude of the two
vectors multiplied by the cosine of the included angle. Note this gives a geometric description
of the dot product which does not depend explicitly on the coordinates of the vectors.
Example 5.2.1 Find the angle between the vectors 2i + j − k and 3i + 4j + k.
√
√ The dot
√ two vectors equals 6+4−1 = 9 and the norms are 4 + 1 + 1 =
√ product of these
6 and 9 + 16 + 1 = 26. Therefore, from 5.11 the cosine of the included angle equals
cos θ = √
9
√ = . 720 58
26 6
Now the cosine is known, the angle can be determines by solving the equation, cos θ = .
720 58. This will involve using a calculator or a table of trigonometric functions. The answer
◦
is θ = . 766 16 radians or in terms of degrees, θ = . 766 16 × 360
2π = 43. 898 . Recall how this
x
360
◦
last computation is done. Set up a proportion, .76616 = 2π because 360 corresponds to 2π
radians. However, in calculus, you should get used to thinking in terms of radians and not
degrees. This is because all the important calculus formulas are defined in terms of radians.
Example 5.2.2 Let u, v be two vectors whose magnitudes are equal to 3 and 4 respectively
and such that if they are placed in standard position with their tails at the origin, the angle
between u and the positive x axis equals 30◦ and the angle between v and the positive x axis
is -30◦ . Find u · v.
From the geometric description of the dot product in 5.11
u · v = 3 × 4 × cos (60◦ ) = 3 × 4 × 1/2 = 6.
Observation 5.2.3 Two vectors are said to be perpendicular if the included angle is π/2
radians (90◦ ). You can tell if two nonzero vectors are perpendicular by simply taking their
dot product. If the answer is zero, this means they are perpendicular because cos θ = 0.
5.2. THE GEOMETRIC SIGNIFICANCE OF THE DOT PRODUCT
63
Example 5.2.4 Determine whether the two vectors, 2i + j − k and 1i + 3j + 5k are perpendicular.
When you take this dot product you get 2 + 3 − 5 = 0 and so these two are indeed
perpendicular.
Definition 5.2.5 When two lines intersect, the angle between the two lines is the smaller
of the two angles determined.
Example 5.2.6 Find the angle between the two lines, (1, 2, 0) + t (1, 2, 3) and (0, 4, −3) +
t (−1, 2, −3) .
These two lines intersect, when t = 0 in the first and t = −1 in the second. It is only a
matter of finding the angle between the direction vectors. One angle determined is given by
cos θ =
−6
−3
=
.
14
7
(5.12)
We don't want this angle because it is obtuse. The angle desired is the acute angle given by
cos θ =
3
.
7
It is obtained by using replacing one of the direction vectors with −1 times it.
5.2.2
Work And Projections
Our first application will be to the concept of work. The physical concept of work does not in
any way correspond to the notion of work employed in ordinary conversation. For example,
if you were to slide a 150 pound weight off a table which is three feet high and shuffle along
the floor for 50 yards, sweating profusely and exerting all your strength to keep the weight
from falling on your feet, keeping the height always three feet and then deposit this weight
on another three foot high table, the physical concept of work would indicate that the force
exerted by your arms did no work during this project even though the muscles in your hands
and arms would likely be very tired. The reason for such an unusual definition is that even
though your arms exerted considerable force on the weight, enough to keep it from falling,
the direction of motion was at right angles to the force they exerted. The only part of a
force which does work in the sense of physics is the component of the force in the direction
of motion (This is made more precise below.). The work is defined to be the magnitude
of the component of this force times the distance over which it acts in the case where this
component of force points in the direction of motion and (−1) times the magnitude of this
component times the distance in case the force tends to impede the motion. Thus the work
done by a force on an object as the object moves from one point to another is a measure of
the extent to which the force contributes to the motion. This is illustrated in the following
picture in the case where the given force contributes to the motion.
µ
»q p
F¡¡
»»» 2
»
»
¡
»
:
θ »»
q »» F||
C»
¡
p1
CO
F⊥ C
In this picture the force, F is applied to an object which moves on the straight line from
p1 to p2 . There are two vectors shown, F|| and F⊥ and the picture is intended to indicate
64
VECTOR PRODUCTS
that when you add these two vectors you get F while F|| acts in the direction of motion and
F⊥ acts perpendicular to the direction of motion. Only F|| contributes to the work done
by F on the object as it moves from p1 to p2 . F|| is called the component of the force
in the direction of motion. From trigonometry, you see the magnitude of F|| should equal
|F| |cos θ| . Thus, since F|| points in the direction of the vector from p1 to p2 , the total work
done should equal
¯ −−→¯
|F| ¯−
p1 p2 ¯ cos θ = |F| |p2 − p1 | cos θ
If the included angle had been obtuse, then the work done by the force, F on the object
would have been negative because in this case, the force tends to impede the motion from
p1 to p2 but in this case, cos θ would also be negative and so it is still the case that the
work done would be given by the above formula. Thus from the geometric description of
the dot product given above, the work equals
|F| |p2 − p1 | cos θ = F· (p2 −p1 ) .
This explains the following definition.
Definition 5.2.7 Let F be a force acting on an object which moves from the point, p1 to
the point p2 . Then the work done on the object by the given force equals F· (p2 − p1 ) .
The concept of writing a given vector, F in terms of two vectors, one which is parallel
to a given vector, D and the other which is perpendicular can also be explained with no
reliance on trigonometry, completely in terms of the algebraic properties of the dot product.
As before, this is mathematically more significant than any approach involving geometry or
trigonometry because it extends to more interesting situations. This is done next.
Theorem 5.2.8 Let F and D be nonzero vectors. Then there exist unique vectors F|| and
F⊥ such that
F = F|| + F⊥
(5.13)
where F|| is a scalar multiple of D, also referred to as
projD (F) ,
and F⊥ · D = 0. The vector projD (F) is called the projection of F onto D.
Proof: Suppose 5.13 and F|| = αD. Taking the dot product of both sides with D and
using F⊥ · D = 0, this yields
2
F · D = α |D|
2
which requires α = F · D/ |D| . Thus there can be no more than one vector, F|| . It follows
F⊥ must equal F − F|| . This verifies there can be no more than one choice for both F|| and
F⊥ .
Now let
F·D
F|| ≡
2 D
|D|
and let
F·D
F⊥ = F − F|| = F−
2 D
|D|
Then F|| = α D where α =
F·D
.
|D|2
It only remains to verify F⊥ · D = 0. But
F·D
2 D·D
|D|
= F · D − F · D = 0.
F⊥ · D = F · D−
This proves the theorem.
5.2. THE GEOMETRIC SIGNIFICANCE OF THE DOT PRODUCT
65
Example 5.2.9 Let F = 2i+7j − 3k Newtons. Find the work done by this force in moving
from the point (1, 2, 3) to the point (−9, −3, 4) along the straight line segment joining these
points where distances are measured in meters.
According to the definition, this work is
(2i+7j − 3k) · (−10i − 5j + k) = −20 + (−35) + (−3)
= −58 Newton meters.
Note that if the force had been given in pounds and the distance had been given in feet,
the units on the work would have been foot pounds. In general, work has units equal to
units of a force times units of a length. Instead of writing Newton meter, people write joule
because a joule is by definition a Newton meter. That word is pronounced "jewel" and it is
the unit of work in the metric system of units. Also be sure you observe that the work done
by the force can be negative as in the above example. In fact, work can be either positive,
negative, or zero. You just have to do the computations to find out.
Example 5.2.10 Find proju (v) if u = 2i + 3j − 4k and v = i − 2j + k.
From the above discussion in Theorem 5.2.8, this is just
Notice how you put the conjugate on the entries of the vector, y. It makes no difference
if the vectors happen to be real vectors but with complex vectors you must do it this way.
The reason for this is that when you take the dot product of a vector with itself, you want
to get the square of the length of the vector, a positive number. Placing the conjugate on
the components of y in the above definition assures this will take place. Thus
X
X
2
x·x=
xj xj =
|xj | ≥ 0.
j
j
If you didn't place a conjugate as in the above definition, things wouldn't work out correctly.
For example,
2
(1 + i) + 22 = 4 + 2i
and this is not a positive number.
The following properties of the dot product follow immediately from the definition and
you should verify each of them.
Properties of the dot product:
1. u · v = v · u.
2. If a, b are numbers and u, v, z are vectors then (au + bv) · z = a (u · z) + b (v · z) .
3. u · u ≥ 0 and it equals 0 if and only if u = 0.
Note this implies (x·αy) = α (x · y) because
(x·αy) = (αy · x) = α (y · x) = α (x · y)
The norm is defined in the usual way.
Definition 5.2.13 For x ∈ Cn ,
Ã
|x| ≡
and zz = |z| .
I will give a proof of this important inequality which depends only on the above list of
properties of the dot product. It will be slightly different than the earlier proof.
Theorem 5.2.15 (Cauchy Schwarz)The following inequality holds for x and y ∈ Cn .
|(x · y)| ≤ (x · x)
1/2
(y · y)
1/2
Equality holds in this inequality if and only if one vector is a multiple of the other.
and this must hold for all t ∈ R. Therefore, if (y · y) = 0 it must be the case that |(x · y)| = 0
also since otherwise the above inequality would be violated. Therefore, in this case,
1/2
|(x · y)| ≤ (x · x)
1/2
(y · y)
.
On the other hand, if (y · y) 6= 0, then p (t) ≥ 0 for all t means the graph of y = p (t) is a
parabola which opens up and it either has exactly one real zero in the case its vertex touches
the t axis or it has no real zeros. From the quadratic formula this happens exactly when
2
4 |(x · y)| − 4 (x · x) (y · y) ≤ 0
which is equivalent to 5.14.
It is clear from a computation that if one vector is a scalar multiple of the other that
equality holds in 5.14. Conversely, suppose equality does hold. Then this is equivalent to
2
saying 4 |(x · y)| − 4 (x · x) (y · y) = 0 and so from the quadratic formula, there exists one
real zero to p (t) = 0. Call it t0 . Then
¯2
¡
¢ ¯
p (t0 ) ≡ x + θt0 y, x + t0 θy = ¯x + θty¯ = 0
and so x = −θt0 y. This proves the theorem.
Note that I only used part of the above properties of the dot product. It was not
necessary to use the one which says that if (x · x) = 0 then x = 0.
By analogy to the case of Rn , length or magnitude of vectors in Cn can be defined.
1/2
Definition 5.2.16 Let z ∈ Cn . Then |z| ≡ (z · z)
.
Theorem 5.2.17 For length defined in Definition 5.2.16, the following hold.
|z| ≥ 0 and |z| = 0 if and only if z = 0
(5.16)
If α is a scalar, |αz| = |α| |z|
(5.17)
|z + w| ≤ |z| + |w| .
(5.18)
Proof: The first two claims are left as exercises. To establish the third, you use the
same argument which was used in Rn .
|z + w|
3. If F is a force and D is a vector, show projD (F) = (|F| cos θ) u where u is the unit
vector in the direction of D, u = D/ |D| and θ is the included angle between the two
vectors, F and D. |F| cos θ is sometimes called the component of the force, F in the
direction, D.
projD (F) =
F·D
D·D D
1
D
= |F| |D| cos θ |D|
2 D = |F| cos θ |D| .
4. A boy drags a sled for 100 feet along the ground by pulling on a rope which is 40
degrees from the horizontal with a force of 10 pounds. How much work does this force
do?
¡ 40 ¢
The component of force is 10 cos 180
π and it acts for 100 feet so the work done is
µ
10 cos
¡
¢
¡ ¢
Recall now that AT ij = Aji . Use this to write a formula for x,AT y Rn .
5.4. THE CROSS PRODUCT
5.4
69
The Cross Product
The cross product is the other way of multiplying two vectors in R3 . It is very different
from the dot product in many ways. First the geometric meaning is discussed and then
a description in terms of coordinates is given. Both descriptions of the cross product are
important. The geometric description is essential in order to understand the applications
to physics and geometry while the coordinate description is the only way to practically
compute the cross product.
Definition 5.4.1 Three vectors, a, b, c form a right handed system if when you extend the
fingers of your right hand along the vector, a and close them in the direction of b, the thumb
points roughly in the direction of c.
For an example of a right handed system of vectors, see the following picture.
In this picture the vector c points upwards from the plane determined by the other two
vectors. You should consider how a right hand system would differ from a left hand system.
Try using your left hand and you will see that the vector, c would need to point in the
opposite direction as it would for a right hand system.
From now on, the vectors, i, j, k will always form a right handed system. To repeat, if
you extend the fingers of our right hand along i and close them in the direction j, the thumb
points in the direction of k.
k
6
¡
-
j
¡
i¡
ª
The following is the geometric description of the cross product. It gives both the direction
and the magnitude and therefore specifies the vector.
Definition 5.4.2 Let a and b be two vectors in R3 . Then a × b is defined by the following
two rules.
1. |a × b| = |a| |b| sin θ where θ is the included angle.
2. a × b · a = 0, a × b · b = 0, and a, b, a × b forms a right hand system.
70
VECTOR PRODUCTS
Note that |a × b| is the area of the parallelogram determined by a and b.
Formula 5.19 follows immediately from the definition. The vectors a × b and b × a have
the same magnitude, |a| |b| sin θ, and an application of the right hand rule shows they have
opposite direction. Formula 5.20 is also fairly clear. If α is a nonnegative scalar, the direction
of (αa) ×b is the same as the direction of a × b,α (a × b) and a× (αb) while the magnitude
is just α times the magnitude of a × b which is the same as the magnitude of α (a × b)
and a× (αb) . Using this yields equality in 5.20. In the case where α < 0, everything works
the same way except the vectors are all pointing in the opposite direction and you must
multiply by |α| when comparing their magnitudes. The distributive laws are much harder
to establish but the second follows from the first quite easily. Thus, assuming the first, and
using 5.19,
(b + c) × a = −a× (b + c)
= − (a × b + a × c)
= b × a + c × a.
A proof of the distributive law is given in a later section for those who are interested.
Now from the definition of the cross product,
i × j = k j × i = −k
k × i = j i × k = −j
j × k = i k × j = −i
With this information, the following gives the coordinate description of the cross product.
Proposition 5.4.3 Let a = a1 i + a2 j + a3 k and b = b1 i + b2 j + b3 k be two vectors. Then
a × b = (a2 b3 − a3 b2 ) i+ (a3 b1 − a1 b3 ) j+
+ (a1 b2 − a2 b1 ) k.
This proves the proposition.
It is probably impossible for most people to remember 5.23. Fortunately, there is a
somewhat easier way to remember it. Define the determinant of a 2 × 2 matrix as follows
¯
¯
¯ a b ¯
¯
¯
¯ c d ¯ ≡ ad − bc
Then
Note that to get the scalar which multiplies i you take the determinant of what is left after
1+1
deleting the first row and the first column and multiply by (−1)
because i is in the first
row and the first column. Then you do the same thing for the j and k. In the case of the j
1+2
there is a minus sign because j is in the first row and the second column and so(−1)
= −1
3+1
while the k is multiplied by (−1)
= 1. The above equals
(a2 b3 − a3 b2 ) i− (a1 b3 − a3 b1 ) j+ (a1 b2 − a2 b1 ) k
(5.26)
which is the same as 5.24. There will be much more presented on determinants later. For
now, consider this an introduction if you have not seen this topic.
Example 5.4.4 Find (i − j + 2k) × (3i − 2j + k) .
Use 5.25 to compute this.
¯
¯ i
j k
¯
¯ 1 −1 2
¯
¯ 3 −2 1
¯
¯ ¯
¯ ¯ −1
¯=¯
¯ ¯ −2
¯
¯ ¯
¯ ¯
2 ¯¯ ¯¯ 1 2 ¯¯ ¯¯ 1
i−
j+
1 ¯ ¯ 3 1 ¯ ¯ 3
¯
−1 ¯¯
k
−2 ¯
= 3i + 5j + k.
Example 5.4.5 Find the area of the parallelogram determined by the vectors, (i − j + 2k)
and (3i − 2j + k) . These are the same two vectors in Example 5.4.4.
From Example 5.4.4 and the geometric description of the cross product,
the area√is just
√
the norm of the vector obtained in Example 5.4.4. Thus the area is 9 + 25 + 1 = 35.
72
VECTOR PRODUCTS
Example 5.4.6 Find the area of the triangle determined by (1, 2, 3) , (0, 2, 5) , and (5, 1, 2) .
This triangle is obtained by connecting the three points with lines. Picking (1, 2, 3) as a
starting point, there are two displacement vectors, (−1, 0, 2) and (4, −1, −1) such that the
given vector added to these displacement vectors gives the other two vectors. The area of
the triangle is half the area of the parallelogram determined by (−1, 0, 2) and
√ (4, −1, −1) .
Thus (−1, 0, 2) × (4, −1, −1) = (2, 7, 1) and so the area of the triangle is 21 4 + 49 + 1 =
√
3
2 6.
Observation 5.4.7 In general, if you have three points (vectors) in R3 , P, Q, R the area
of the triangle is given by
1
|(Q − P) × (R − P)| .
2
P
5.4.1
Q
r
Á
r
-r
R
The Distributive Law For The Cross Product
This section gives a proof for 5.21, a fairly difficult topic. It is included here for the interested
student. If you are satisfied with taking the distributive law on faith, it is not necessary
to read this section. The proof given here is quite clever and follows the one given in [3].
Another approach, based on volumes of parallelepipeds is found in [12] and is discussed a
little later.
Lemma 5.4.8 Let b and c be two vectors. Then b × c = b × c⊥ where c|| + c⊥ = c and
c⊥ · b = 0.
Proof: Consider the following picture.
c⊥ 6c
µ
¡
¡
¡θ
b
-
b b
Now c⊥ = c − c· |b|
|b| and so c⊥ is in the plane determined by c and b. Therefore, from
the geometric definition of the cross product, b × c and b × c⊥ have the same direction.
Now, referring to the picture,
|b × c⊥ | = |b| |c⊥ |
= |b| |c| sin θ
= |b × c| .
Therefore, b × c and b × c⊥ also have the same magnitude and so they are the same vector.
With this, the proof of the distributive law is in the following theorem.
Proof: Suppose first that a · b = a · c = 0. Now imagine a is a vector coming out of the
page and let b, c and b + c be as shown in the following picture.
a × (b + c)
BM
B
B
B
B
B
B
B
B
B
I
@
B
@
a×c @ B
6
a×b
1
B
³³
c¡
µ
³³HH
@B ¡
³
b+c
³
@B³
¡³
b
Then a × b, a× (b + c) , and a × c are each vectors in the same plane, perpendicular to a
as shown. Thus a × c · c = 0, a× (b + c) · (b + c) = 0, and a × b · b = 0. This implies that
to get a × b you move counterclockwise through an angle of π/2 radians from the vector, b.
Similar relationships exist between the vectors a× (b + c) and b + c and the vectors a × c
and c. Thus the angle between a × b and a× (b + c) is the same as the angle between b + c
and b and the angle between a × c and a× (b + c) is the same as the angle between c and
b + c. In addition to this, since a is perpendicular to these vectors,
|a × b| = |a| |b| , |a× (b + c)| = |a| |b + c| , and
|a × c| = |a| |c| .
Therefore,
|a × c|
|a × b|
|a× (b + c)|
=
=
= |a|
|b + c|
|c|
|b|
and so
|b + c| |a× (b + c)|
|b + c|
|a× (b + c)|
=
,
=
|a × c|
|c|
|a × b|
|b|
showing the triangles making up the parallelogram on the right and the four sided figure on
the left in the above picture are similar. It follows the four sided figure on the left is in fact
a parallelogram and this implies the diagonal is the vector sum of the vectors on the sides,
yielding 5.27.
Now suppose it is not necessarily the case that a · b = a · c = 0. Then write b = b|| + b⊥
where b⊥ · a = 0. Similarly c = c|| + c⊥ . By the above lemma and what was just shown,
a× (b + c) = a× (b + c)⊥
= a× (b⊥ + c⊥ )
= a × b⊥ + a × c⊥
= a × b + a × c.
This proves the theorem.
The result of Problem ?? of the exercises ?? is used to go from the first to the second
line.
74
VECTOR PRODUCTS
5.4.2
The Box Product
Definition 5.4.10 A parallelepiped determined by the three vectors, a, b, and c consists of
{ra+sb + tc : r, s, t ∈ [0, 1]} .
That is, if you pick three numbers, r, s, and t each in [0, 1] and form ra+sb + tc, then the
collection of all such points is what is meant by the parallelepiped determined by these three
vectors.
The following is a picture of such a thing.
6
the area of the base of the parallelepiped, the parallelogram determined by
and b has area equal to |a × b| while the altitude of the parallelepiped is
θ is the angle shown in the picture between c and a × b. Therefore, the
parallelepiped is the area of the base times the altitude which is just
|a × b| |c| cos θ = a × b · c.
This expression is known as the box product and is sometimes written as [a, b, c] . You
should consider what happens if you interchange the b with the c or the a with the c. You
can see geometrically from drawing pictures that this merely introduces a minus sign. In any
case the box product of three vectors always equals either the volume of the parallelepiped
determined by the three vectors or else minus this volume.
Example 5.4.11 Find the volume of the parallelepiped determined by the vectors, i + 2j −
5k, i + 3j − 6k,3i + 2j + 3k.
According to the above discussion, pick any two of these, take the cross product and
then take the dot product of this with the third of these vectors. The result will be either
the desired volume or minus the desired volume.
¯
¯
¯ i j k ¯
¯
¯
(i + 2j − 5k) × (i + 3j − 6k) = ¯¯ 1 2 −5 ¯¯
¯ 1 3 −6 ¯
= 3i + j + k
Now take the dot product of this vector with the third which yields
(3i + j + k) · (3i + 2j + 3k) = 9 + 2 + 3 = 14.
This shows the volume of this parallelepiped is 14 cubic units.
There is a fundamental observation which comes directly from the geometric definitions
of the cross product and the dot product.
5.4. THE CROSS PRODUCT
75
Lemma 5.4.12 Let a, b, and c be vectors. Then (a × b) ·c = a· (b × c) .
Proof: This follows from observing that either (a × b) ·c and a· (b × c) both give the
volume of the parallellepiped or they both give −1 times the volume.
A. Evaluate the determinant of a square matrix using by applying
(a) the cofactor formula or
(b) row operations.
B. Recall the effects that row operations have on determinants.
C. Recall
1. and verify the following:
(a) The determinant of a product of matrices is the product of the determinants.
(b) The determinant of a matrix is equal to the determinant of its transpose.
D. Apply Cramer's Rule to solve a 2 × 2 or a 3 × 3 linear system.
E. Use determinants to determine whether a matrix has an inverse.
F. Evaluate the inverse of a matrix using cofactors.
6.1
6.1.1
Basic Techniques And Properties
Cofactors And 2 × 2 Determinants
Let A be an n × n matrix. The determinant of A, denoted as det (A) is a number. If the
matrix is a 2×2 matrix, this number is very easy to find.
µ
¶
a b
Definition 6.1.1 Let A =
. Then
c d
det (A) ≡ ad − cb.
The determinant is also often denoted by enclosing the matrix with two vertical lines. Thus
¯
µ
¶ ¯
¯ a b ¯
a b
¯.
det
= ¯¯
c d
c d ¯
µ
Example 6.1.2 Find det
2 4
−1 6
¶
.
77
78
DETERMINANTS
From the definition this is just (2) (6) − (−1) (4) = 16.
Having defined what is meant by the determinant of a 2 × 2 matrix, what about a 3 × 3
matrix?
Definition 6.1.3 Suppose A is a 3 × 3 matrix. The ij th minor, denoted as minor(A)ij ,
is the determinant of the 2 × 2 matrix which results from deleting the ith row and the j th
column.
Example 6.1.4 Consider the matrix,
1
4
3
2
3
2i+j
Definition 6.1.5 Suppose A is a 3×3 matrix. The ij th cofactor is defined to be (−1) ×
¡ th
¢
i+j
ij minor . In words, you multiply (−1)
times the ij th minor to get the ij th cofactor.
The cofactors of a matrix are so important that special notation is appropriate when referring
to them. The ij th cofactor of a matrix, A will be denoted by cof (A)ij . It is also convenient
to refer to the cofactor of an entry of a matrix as follows. For aij an entry of the matrix,
its cofactor is just cof (A)ij . Thus the cofactor of the ij th entry is just the ij th cofactor.
Example 6.1.6 Consider the matrix,
1 2
A= 4 3
3 2It follows
µ
cof (A)12 = (−1)
1+2
det
4 2
3 1
¶
1+2
= (−1)
(−2) = 2
6.1. BASIC TECHNIQUES AND PROPERTIES
Therefore,
µ
2+3
cof (A)23 = (−1)
det
1
3
79
2
2
Similarly,
¶
= (−1)
µ
2+2
cof (A)22 = (−1)
det
1 3
3 1
2+3
(−4) = 4.
¶
= −8.
Definition 6.1.7 The determinant of a 3 × 3 matrix, A, is obtained by picking a row (column) and taking the product of each entry in that row (column) with its cofactor and adding
these up. This process when applied to the ith row (column) is known as expanding the
determinant along the ith row (column).
Example 6.1.8 Find the determinant of
3
2 .
1
1 2
A= 4 3
3 2
Here is how it is done by "expanding along the first column".
z
cof(A)11
}|¯
1+1 ¯¯ 3 2
1(−1)
¯ 2 1
cof(A)21
{¯
}|¯
z
¯ 2
¯
2+1
¯
¯ + 4(−1)
¯ 2
¯
cof(A)31
{¯
}|¯
z
3 ¯¯
3+1 ¯¯ 2
+ 3(−1)
¯ 3
1 ¯
{¯
3 ¯¯
= 0.
2 ¯
You see, we just followed the rule in the above definition. We took the 1 in the first column
and multiplied it by its cofactor, the 4 in the first column and multiplied it by its cofactor,
and the 3 in the first column and multiplied it by its cofactor. Then we added these numbers
together.
You could also expand the determinant along the second row as follows.
z
cof(A)21
}|¯
2+1 ¯¯ 2 3
4(−1)
¯ 2 1
cof(A)22
{¯
}|¯
z
¯ 1
¯
2+2
¯
¯ + 3(−1)
¯ 3
¯
cof(A)23
{¯
}|¯
z
3 ¯¯
2+3 ¯¯ 1
+ 2(−1)
¯ 3
1 ¯
{¯
2 ¯¯
= 0.
2 ¯
Observe this gives the same number. You should try expanding along other rows and
columns. If you don't make any mistakes, you will always get the same answer.
What about a 4 × 4 matrix? You know now how to find the determinant of a 3 × 3
matrix. The pattern is the same.
Definition 6.1.9 Suppose A is a 4 × 4 matrix. The ij th minor is the determinant of the
th
th
th
3 × 3 matrix you obtain when you delete
¡ the i row
¢ and the j column. The iji+jcofactor,
i+j
cof (A)ij is defined to be (−1)
× ij th minor . In words, you multiply (−1)
times the
th
th
ij minor to get the ij cofactor.
Definition 6.1.10 The determinant of a 4 × 4 matrix, A, is obtained by picking a row
(column) and taking the product of each entry in that row (column) with its cofactor and
adding these up. This process when applied to the ith row (column) is known as expanding
the determinant along the ith row (column).
Example 6.1.11 Find det (A) where
Now you know how to expand each of these 3×3 matrices along a row or a column. If you do
so, you will get −12 assuming you make no mistakes. You could expand this matrix along
any row or any column and assuming you make no mistakes, you will always get the same
thing which is defined to be the determinant of the matrix, A. This method of evaluating
a determinant by expanding along a row or a column is called the method of Laplace
expansion.
Note that each of the four terms above involves three terms consisting of determinants
of 2 × 2 matrices and each of these will need 2 terms. Therefore, there will be 4 × 3 × 2 = 24
terms to evaluate in order to find the determinant using the method of Laplace expansion.
Suppose now you have a 10 × 10 matrix and you follow the above pattern for evaluating
determinants. By analogy to the above, there will be 10! = 3, 628 , 800 terms involved in
the evaluation of such a determinant by Laplace expansion along a row or column. This is
a lot of terms.
In addition to the difficulties just discussed, you should regard the above claim that you
always get the same answer by picking any row or column with considerable skepticism. It
is incredible and not at all obvious. However, it requires a little effort to establish it. This
is done in the section on the theory of the determinant.
Definition 6.1.12 Let A = (aij ) be an n × n matrix and suppose the determinant of a
(n − 1) × (n − 1) matrix has been defined. Then a new matrix called the cofactor matrix, cof (A) is defined by cof (A) = (cij ) where to obtain cij delete the ith row and
the j th column of A, take the determinant of the (n − 1) × (n − 1) matrix which results,
i+j
(This is called the ij th minor of A. ) and then multiply this number by (−1) . Thus
¡
¢
i+j
(−1) × the ij th minor equals the ij th cofactor. To make the formulas easier to remember, cof (A)ij will denote the ij th entry of the cofactor matrix.
With this definition of the cofactor matrix, here is how to define the determinant of an
n × n matrix.
Definition 6.1.13 Let A be an n × n matrix where n ≥ 2 and suppose the determinant of
an (n − 1) × (n − 1) has been defined. Then
det (A) =
n
X
j=1
aij cof (A)ij =
n
X
aij cof (A)ij .
(6.1)
i=1
The first formula consists of expanding the determinant along the ith row and the second
expands the determinant along the j th column.
Theorem 6.1.14 Expanding the n × n matrix along any row or column always gives the
same answer so the above definition is a good definition.
6.1. BASIC TECHNIQUES AND PROPERTIES
6.1.2
81
The Determinant Of A Triangular Matrix
Notwithstanding the difficulties involved in using the method of Laplace expansion, certain
types of matrices are very easy to deal with.
Definition 6.1.15 A matrix M , is upper triangular if Mij = 0 whenever i > j. Thus such
a matrix equals zero below the main diagonal, the entries of the form Mii ,You should verify the following using the above theorem on Laplace expansion.
Corollary 6.1.16 Let M be an upper (lower) triangular matrix. Then det (M ) is obtained
by taking the product of the entries on the main diagonal.
Example 6.1.17 Let
Next expand this last determinant along the first column to obtain the above equals
1 × 2 × 3 × (−1) = −6
which is just the product of the entries down the main diagonal of the original matrix.
82
DETERMINANTS
6.1.3
Properties Of Determinants
There are many properties satisfied by determinants. Some of these properties have to do
with row operations. Recall the row operations.
Definition 6.1.18 The row operations consist of the following
1. Switch two rows.
2. Multiply a row by a nonzero number.
3. Replace a row by a multiple of another row added to itself.
Theorem 6.1.19 Let A be an n × n matrix and let A1 be a matrix which results from
multiplying some row of A by a scalar, c. Then c det (A) = det (A1 ).
µ
Example 6.1.20 Let A =
1 2
3 4
¶
µ
, A1 =
2 4
3 4
¶
. det (A) = −2, det (A1 ) = −4.
Theorem 6.1.21 Let A be an n × n matrix and let A1 be a matrix which results from
switching two rows of A. Then det (A) = − det (A1 ) . Also, if one row of A is a multiple of
another row of A, then det (A) = 0.
µ
Example 6.1.22 Let A =
1 2
3 4
¶
µ
and let A1 =
3 4
1 2
¶
. det A = −2, det (A1 ) = 2.
Theorem 6.1.23 Let A be an n × n matrix and let A1 be a matrix which results from
applying row operation 3. That is you replace some row by a multiple of another row added
to itself. Then det (A) = det (A1 ).
µ
¶
µ
¶
1 2
1 2
Example 6.1.24 Let A =
and let A1 =
. Thus the second row of A1
3 4
4 6
is one times the first row added to the second row. det (A) = −2 and det (A1 ) = −2.
Theorem 6.1.25 In Theorems 6.1.19 - 6.1.23 you can replace the word, "row" with the
word "column".
There are two other major properties of determinants which do not involve row operations.
Theorem 6.1.26 Let A and B be two n × n matrices. Then
det (AB) = det (A) det (B).
Also,
Theorems 6.1.23 - 6.1.25 can be used to find determinants using row operations. As pointed
out above, the method of Laplace expansion will not be practical for any matrix of large
size. Here is an example in which all the row operations are used.
Example 6.1.28 Find the determinant of
1
5
A=
4
2
the matrix,
2
1
5
2
3
2
4
−4
4
3
3
5
Replace the second row by (−5) times the first row added to it. Then replace the third
row by (−4) times the first row added to it. Finally, replace the fourth row by (−2) times
the first row added to it. This yields the matrix,
1 2
3
4
0 −9 −13 −17
B=
0 −3 −8 −13
0 −2 −10 −3
and from Theorem
6.1.23, it has the same determinant as A. Now using other row operations,
¡ ¢
det (B) = −1
det (C) where
3
1
0
C=
0
0
2
0
−3
6
3
11
−8
30
4
22
.
−13
9
The second row was replaced by (−3) times the third row added to the second row. By
Theorem 6.1.23 this didn't change the value of the determinant. Then the last row was
multiplied by (−3) . By Theorem 6.1.19 the resulting matrix has a determinant which is
(−3) times the determinant of the unmultiplied matrix. Therefore, we multiplied by −1/3
to retain the correct value. Now replace the last row with 2 times the third added to
it. This does not change the value of the determinant by Theorem 6.1.23. Finally switch
Replace the second row by (−1) times the first row added to it. Next take −2 times the
first row and add to the third and finally take −3 times the first row and add to the last
row. This yields
1
2
3
2
0 −5 −1 −1
0 −3 −4 1 .
0 −10 −8 −4
By Theorem 6.1.23 this matrix has the same determinant as the original matrix. Remember
you can work with the columns also. Take −5 times the last column and add to the second
column. This yields
1 −8 3
2
0 0 −1 −1
0 −8 −4 1
0 10 −8 −4
By Theorem 6.1.25 this matrix has the same
(−1) times the third row and add to the top
1 0
0 0
0 −8
0 10
which by Theorem 6.1.23 has the same determinant as the original matrix. Lets expand it
now along the first column. This yields the following for the determinant of the original
matrix.
0 −1 −1
det −8 −4 1
10 −8 −4
which equals
µ
8 det
−1 −1
−8 −4
¶
µ
+ 10 det
−1
−4
−1
1
¶
= −82
6.2. APPLICATIONS
85
We suggest you do not try to be fancy in using row operations. That is, stick mostly to
the one which replaces a row or column with a multiple of another row or column added to
it. Also note there is no way to check your answer other than working the problem more
than one way. To be sure you have gotten it right you must do this.
6.2
6.2.1
Applications
A Formula For The Inverse
The definition of the determinant in terms of Laplace expansion along a row or column also
provides a way to give a formula for the inverse of a matrix. Recall the definition of the
inverse of a matrix in Definition 4.1.28 on Page 52. Also recall the definition of the cofactor
matrix given in Definition 6.1.12 on Page 80. This cofactor matrix was just the matrix
which results from replacing the ij th entry of the matrix with the ij th cofactor.
The following theorem says that to find the inverse, take the transpose of the cofactor
matrix and divide by the determinant. The transpose of the cofactor matrix is called the
adjugate or sometimes the classical adjoint of the matrix A. In other words, A−1 is
equal to one divided by the determinant of A times the adjugate matrix of A. This is what
the following theorem says with more precision.
¡
¢
Theorem 6.2.1Example 6.2.2 Find the inverse of the matrix,
1 2
A= 3 0
1 2
3
1
1
First find the determinant of this matrix. Using Theorems 6.1.23 - 6.1.25 on Page 82,
the determinant of this matrix equals the determinant of the matrix,
1 2
3
0 −6 −8
0 0 −2
which equals 12. The cofactor matrix of A
−2
4
2
and so we got it right. If the result of multiplying these matrices had been something other
than the identity matrix, you would know there was an error. When this happens, you
need to search for the mistake if you am interested in getting the right answer. A common
mistake is to forget to take the transpose of the cofactor matrix.
6.2. APPLICATIONS
87
Proof of Theorem 6.2.1: From the definition of the determinant in terms of expansion
along a column, and letting (air ) = A, if det (A) 6= 0,
n
X
air cof (A)ir det(A)−1 = det(A) det(A)−1 = 1.
i=1
Now consider
n
X
air cof (A)ik det(A)−1
i=1
th
when k 6= r. Replace the k column with the rth column to obtain a matrix, Bk whose
determinant equals zero by Theorem 6.1.21. However, expanding this matrix, Bk along the
k th column yields
−1
0 = det (Bk ) det (A)
=
n
X
−1
air cof (A)ik det (A)
i=1
Summarizing,
n
X
½
−1
air cof (A)ik det (A)
= δ rk ≡
i=1
Now
n
X
air cof (A)ik =
i=1
n
X
1 if r = k
.
0 if r 6= k
T
air cof (A)ki
i=1
T
which is the krth entry of cof (A) A. Therefore,
T
cof (A)
A = I.
det (A)
(6.2)
Using the other formula in Definition 6.1.13, and similar reasoning,
n
X
arj cof (A)kj det (A)
−1
= δ rk
j=1
Now
n
X
arj cof (A)kj =
j=1
n
X
T
arj cof (A)jk
j=1
T
which is the rk th entry of A cof (A) . Therefore,
T
A
cof (A)
= I,
det (A)
(6.3)
¡
¢
and it follows from 6.2 and 6.3 that A−1 = a−1
ij , where
a−1
ij = cof (A)ji det (A)
In other words,
T
A−1 =
cof (A)
.
det (A)
−1
.
88
DETERMINANTS
Now suppose A−1 exists. Then by Theorem 6.1.26,
¡
¢
¡
¢
1 = det (I) = det AA−1 = det (A) det A−1
so det (A) 6= 0. This proves the theorem.
This way of finding inverses is especially useful in the case where it is desired to find the
inverse of a matrix whose entries are functions.
Example 6.2.4 Suppose
This formula for the inverse also implies a famous procedure known as Cramer's rule.
Cramer's rule gives a formula for the solutions, x, to a system of equations, Ax = y in the
special case that A is a square matrix. Note this rule does not apply if you have a system
of equations in which there is a different number of equations than variables
You see the pattern. For large systems Cramer's rule is less than useful if you want to find
an answer. This is because to use it you must evaluate determinants. However, you have no
practical way to evaluate determinants for large matrices other than row operations and if
you are using row operations, you might just as well use them to solve the system to begin
with. It will be a lot less trouble. Nevertheless, there are situations in which Cramer's rule
is useful.
Example 6.2.7 Solve for z if
Finally, take −6/8 times the second row and add to the third.
¯
¯
¯ −1 2
¯
0
3
¯
¯
¯ 0 8
¯
1
12
¯
¯.
−¯
¯
0
0
−6/8
5
+
(−6/8)
(12)
¯
¯
¯ 0 0
0
−44 + (9/2) 5 ¯
92
DETERMINANTS
This material is definitely not for the faint of heart. It is only for people who want
to see everything proved. It is a fairly complete and unusually elementary treatment of
the subject. There will be some repetition between this section and the earlier section on
determinants. The main purpose is to give all the missing proofs. Two books which give
a good introduction to determinants are Apostol [1] and Rudin [11]. A recent book which
also has a good introduction is Baker [2]. Most linear algebra books do not do an honest
job presenting this topic.
It is easiest to give a different definition of the determinant which is clearly well defined
and then prove the earlier one in terms of Laplace expansion. Let (i1 , · · · , in ) be an ordered
list of numbers from {1, · · · , n} . This means the order is important so (1, 2, 3) and (2, 1, 3)
are different.
The following Lemma will be essential in the definition of the determinant.
Lemma 6.4.1 There exists a unique function, sgnn which maps each list of numbers from
{1, · · · , n} to one of the three numbers, 0, 1, or −1 which also has the following properties.
sgnn (1, · · · , n) = 1
This proves the existence of the desired function.
To see this function is unique, note that you can obtain any ordered list of distinct
numbers from a sequence of switches. If there exist two functions, f and g both satisfying
6.5 and 6.6, you could start with f (1, · · · , n) = g (1, · · · , n) and applying the same sequence
of switches, eventually arrive at f (i1 , · · · , in ) = g (i1 , · · · , in ) . If any numbers are repeated,
then 6.6 gives both functions are equal to zero for that ordered list. This proves the lemma.
In what follows sgn will often be used rather than sgnn because the context supplies the
appropriate n.
Definition 6.4.2 Let f be a real valued function which has the set of ordered lists of numbers
from {1, · · · , n} as its domain. Define
X
f (k1 · · · kn )
(k1 ,··· ,kn )
det (A (r1 , · · · , rn )) = (−1) det (A) = sgn (r1 , · · · , rn ) det (A)
and proves the proposition in the case when there are no repeated numbers in the ordered
list, (r1 , · · · , rn ). However, if there is a repeat, say the rth row equals the sth row, then the
reasoning of 6.13 -6.14 shows that A (r1 , · · · , rn ) = 0 and also sgn (r1 , · · · , rn ) = 0 so the
formula holds in this case also.
Observation 6.4.5 There are n! ordered lists of distinct numbers from {1, · · · , n} .
To see this, consider n slots placed in order. There are n choices for the first slot. For
each of these choices, there are n − 1 choices for the second. Thus there are n (n − 1) ways
to fill the first two slots. Then for each of these ways there are n − 2 choices left for the third
slot. Continuing this way, there are n! ordered lists of distinct numbers from {1, · · · , n} as
stated in the observation.
With the above, it is possible to give a more
¡ ¢ symmetric description of the determinant
from which it will follow that det (A) = det AT .
Summing over all ordered lists, (r1 , · · · , rn ) where the ri are distinct, (If the ri are not
distinct, sgn (r1 , · · · , rn ) = 0 and so there is no contribution to the sum.)
n! det (A) =
X
X
sgn (r1 , · · · , rn ) sgn (k1 , · · · , kn ) ar1 k1 · · · arn kn .
(r1 ,··· ,rn ) (k1 ,··· ,kn )
This proves the corollary since the formula gives the same number for A as it does for AT .
Corollary 6.4.7 If two rows or two columns in an n × n matrix, A, are switched, the
determinant of the resulting matrix equals (−1) times the determinant of the original matrix.
If A is an n×n matrix in which two rows are equal or two columns are equal then det (A) = 0.
Suppose the ith row of A equals (xa1 + yb1 , · · · , xan + ybn ). Then
det (A) = x det (A1 ) + y det (A2 )
where the ith row of A1 is (a1 , · · · , an ) and the ith row of A2 is (b1 , · · · , bn ) , all other rows
of A1 and A2 coinciding with those of A. In other words, det is a linear function of each
row A. The same is true with the word "row" replaced with the word "column".
Proof: By Proposition 6.4.4 when two rows are switched, the determinant of the resulting matrix is (−1) times the determinant of the original matrix. By Corollary 6.4.6 the
same holds for columns because the columns of the matrix equal the rows of the transposed
matrix. Thus if A1 is the matrix obtained from A by switching two columns,
¡ ¢
¡ ¢
det (A) = det AT = − det AT1 = − det (A1 ) .
If A has two equal columns or two equal rows, then switching them results in the same
matrix. Therefore, det (A) = − det (A) and so det (A) = 0.
It remains to verify the last assertion.
X
det (A) ≡
sgn (k1 , · · · , kn ) a1k1 · · · (xaki + ybki ) · · · ankn
(k1 ,··· ,kn )
=x
X
sgn (k1 , · · · , kn ) a1k1 · · · aki · · · ankn
(k1 ,··· ,kn )
+y
X
sgn (k1 , · · · , kn ) a1k1 · · · bki · · · ankn
(k1 ,··· ,kn )
≡ x det (A1 ) + y det (A2 ) .
¡ ¢
The same is true of columns because det AT = det (A) and the rows of AT are the columns
of A.
The following corollary is also of great use.
Corollary 6.4.9 Suppose A is an n × n matrix and some column (row) is a linear combination of r other columns (rows). Then det (A) = 0.
¡
¢
Proof: Let A = a1 · · · an be the columns of A and suppose the condition that
one column is a linear combination of r of the others is satisfied. Then by using Corollary
6.4.7 you may rearrange thePcolumns to have the nth column a linear combination of the
r
first r columns. Thus an = k=1 ck ak and so
¡
¢
Pr
det (A) = det a1 · · · ar · · · an−1
.
k=1 ck ak
By Corollary 6.4.7
r
X
One of the most important rules about determinants is that the determinant of a product
equals the product of the determinants.
Theorem 6.4.11 Let A and B be n × n matrices. Then
det (AB) = det (A) det (B) .
Proof: Let cij be the ij th entry of AB. Then by Proposition 6.4.4,
det (AB) =
X
sgn (k1 , · · · , kn ) c1k1 · · · cnkn
(k1 ,··· ,kn )
=
X
sgn (k1 , · · · , kn )
(k1 ,··· ,kn )
=
X
X
Ã
X
r1
!
a1r1 br1 k1
Ã
···
X
!
anrn brn kn
rn
sgn (k1 , · · · , kn ) br1 k1 · · · brn kn (a1r1 · · · anrn )
(r1 ··· ,rn ) (k1 ,··· ,kn )
=
X
(r1 ··· ,rn )
This proves the theorem.
sgn (r1 · · · rn ) a1r1 · · · anrn det (B) = det (A) det (B) .
6.4. THE MATHEMATICAL THEORY OF DETERMINANTS∗
Lemma 6.4.12 Suppose a matrix is of the form
µ
¶
A ∗
M=
0 a
or
µ
M=
A
∗
0
a
99
(6.16)
¶
(6.17)
where a is a number and A is an (n − 1) × (n − 1) matrix and ∗ denotes either a column
or a row having length n − 1 and the 0 denotes either a column or a row of length n − 1
consisting entirely of zeros. Then
det (M ) = a det (A) .
Proof: Denote M by (mij ) . Thus in the first case, mnn = a and mni = 0 if i 6= n while
in the second case, mnn = a and min = 0 if i 6= n. From the definition of the determinant,
X
sgnn (k1 , · · · , kn ) m1k1 · · · mnkn
det (M ) ≡
(k1 ,··· ,kn )
Now suppose 6.17. Then if kn 6= n, the term involving mnkn in the above expression equals
zero. Therefore, the only terms which survive are those for which θ = n or in other words,
those for which kn = n. Therefore, the above expression reduces to
X
a
sgnn−1 (k1 , · · · kn−1 ) m1k1 · · · m(n−1)kn−1 = a det (A) .
(k1 ,··· ,kn−1 )
To get the assertion in the situation of 6.16 use Corollary 6.4.6 and 6.17 to write
µµ T
¶¶
¡
¢
¡ ¢
A
0
det (M ) = det M T = det
= a det AT = a det (A) .
∗ a
This proves the lemma.
In terms of the theory of determinants, arguably the most important idea is that of
Laplace expansion along a row or a column. This will follow from the above definition of a
determinant.
Definition 6.4.13 Let A = (aij ) be an n×n matrix. Then a new matrix called the cofactor
matrix, cof (A) is defined by cof (A) = (cij ) where to obtain cij delete the ith row and the
j th column of A, take the determinant of the (n − 1) × (n − 1) matrix which results, (This
i+j
is called the ij th minor of A. ) and then multiply this number by (−1) . To make the
th
formulas easier to remember, cof (A)ij will denote the ij entry of the cofactor matrix.
The following is the main result. Earlier this was given as a definition and the outrageous
totally unjustified assertion was made that the same number would be obtained by expanding
the determinant along any row or column. The following theorem proves this assertion.
100
DETERMINANTS
Theorem 6.4.14 Let A be an n × n matrix where n ≥ 2. Then
det (A) =
n
X
aij cof (A)ij =
j=1
n
X
aij cof (A)ij .
(6.18)
i=1
The first formula consists of expanding the determinant along the ith row and the second
expands the determinant along the j th column.
Proof: Let (ai1 , · · · , ain ) be the ith row of A. Let Bj be the matrix obtained from A by
leaving every row the same except the ith row which in Bj equals (0, · · · , 0, aij , 0, · · · , 0) .
Then by Corollary 6.4.7,
n
X
det (A) =
det (Bj )
j=1
which is the formula for expanding det (A) along the ith row. Also,
det (A)
=
n
¡ ¢ X
¡ ¢
det AT =
aTij cof AT ij
j=1
=
n
X
aji cof (A)ji
j=1
which is the formula for expanding det (A) along the ith column. This proves the theorem.
Note that this gives an easy way to write a formula for the inverse of an n × n matrix.
¡
¢
Theorem 6.4.15Proof: By Theorem 6.4.14 and letting (air ) = A, if det (A) 6= 0,
n
X
i=1
Now suppose A−1 exists. Then by Theorem 6.4.11,
¡
¢
¡
¢
1 = det (I) = det AA−1 = det (A) det A−1
so det (A) 6= 0. This proves the theorem.
The next corollary points out that if an n × n matrix, A has a right or a left inverse,
then it has an inverse.
Corollary 6.4.16 Let A be an n × n matrix and suppose there exists an n × n matrix, B
such that BA = I. Then A−1 exists and A−1 = B. Also, if there exists C an n × n matrix
such that AC = I, then A−1 exists and A−1 = C.
Proof: Since BA = I, Theorem 6.4.11 implies
det B det A = 1
and so det A 6= 0. Therefore from Theorem 6.4.15, A−1 exists. Therefore,
¡
¢
A−1 = (BA) A−1 = B AA−1 = BI = B.
The case where CA = I is handled similarly.
The conclusion of this corollary is that left inverses, right inverses and inverses are all
the same in the context of n × n matrices.
Theorem 6.4.15 says that to find the inverse, take the transpose of the cofactor matrix
and divide by the determinant. The transpose of the cofactor matrix is called the adjugate
or sometimes the classical adjoint of the matrix A. It is an abomination to call it the adjoint
although you do sometimes see it referred to in this way. In words, A−1 is equal to one over
the determinant of A times the adjugate matrix of A
102
DETERMINANTS
Definition 6.4.17 A matrix M , is upper triangular if Mij = 0 whenever i > j. Thus such
a matrix equals zero below the main diagonal, the entries of the form MiiWith this definition, here is a simple corollary of Theorem 6.4.14.
Corollary 6.4.18 Let M be an upper (lower) triangular matrix. Then det (M ) is obtained
by taking the product of the entries on the main diagonal.
6.5
The Cayley Hamilton Theorem∗
Definition 6.5.1 Let A be an n × n matrix. The characteristic polynomial is defined as
pA (t) ≡ det (tI − A)
and the solutions to pA (t) = 0 are called eigenvalues. For A a matrix and p (t) = tn +
an−1 tn−1 + · · · + a1 t + a0 , denote by p (A) the matrix defined by
p (A) ≡ An + an−1 An−1 + · · · + a1 A + a0 I.
The explanation for the last term is that A0 is interpreted as I, the identity matrix.
The Cayley Hamilton theorem states that every matrix satisfies its characteristic equation, that equation defined by PA (t) = 0. It is one of the most important theorems in linear
algebra1 . The following lemma will help with its proof.
1 A special case was first proved by Hamilton in 1853. The general case was announced by Cayley some
time later and a proof was given by Frobenius in 1878.
A. Recognize and find the row reduced echelon form of a matrix.
B. Determine the rank of a matrix.
C. Describe the row space, column space and null space of a matrix.
D. Define the span of a set of vectors. Recall that a span of vectors in a vector space is
a subspace.
E. Determine whether a set of vectors is a subspace.
F. Define linear independence.
G. Determine whether a set of vectors is linearly independent or linearly dependent.
H. Determine a basis and the dimension of a vector space.
I. Characterize the solution set to a matrix equation using rank.
J. Argue that a homogeneous linear system always has a solution and find the solutions.
K. Understand and use the Fredholm alternative.
7.1
Elementary Matrices
The elementary matrices result from doing a row operation to the identity matrix.
Definition 7.1.1 The row operations consist of the following
1. Switch two rows.
2. Multiply a row by a nonzero number.
3. Replace a row by a multiple of another row added to it.
The elementary matrices are given in the following definition.
Definition 7.1.2 The elementary matrices consist of those matrices which result by applying a row operation to an identity matrix. Those which involve switching rows of the identity
are called permutation matrices1 .
1 More generally, a permutation matrix is a matrix which comes by permuting the rows of the identity
matrix, not just switching two rows.
105
106
RANK OF A MATRIX
As an example of why these elementary matrices are interesting, consider the following.
0
1
0
1
0
0
0
a
0 x
1
f
b
y
g
d
x
w = a
i
f
c
z
h
y
b
g
w
d
i
z
c
h
A 3 × 4 matrix was multiplied on the left by an elementary matrix which was obtained from
row operation 1 applied to the identity matrix. This resulted in applying the operation 1
to the given matrix. This is what happens in general.
Now consider what these elementary matrices look like. First consider the one which
involves switching row i and row j where i < j. This matrix is1
0
..
.
0
..
.
0 ···
..
.
..
.
0 ···
0 ···
..
.
..
.
0 ···
···
1
0
···
0
···
0
..
.
..
.
0
1
···
···
···
···
···
···
0
1
1
0
..
.
···
0
..
.
···
0
0
···
1
···
0
0
···
···
···
···
···
···
···
···
0
..
.
..
.
0
..
.
..
.
0
0
..
.
0
1
1
..
···
···
···
···
···
···
···
.
0
The two exceptional rows are shown. The ith row was the j th and the j th row was the ith
in the identity matrix. Now consider what this does to a column vector.This has established the following lemma.
Lemma 7.1.3 Let P ij denote the elementary matrix which involves switching the ith and
the j th rows. Then
P ij A = B
where B is obtained from A by switching the ith and the j th rows.
Next consider the row operation which involves multiplying the ith row by a nonzero
constant, c. The elementary matrix which results from applying this operation to the ith
Lemma 7.1.4 Let E (c, i) denote the elementary matrix corresponding to the row operation in which the ith row is multiplied by the nonzero constant, c. Thus E (c, i) involves
multiplying the ith row of the identity matrix by c. Then
E (c, i) A = B
where B is obtained from A by multiplying the ith row of A by c.
Finally consider the third of these row operations. Denote by E (c × i + j) the elementary
matrix which replaces the j th row with itself added to c times the ith row added to it. In
case i < j this will beThe case where i > j is handled similarly. This proves the following lemma.
Lemma 7.1.5 Let E (c × i + j) denote the elementary matrix obtained from I by replacing
the j th row with c times the ith row added to it. Then
E (c × i + j) A = B
where B is obtained from A by replacing the j th row of A with itself added to c times the
ith row of A.
7.2. THE ROW REDUCED ECHELON FORM OF A MATRIX
111
The next theorem is the main result.
Theorem 7.1.6 To perform any of the three row operations on a matrix, A it suffices to
do the row operation on the identity matrix obtaining an elementary matrix, E and then
take the product, EA. Furthermore, each elementary matrix is invertible and its inverse is
an elementary matrix.
Proof: The first part of this theorem has been proved in Lemmas 7.1.3 - 7.1.5. It
only remains to verify the claim about the inverses. Consider first the elementary matrices
corresponding to row operation of type three.
E (−c × i + j) E (c × i + j) = I
This follows because the first matrix takes c times row i in the identity and adds it to row j.
When multiplied on the left by E (−c × i + j) it follows from the first part of this theorem
that you take the ith row of E (c × i + j) which coincides with the ith row of I since that
row was not changed, multiply it by −c and add to the j th row of E (c × i + j) which was
the j th row of I added to c times the ith row of I. Thus E (−c × i + j) multiplied on the
left, undoes the row operation which resulted in E (c × i + j). The same argument applied
to the product
E (c × i + j) E (−c × i + j)
replacing c with −c in the argument yields that this product is also equal to I. Therefore,
−1
E (c × i + j) = E (−c × i + j) .
Similar reasoning shows that for E (c, i) the elementary matrix which comes from multiplying the ith row by the nonzero constant, c,
¡
¢
−1
E (c, i) = E c−1 , i .
Finally, consider P ij which involves switching the ith and the j th rows.
P ij P ij = I
because by the first part of this theorem, multiplying on the left by P ij switches the ith
and j th rows of P ij which was obtained from switching the ith and j th rows of the identity.
First you switch them to get P ij and then you multiply on the left by P ij which switches
¡ ¢−1
these rows again and restores the identity matrix. Thus P ij
= P ij .
7.2
The Row Reduced Echelon Form Of A Matrix
Recall that putting a matrix in row reduced echelon form involves doing row operations as
described on Page 33. In this section we review the description of the row reduced echelon
form and prove the row reduced echelon form for a given matrix is unique. That is, every
matrix can be row reduced to a unique row reduced echelon form. Of course this is not true
of the echelon form. The significance of this is that it becomes possible to use the definite
article in referring to the row reduced echelon form and hence important conclusions about
the original matrix may be logically deduced from an examination of its unique row reduced
echelon form. First we need the following definition of some terminology.
Definition 7.2.1 Let v1 , · · · , vk , u be vectors. Then u is said to be a linear combination
of the vectors {v1 , · · · , vk } if there exist scalars, c1 , · · · , ck such that
u=
k
X
i=1
ci vi .
112
RANK OF A MATRIX
The collection of all linear combinations of the vectors, {v1 , · · · , vk } is known as the span
of these vectors and is written as span (v1 , · · · , vk ).
Another way to say the same thing as expressed in the earlier definition of row reduced
echelon form found on Page 32 is the following which is a more useful description when
proving the major assertions about the row reduced echelon form.
Definition 7.2.2 Let ei denote the column vector which has all zero entries except for the
ith slot which is one. An m × n matrix is said to be in row reduced echelon form if,
in viewing successive columns from left to right, the first nonzero column encountered is e1
and if you have encountered e1 , e2 , · · · , ek , the next column is either ek+1 or is a linear
combination of the vectors, e1 , e2 , · · · , ek .
Theorem 7.2.3 Let A be an m × n matrix. Then A has a row reduced echelon form
determined by a simple process.
Proof: Viewing the columns of A from left to right take the first nonzero column. Pick
a nonzero entry in this column and switch the row containing this entry with the top row
of A. Now divide this new top row by the value of this nonzero entry to get a 1 in this
position and then use row operations to make all entries below this element equal to zero.
Thus the first nonzero column is now e1 . Denote the resulting matrix by A1 . Consider
the sub-matrix of A1 to the right of this column and below the first row. Do exactly the
same thing for this sub-matrix that was done for A. This time the e1 will refer to Fm−1 .
Use the first 1 obtained by the above process which is in the top row of this sub-matrix
and row operations to zero out every element above it in the rows of A1 . Call the resulting
matrix, A2 . Thus A2 satisfies the conditions of the above definition up to the column just
encountered. Continue this way till every column has been dealt with and the result must
be in row reduced echelon form.
The following diagram illustrates the above procedure. Say the matrix looked something
like the following.
0 ∗ ∗ ∗ ∗ ∗ ∗
0 ∗ ∗ ∗ ∗ ∗ ∗
.. .. .. .. .. .. ..
. . . . . . .
0
First step would yield something like
0
0
..
.
Thus, after zeroing out the term in the top row above the 1, you get the following for the
next step in the computation of the row reduced echelon form for the original matrix.
0 1 ∗ 0 ∗ ∗ ∗
0 0 0 1 ∗ ∗ ∗
.. .. .. .. .. .. .. .
. . . . . . .
0
0
0
0
∗ ∗
∗
Next you look at the lower right matrix below the top two rows and to the right of the first
four columns and repeat the process.
Recall the following definition which was discussed earlier.
Definition 7.2.4 The first pivot column of A is the first nonzero column of A. The next
pivot column is the first column after this which becomes e2 in the row reduced echelon form.
The third is the next column which becomes e3 in the row reduced echelon form and so forth.
There are three choices for row operations at each step in the above theorem. A natural
question is whether the same row reduced echelon matrix always results in the end from
following the above algorithm applied in any way. The next corollary says this is the case.
In rough terms, the following lemma states that linear relationships between columns
in a matrix are preserved by row operations. This simple lemma is the main result in
understanding all the major questions related to the row reduced echelon form as well as
many other topics.
Lemma 7.2.5 Let A and B be two m × n matrices and suppose B results from a row
operation applied to A. Then the k th column of B is a linear combination of the i1 , · · · , ir
columns of B if and only if the k th column of A is a linear combination of the i1 , · · · , ir
columns of A. Furthermore, the scalars in the linear combination are the same. (The linear
relationship between the k th column of A and the i1 , · · · , ir columns of A is the same as the
linear relationship between the k th column of B and the i1 , · · · , ir columns of B.)
Proof: Let A equal the following matrix in which the ak are the columns
¡
¢
a1 a2 · · · an
and let B equal the following matrix in which the columns are given by the bk
¡
¢
b1 b2 · · · bn
Then by Theorem 7.1.6 on Page 111 bk = Eak where E is an elementary matrix. Suppose
then that one of the columns of A is a linear combination of some other columns of A. Say
X
ak =
cr ar .
r∈S
Then multiplying by E,
bk = Eak =
X
r∈S
cr Ear =
X
cr br .
r∈S
This proves the lemma.
Definition 7.2.6 Two matrices are said to be row equivalent if one can be obtained from
the other by a sequence of row operations.
114
RANK OF A MATRIX
It has been shown above that every matrix is row equivalent to one which is in row
reduced echelon form.
Corollary 7.2.7 The row reduced echelon form is unique. That is if B, C are two matrices
in row reduced echelon form and both are row equivalent to A, then B = C.
Proof: Suppose B and C are both row reduced echelon forms for the matrix, A. Then
they clearly have the same zero columns since row operations leave zero columns unchanged.
If B has the sequence e1 , e2 , · · · , er occurring for the first time in the positions, i1 , i2 , · · · , ir ,
the description of the row reduced echelon form means that each of these columns is not
a linear combination of the preceding columns. Therefore, by Lemma 7.2.5, the same is
true of the columns in positions i1 , i2 , · · · , ir for C. It follows from the description of the
row reduced echelon form that e1 , · · · , er occur respectively for the first time in columns
i1 , i2 , · · · , ir for C. Therefore, both B and C have the sequence e1 , e2 , · · · , er occurring for
the first time in the positions, i1 , i2 , · · · , ir . By Lemma 7.2.5, the columns between the ik
and ik+1 position in the two matrices are linear combinations involving the same scalars
of the columns in the i1 , · · · , ik position. Also the columns after the ir position are linear
combinations of the columns in the i1 , · · · , ir positions involving the same scalars in both
matrices. This is equivalent to the assertion that each of these columns is identical and this
proves the corollary.
Now with the above corollary, here is a very fundamental observation. It concerns a
matrix which looks like this: (More columns than rows.)
Corollary 7.2.8 Suppose A is an m×n matrix and that m < n. That is, the number of rows
is less than the number of columns. Then one of the columns of A is a linear combination
of the preceding columns of A.
Proof: Since m < n, not all the columns of A can be pivot columns. In reading from
left to right, pick the first one which is not a pivot column. Then from the description of
the row reduced echelon form, this column is a linear combination of the preceding columns.
This proves the corollary.
Example 7.2.9 Find the row reduced echelon form of the matrix,
0 0 2 3
0 2 0 1
0 1 1 5
The first nonzero column is the second in the matrix. We switch the third and first rows
to obtain
0 1 1 5
0 2 0 1
0 0 2 3
Now we multiply the top row by −2 and
0
0
0
add to the second.
1 1
5
0 −2 −9
0 2
3
7.3. THE RANK OF A MATRIX
Next, add the second row to the bottom
0
0
0
To begin, here is a definition to introduce some terminology.
Definition 7.3.1 Let A be an m × n matrix. The column space of A is the span of the
columns. The row space is the span of the rows.
There are three definitions of the rank of a matrix which are useful. These are given
in the following definition. It turns out that the concept of determinant rank is the one
most useful in applications to analysis but is virtually impossible to find directly. The other
two concepts of rank are very easily determined and it is a happy fact that all three yield
the same number. This is shown later.
Definition 7.3.2 A sub-matrix of a matrix A is a space of a matrix is the span of the rows and
the column space of a matrix is the span of the columns. The row rank of a matrix is
the number of nonzero rows in the row reduced echelon form and the column rank is the
number columns in the row reduced echelon form which are one of the ek vectors. Thus the
column rank equals the number of pivot columns. It follows the row rank equals the column
rank. This is also called the rank of the matrix. The rank of a matrix, A is denoted by
rank (A) .
Each has determinant equal to 0. Therefore, the rank is less than 2. Now look at the 1 × 1
submatrices. There exists one of these which has nonzero determinant. For example (1) has
determinant equal to 1 and so the rank of this matrix equals 1.
Of course this example was pretty easy but what if you had a 4 × 7 matrix? You would
have to consider all the 4 × 4 submatrices and then all the 3 × 3 submatrices and then all
the 2 × 2 matrices and finally all the 1 × 1 matrices in order to compute the rank. Clearly
this is not practical. The following theorem will remove the difficulties just indicated.
The following theorem is proved later.
Theorem 7.3.4 Let A be an m × n matrix. Then the row rank, column rank and determinant rank are all the same.
Example 7.3.5 Find the rank of the matrix,
1
2
1 3 0
−4 3
2 1 2
.
3
2
1 6 5
4 −3 −2 1 7
From the above definition, all you have to
then count up the number of nonzero rows.
matrix is
1 0 0
0 1 0
0 0 1
0 0 0
and so the rank of this matrix is 4.
Find the rank of the matrix
The row reduced echelon form also can be used to obtain an efficient description of the row
and column space of a matrix. Of course you can get the column space by simply saying
that it equals the span of all the columns but often you can get the column space as the
span of fewer columns than this. This is what we mean by an "efficient description". This
is illustrated in the next example.
Example 7.3.6 Find the rank of the following matrix and describe the column and row
spaces efficiently.
1 2 1 3 2
1 3 6 0 2
(7.1)
3 7 8 6 6
The row reduced echelon form is
1 0
0 1
0 0
By Lemma 7.2.5, all columns of the original matrix, are similarly contained in the span of
the first two columns of that matrix. For example, consider the third column of the original
matrix.
1
1
2
6 = −9 1 + 5 3 .
8
3
7
How did I know to use −9 and 5 for the coefficients? This is what Lemma 7.2.5 says! It says
linear relationships are all preserved. Therefore, the column space of the original matrix
equals the span of the first two columns. This is the desired efficient description of the
column space.
What about an efficient description of the row space? When row operations are used, the
resulting vectors remain in the row space. Thus the rows in the row reduced echelon form
are in the row space of the original matrix. Furthermore, by reversing the row operations,
each row of the original matrix can be obtained as a linear combination of the rows in the
row reduced echelon form. It follows that the span of the nonzero rows in the row reduced
echelon equals the span of the
example, the row
¡ original rows. In¢the above
¡
¢ space equals
the span of the two vectors, 1 0 −9 9 2 and 0 1 5 −3 0 .
Example 7.3.7 Find the rank of the following matrix and describe the column and row
spaces efficiently.
1 2 1 3 2
1 3 6 0 2
(7.2)
1 2 1 3 2
1 3 2 4 0
It follows the rank is three and the column space is the span of the first, second and fourth
columns of the original matrix.
1
2
0
span 2 , 1 , 2
−1
2
3
¡
¢ ¡
¢
2
0 0 0 1 14
0 1 1 0 − 17
while¡ the row space is the
span
of
the
vectors
,
,
17
¢
21
and 1 0 1 0 17 .
Procedure 7.3.9 To find the rank of a matrix, obtain the row reduced echelon form for
the matrix. Then count the number of nonzero rows or equivalently the number of pivot
columns. This is the rank. The row space is the span of the nonzero rows in the row reduced
echelon form and the column space is the span of the pivot columns of the original matrix.
7.4
7.4.1
Linear Independence And Bases
Linear Independence And Dependence
First we consider the concept of linear independence. We define what it means for vectors
in Fn to be linearly independent and then give equivalent descriptions. In the following
definition, the symbol,
¡
¢
v1 v2 · · · vk
denotes the matrix which has the vector, v1 as the first column, v2 as the second column
and so forth until vk is the k th column.
7.4. LINEAR INDEPENDENCE AND BASES
119
Definition 7.4.1 Let {v1 , · · · , vk } be vectors in Fn . Then this collection
of vectors is said¢
¡
to be linearly independent if each of the columns of the n×k matrix ¡ v1 v2 · · · vk ¢
is a pivot column. Thus the row reduced echelon form for this matrix is e1 e2 · · · ek .
The question whether any vector in the first k columns in a matrix is a pivot column is
independent of the presence of later columns. Thus each of {v1 , · · · , vk } is a pivot column
in
¡
¢
v1 v2 · · · vk
if and only if these vectors are each pivot columns in
¡
v1 v2 · · · vk w1 · · ·
wr
¢
Here is what the above means in terms of linear relationships.
Corollary 7.4.2 The collection of vectors, {v1 , · · · , vk } is linearly independent if and only
if none of these vectors is a linear combination of the others.
Proof: If {v1 , · · · , vk } is linearly independent, then every column in
¡
¢
v1 v2 · · · vk
is a pivot column which requires that the row reduced echelon form is
¡
¢
e1 e2 · · · ek .
Now none of the ei vectors is a linear combination of the others. By Lemma 7.2.5 on Page
113 none of the vi is a linear combination of the others. Recall this lemma says linear
relationships between the columns are preserved under row operations.
Next suppose none of the vectors {v1 , · · · , vk } is a linear combination of the others.
Then none of the columns in
¡
¢
v1 v2 · · · vk
is a linear combination of the others. By Lemma 7.2.5 the same is true of the row reduced
echelon form for this matrix. From the description of the row reduced echelon form, it
follows that the ith column of the row reduced echelon form must be ei since otherwise, it
would be a linear combination of the first i − 1 vectors e1 ,· · · , ei−1 and by Lemma 7.2.5, it
follows vi would be the same linear
combination of ¢v1 , · · · , vi−1 contrary to the assumption
¡
that none of the columns in v1 v2 · · · v¡k is a linear combination
of the others.
¢
Therefore, each of the k columns in columns in v1 v2 · · · vk is a pivot column and
so {v1 , · · · , vk } is linearly independent.
Corollary 7.4.3 The collection of vectors, {v1 , · · · , vk } is linearly independent if and only
if whenever
n
X
ci vi = 0
i=1
it follows each ci = 0.
Proof: Suppose first {v1 , · · · , vk } is linearly independent. Then by Corollary 7.4.2,
none of the vectors is a linear combination of the others. Now suppose
n
X
i=1
ci vi = 0
120
RANK OF A MATRIX
and not all the ci = 0. Then pick ci which is not zero, divide by it and solve for vi in terms
of the other vj , contradicting the fact that none of the vi equals a linear combination of the
others.
Now suppose the condition about the sum holds. If vi is a linear combination of the
other vectors in the list, then you could obtain an equation of the form
X
vi =
cj vj
j6=i
and so
0=
X
cj vj + (−1) vi ,
j6=i
contradicting the condition about the sum.
Sometimes we refer to this last condition about sums as follows: The set of vectors,
{v1 , · · · , vk } is linearly independent if and only if there is no nontrivial linear combination
which equals zero. (A nontrivial linear combination is one in which not all the scalars equal
zero.)
We give the following equivalent definition of linear independence which follows from the
above corollaries.
Definition 7.4.4 A set of vectors, {v1 , · · · , vk } is linearly independent if and only if none
of the vectors is a linear combination of the others or equivalently if there is no nontrivial
linear combination of the vectors which equals 0. It is said to be linearly dependent if at
least one of the vectors is a linear combination of the others or equivalently there exists a
nontrivial linear combination which equals zero.
Note the meaning of the words. To say a set of vectors is linearly dependent means at
least one is a linear combination of the others. In other words, it is in a sense "dependent"
on these other vectors.
The following corollary follows right away from the row reduced echelon form. It concerns
a matrix which looks like this: (More columns than rows.)
Corollary 7.4.5 Let {v1 , · · · , vk } be a set of vectors in Fn . Then if k > n, it must be
the case that {v1 , · · · , vk } is not linearly independent. In other words, if k > n, then
{v1 , · · · , vk } is dependent.
¡
¢
Proof: If k > n, then the columns of v1 v2 · · · vk cannot each be a pivot
column because there are at most n pivot columns due to the fact the matrix has only n
rows. In reading from left to right, pick the first column which is not a pivot column. Then
from the description of row reduced echelon form, this column is a linear combination of the
preceding columns and so the given vectors are dependent by Corollary 7.4.2.
1
2
0
3
1 2
2
1
Example 7.4.6 Determine whether the vectors,
are
3
0 1 2
0
1
2
−1
linearly independent. If they are linearly dependent, exhibit one of the vectors as a linear
combination of the others.
and so every column is a pivot column. Therefore, these vectors are linearly independent
and there is no way to obtain one of the vectors as a linear combination of the others.
122
RANK OF A MATRIX
7.4.2
Subspaces
It turns out that the span of a set of vectors is something called a subspace. We will now
give a different, easier to remember description of subspaces and will then show that every
subspace is the span of a set of vectors.
Definition 7.4.8 Let V be a nonempty collection of vectors in Fn . Then V is called a
subspace if whenever α, β are scalars and u, v are vectors in V, the linear combination,
αu + βv is also in V .
Theorem 7.4.9 V is a subspace of Fn if and only if there exist vectors of V, {u1 , · · · , uk }
such that V = span (u1 , · · · , uk ) .
Proof: Pick a vector of V, u1 . If V = span {u1 } , then stop. You have found your
list of vectors. If V 6= span (u1 ) , then there exists u2 a vector of V which is not a vector
in span (u1 ) . Consider span (u1 , u2 ) . If V = span (u1 , u2 ) , stop. Otherwise, pick u3 ∈
/
span (u1 , u2 ) . Continue this way. Note that since V is a subspace, these spans are each
contained in V . The process must stop with uk for some k ≤ n since otherwise, the matrix
¡
¢
u1 · · · uk
having these vectors as columns would have n rows and k > n columns. Consequently, it
can have no more than n pivot columns and so the first column which is not a pivot column
would be a linear combination of the preceding columns contrary
Pk
Pk to the construction.
For the other half, suppose V = span (u1 , · · · , uk ) and let i=1 ci ui and i=1 di ui be
two vectors in V. Now let α and β be two scalars. Then
α
Suppose s < r. Then the matrix C whose ij th entry is cij has fewer rows, s than columns,
r. By Corollary 7.4.5 the columns of this matrix are linearly dependent. Thus there exist
scalars bj not all zero such that for c1 , · · · , cr the columns of C
r
X
dj cj = 0.
j=1
In other words,
r
X
cij dj = 0, i = 1, 2, · · · , s
j=1
Therefore,
r
X
dj xj
=
j=1
r
X
j=1
=
s
X
i=1
dj
s
X
i=1
r
X
cij yi
cij dj yi =
j=1
s
X
0yi = 0
i=1
which contradicts {x1 , · · · , xr } is linearly independent because not all the dj = 0. Thus
s ≥ r and this proves the lemma.
7.4.3
Basis Of A Subspace
It was just shown in Corollary 7.4.10 that every subspace of Fn is equal to the span of a
linearly independent collection of vectors of Fn . Such a collection of vectors is called a basis.
Definition 7.4.12 Let V be a subspace of Fn . Then {u1 , · · · , uk } is a basis for V if the
following two conditions hold.
1. span (u1 , · · · , uk ) = V.
2. {u1 , · · · , uk } is linearly independent.
The plural of basis is bases.
The main theorem about bases is the following.
Theorem 7.4.13 Let V be a subspace of Fn and suppose {u1 , · · · , uk } and {v1 , · · · , vm }
are two bases for V . Then k = m.
Proof: This follows right away from Lemma 7.4.11. {u1 , · · · , uk } is a spanning set
while {v1 , · · · , vm } is linearly independent so k ≥ m. Also {v1 , · · · , vm } is a spanning set
while {u1 , · · · , uk } is linearly independent so m ≥ k.
Now here is another proof. Suppose k < m. Then since {u1 , · · · , uk } is a basis for V,
each vi is a linear combination of the vectors of {u1 , · · · , uk } . Consider the matrix
¡
¢
u1 · · · uk v1 · · · vm
in which each of the ui is a pivot column because the {u1 , · · · , uk } are linearly independent.
Therefore, the row reduced echelon form of this matrix is
¡
¢
e1 · · · ek w1 · · · wm
(7.3)
124
RANK OF A MATRIX
where each wj has zeroes below the k th row. This is because of Lemma 7.2.5 which implies
each wi is a linear combination of the e1 , · · · , ek . Discarding the bottom n−k rows of zeroes
in the above, yields the matrix,
¡ 0
¢
0
e1 · · · e0k w10 · · · wm
in which all vectors are in Fk . Since m > k, it follows from Corollary 7.4.5 that the vectors,
0
{w10 , · · · , wm
} are dependent. Therefore, some wj0 is a linear combination of the other wi0 .
Therefore, wj is a linear combination of the other wi in 7.3. By Lemma 7.2.5 again, the
same linear relationship exists between the {v1 , · · · , vm } showing that {v1 , · · · , vm } is not
linearly independent and contradicting the assumption that {v1 , · · · , vm } is a basis. It
follows m ≤ k. Similarly, k ≤ m. This proves the theorem.
This is a very important theorem so here is yet another proof of it.
Theorem 7.4.14 Let V be a subspace and suppose {u1 , · · · , uk } and {v1 , · · · , vm } are two
bases for V . Then k = m.
Proof: Suppose k > m. Then since the vectors, {u1 , · · · , uk } span V, there exist scalars,
cij such that
m
X
cij vi = uj .
i=1
Therefore,
k
X
dj uj = 0 if and only if
j=1
k X
m
X
cij dj vi = 0
j=1 i=1
if and only if
m
X
i=1
k
X
cij dj vi = 0
j=1
Now since{v1 , · ·The following definition can now be stated.
Definition 7.4.15 Let V be a subspace of Fn . Then the dimension of V is defined to be
the number of vectors in a basis.
Corollary 7.4.16 The dimension of Fn is n.
Proof: You only need to exhibit a basis for Fn which has n vectors. Such a basis is
{e1 , · · · , en }.
7.4. LINEAR INDEPENDENCE AND BASES
125
Corollary 7.4.17 Suppose {v1 , · · · , vn } is linearly independent and each vi is a vector in
Fn . Then {v1 , · · · , vn } is a basis for Fn . Suppose {v1 , · · · , vm } spans Fn . Then m ≥ n. If
{v1 , · · · , vn } spans Fn , then {v1 , · · · , vn } is linearly independent.
Proof: Let u be a vector of Fn and consider the matrix,
¡
¢
v1 · · · vn u .
Since each vi is a pivot column, the row reduced echelon form is
¡
¢
e1 · · · en w
and so, since w is in span (e1 , · · · , en ) , it follows from Lemma 7.2.5 that u is one of the
vectors in span (v1 , · · · , vn ) . Therefore, {v1 , · · · , vn } is a basis as claimed.
To establish the second claim, suppose that m < n. Then letting vi1 , · · · , vik be the
pivot columns of the matrix
¡
¢
v1 · · · vm
it follows k ≤ m < n and these k pivot columns would be a basis for Fn having fewer than
n vectors, contrary to Theorem 7.4.13 which states every two bases have the same number
of vectors in them.
Finally consider the third claim. If {v1 , · · · , vn } is not linearly independent, then replace
this list with {vi1 , · · · , vik } where these are the pivot columns of the matrix,
¡
¢
v1 · · · vn
Then {vi1 , · · · , vik } spans Fn and is linearly independent so it is a basis having less than n
vectors contrary to Theorem 7.4.13 which states every two bases have the same number of
vectors in them. This proves the corollary.
Example 7.4.18 Find the rank of the following matrix. If the rank is r, identify r columns
in the original matrix which have the property that every other column may be written
as a linear combination of these. Also find a basis for the row and column spaces of the
matrices.
1 2 3
2
1 5 −4 −1
−2 3 1
0
The row reduced echelon form is
1 0 0
0 1 0
0 0 1
27
70
1
10
33
70
and so the rank of the matrix is 3. A basis for the column space is the first three columns of
the original matrix. I know they span because the first three columns of the row reduced
echelon form above span the column space of that matrix. They are linearly independent
because the first three columns of the row reduced echelon form are linearly independent.
By Lemma 7.2.5 all linear relationships are preserved and so these first three vectors form
a basis for the column space. The four rows of the row reduced echelon form form a basis
for the row space of the original matrix.
Example 7.4.19 Find the rank of the following matrix. If the rank is r, identify r columns
in the original matrix which have the property that every other column may be written
A basis for the column space of this row reduced echelon form is the first second and fourth
columns. Therefore, a basis for the column space in the original matrix is the first second
and fourth columns. The rank of the matrix is 3. A basis for the row space of the original
matrix is the columns of the row reduced echelon form.
7.4.4
Extending An Independent Set To Form A Basis
Suppose {v1 , · · · , vm } is a linearly independent set of vectors in Fn . It turns out there is
a larger set of vectors, {v1 , · · · , vm , vm+1 , · · · , vn } which is a basis for Fn . It is easy to do
this using the row reduced echelon form. Consider the following matrix having rank n in
which the columns are shown.
¡
Let A be an m × n matrix.
Definition 7.4.22 ker (A), also referred to as the null space of A is defined as follows.
ker (A) = {x : Ax = 0}
and to find ker (A) one must solve the system of equations Ax = 0.
This is not new! There is just some new terminology being used. To repeat, ker (A) is
the solution to the system Ax = 0.
Example 7.4.23 Let
Notice also that the three vectors above are linearly independent and so the dimension of
ker (A) is 3. This is generally the way it works. The number of free variables equals the
dimension of the null space while the number of basic variables equals the number of pivot
columns which equals the rank. We state this in the following theorem.
7.5. FREDHOLM ALTERNATIVE
129
Definition 7.4.25 The dimension of the null space of a matrix is called the nullity2 and
written as null (A) .
Theorem 7.4.26 Let A be an m × n matrix. Then rank (A) + null (A) = n.
7.4.6
Rank And Existence Of Solutions To Linear Systems
Consider the linear system of equations,
Ax = b
(7.4)
where A is an m × n matrix, x is a n × 1 column vector, and b is an m × 1 column vector.
Suppose
¡
¢
A = a1 · · · an
T
where the ak denote the columns of A. Then x = (x1 , · · · , xn ) is a solution of the system
7.4, if and only if
x1 a1 + · · · + xn an = b
which says that b is a vector in span (a1 , · · · , an ) . This shows that there exists a solution
to the system, 7.4 if and only if b is contained in span (a1 , · · · , an ) . In words, there is a
solution to 7.4 if and only if b is in the column space of A. In terms of rank, the following
proposition describes the situation.
Proposition 7.4.27 Let A be an m × n matrix and let b be an m × 1 column vector. Then
there exists a solution to 7.4 if and only if
¡
¢
rank A | b = rank (A) .
(7.5)
¡
¢
Proof: Place A | b and A in row reduced echelon form, respectively B and C. If
the above condition on rank is true, then both B and C have the same number of nonzero
rows. In particular, you cannot have a row of the form
¡
¢
0 ··· 0 ¥
where ¥ 6= 0 in B. Therefore, there will exist a solution to the system 7.4.
Conversely, suppose there exists a solution. This means there cannot be such a row in
B described above. Therefore, B and C must have the same number of zero rows and so
they have the same number of nonzero rows. Therefore, the rank of the two matrices in 7.5
is the same. This proves the proposition.
7.5
Fredholm Alternative
There is a very useful version of Proposition 7.4.27 known as the Fredholm alternative.
I will only present this for the case of real matrices here. Later a much more elegant and
general approach is presented which allows for the general case of complex matrices.
The following definition is used to state the Fredholm alternative.
Definition 7.5.1 Let S ⊆ Rm . Then S ⊥ ≡ {z ∈ Rm : z · s = 0 for every s ∈ S} . The funny
exponent, ⊥ is called "perp".
2 Isn't
it amazing how many different words are available for use in linear algebra?
This proves the lemma.
Now it is time to state the Fredholm alternative. The first version of this is the following
theorem.
Theorem 7.5.3 Let A be a real m × n matrix and let b ∈ Rm . There exists a solution, x
¡ ¢⊥
to the equation Ax = b if and only if b ∈ ker AT .
¡ ¢⊥
Proof: First suppose b ∈ ker AT . Then this says that if AT x = 0, it follows that
b · x = 0. In other words, taking the transpose, if
xT A = 0, then b · x = 0.
T
In other words, if you get a row of zeros in row reduced echelon form for A then you the
same row operations produce a zero in the m × 1 matrix b.
Consequently
¡
¢
rank A | b = rank (A)
and so by Proposition 7.4.27, there exists a solution, x to the system Ax = b. It remains
to go the other¡direction.
¢
Let z ∈ ker AT and suppose Ax = b. I need to verify b · z = 0. By Lemma 7.5.2,
b · z = Ax · z = x · AT z = x · 0 = 0
This proves the theorem.
This implies the following corollary which is also called the Fredholm alternative. The
"alternative" becomes more clear in this corollary.
7.5. FREDHOLM ALTERNATIVE
131
Corollary 7.5.4 Let A be an m × n matrix. Then A maps Rn onto Rm if and only if the
only solution to AT x = 0 is x = 0.
¡ ¢
¡ ¢⊥
Proof: If the only solution to AT x = 0 is x = 0, then ker AT = {0} and so ker AT
=
Rm because every b ∈ Rm has the property that b · 0 = 0. Therefore, Ax = b has a solu¡ ¢⊥
tion for any b ∈ Rm because the b for which there is a solution are those in ker AT
by
Theorem 7.5.3. In other words, A maps Rn onto Rm .
¡ ¢⊥
Conversely if A is onto, then by Theorem 7.5.3 every b ∈ Rm is in ker AT
and so if
AT x = 0, then b · x = 0 for every b. In particular, this holds for b = x. Hence if AT x = 0,
then x = 0. This proves the corollary.
Here is an amusing example.
Example 7.5.5 Let A be an m × n matrix in which m > n. Then A cannot map onto Rm .
The reason for this is that AT is an n × m where m > n and so in the augmented matrix,
¡ T ¢
A |0
there must be some free variables. Thus there exists a nonzero vector x such that AT x = 0.
7.5.1
Row, Column, And Determinant Rank
I will now present a review of earlier topics and prove Theorem 7.3.4.
Definition 7.5.6 A sub-matrix of a matrix A is the rank is defined to be the dimension of the span
of the rows. The column rank is defined to be the dimension of the span of the columns.
Theorem 7.5.7 If A, an m × n matrix has determinant rank, r, then there exist r rows of
the matrix such that every other row is a linear combination of these r rows.
Proof: Suppose the determinant rank of A = (aij ) equals r. Thus some r × r submatrix
has non zero determinant and there is no larger square submatrix which has non zero
determinant. Suppose such a submatrix is determined by the r columns whose indices are
j1 < · · · < j r
and the r rows whose indices are
i 1 < · · · < ir
I want to show that every row is a linear combination of these rows. Consider the lth row
and let p be an index between 1 and n. Form the following (r + 1) × (r + 1) matrix
ai1 j1 · · · ai1 jr ai1 p
..
..
..
.
.
.
air j1 · · · air jr air p
alj1
· · · aljr
alp
Of course you can assume l ∈
/ {i1 , · · · , ir } because there is nothing to prove if the lth
row is one of the chosen ones. The above matrix has determinant 0. This is because if
p∈
/ {j1 , · · · , jr } then the above would be a submatrix of A which is too large to have non
132
RANK OF A MATRIX
zero determinant. On the other hand, if p ∈ {j1 , · · · , jr } then the above matrix has two
columns which are equal so its determinant is still 0.
Expand the determinant of the above matrix along the last column. Let Ck denote the
cofactor associated with the entry aik p . This is not dependent on the choice of p. Remember,
you delete the column and the row the entry is in and take the determinant of what is left
and multiply by −1 raised to an appropriate power. Let C denote the cofactor associated
with alp . This is given to be nonzero, it being the determinant of the matrix
a i1 j 1
..
.
a ir j 1
···
···
Thus
0 = alp C +
ai1 jr
..
.
air jr
r
X
Ck a i k p
k=1
which implies
alp =
r
X
−Ck
k=1
C
a ik p ≡
r
X
mk aik p
k=1
Since this is true for every p and since mk does not depend on p, this has shown the lth row
is a linear combination of the i1 , i2 , · · · , ir rows. This proves the theorem.
Corollary 7.5.8 The determinant rank equals the row rank.
Proof: From Theorem 7.5.7, the row rank is no larger than the determinant rank. Could
the row rank be smaller than the determinant rank? If so, there exist p rows for p < r such
that the span of these p rows equals the row space. But this implies that the r×r sub-matrix
whose determinant is nonzero also has row rank no larger than p which is impossible if its
determinant is to be nonzero because at least one row is a linear combination of the others.
Corollary 7.5.9 If A has determinant rank, r, then there exist r columns of the matrix
such that every other column is a linear combination of these r columns. Also the column
rank equals the determinant rank.
Proof: This follows from the above by considering AT . The rows of AT are the columns
of A and the determinant rank of AT and A are the same. Therefore, from Corollary 7.5.8,
column rank of A = row rank of AT = determinant rank of AT = determinant rank of A.
The following theorem is of fundamental importance and ties together many of the ideas
presented above.
Theorem 7.5.10 Let A be an n × n matrix. Then the following are equivalent.
1. det (A) = 0.
2. A, AT are not one to one.
3. A is not onto.
Proof: Suppose det (A) = 0. Then the determinant rank of A = r < n. Therefore,
there exist r columns such that every other column is a linear combination of these columns
by Theorem 7.5.7. In particular, it follows that for some m, the mth column is a linear
Since also A0 = 0, it follows A is not one to one. Similarly, AT is not one to one by the
same argument applied to AT . This verifies that 1.) implies 2.).
Now suppose 2.). Then since AT is not one to one, it follows there exists x 6= 0 such that
AT x = 0.
Taking the transpose of both sides yields
xT A = 0
where the 0 is a 1 × n matrix or row vector. Now if Ay = x, then
¡
¢
2
|x| = xT (Ay) = xT A y = 0y = 0
contrary to x 6= 0. Consequently there can be no y such that Ay = x and so A is not onto.
This shows that 2.) implies 3.).
Finally, suppose 3.). If 1.) does not hold, then det (A) 6= 0 but then from Theorem 6.4.15
A−1 exists and so for every y ∈ Fn there exists a unique x ∈ Fn such that Ax = y. In fact
x = A−1 y. Thus A would be onto contrary to 3.). This shows 3.) implies 1.) and proves the
theorem.
Corollary 7.5.11 Let A be an n × n matrix. Then the following are equivalent.
1. det(A) 6= 0.
2. A and AT are one to one.
3. A is onto.
Proof: This follows immediately from the above theorem.
Corollary 7.5.12 Let A be an invertible n × n matrix. Then A equals a finite product of
elementary matrices.
Proof: Since A−1 is given to exist, det (A) 6= 0 and it follows A must have rank n and
so the row reduced echelon form of A is I. Therefore, by Theorem 7.1.6 there is a sequence
of elementary matrices, E1 , · · · , Ep which accomplish successive row operations such that
(Ep Ep−1 · · · E1 ) A = I.
−1
−1
But now multiply on the left on both sides by Ep−1 then by Ep−1
and then by Ep−2
etc.
until you get
−1
A = E1−1 E2−1 · · · Ep−1
Ep−1
and by Theorem 7.1.6 each of these in this product is an elementary matrix.
134
RANK OF A MATRIX
Linear Transformations
8.0.2
Outcomes
A. Define linear transformation. Interpret a matrix as a linear transformation.
B. Find a matrix that represents a linear transformation given by a geometric description.
C. Write the solution space of a homogeneous system as the span of a set of basis vectors.
Determine the dimension of the solution space.
D. Relate the solutions of a non-homogeneous system to the solutions of a homogeneous
system.
The idea is to define a function which takes vectors in F3 and delivers new vectors in F2 .
This is an example of something called a linear transformation.
135
136
LINEAR TRANSFORMATIONS
Definition 8.1.2 Let T : Fn → Fm be a function. Thus for each x ∈ Fn , T x ∈ Fm . Then T
is a linear transformation if whenever α, β are scalars and x1 and x2 are vectors in Fn ,
T (αx1 + βx2 ) = α1 T x1 + βT x2 .
In words, linear transformations distribute across + and allow you to factor out scalars.
At this point, recall the properties of matrix multiplication. The pertinent property is 4.14
on Page 49. Recall it states that for a and b scalars,
A (aB + bC) = aAB + bAC
In particular, for A an m × n matrix and B and C, n × 1 matrices (column vectors) the
above formula holds which is nothing more than the statement that matrix multiplication
gives an example of a linear transformation.
Definition 8.1.3 A linear transformation is called one to one (often written as 1 − 1) if
it never takes two different vectors to the same vector. Thus T is one to one if whenever
x 6= y
T x 6= T y.
Equivalently, if T (x) = T (y) , then x = y.
In the case that a linear transformation comes from matrix multiplication, it is common
usage to refer to the matrix as a one to one matrix when the linear transformation it
determines is one to one.
Definition 8.1.4 A linear transformation mapping Fn to Fm is called onto if whenever
y ∈ Fm there exists x ∈ Fn such that T (x) = y.
Thus T is onto if everything in Fm gets hit. In the case that a linear transformation
comes from matrix multiplication, it is common to refer to the matrix as onto when the
linear transformation it determines is onto. Also it is common usage to write T Fn , T (Fn ) ,or
Im (T ) as the set of vectors of Fm which are of the form T x for some x ∈ Fn . In the case
that T is obtained from multiplication by an m × n matrix, A, it is standard to simply write
A (Fn ) AFn , or Im (A) to denote those vectors in Fm which are obtained in the form Ax for
some x ∈ Fn .
8.2
Constructing The Matrix Of A Linear Transformation
It turns out that if T is any linear transformation which maps Fn to Fm , there is always an
m × n matrix, A with the property that
Ax = T x
(8.1)
for all x ∈ Fn . Here is why. Suppose T : Fn → Fm is a linear transformation and you want
to find the matrix defined by this linear transformation as described in 8.1. Then if x ∈ Fn
it follows
n
X
x=
x i ei
i=1
8.2. CONSTRUCTING THE MATRIX OF A LINEAR TRANSFORMATION
137We state this as the following theorem.
Theorem 8.2.1 Let T be a linear transformation from Fn to Fm . Then the matrix, A
satisfying 8.1
8.2.1
Rotations of R2
Sometimes you need to find a matrix which represents a given linear transformation which
is described in geometrical terms. The idea is to produce a matrix which you can multiply
a vector by to get the same thing as some geometrical description. A good example of this
is the problem of rotation of vectors.
Example 8.2.2 Determine the matrix which represents the linear transformation defined
by rotating every vector through an angle of θ.
µ
From the above, you only need to find T e1 and T e2 , the first being the first column of
the desired matrix, A and the second being the second column. From drawing a picture and
doing a little geometry, you see that
¶
¶
µ
µ
cos θ
− sin θ
T e1 =
, T e2 =
.
sin θ
cos θ
Therefore, from Theorem 8.2.1,
µ
A=
cos θ
sin θ
− sin θ
cos θ
¶
Example 8.2.3You know how to multiply matrices. Do so to the pair on the right. This yields
¶
¶ µ
µ
cos θ cos φ − sin θ sin φ − cos θ sin φ − sin θ cos φ
cos (θ + φ) − sin (θ + φ)
.
=
sin θ cos φ + cos θ sin φ cos θ cos φ − sin θ sin φ
sin (θ + φ) cos (θ + φ)
Don't these look familiar? They are the usual trig. identities for the sum of two angles
derived here using linear algebra concepts.
You do not have to stop with two dimensions. You can consider rotations and other
geometric concepts in any number of dimensions. This is one of the major advantages
of linear algebra. You can break down a difficult geometrical procedure into small steps,
each corresponding to multiplication by an appropriate matrix. Then by multiplying the
matrices, you can obtain a single matrix which can give you numerical information on the
results of applying the given sequence of simple procedures. That which you could never
visualize can still be understood to the extent of finding exact numerical answers. Another
example follows.
Example 8.2.4 Find the matrix of the linear transformation which is obtained by first
rotating all vectors through an angle of π/6 and then reflecting through the x axis.
As shown in Example 8.2.3, the matrix of the transformation which involves rotating
through an angle of π/6 is
¶
µ
¶ µ 1√
3 −√12
cos (π/6) − sin (π/6)
2
=
1
1
sin (π/6) cos (π/6)
2
2 3
In Physics it is important to consider the work done by a force field on an object. This
involves the concept of projection onto a vector. Suppose you want to find the projection
of a vector, v onto the given vector, u, denoted by proju (v) This is done using the dot
product as follows.
³v · u´
proju (v) =
u
u·u
Because of properties of the dot product, the map v → proju (v) is linear,
µ
¶
³v · u´
³w · u´
αv+βw · u
proju (αv+βw) =
u=α
u+β
u
u·u
u·u
u·u
= α proju (v) + β proju (w) .
T
Example 8.2.5 Let the projection map be defined above and let u = (1, 2, 3) . Does this
linear transformation come from multiplication by a matrix? If so, what is the matrix?
You can find this matrix in the same way as in the previous example. Let ei denote the
vector in Rn which has a 1 in the ith position and a zero everywhere else. Thus a typical
T
vector, x = (x1 , · · · , xn ) can be written in a unique way as
x=
n
X
xj ej .
j=1
From the way you multiply a matrix by a vector, it follows that proju (ei ) gives the ith
column of the desired matrix. Therefore, it is only necessary to find
³ e ·u ´
i
proju (ei ) ≡
u
u·u
For the given vector in the example, this implies the columns of the desired matrix are
1
1
1
3
1 2
2 .
2
2 ,
,
14
14
14
3
3
3
Hence the matrix is
Proof: This follows from the definition of matrix multiplication in Definition 4.1.9 on
Page 44.
The following is a theorem of major significance. First here is an interesting observation.
Observation 8.2.7 Let A be an m × n matrix. Then A is one to one if and only if Ax = 0
implies x = 0.
Here is why: A0 = A (0 + 0) = A0 + A0 and so A0 = 0.
Now suppose A is one to one and Ax = 0. Then since A0 = 0, it follows x = 0. Thus if
A is one to one and Ax = 0, then x = 0.
Next suppose the condition that Ax = 0 implies x = 0 is valid. Then if Ax = Ay, then
A (x − y) = 0 and so from the condition, x − y = 0 so that x = y. Thus A is one to one.
Theorem 8.2.8 Suppose A is an n × n matrix. Then A is one to one if and only if A is
onto. Also, if B is an n × n matrix and AB = I, then it follows BA = I.
Proof: First suppose A is one to one. Consider the vectors, {Ae1 , · · · , Aen } where ek
is the column vector which is all zeros except for a 1 in the k th position. This set of vectors
is linearly independent because if
n
X
ck Aek = 0,
k=1
showing that, since y was arbitrary, A is onto.
Next suppose A is onto. This implies the span of the columns of A equals Fn and by
Corollary 7.4.17 this implies the columns of A are independent. If Ax = 0, then letting
T
x = (x1 , · · · , xn ) , it follows
n
X
xi ai = 0
i=1
8.2. CONSTRUCTING THE MATRIX OF A LINEAR TRANSFORMATION
141
and so each xi = 0. If Ax = Ay, then A (x − y) = 0 and so x = y. This shows A is one to
one.
Now suppose AB = I. Why is BA = I? Since AB = I it follows B is one to one since
otherwise, there would exist, x 6= 0 such that Bx = 0 and then ABx = A0 = 0 6= Ix.
Therefore, from what was just shown, B is also onto. In addition to this, A must be one
to one because if Ay = 0, then y = Bx for some x and then x = ABx = Ay = 0 showing
y = 0. Now from what is given to be so, it follows (AB) A = A and so using the associative
law for matrix multiplication,
A (BA) − A = A (BA − I) = 0.
But this means (BA − I) x = 0 for all x since otherwise, A would not be one to one. Hence
BA = I as claimed. This proves the theorem.
This theorem shows that if an n × n matrix, B acts like an inverse when multiplied on
one side of A it follows that B = A−1 and it will act like an inverse on both sides of A.
The conclusion of this theorem pertains to square matrices only. For example, let
µ
¶
1 0
1 0 0
0
1
, B=
(8.2)
A=
1 1 −1
1 0
Then
µ
BA =
but
1
AB = 1
1
1 0
0 1
0
1
0
¶
0
−1 .
0
There is also an important characterization in terms of determinants. This is proved
completely in the section on the mathematical theory of the determinant.
Theorem 8.2.9 Let A be an n × n matrix and let TA denote the linear transformation
determined by A. Then the following are equivalent.
1. TA is one to one.
2. TA is onto.
3. det (A) 6= 0.
8.2.4
The General Solution Of A Linear System
Recall the following definition which was discussed above.
Definition 8.2.10 T is a linear transformation if whenever x, y are vectors and a, b
scalars,
T (ax + by) = aT x + bT y.
(8.3)
Thus linear transformations distribute across addition and pass scalars to the outside. A
linear system is one which is of the form
T x = b.
If T xp = b, then xp is called a particular solution to the linear system.
142
LINEAR TRANSFORMATIONS
For example, if A is an m × n matrix and TA is determined by
TA (x) = Ax,
then from the properties of matrix multiplication, TA is a linear transformation. In this
setting, we will usually write A for the linear transformation as well as the matrix. There
are many other examples of linear transformations other than this. In differential equations,
you will encounter linear transformations which act on functions to give new functions. In
this case, the functions are considered as vectors. Don't worry too much about this at this
time. It will happen later. The fundamental idea is that something is linear if 8.3 holds and
if whenever a, b are scalars and x, y are vectors ax + by is a vector. That is you can add
vectors and multiply by scalars.
Definition 8.2.11 Let T be a linear transformation. Define
ker (T ) ≡ {x : T x = 0} .
In words, ker (T ) is called the kernel of T . As just described, ker (T ) consists of the set of
all vectors which T sends to 0. This is also called the null space of T . It is also called the
solution space of the equation T x = 0.
The above definition states that ker (T ) is the set of solutions to the equation,
T x = 0.
In the case where T is really a matrix, you have been solving such equations for quite some
time. However, sometimes linear transformations act on vectors which are not in Fn . There
is more on this in Chapter 15 on Page 15 and this is discussed more carefully then. However,
consider the following familiar example.
d
denote the linear transformation defined on X, the functions which
Example 8.2.12 Let dx
¡d¢
are defined on R and have a continuous derivative. Find ker dx
.
df
The example asks for functions, f which the property that dx
= 0. As you know from
¡d¢
calculus, these functions are the constant functions. Thus ker dx = constant functions.
When T is a linear transformation, systems of the form T x = 0 are called homogeneous
systems. Thus the solution to the homogeneous system is known as ker (T ) .
Systems of the form T x = b where b 6= 0 are called nonhomogeneous systems. It
turns out there is a very interesting and important relation between the solutions to the
homogeneous systems and the solutions to the nonhomogeneous systems.
Theorem 8.2.13 Suppose xp is a solution to the linear system,
Tx = b
Then if y is any other solution to the linear system, there exists x ∈ ker (T ) such that
y = xp + x.
¡
¢
Proof: Consider y − xp ≡ y+ (−1) xp . Then T y − xp = T y − T xp = b − b = 0. Let
x ≡ y − xp . This proves the theorem.
Sometimes people remember the above theorem in the following form. The solutions
to the nonhomogeneous system, T x = b are given by xp + ker (T ) where xp is a particular
solution to T x = b.
We have been vague about what T is and what x is on purpose. This theorem is
completely algebraic in nature and will work whenever you have linear transformations. In
particular, it will be important in differential equations. For now, here is a familiar example.
A. Determine LU factorizations when possible.
B. Solve a linear system of equations using the LU factorization.
9.1
Definition Of An LU factorization
An LU factorization of a matrix involves writing the given matrix as the product of a lower
triangular matrix which has the main diagonal consisting entirely of ones L, and an upper
triangular matrix U in the indicated order. This is the version discussed here but it is
sometimes the case that the L has numbers other than 1 down the main diagonal. It is still
a useful concept. The L goes with "lower" and the U with "upper". It turns out many
matrices can be written in this way and when this is possible, people get excited about slick
ways of solving the system of equations, Ax = y. It is for this reason that you want to study
the LU factorization. It allows you to work only with triangular matrices. It turns out that
it takes about 2n3 /3 operations to use Gauss elimination but only n3 /3 to obtain an LU
factorization.
First it should be noted not all matrices have an LU factorization and so we will emphasize the techniques for achieving it rather than formal proofs.
µ
Example 9.1.1 Can you write
0
1
1
0
¶
in the form LU as just described?
To do so you would need
µ
¶µ
¶ µ
1 0
a b
a
=
x 1
0 c
xa
b
xb + c
¶
µ
=
0
1
1
0
¶
.
Therefore, b = 1 and a = 0. Also, from the bottom rows, xa = 1 which can't happen and
have a = 0. Therefore, you can't write this matrix in the form LU. It has no LU factorization.
This is what we mean above by saying the method lacks generality.
9.2
Finding An LU Factorization By Inspection
Which matrices have an LU factorization? It turns out it is those whose row reduced echelon
form can be achieved without switching rows and which only involve row operations of type
3 in which row j is replaced with a multiple of row i added to row j for i < j.
145
146
THE LU FACTORIZATION
1
Example 9.2.1 Find an LU factorization of A = 1
2
One way to find the
1
1
2
There is also a convenient procedure for finding an LU factorization. It turns out that it
is only necessary to keep track of the multipliers which are used to row reduce to upper
triangular form. This procedure is described in the following examples.
1 2 3
Example 9.3.1 Find an LU factorization for A = 2 1 −4
1 5 2
Write the matrix next to the identity matrix as shown.
1 0 0
1 2 3
0 1 0 2 1 −4 .
0 0 1
1 5 2
The process involves doing row operations to the matrix on the right while simultaneously
updating successive columns of the matrix on the left. First take −2 times the first row and
add to the second in the matrix on the right.
1 0 0
1 2
3
2 1 0 0 −3 −10
0 0 1
1 5
2
Note the way we updated the matrix on the left. We put a 2 in the second entry of the first
column because we used −2 times the first row added to the second row. Now replace the
9.4. SOLVING SYSTEMS USING THE LU FACTORIZATION
third row in the matrix on the right by −1 times
next step is
1 0 0
1
2 1 0 0
1 0 1
0
Finally, we will add the second row to
1 0
2 1
1 −1
At this point, we stop because the matrix on the right is
ization is the above.
The justification for this gimmick will be given later.
1
2
Example 9.3.2 Find an LU factorization for A =
2
1
upper triangular. An LU factor-
2
0
3
0
1
2
1
1
2
1
3
1
1
1
.
2
2
We will use the same procedure as above. However, this time we will do everything for
one column at a time. First multiply the first row by (−1) and then add to the last row.
Next take (−2) times the first and add to the second and then (−2) times the first and add
to the third.
1 0 0 0
1 2
1
2
1
2 1 0 0 0 −4 0 −3 −1
2 0 1 0 0 −1 −1 −1 0 .
1 0 0 1
0 −2 0 −1 1
This finishes the first column of L and the first column of U. Now take − (1/4) times the
second row in the matrix on the right and add to the third followed by − (1/2) times the
second added to the last.
1 0 0 0
1 2
1
2
1
2 1 0 0 0 −4 0
−3
−1
2 1/4 1 0 0 0 −1 −1/4 1/4
1 1/2 0 1
0 0
0
1/2 3/2
This finishes the second column of L as well as the second column of U . Since the matrix
on the right is upper triangular, stop. The LU factorization has now been obtained. This
technique is called Dolittle's method.
This process is entirely typical of the general case. The matrix U is just the first upper
triangular matrix you come to in your quest for the row reduced echelon form using only
the row operation which involves replacing a row by itself added to a multiple of another
row. The matrix, L is what you get by updating the identity matrix as illustrated above.
You should note that for a square matrix, the number of row operations necessary to
reduce to LU form is about half the number needed to place the matrix in row reduced
echelon form. This is why an LU factorization is of interest in solving systems of equations.
9.4
Solving Systems Using The LU Factorization
The reason people care about the LU factorization is it allows the quick solution of systems
of equations. Here is an example.
Why does the multiplier method work for finding the LU factorization? Suppose A is a
matrix which has the property that the row reduced echelon form for A may be achieved
using only the row operations which involve replacing a row with itself added to a multiple
of another row. It is not ever necessary to switch rows. Thus every row which is replaced
using this row operation in obtaining the echelon form may be modified by using a row
which is above it. Furthermore, in the multiplier method for finding the LU factorization,
we zero out the elements below the pivot element in first column and then the next and
so on when scanning from the left. In terms of elementary matrices, this means the row
operations used to reduce A to upper triangular form correspond to multiplication on the
left by lower triangular matrices having all ones down the main diagonal.and the sequence
of elementary matrices which row reduces A has the property that in scanning the list of
9.5. JUSTIFICATION FOR THE MULTIPLIER METHOD
149
elementary matrices from the right to the left, this list consists of several matrices which
involve only changes from the identity in the first column, then several which involve only
changes from the identity in the second column and so forth. More precisely, Ep · · · E1 A = U
where U is upper triangular, each Ei is a lower triangular elementary matrix having all ones
down the main diagonal, for some ri , each of Er1 · · · E1 differs from the identity only in the
first column, each of Er2 · · · Er1 +1 differs from the identity only in the second column and
Will be L
z
}|
{
−1
so forth. Therefore, A = E1−1 · · · Ep−1
Ep−1 U. You multiply the inverses in the reverse order.
Now each of the Ei−1 is also lower triangular with 1 down the main diagonal. Therefore
their product has this property. Recall also that if Ei equals the identity matrix except
for having an a in the j th column somewhere below the main diagonal, Ei−1 is obtained by
replacing the a in Ei with −a thus explaining why we replace with −1 times the multiplier
−1
in computing L. In the case where A is a 3 × m matrix, E1−1 · · · Ep−1
Ep−1 is of the form
1 0
1 0 0
1 0 0
a 1 0 0 1 0 0 1
b 0 1
0 c
0 0 1
1 0 0
0
0 = a 1 0 .
1
b c 1
Note that scanning from left to right, the first two in the product involve changes in the
identity only in the first column while in the third matrix, the change is only in the second.
If we had zeroed out the elements of the first column in a different order, we would have
obtained.
1 0 0
1 0 0
1 0 0
1 0 0
0 1 0 a 1 0 0 1 0 = a 1 0
b 0 1
0 0 1
0 c 1
b c 1
However, it is important to be working from the left to the right, one column at a time.
A similar observation holds in any dimension. Multiplying the elementary matrices which
involve a change only in the j th column you obtain A equal to an upper triangular, n × m
matrix, U multiplied on its left by a sequence of lower triangular matrices which is of the
following form, the aij being negatives of multipliers used in row reducing A to an upper
triangular matrix.
Notice how the end result of the matrix multiplication made no change in the aij . It just
filled in the empty spaces with the aij which occurred in one of the matrices in the product.
This is why, in computing L, it is sufficient to begin with the left column and work column
by column toward the right, replacing entries with the negative of the multiplier used in the
row operation which produces a zero in that entry.
9.6
The P LU Factorization
As indicated above, some matrices don't have an LU factorization. Here is an example.
1 2 3 2
M = 1 2 3 0
(9.1)
4 3 1 1
In this case, there is another factorization which is useful called a P LU factorization. Here
P is a permutation matrix.
Example 9.6.1 Find a P LU factorization for the above matrix in 9.1.
Proceed as before trying to find the row echelon form of the matrix. First add −1 times
the first row to the second row and then add −4 times the first to the third. This yields
1 0 0
1 2
3
2
1 1 0 0 0
0
−2
4 0 1
0 −5 −11 −7
There is no way to do only row operations involving replacing a row with itself added
to a multiple of another row to the matrix on the right in such a way as to obtain an
upper triangular matrix. Therefore, consider the original matrix with the bottom two rows
switched.
1 2 3 2
1 0 0
1 2 3 2
M0 = 4 3 1 1 = 0 0 1 1 2 3 0
1 2 3 0
0 1 0
4 3 1 1
= PM
Now try again with this matrix. First take −1 times the first row and add to the bottom
row and then take −4 times the first row and add to the second row. This yields
1 0 0
1 2
3
2
4 1 0 0 −5 −11 −7
1 0 1
0 0
0
−2
As pointed out above, the LU factorization is not a mathematically respectable thing because it does not always exist. There is another factorization which does always exist. Much
more can be said about it than I will say here. I will only deal with real matrices and so
the dot product will be the usual real dot product.
152
THE LU FACTORIZATION
Definition 9.7.1 An n × n real matrix Q is called an orthogonal matrix if
QQT = QT Q = I.
Thus an orthogonal matrix is one whose inverse is equal to its transpose.
First note that if a matrix is orthogonal this says
X
X
QTij Qjk =
Qji Qjk = δ ik
j
Thus
2
|Qx| =
X
j
2
XXX
X
Qis xs Qir xr
Qij xj =
i
=
r
i
Qis Qir xs xr =
XX
r
s
XXX
s
=
r
i
j
XXX
r
δ sr xs xr =
X
s
s
Qis Qir xs xr
i
x2r = |x|
2
r
This shows that orthogonal transformations preserve distances. You can show that if you
have a matrix which does preserve distances, then it must be orthogonal also.
Example 9.7.2 One of the most important examples of an orthogonal matrix is the so
called Householder matrix. You have v a unit vector and you form the matrix,
I − 2vvT
This is an orthogonal matrix which is also symmetric. To see this, you use the rules of
matrix operations.
¡
¢T
¡
¢T
I − 2vvT
= I T − 2vvT
=
I − 2vvT
so it is symmetric. Now to show it is orthogonal,
¡
¢¡
¢
I − 2vvT I − 2vvT
= I − 2vvT − 2vvT + 4vvT vvT
=
The orthogonal matrix Q reflects across the dotted line taking x to y and y to x.
Definition 9.7.4 Let A be an m × n matrix. Then a QR factorization of A consists of two
matrices, Q orthogonal and R upper triangular or in other words equal to zero below the
main diagonal such that A = QR.
With the solution to this simple problem, here is how to obtain a QR factorization for
any matrix A. Let
A = (a1 , a2 , · · · , an )
where the ai are the columns. If a1 = 0, let Q1 = I. If a1 6= 0, let
|a1 |
0
b ≡ .
..
0
and form the Householder matrix,
Q1 ≡ I − 2
(a1 − b)
2
|a1 − b|
As in the above problem Q1 a1 = b and so
µ
|a1 |
Q1 A =
0
(a1 − b)
∗
A2
T
¶
where A2 is a m−1×n−1 matrix. Now find in the same way as was just done a n−1×n−1
b 2 such that
matrix Q
µ
¶
∗ ∗
b
Q2 A2 =
0 A3
154
THE LU FACTORIZATION
Let
µ
1
0
Q2 ≡
Then
µ
Q2 Q1 A =
=
1
0
0
b2
Q
¶µ
0
b2
Q
|a1 |
..
.
0
∗
∗
0
∗
¶
.
|a1 | ∗
0 A2
¶
∗
A3
Continuing this way untill the result is upper triangular, you get a sequence of orthogonal
matrices Qp Qp−1 · · · Q1 such that
Qp Qp−1 · · · Q1 A = R
(9.2)
where R is upper triangular.
Now if Q1 and Q2 are orthogonal, then from properties of matrix multiplication,
T
Q1 Q2 (Q1 Q2 ) = Q1 Q2 QT2 QT1 = Q1 IQT1 = I
and similarly
T
(Q1 Q2 ) Q1 Q2 = I.
Thus the product of orthogonal matrices is orthogonal. Also the transpose of an orthogonal
matrix is orthogonal directly from the definition. Therefore, from 9.2
T
A = (Qp Qp−1 · · · Q1 ) R ≡ QR.
This proves the following theorem.
Theorem 9.7.5 Let A be any real m × n matrix. Then there exists an orthogonal matrix,
Q and an upper triangular matrix R such that
A = QR
and this factorization can be accomplished in a systematic manner.
Linear Programming
10.1
Simple Geometric Considerations
One of the most important uses of row operations is in solving linear program problems
which involve maximizing a linear function subject to inequality constraints determined
from linear equations. Here is an example. A certain hamburger store has 9000 hamburger
patties to use in one week and a limitless supply of special sauce, lettuce, tomatoes, onions,
and buns. They sell two types of hamburgers, the big stack and the basic burger. It has also
been determined that the employees cannot prepare more than 9000 of either type in one
week. The big stack, popular with the teen agers from the local high school, involves two
patties, lots of delicious sauce, condiments galore, and a divider between the two patties.
The basic burger, very popular with children, involves only one patty and some pickles
and ketchup. Demand for the basic burger is twice what it is for the big stack. What
is the maximum number of hamburgers which could be sold in one week given the above
limitations?
Let x be the number of basic burgers and y the number of big stacks which could be sold
in a week. Thus it is desired to maximize z = x + y subject to the above constraints. The
total number of patties is 9000 and so the number of patty used is x+2y. This number must
satisfy x + 2y ≤ 9000 because there are only 9000 patty available. Because of the limitation
on the number the employees can prepare and the demand, it follows 2x + y ≤ 9000.
You never sell a negative number of hamburgers and so x, y ≥ 0. In simpler terms the
problem reduces to maximizing z = x + y subject to the two constraints, x + 2y ≤ 9000 and
2x + y ≤ 9000. This problem is pretty easy to solve geometrically. Consider the following
picture in which R labels the region described by the above inequalities and the line z = x+y
is shown for a particular value of z.
A
x+y =z
A
A2x + y = 4
@ A
@ A
HH
@H
H
A
@
AH
R A@HHx + 2y = 4
A@ HHH
A @
As you make z larger this line moves away from the origin, always having the same slope
155
156
LINEAR PROGRAMMING
and the desired solution would consist of a point in the region, R which makes z as large as
possible or equivalently one for which the line is as far as possible from the origin. Clearly
this point is the point of intersection of the two lines, (3000, 3000) and so the maximum
value of the given function is 6000. Of course this type of procedure is fine for a situation in
which there are only two variables but what about a similar problem in which there are very
many variables. In reality, this hamburger store makes many more types of burgers than
those two and there are many considerations other than demand and available patty. Each
will likely give you a constraint which must be considered in order to solve a more realistic
problem and the end result will likely be a problem in many dimensions, probably many
more than three so your ability to draw a picture will get you nowhere for such a problem.
Another method is needed. This method is the topic of this section. I will illustrate with
this particular problem. Let x1 = x and y = x2 . Also let x3 and x4 be nonnegative variables
such that
x1 + 2x2 + x3 = 9000, 2x1 + x2 + x4 = 9000.
To say that x3 and x4 are nonnegative is the same as saying x1 + 2x2 ≤ 9000 and 2x1 + x2 ≤
9000 and these variables are called slack variables at this point. They are called this because
they "take up the slack". I will discuss these more later. First a general situation is
considered.
10.2
The Simplex Tableau
Here is some notation.
Definition 10.2.1 Let x, y be vectors in Rq . Then x ≤ y means for each i, xi ≤ yi .
The problem is as follows:
Let A be an m × (m + n) real matrix of rank m. It is desired to find x ∈ Rn+m such
that x satisfies the constraints,
x ≥ 0, Ax = b
(10.1)
and out of all such x,
z≡
m+n
X
ci xi
i=1
is as large (or small) as possible. This is usually referred to as maximizing or minimizing z
subject to the¡ above constraints.¢ First I will consider the constraints.
Let A = a1 · · · an+m . First you find a vector, x0 ≥ 0, Ax0 = b such that n of
the components of this vector equal 0. Letting i1 , · · · , in be the positions of x0 for which
x0i = 0, suppose also that {aj1 , · · · , ajm } is linearly independent for ji the other positions
of x0 . Geometrically, this means that x0 is a corner of the feasible region, those x which
satisfy the constraints. This is called a basic feasible solution. Also define
and
cB
≡
(cj1 . · · · , cjm ) , cF ≡ (ci1 , · · · , cin )
xB
≡
(xj1 , · · · , xjm ) , xF ≡ (xi1 , · · · , xin ) .
¡ ¢ ¡
z ≡ z x0 = cB
0
cF
¢
µ
x0B
x0F
¶
= cB x0B
since x0F = 0. The variables which are the components of the vector xB are called the basic
variables and the variables which are the entries of xF are called the free variables. You
¡
¢T
The reason there is a z 0 on the bottom right corner is that xF = 0 and x0B , x0F , z 0 is a
solution of the system of equations represented by the above augmented matrix because it is
158
LINEAR PROGRAMMING
a solution to the system of equations corresponding to the system of equations represented
by 10.6 and row operations leave solution sets unchanged. Note how attractive this is. The
z0 is the value of z at the point x0 . The augmented matrix of 10.9 is called the simplex
tableau and it is the beginning point for the simplex algorithm to be described a little
later. It is very convenient to express the simplexµtableau
¶ in the above form in which the
I
on the left side. However, as far
variables are possibly permuted in order to have
0
as the simplex algorithm is concerned it is not necessary to be permuting the variables in
this manner. Starting with 10.9 you could permute the variables and columns to obtain an
augmented matrix in which the variables are in their original order. What is really required
for the simplex tableau?
It is an augmented m + 1 × m + n + 2 matrix which represents a system of equations
T
which has the same set of solutions, (x,z) as the system whose augmented matrix is
µ
¶
A 0 b
−c 1 0
(Possibly the variables for x are taken in another order.) There are m linearly independent
columns in the first m + n columns for which there is only one nonzero entry, a 1 in one of
the first m rows, the "simple columns", the other first m + n columns being the "nonsimple
columns". As in the above, the variables corresponding to the simple columns are xB ,
the basic variables and those corresponding to the nonsimple columns are xF , the free
variables. Also, the top m entries of the last column on the right are nonnegative. This is
the description of a simplex tableau.
In a simplex tableau it is easy to spot a basic feasible solution. You can see one quickly
by setting the variables, xF corresponding to the nonsimple columns equal to zero. Then the
other variables, corresponding to the simple columns are each equal to a nonnegative entry
in the far right column. Lets call this an "obvious basic feasible solution". If a solution is
obtained by setting the variables corresponding to the nonsimple columns equal to zero and
the variables corresponding to the simple columns equal to zero this will be referred to as
an "obvious" solution. Lets also call the first m + n entries in the bottom row the "bottom
left row". In a simplex tableau, the entry in the bottom right corner gives the value of the
variable being maximized or minimized when the obvious basic feasible solution is chosen.
The following is a special case of the general theory presented above and shows how such
a special case can be fit into the above framework. The following example is rather typical
of the sorts of problems considered. It involves inequality constraints instead of Ax = b.
This is handled by adding in "slack variables" as explained below.
Example 10.2.2 Consider z = x1 − x2 subject to the constraints, x1 + 2x2 ≤ 10, x1 + 2x2 ≥
2, and 2x1 + x2 ≤ 6, xi ≥ 0. Find a simplex tableau for a problem of the form x ≥ 0,Ax = b
which is equivalent to the above problem.
You add in slack variables. These are positive variables, one for each of the first three constraints, which change the first three inequalities into equations. Thus the first three inequalities become x1 +2x2 +x3 = 10, x1 +2x2 −x4 = 2, and 2x1 +x2 +x5 = 6, x1 , x2 , x3 , x4 , x5 ≥ 0.
Now it is necessary to find a basic feasible solution. You mainly need to find a positive solution to the equations,
x1 + 2x2 + x3 = 10
x1 + 2x2 − x4 = 2 .
2x1 + x2 + x5 = 6
the solution set for the above system is given by
x2 =
The obvious solution is not feasible because of that -1 in the fourth column. Consider the
second column and select the 2 as a pivot to zero out that which is above and below the 2.
This is because that 2 satisfies the criterion for being chosen as a pivot.
0 0 1 1 0 4
1 1 0 −1 0 3
2
2
3
1
0 0
1 3
2
2
This one is good. The obvious solution is now feasible. You can now assemble the simplex
tableau. The first step is to include a column and row for z. This yields
0 0 1 1 0 0 4
1 1 0 −1 0 0 3
2
23
1
0 0
1 0 3
2
2
−1 0 1 0 0 1 0
Now you need to get zeros in the right places so the simple columns will be preserved as
simple columns. This means you need to zero out the 1 in the third column on the bottom.
A simplex tableau is now
0 0 1 1 0 0 4
1 1 0 −1 0 0 3
2
32
.
1
0 0
1 0 3
2
2
−1 0 0 −1 0 1 −4
Note it is not the same one obtained earlier. There is no reason a simplex tableau should
be unique. In fact, it follows from the above general description that you have one for each
basic feasible point of the region determined by the constraints.
10.3
The Simplex Algorithm
10.3.1
Maximums
The simplex algorithm takes you from one basic feasible solution to another while maximizing or minimizing the function you are trying to maximize or minimize. Algebraically,
it takes you from one simplex tableau to another in which the lower right corner either
increases in the case of maximization or decreases in the case of minimization.
I will continue writing the simplex tableau in such a way that the simple columns having
only one entry nonzero are on the left. As explained above, this amounts to permuting the
variables. I will do this because it is possible to describe what is going on without onerous
notation. However, in the examples, I won't worry so much about it. Thus, from a basic
feasible solution, a simplex tableau of the following form has been obtained in which the
columns for the basic variables, xB are listed first and b ≥ 0.
µ
¶
I F 0 b
(10.10)
0 c 1 z0
¡
¢
Let x0i = bi for i = 1, · · · , m and x0i¡ = 0 for
i > m. Then x0 , z 0 is a solution to the above
¢
system and since b ≥ 0, it follows x0 , z 0 is a basic feasible solution.
µ
¶
F
If ci < 0 for some i, and if Fji ≤ 0 so that a whole column of
is ≤ 0 with the
c
bottom entry < 0, then letting xi be the variable corresponding to that column, you could
10.3. THE SIMPLEX ALGORITHM
161
leave all the other entries of xF equal to zero but change xi to be positive. Let the new
vector be denoted by x0F and letting x0B = b − F x0F it follows
X
(x0B )k = bk −
Fkj (xF )j
j
=
bk − Fki xi ≥ 0
(x0B , x0F )
Now this shows
is feasible whenever xi > 0 and so you could let xi become
arbitrarily large and positive and conclude there is no maximum for z because
z = −cx0F + z 0 = (−ci ) xi + z 0
(10.11)
If this happens in a simplex tableau, you can say there is no maximum and stop.
What if c ≥ 0? Then z = z 0 − cxF and to satisfy the constraints, xF ≥ 0. Therefore,
in this case, z 0 is the largest possible value of z and so the maximum has been found. You
stop when this occurs. Next I explain what to do if neither of the above stopping conditions
hold.
µThe only
¶ case which remains is that some ci < 0 and some Fji > 0. You pick a column
F
in
in which ci < 0, usually the one for which ci is the largest in absolute value.
c
You pick Fji > 0 as a pivot element, divide the j th row by Fji and then use to obtain
zeros above Fji and below Fji , thus obtaining a new simple column. This row operation
also makes exactly one of the other simple columns into a nonsimple column. (In terms of
variables, it is said that a free variable becomes a basic variable and a basic variable becomes
a free variable.) Now permuting the columns and variables, yields
¶
µ
I F 0 0 b0
0 c0 1 z 00
³ ´
bj
and ci < 0. If b0 ≥ 0, you are in the same
where z 00 ≥ z 0 because z 00 = z 0 − ci Fji
position you were at the beginning but now z 0 is larger. Now here is the important thing.
You don't pick just any Fji when you do these row operations. You pick the positive one
for which the row operation results in b0 ≥ 0. Otherwise the obvious basic feasible
solution obtained by letting x0F = 0 will fail to satisfy the constraint that x ≥ 0.
How is this done? You need
Fpi bj
b0p ≡ bp −
≥0
(10.12)
Fji
for each p = 1, · · · , m or equivalently,
bp ≥
Fpi bj
.
Fji
(10.13)
Now if Fpi ≤ 0 the above holds. Therefore, you only need to check Fpi for Fpi > 0. The
pivot, Fji is the one which makes the quotients of the form
bp
Fpi
for all positive Fpi the smallest. Having gotten a new simplex tableau, you do the same
thing to it which was just done and continue. As long as b > 0, so you don't encounter the
degenerate case, the values for z associated with setting xF = 0 keep getting strictly larger
every time the process is repeated. You keep going until you find c ≥ 0. Then you stop.
You are at a maximum. Problems can occur in the process in the so called degenerate case
when at some stage of the process some bj = 0. In this case you can cycle through different
values for x with no improvement in z. This case will not be discussed here.
162
LINEAR PROGRAMMING
10.3.2
Minimums
How does it differ if you are finding a minimum? From a basic feasible solution, a simplex
tableau of the following form has been obtained in which the simple columns for the basic
variables, xB are listed first and b ≥ 0.
µ
¶
I F 0 b
(10.14)
0 c 1 z0
¢
¡
Let x0i = bi for i = 1, · · · , m and x0i ¡= 0 for¢ i > m. Then x0 , z 0 is a solution to the above
system and since b ≥ 0, it follows x0 , z 0 is a basic feasible solution. So far, there is no
change.
Suppose first that some ci > 0 and Fji ≤ 0 for each j. Then let x0F consist of changing xi
by making it positive but leaving the other entries of xF equal to 0. Then from the bottom
row,
z = −cx0F + z 0 = −ci xi + z 0
and you let x0B = b − F x0F ≥ 0. Thus the constraints continue to hold when xi is made
increasingly positive and it follows from the above equation that there is no minimum for
z. You stop when this happens.
Next suppose c ≤ 0. Then in this case, z = z 0 − cxF and from the constraints, xF ≥ 0
and so −cxF ≥ 0 and so z 0 is the minimum value and you stop since this is what you are
looking for.
What do you do in the case where some ci > 0 and some Fji > 0? In this case, you use
the simplex algorithm as in the case of maximums to obtain a new simplex tableau in which
z 00 is smaller. You choose Fji the same way to be the positive entry of the ith column such
that bp /Fpi ≥ bj /Fji for all positive entries, Fpi and do the same row operations. Now this
time,
µ
¶
bj
00
0
z = z − ci
< z0
Fji
As in the case of maximums no problem can occur and the process will converge unless
you have the degenerate case in which some bj = 0. As in the earlier case, this is most
unfortunate when it occurs. You see what happens of course. z 0 does not change and the
algorithm just delivers different values of the variables forever with no improvement.
To summarize the geometrical significance of the simplex algorithm, it takes you from
one corner of the feasible region to another. You go in one direction to find the maximum
and in another to find the minimum. For the maximum you try to get rid of negative
entries of c and for minimums you try to eliminate positive entries of c where the method of
elimination involves the auspicious use of an appropriate pivot element and row operations.
Now return to Example 10.2.2. It will be modified to be a maximization problem.
Example 10.3.1 Maximize z = x1 −x2 subject to the constraints, x1 +2x2 ≤ 10, x1 +2x2 ≥
2, and 2x1 + x2 ≤ 6, xi ≥ 0.
Recall this is the same as maximizing z = x1 − x2 subject to
x1
x2
1 2 1 0 0
10
1 2 0 −1 0 x3 = 2 , x ≥ 0,
x4
2 1 0 0 1
6
x5
This information is available to a pig farmer and Fi denotes a particular feed. The numbers
in the table contain the number of units of a particular nutrient contained in one pound of
the given feed. Thus F2 has 2 units of iron in one pound. Now suppose the cost of each feed
in cents per pound is given in the following table.
F1
2
F2
3
F3
2
F4
3
A typical pig needs 5 units of iron, 8 of protein, 6 of folic acid, 7 of copper and 4 of calcium.
(The units may change from nutrient to nutrient.) How many pounds of each feed per pig
should the pig farmer use in order to minimize his cost?
How in the world can you find a basic feasible solution? Remember the simplex algorithm
is designed to keep the entries in the right column nonnegative so you use this algorithm a
few times till the obvious solution is a basic feasible solution.
Consider the first column. The pivot is the 5. Using the row operations described in the
algorithm, you get
7
3
14
1
17
0
−1
0
0
0
5
5
5
5
5
3
2
1
8
1
0 − 15
0
0
0
5
5
5
5
7
8
4
1
22
0
0
−1 0
0
5
5
5
5
5
2
19
0 −1 1 3
0
0
−1
0
5
5
5
5
5
2
3
4
1
0
0
0
0 −1 12
5
5
5
5
5
Now go to the second column. The pivot in this column is the 7/5. This is in a different
row than the pivot in the first column so I will use it to zero out everything below it. This
will get rid of the zeros in the fifth column and introduce zeros in the second. This yields
1
17
2 − 57
0
0
0
0 1 37
7
7
3
1
1 0 1 −1
− 27
0
0
0
7
7
7
0 0 1 −2 1
0 −1 0
0
1
3
1
30
0 0 2
1 −7
0 −1 0
7
7
7
1
2
0 0 37
0
0
0 −1 10
7
7
7
Now consider another column, this time the fourth. I will pick this one because it has
some negative numbers in it so there are fewer entries to check in looking for a pivot.
Unfortunately, the pivot is the top 2 and I don't want to pivot on this because it would
destroy the zeros in the second column. Consider the fifth column. It is also not a good
choice because the pivot is the second element from the top and this would destroy the zeros
10.3. THE SIMPLEX ALGORITHM
165
in the first column. Consider the sixth column. I can use either of the two bottom entries
as the pivot. The matrix is
0
1
0
0
0
1
0
0
0
0
0
1
1
−1
3
2
−1
−2
1
0
−1
1
1
−1
2
0 0
0 0
0 −1
0 0
1 0
0
0
0
−1
0
1
−2
0
3
−7
1
3
1
0
10
Next consider the third column. The pivot is the 1 in the third row. This yields
0
1
0
0
0
1
0
0
0
0
0
0
1
0
0
2
1
−2
−1
6
−1
0
1
0
−1
0
0
0
0
1
0
1
−1
−1
3
0
0
0
−1
0
1
−2
0
3
−7
1
2
1
1
7
.
There are still 5 columns which consist entirely of zeros except for one entry. Four of them
have that entry equal to 1 but one still has a -1 in it, the -1 being in the fourth column.
I need to do the row operations on a nonsimple column which has the pivot in the fourth
row. Such a column is the second to the last. The pivot is the 3. The new matrix is
Now you do row operations to keep the simple columns of 10.15 simple in 10.16. Of course
you could permute the columns if you wanted but this is not necessary.
This yields the following for a simplex tableau. Now it is a matter of getting rid of the
positive entries in the bottom row because you are trying to minimize.
and the process stops. The maximum for z is 7 and it occurs when x1 = 13/2, x2 = 0, x3 =
1/2.
10.4. FINDING A BASIC FEASIBLE SOLUTION
10.4
169
Finding A Basic Feasible Solution
By now it should be fairly clear that finding a basic feasible solution can create considerable
difficulty. Indeed, given a system of linear inequalities along with the requirement that each
variable be nonnegative, do there even exist points satisfying all these inequalities? If you
have many variables, you can't answer this by drawing a picture. Is there some other way
to do this which is more systematic than what was presented above? The answer is yes. It
is called the method of artificial variables. I will illustrate this method with an example.
Example 10.4.1 Find a basic feasible solution to the system 2x1 +x2 −x3 ≥ 3, x1 +x2 +x3 ≥
2, x1 + x2 + x3 ≤ 7 and x ≥ 0.
If you write the appropriate augmented
2 1 −1
1 1 1
1 1 1
matrix with the slack variables,
−1 0 0 3
0 −1 0 2
0
0 1 7
(10.18)
The obvious solution is not feasible. This is why it would be hard to get started with
the simplex method. What is the problem? It is those −1 entries in the fourth and fifth
columns. To get around this, you add in artificial variables to get an augmented matrix of
the form
2 1 −1 −1 0 0 1 0 3
1 1 1
0 −1 0 0 1 2
(10.19)
1 1 1
0
0 1 0 0 7
Thus the variables are x1 , x2 , x3 , x4 , x5 , x6 , x7 , x8 . Suppose you can find a feasible solution
to the system of equations represented by the above augmented matrix. Thus all variables
are nonnegative. Suppose also that it can be done in such a way that x8 and x7 happen
to be 0. Then it will follow that x1 , · · · , x6 is a feasible solution for 10.18. Conversely, if
you can find a feasible solution for 10.18, then letting x7 and x8 both equal zero, you have
obtained a feasible solution to 10.19. Since all variables are nonnegative, x7 and x8 both
equalling zero is equivalent to saying the minimum of z = x7 + x8 subject to the constraints
represented by the above augmented matrix equals zero. This has proved the following
simple observation.
Observation 10.4.2 There exists a feasible solution to the constraints represented by the
augmented matrix of 10.18 and x ≥ 0 if and only if the minimum of x7 + x8 subject to the
constraints of 10.19 and x ≥ 0 exists and equals 0.
Of course a similar observation would hold in other similar situations. Now the point of
all this is that it is trivial to see a feasible solution to 10.19, namely x6 = 7, x7 = 3, x8 = 2
and all the other variables may be set to equal zero. Therefore, it is easy to find an initial
simplex tableau for the minimization problem just described. First add the column and row
for z
2 1 −1 −1 0 0 1
0 0 3
1 1 1
0 −1 0 0
1 0 2
1 1 1
0
0 1 0
0 0 7
0 0 0
0
0 0 −1 −1 1 0
Next it is necessary to make the last two columns on the bottom left row into simple columns.
Performing the row operation, this yields an initial simplex tableau,
2 1 −1 −1 0 0 1 0 0 3
1 1 1
0 −1 0 0 1 0 2
1 1 1
0
0 1 0 0 0 7
3 2 0 −1 −1 0 0 0 1 5
and you see there are only nonpositive numbers on the bottom left column so the process
stops and yields 0 for the minimum of z = x7 +x8 . As for the other variables, x1 = 5/3, x2 =
0, x3 = 1/3, x4 = 0, x5 = 0, x6 = 5. Now as explained in the above observation, this is a
basic feasible solution for the original system 10.18.
Now consider a maximization problem associated with the above constraints.
Example 10.4.3 Maximize x1 − x2 + 2x3 subject to the constraints, 2x1 + x2 − x3 ≥
3, x1 + x2 + x3 ≥ 2, x1 + x2 + x3 ≤ 7 and x ≥ 0.
From 10.20 you can immediately assemble an initial simplex tableau. You begin with
the first 6 columns and top 3 rows in 10.20. Then add in the column and row for z. This
yields
2
0 − 13 − 13 0 0 53
1
3
1
1
0
1
− 23 0 0 13
3
3
0 0 0
0
1 1 0 5
−1 1 −2 0
0 0 1 0
and you first do row operations to make the first and third columns simple columns. Thus
the next simplex tableau is− 13
− 23
1
− 53
0 0
0 0
1 0
0 1
5
3
1
3
5
7
3
You are trying to get rid of negative entries in the bottom left row. There is only one, the
−5/3. The pivot is the 1. The next simplex tableau is then0
0
1
0
1
3
2
3
1
5
3
0
0
0
1
10
3
11
3
5
32
3
and so the maximum value of z is 32/3 and it occurs when x1 = 10/3, x2 = 0 and x3 = 11/3.
10.5. DUALITY
10.5
171
Duality
You can solve minimization problems by solving maximization problems. You can also go
the other direction and solve maximization problems by minimization problems. Sometimes
this makes things much easier. To be more specific, the two problems to be considered are
A.) Minimize z = cx subject to x ≥ 0 and Ax ≥ b and
B.) Maximize w = yb such that y ≥ 0 and yA ≤ c,
¡
¢
equivalently AT yT ≥ cT and w = bT yT .
In these problems it is assumed A is an m × p matrix.
I will show how a solution of the first yields a solution of the second and then show how
a solution of the second yields a solution of the first. The problems, A.) and B.) are called
dual problems.
Lemma 10.5.1 Let x be a solution of the inequalities of A.) and let y be a solution of the
inequalities of B.). Then
cx ≥ yb.
and if equality holds in the above, then x is the solution to A.) and y is a solution to B.).
Proof: This follows immediately. Since c ≥ yA,
cx ≥ yAx ≥ yb.
It follows from this lemma that if y satisfies the inequalities of B.) and x satisfies the
inequalities of A.) then if equality holds in the above lemma, it must be that x is a solution
of A.) and y is a solution of B.). This proves the lemma.
Now recall that to solve either of these problems using the simplex method, you first
add in slack variables. Denote by x0 and y0 the enlarged list of variables. Thus x0 has at
least m entries and so does y0 and the inequalities involving A were replaced by equalities
whose augmented matrices were of the form
¡
¢
¡
¢
A −I b , and AT I cT
Then you included the row and column for z and w to obtain
µ
¶
µ
¶
A −I 0 b
AT
I 0 cT
and
.
−c 0 1 0
−bT 0 1 0
Similar considerations apply to the second problem. Thus as just described, a basic feasible
solution is one which determines a simplex tableau like the above in which you get a feasible
solution by setting all but the first m variables equal to zero. The simplex algorithm takes
you from one basic feasible solution to another till eventually, if there is no degeneracy, you
obtain a basic feasible solution which yields the solution of the problem of interest.
Theorem 10.5.2 Suppose there exists a solution, x to A.) where x is a basic feasible solution of the inequalities of A.). Then there exists a solution, y to B.) and cx = by. It is
also possible to find y from x using a simple formula.
Proof: Since the solution to A.) is basic and feasible, there exists a simplex tableau
like 10.23 such that x0 can be split into xB and xF such that xF = 0 and xB = B −1 b.
Now since it is a minimizer, it follows cB B −1 F − cF ≤ 0 and the minimum value for cx is
cB B −1 b. Stating this again, cx = cB B −1 b. Is it possible you can take y = cB B −1 ? From
−1
Lemma 10.5.1 this will be so if cB B −1 solves the constraints of problem B.). Is
¡ cB B ≥¢ 0?
−1
−1
A −I
Is
≤
¡ cB B ¢ A ≤ c? These two conditions are satisfied if and only if cB B
c 0 . Referring to the process of permuting the columns of the first augmented
matrix
¡
¢
A −I¢ and
the columns
of
¡of 10.21 ¢to get 10.22 and doing the same permutations on
¡
¢
¡
−1
c 0 , the desired inequality
B F ≤ cB cF which
holds if¢and¡ only if cB B
¡
¢
−1
is equivalent to saying cB cB B F ≤ cB cF and this is true because cB B −1 F −
cF ≤ 0 due to the assumption that x is a minimizer. The simple formula is just
y = cB B −1 .
This proves the theorem.
The proof of the following corollary is similar.
Corollary 10.5.3 Suppose there exists a solution, y to B.) where y is a basic feasible
solution of the inequalities of B.). Then there exists a solution, x to A.) and cx = by. It is
also possible to find x from y using a simple formula. In this case, and referring to 10.23,
the simple formula is x = B1−T bB1 .
As an example, consider the pig farmers problem. The main difficulty in this problem
was finding an initial simplex tableau. Now consider the following example and marvel at
how all the difficulties disappear.
Example 10.5.4 minimize C ≡ 2x1 + 3x2 + 2x3 + 3x4 subject to the constraints
x1 + 2x2 + x3 + 3x4
5x1 + 3x2 + 2x3 + x4
x1 + 2x2 + 2x3 + x4
which agrees with the original way of doing the problem.
Two good books which give more discussion of linear programming are Strang [13] and
Nobel and Daniels [10]. Also listed in these books are other references which may prove
useful if you are interested in seeing more on these topics. There is a great deal more which
can be said about linear programming.
Spectral Theory
11.0.1
Outcomes
A. Describe the eigenvalue problem geometrically and algebraically.
B. Evaluate the spectrum and eigenvectors for a square matrix.
C. Find the principle directions of a deformation matrix.
D. Model a Markov process.
(a) Find the limit state.
(b) Determine comparisons of population after a long period of time.
11.1
Eigenvalues And Eigenvectors Of A Matrix
Spectral Theory refers to the study of eigenvalues and eigenvectors of a matrix. It is of
fundamental importance in many areas. Row operations will no longer be such a useful tool
in this subject.
11.1.1
Definition Of Eigenvectors And Eigenvalues
In this section, F = C.
To illustrate the idea behind what will be discussed, consider the following example.
Example 11.1.1 Here is a matrix.
0
0
0
Multiply this matrix by the vector
5
22
−9
−10
16 .
−2
−5
−4
3
and see what happens. Then multiply it by
1
0
0
and see what happens. Does this matrix act this way for some other vector?
175
When you multiply the first vector by the given matrix, it stretched the vector, multiplying
it by 10. When you multiplied the matrix by the second vector it sent it to the zero vector.
Now consider
−5
1
0 5 −10
0 22
16 1 = 38 .
1
−11
0 −9 −2
In this case, multiplication by the matrix did not result in merely multiplying the vector by
a number.
In the above example, the first two vectors were called eigenvectors and the numbers, 10
and 0 are called eigenvalues. Not every number is an eigenvalue and not every vector is an
eigenvector.
Definition 11.1.2 Let M be an n × n matrix and let x ∈ Cn be a nonzero vector for
which
M x = λx
(11.1)
for some scalar, λ. Then x is called an eigenvector and λ is called an eigenvalue (characteristic value) of the matrix, M.
Note: Eigenvectors are never equal to zero!
The set of all eigenvalues of an n × n matrix, M, is denoted by σ (M ) and is referred to as
the spectrum of M.
The eigenvectors of a matrix M are those vectors, x for which multiplication by M
results in a vector in the same direction or opposite direction to x. Since the zero vector, 0
has no direction this would make no sense for the zero vector. As noted above, 0 is never
allowed to be an eigenvector. How can eigenvectors be identified? Suppose x satisfies 11.1.
Then
(M − λI) x = 0
for some x 6= 0. (Equivalently, you could write (λI − M ) x = 0.) Sometimes we will use
(λI − M ) x = 0
and sometimes (M − λI) x = 0. It makes absolutely no difference and you should use
whichever you like better. Therefore, the matrix M − λI cannot have an inverse because if
it did, the equation could be solved,
³
´
−1
−1
−1
x = (M − λI) (M − λI) x = (M − λI) ((M − λI) x) = (M − λI) 0 = 0,
and this would require x = 0, contrary to the requirement that x 6= 0. By Theorem 6.2.1
on Page 85,
det (M − λI) = 0.
(11.2)
11.1. EIGENVALUES AND EIGENVECTORS OF A MATRIX
177
(Equivalently you could write det (λI − M ) = 0.) The expression, det (λI − M ) or equivalently, det (M − λI) is a polynomial called the characteristic polynomial and the above
equation is called the characteristic equation. For M an n × n matrix, it follows from the
theorem on expanding a matrix by its cofactor that det (M − λI) is a polynomial of degree n. As such, the equation, 11.2 has a solution, λ ∈ C by the fundamental theorem of
algebra. Is it actually an eigenvalue? The answer is yes and this follows from Observation
8.2.7 on Page 140 along with Theorem 6.2.1 on Page 85. Since det (M − λI) = 0 the matrix, det (M − λI) cannot be one to one and so there exists a nonzero vector, x such that
(M − λI) x = 0. This proves the following corollary.
Corollary 11.1.3 Let M be an n × n matrix and det (M − λI) = 0. Then there exists a
nonzero vector, x ∈ Cn such that (M − λI) x = 0.
11.1.2
Finding Eigenvectors And Eigenvalues
As an example, consider the following.
Example 11.1.4 Find the eigenvalues and eigenvectors for the matrix,
5 −10 −5
14
2 .
A= 2
−4 −8
6
You first need to identify the eigenvalues. Recall this requires the solution of the equation
det (A − λI) = 0.
In this case this equation is
5
det 2
−4
−10
14
−8
−5
1 0
2 − λ 0 1
6
0 0
0
0 = 0
1
When you expand this determinant and simplify, you find the equation you need to solve is
¡
¢
(λ − 5) λ2 − 20λ + 100 = 0
and so the eigenvalues are
5, 10, 10.
We have listed 10 twice because it is a zero of multiplicity two due to
2
By now this is an old problem. You set up the augmented matrix and row reduce to get the
solution. Thus the matrix you must row reduce is
0 −10 −5 | 0
2
9
2 | 0 .
(11.3)
−4 −8
1 | 0
The row reduced echelon form is
1 0
0 1
0 0
− 54
1
2
0
| 0
| 0
| 0
and so the solution is any vector of the form
5
4t
−1
2 t = t
t
5
4
−1
2
1
where t ∈ F. You would obtain the same collection of vectors if you replaced t with 4t. Thus
a simpler description for the solutions to this system of equations whose augmented matrix
is in 11.3 is
5
t −2
(11.4)
4
where t ∈ F. Now you need to remember that you can't take t = 0 because this would result
in the zero vector and
Eigenvectors are never equal to zero!
Other than this value, every other choice of z in 11.4 results in an eigenvector. It is a good
idea to check your work! To do so, we will take the original matrix and multiply by this
vector and see if we get 5 times this vector.
5 −10 −5
5
25
5
2
14
2 −2 = −10 = 5 −2
−4 −8
6
4
20
4
so it appears this is correct. Always check your work on these problems if you care about
getting the answer right.
The parameter, t is sometimes called a free variable. The set of vectors in 11.4 is
called the eigenspace and it equals ker (A − λI) . You should observe that in this case
the eigenspace has dimension 1 because the eigenspace is the span of a single vector. In
general, you obtain the solution from the row echelon form and the number of different free
variables gives you the dimension of the eigenspace. Just remember that not every vector
in the eigenspace is an eigenvector. The vector, 0 is not an eigenvector although it is in the
eigenspace because
Eigenvectors are never equal to zero!
Next consider the eigenvectors for λ = 10. These vectors are solutions to the equation,
5 −10 −5
1 0 0
x
0
2
14
2 − 10 0 1 0 y = 0
−4 −8
6
0 0 1
z
0
11.1. EIGENVALUES AND EIGENVECTORS OF A MATRIX
That is you must find the solutions to
−5 −10
2
4
−4 −8
and so the eigenvectors are of the form
−1
−2s − t
−2
= s 1 + t 0 .
s
1
0
t
You can't pick t and s both equal to zero because this would result in the zero vector and
Eigenvectors are never equal to zero!
However, every other choice of t and s does result in an eigenvector for the eigenvalue λ = 10.
As in the case for λ = 5 you should check your work if you care about getting it right.
5 −10 −5
−1
−10
−1
2
14
2 0 = 0 = 10 0
−4 −8
6
1
10
1
so it worked. The other vector will also work. Check it.
11.1.3
A Warning
The above example shows how to find eigenvectors and eigenvalues algebraically. You may
have noticed it is a bit long. Sometimes students try to first row reduce the matrix before looking for eigenvalues. This is a terrible idea because row operations destroy the
eigenvalues. The eigenvalue problem is really not about row operations.
The general eigenvalue problem is the hardest problem in algebra and people still do
research on ways to find eigenvalues and their eigenvectors. If you are doing anything which
would yield a way to find eigenvalues and eigenvectors for general matrices without too
much trouble, the thing you are doing will certainly be wrong. The problems you will see
in these notes are not too hard because they are cooked up by us to be easy. Later we
will describe general methods to compute eigenvalues and eigenvectors numerically. These
methods work even when the problem is not cooked up to be easy.
If you are so fortunate as to find the eigenvalues as in the above example, then finding the
eigenvectors does reduce to row operations and this part of the problem is easy. However,
finding the eigenvalues along with the eigenvectors is anything but easy because for an
n × n matrix, it involves solving a polynomial equation of degree n. If you only find a good
180
SPECTRAL THEORY
approximation to the eigenvalue, it won't work. It either is or is not an eigenvalue and
if it is not, the only solution to the equation, (M − λI) x = 0 will be the zero solution as
explained above and
Eigenvectors are never equal to zero!
Here is another example.
Example 11.1.5 Let
2 2
A= 1 3
−1 1
First find the eigenvalues.
2 2
det 1 3
−1 1
−2
−1
1
−2
1
−1 − λ 0
1
0
0
1
0
0
0 = 0
1
This reduces to λ3 − 6λ2 + 8λ = 0 and the solutions are 0, 2, and 4.
0 Can be an Eigenvalue!
Now find the eigenvectors. For λ = 0 the
2
1
−1
and the row reduced echelon form is
Although it is usually hard to solve the eigenvalue problem, there is a kind of matrix for
which this is not the case. These are the upper or lower triangular matrices. I will illustrate
by a examples.
1 2
Example 11.1.6 Let A = 0 4
0 0
4
7 . Find its eigenvalues.
6
You need to solve
1 2 4
1 0 0
0 = det 0 4 7 − λ 0 1 0
0 0 6
0 0 1
1−λ
2
4
4−λ
7 = (1 − λ) (4 − λ) (6 − λ) .
= det 0
0
0
6−λ
Thus the eigenvalues are just the diagonal entries of the original matrix. You can see it
would work this way with any such matrix. These matrices are called upper triangular.
Stated precisely, a matrix A is upper triangular if Aij = 0 for all i > j. Similarly, it is easy
to find the eigenvalues for a lower triangular matrix, on which has all zeros above the main
diagonal.
182
SPECTRAL THEORY
11.1.5
Defective And Nondefective Matrices
Definition 11.1.7 By the fundamental theorem of algebra, it is possible to write the characteristic equation in the form
r
r
rm
(λ − λ1 ) 1 (λ − λ2 ) 2 · · · (λ − λm )
=0
where ri is some integer no smaller than 1. Thus the eigenvalues are λ1 , λ2 , · · · , λm . The
algebraic multiplicity of λj is defined to be rj .
Example 11.1.8 Consider the matrix,
1
A= 0
0
1
1
0
0
1
1
(11.5)
What is the algebraic multiplicity of the eigenvalue λ = 1?
In this case the characteristic equation is
3
The augmented matrix which must be row reduced to get this solution is therefore,
0 1 0 | 0
0 0 1 | 0
0 0 0 | 0
This requires z = y = 0 and x is arbitrary. Thus the eigenspace is
1
t 0 , t ∈ F.
0
It follows the geometric multiplicity of λ = 1 is 1.
Definition 11.1.11 An n × n matrix is called defective if the geometric multiplicity is not
equal to the algebraic multiplicity for some eigenvalue. Sometimes such an eigenvalue for
which the geometric multiplicity is not equal to the algebraic multiplicity is called a defective
eigenvalue. If the geometric multiplicity for an eigenvalue equals the algebraic multiplicity,
the eigenvalue is sometimes referred to as nondefective.
11.1. EIGENVALUES AND EIGENVECTORS OF A MATRIX
183
Here is another more interesting example of a defective matrix.
Example 11.1.12 Let
2
A = −2
14
−1
−2 .
14
−2
−1
25
Find the eigenvectors and eigenvalues.
In this case the eigenvalues are 3, 6, 6 where we have listed 6 twice because it is a zero
of algebraic multiplicity two, the characteristic equation being
2
where t ∈ F.
Note that in this example the eigenspace for the eigenvalue, λ = 6 is of dimension 1
because there is only one parameter. However, this eigenvalue is of multiplicity two as a
root to the characteristic equation. Thus this eigenvalue is a defective eigenvalue. However,
the eigenvalue 3 is nondefective. The matrix is defective because it has a defective eigenvalue.
The word, defective, seems to suggest there is something wrong with the matrix. This
is in fact the case. Defective matrices are a lot of trouble in applications and we may wish
they never occurred. However, they do occur as the above example shows. When you study
linear systems of differential equations, you will have to deal with the case of defective
matrices and you will see how awful they are. The reason these matrices are so horrible
to work with is that it is impossible to obtain a basis of eigenvectors. When you study
differential equations, solutions to first order systems are expressed in terms of eigenvectors
of a certain matrix times eλt where λ is an eigenvalue. In order to obtain a general solution
of this sort, you must have a basis of eigenvectors. For a defective matrix, such a basis does
not exist and so you have to go to something called generalized eigenvectors. Unfortunately,
it is never explained in beginning differential equations courses why there are enough
generalized eigenvectors and eigenvectors to represent the general solution. In fact, this
reduces to a difficult question in linear algebra equivalent to the existence of something
called the Jordan Canonical form which is much more difficult than everything discussed in
the entire differential equations course. If you become interested in this, see [9]or Appendix
A.
Ultimately, the algebraic issues which will occur in differential equations are a red herring
anyway. The real issues relative to existence of solutions to systems of ordinary differential
equations are analytical, having much more to do with calculus than with linear algebra
although this will likely not be made clear when you take a beginning differential equations
class.
In terms of algebra, this lack of a basis of eigenvectors says that it is impossible to obtain
a diagonal matrix which is similar to the given matrix.
Although there may be repeated roots to the characteristic equation, 11.2 and it is not
known whether the matrix is defective in this case, there is an important theorem which
holds when considering eigenvectors which correspond to distinct eigenvalues.
Theorem 11.1.13 of this theorem is not true, then there exist non zero scalars,
ckj such that
m
X
ckj vkj = 0.
(11.6)
j=1
11.1. EIGENVALUES AND EIGENVECTORS OF A MATRIX
185
Take m to be the smallest number possible for an expression of the form 11.6 to hold. Then
solving for vk1
X
vk1 =
dkj vkj
(11.7)
kj 6=k1
a sum having fewer than m terms. However, from the assumption that m is as small as
possible for 11.6 to hold with all the scalars, ckj non zero, it follows that for some j 6= 1,
¡
¢
dkj λk1 − λkj = 0
which implies λk1 = λkj , a contradiction.
Here is another proof in case you did not follow the above.
Theorem 11.1.14 is not true, there exists a basis for span (v1 , · · · , vr ) , consisting
of some vectors from the list {v1 , · · · , vr } which has fewer than r vectors. Let {w1 , · · · , wk }
be this list of vectors. Thus {w1 , · · · , wk } are the pivot columns in the matrix
¡
¢
v1 v2 · · · vk .
Then there exists v ∈ {v1 , · · · , vr } which is a linear combination of the {w1 , · · · , wk }. Thus
v=
k
X
ci wi .
(11.8)
ci λwi wi
(11.9)
i=1
Then doing M to both sides yields
λv v =
k
X
i=1
where λv denotes the eigenvalue for v. But also you could multiply both sides of 11.8 by
λv to get
k
X
λv v =
ci λv wi .
i=1
And now subtracting this from 11.9 yields
0=
k
X
i=1
ci (λv − λwi ) wi
186
SPECTRAL THEORY
and by independence of the {w1 , · · · , wk } , this requires ci (λv − λwi ) = 0 for each i. Since
the eigenvalues are distinct, λv − λwi 6= 0 and so each ci = 0. But from 11.8, this requires
v = 0 which is impossible because v is an eigenvector and
Eigenvectors are never equal to zero!
This proves the theorem.
11.1.6
Complex Eigenvalues
Sometimes you have to consider eigenvalues which are complex numbers. This occurs in
differential equations for example. You do these problems exactly the same way as you do
the ones in which the eigenvalues are real. Here is an example.
Example 11.1.15 Find the eigenvalues and
1
A= 0
0
Recall that n × n matrices can be considered as linear transformations. If F is a 3 × 3 real
matrix having positive determinant, it can be shown that F = RU where R is a rotation
matrix and U is a symmetric real matrix having positive eigenvalues. An application of
this wonderful result, known to mathematicians as the right polar factorization, is to
continuum mechanics where a chunk of material is identified with a set of points in three
dimensional space.
The linear transformation, F in this context is called the deformation gradient and
it describes the local deformation of the material. Thus it is possible to consider this
deformation in terms of two processes, one which distorts the material and the other which
just rotates it. It is the matrix, U which is responsible for stretching and compressing.
This is why in elasticity, the stress is often taken to depend on U which is known in this
context as the right Cauchy Green strain tensor. In this context, the eigenvalues will
always be positive. The symmetry of U allows the proof of a theorem which says that if
λM is the largest eigenvalue, then in every other direction other than the one corresponding
to the eigenvector for λM the material is stretched less than λM and if λm is the smallest
eigenvalue, then in every other direction other than the one corresponding to an eigenvector
of λm the material is stretched more than λm . This process of writing a matrix as a product
of two such matrices, one of which preserves distance and the other which distorts is
also important in applications to geometric measure theory an interesting field of study in
mathematics and to the study of quadratic forms which occur in many applications such as
statistics. Here we are emphasizing the application to mechanics in which the eigenvectors
of the symmetric matrix U determine the principle directions, those directions in which
the material is stretched the most or the least.
Example 11.2.1 Find the principle directions determined by the matrix,
29 6
6
The eigenvalues are 3, 1, and 12 .
11
11
11
6
11
41
44
19
44
6
11
19
44
41
44
188
SPECTRAL THEORY
It is nice to be given the eigenvalues. The largest eigenvalue is 3 which means that in
the direction determined by the eigenvector associated with 3 the stretch is three times as
large. The smallest eigenvalue is 1/2 and so in the direction determined by the eigenvector
for 1/2 the material is stretched by a factor of 1/2, becoming locally half as long. It remains
to find these directions. First consider the eigenvector for 3. It is necessary to solve
29 6
6
11
11
11
0
1 0 0
x
6 41 19
0 1 0 − 11 44 44 y = 0
3
0
0 0 1
z
6 19 41
11
and so the principle direction for the eigenvalue, 3 in which the material is stretched to the
maximum extent is
3
1 .
1
A direction vector (or unit vector) in this direction is
√
3/√11
1/ 11 .
√
1/ 11
You should show that the direction in which the material is compressed the most is in the
direction
0√
−1/ 2
√
1/ 2
Note this is meaningful information which you would have a hard time finding without
the theory of eigenvectors and eigenvalues.
11.2.2
Migration Matrices
There are applications which are of great importance which feature only one eigenvalue.
Definition 11.2.2 Let n locations be denoted by the numbers 1, 2, · · · , n. Also suppose it is
the case that each year aij denotes the proportion of residents in location j which move to
location i. Also suppose no one escapes or emigrates from without these n locations. This last
11.2. SOME APPLICATIONS OF EIGENVALUES AND EIGENVECTORS
189
P
assumption requires i aij = 1. Such matrices in which the columns are nonnegative numbers which sum to one are called Markov matrices. In this context describing migration,
they are also called migration matrices.
Example 11.2.3 Here is an example of one of these matrices.
µ
¶
.4 .2
.6 .8
Thus if it is considered as a migration matrix, .4 is the proportion of residents in location
1 which stay in location one in a given time period while .6 is the proportion of residents in
location 1 which move to location 2 and .2 is the proportion of residents in location 2 which
move to location 1. Considered as a Markov matrix, these numbers are usually identified
with probabilities.
T
If v = (x1 , · · · , xn ) where xi is the population of location
P i at a given instant, you obtain
the population of location i one year later by computing j aij xj = (Av)i . Therefore, the
¡
¢
population of location i after k years is Ak v i . An obvious application of this would be to
a situation in which you rent trailers which can go to various parts of a city and you observe
through experiments the proportion of trailers which go from point i to point j in a single
day. Then you might want to find how many trailers would be in all the locations after 8
days.
Proposition 11.2.4 Let A = (aij ) be a migration matrix. Then 1 is always an eigenvalue
for A.
¡ ¢
Proof: Remember that det B T = det (B) . Therefore,
³
´
¡
¢
T
det (A − λI) = det (A − λI)
= det AT − λI
because I T = I. Thus the characteristic equation for A is the same as the characteristic
equation for AT and so A and AT have the same eigenvalues. We will show that 1 is an
eigenvalue for AT and then it will follow that
P 1 is an eigenvalue for A.
Remember that for a migration matrix, i aij = 1. Therefore, if AT = (bij ) so bij = aji ,
it follows that
X
X
bij =
aji = 1.
j
1
As explained above, this shows that λ = 1 is an eigenvalue for A because A and AT have
the same eigenvalues.
.6 0 .1
Example 11.2.5 10 units of time.
190
SPECTRAL THEORY
From the above, it suffices to consider
10
.6 0 .1
100
115. 085 829 22
.2 .8 0 200 = 120. 130 672 44
.2 .2 .9
400
464. 783 498 34
Of course you would need to round these numbers off.
A related problem asks for how many there will be in the various locations after a long
time. It turns out that if some power of the migration matrix has all positive entries, then
there is a limiting vector, x = limk→∞ Ak x0 where x0 is the initial vector describing the
number of inhabitants in the various locations initially. This vector will be an eigenvector
for the eigenvalue 1 because
x = lim Ak x0 = lim Ak+1 x0 = A lim Ak x = Ax,
k→∞
k→∞
k→∞
and the sum of its entries will equal the sum of the entries of the initial vector, x0 because
this sum is preserved for every multiplication by A since
Ã
!
X
X
X
XX
aij xj =
xj
aij =
xj .
i
j
j
i
j
Here is an example. It is the same example as the one above but here it will involve the
long time limit.
.6 0 .1
Example 11.2.6 a long time.
You just need to find the eigenvector which goes with the eigenvalue 1 and then normalize
it so the sum of its entries equals the sum of the entries of the initial vector. Thus you need
to find a solution to
1 0 0
.6 0 .1
x
0
0 1 0 − .2 .8 0 y = 0
0 0 1
.2 .2 .9
z
0
The augmented matrix is
.4
−. 2
−. 2
0
−. 1
.2
0
−. 2 . 1
and its row reduced echelon form is
1 0
0 1
0 0
Therefore, the eigenvectors are
−. 25
−. 25
0
(1/4)
s (1/4)
1
|
|
|
0
0
0
0
0
0
11.2. SOME APPLICATIONS OF EIGENVALUES AND EIGENVECTORS
191
and all that remains is to choose the value of s such that
1
1
s + s + s = 100 + 200 + 400
4
4
This yields s =
You would of course need to round these numbers off. You see that you are not far off after
just 10 units of time. Therefore, you might consider this as a useful procedure because it is
probably easier to solve a simple system of equations than it is to raise a matrix to a large
power.
1 1 1
Example 11.2.7 Suppose a migration matrix is
location 3 than in location 2.
You see the eigenvalue problem makes these sorts of determinations fairly simple.
192
SPECTRAL THEORY
There are many other things which can be said about these sorts of migration problems. They include things like the gambler's ruin problem which asks for the probability
that a compulsive gambler will eventually lose all his money. However those problems are
not so easy although they still involve eigenvalues and eigenvectors.
There are many other important applications of eigenvalue problems. We have just given
a few such applications here. As pointed out, this is a very hard problem but sometimes
you don't need to find the eigenvalues exactly.
11.3
The Estimation Of Eigenvalues
There are ways to estimate the eigenvalues for matrices from just looking at the matrix.
The most famous is known as Gerschgorin's theorem. This theorem gives a rough idea
where the eigenvalues are just from looking at the matrix.
Theorem 11.3.1 Let A be an n × n matrix. Consider the n Gerschgorin discs defined
as
X
Di ≡ λ ∈ C : |λ − aii | ≤
|aij | .
j6=i
Then every eigenvalue is contained in some Gerschgorin disc.
This theorem says to add up the absolute values of the entries of the ith row which are
off the main diagonal and form the disc centered at aii having this radius. The union of
these discs contains σ (A) , the spectrum of A.
Proof: Suppose Ax = λx where x 6= 0. Then for A = (aij )
X
aij xj = (λ − aii ) xi .
j6=i
Estimate the eigenvalues.
The exact eigenvalues are 35, 56, and 28. The Gerschgorin disks are
D1 = {λ ∈ C : |λ − 21| ≤ 22} ,
D2 = {λ ∈ C : |λ − 60| ≤ 26} ,
and
D3 = {λ ∈ C : |λ − 38| ≤ 15} .
Gerschgorin's theorem says these three disks contain the eigenvalues. Now 35 is in D3 , 56
is in D2 and 28 is in D1 .
More can be said when the Gerschgorin disks are disjoint but this is an advanced topic
which requires the theory of functions of a complex variable. If you are interested and have
a background in complex variable techniques, this is in [9]
Next you need to find the solutions to this equation. Of course this is a real joy. If
there are any rational zeros they are
±
factor of 500
factor of 1
I hope to find a rational zero. If there are none, then I don't know what to do at this
point. This is a really lousy method for finding eigenvalues and eigenvectors. It only
works if things work out well. Lets try 10. You can plug it in and see if it works or
you can use synthetic division.
0
10
1
−25
10
1
−15
200 −500
−150 500
50
0
¡
¢
Yes, it appears 10 works and you can factor the polynomial as (λ − 10) λ2 − 15λ + 50
which factors further to (λ − 10) (λ − 5) (λ − 10) so you find the eigenvalues are 5, 10,
and 10. It remains to find the eigenvectors. First find an eigenvector for λ = 5. To do
this, you find a vector which is sent to 0 by the matrix on the right in 11.10 in which
you let λ = 5. Thus the augmented matrix of the system of equations you need to
solve to get the eigenvector is
5−8
3
−1
| 0
2
5−7
−1
| 0
0
0
5 − 10 | 0
Now the row reduced echelon form is
1 −1 0 |
0 0 1 |
0 0 0 |
0
0
0
T
and so you need x = y and z = 0. An eigenvector is (1, 1, 0) . Now you have the
glorious opportunity to solve for the eigenvectors associated with λ = 10. You do it
the same way. The augmented matrix for the system of equations you solve to find
the eigenvectors is
10 − 8
3
−1
| 0
2
10 − 7
−1
| 0
0
0
10 − 10 | 0
194
SPECTRAL THEORY
The row reduced echelon form is
1
0
0
3
2
0
0
− 12
0
0
| 0
| 0
| 0
and so you need x = − 32 y + 12 z. It follows the eigenvectors for λ = 10 are
¶T
µ
1
3
− y + z, y, z
2
2
where x, y ∈ R, not both equal to zero. Why? Let y = 2 and z = 0. This gives the
vector,
T
(−3, 2, 0)
as one of the eigenvectors. You could also let y = 0 and z = 2 to obtain another
eigenvector,
T
(1, 0, 2) .
If there exists a basis of eigenvectors, then the matrix is nondefective and as discussed
above, the matrix can be diagonalized by considering S −1 AS where the columns of S
are the eigenvectors. In this case, I have found three eigenvectors and so it remains
to determine whether these form a basis. Remember how to do this. You let them be
the columns of a matrix and then find the rank of this matrix. If it is three, then they
are a basis because they are linearly independent and the vectors are in R3 . This is
equivalent to the following matrix has an inverse.
1 −3 1
1 2 0
0 0 2
−1 2
1 −3 1
5
1 2 0 = −1
5
0 0 2
0
Then to diagonalize
2
5
− 15
0
3
5
1
5
0
− 15
1
10
1
2
8 −3
−2 7
0
0
5 0
0 10
0 0
3
5
1
5
− 15
0
1
1
1 1
10
0
0
0
10
1
10
1
2
−3 1
2 0 =
0 2
Isn't this stuff marvelous! You can know this matrix is nondefective at the point when
you find the eigenvectors for the repeated eigenvalue. This eigenvalue was repeated
with multiplicity 2 and there were two parameters, y and z in the description of
the eigenvectors. Therefore, the matrix is nondefective. Also note that there is no
uniqueness for the similarity transformation.
2 1 0
2. Now consider the matrix, 0 1 0 . Find its eigenvectors and eigenvalues and
−1 0 1
determine whether it is defective.
Therefore the eigenvectors are of the form (x, 0, −x) . One such eigenvector is
T
(1, 0, −1) .
3
and so λ = λ showing that λ must be real.
45A
are λ1 , · · · , λn and that A is nondefective. Show that
−1
S where S is the matrix which satisfies S −1 AS = D.
eλ n t
The diagonal matrix, D has the same characteristic equation as A why? and so it has
the same eigenvalues. However the eigenvalues of D are the diagonal entries and so
the diagonal entries of D are the eigenvalues of A. Now
S −1 tAS = tD
198
SPECTRAL THEORY
and
n
(λ1 t)
n
..
(tD) =
.
0
···
..
.
···
0
..
.
(λn t)
n
Therefore,
¢n
∞ ¡ −1
X
S tAS
=
n!
n=0
∞
X
1
n
(tD)
n!
n=0
= S −1
∞
n
X
(tA)
S.
n!
n=0
Now the left side equals
∞
X
1
n
(tD)
n!
n=0
=
n=0
..
.
0
eλ1 t
= ...
0
Therefore,
eλ 1 t
(tA)
≡
= S ...
n!
n=0
0
∞
X
···
..
.
···
n!
···
..
.
···
etA
n
(λ1 t)
1
..
=
.
n!
n=0
0
P∞ (λ1 t)n
∞
X
n
0
..
.
0
..
.
(λn t)
n
···
0
..
..
.
.
P∞ (λn t)n
···
n=0
n!
.
eλn t
···
..
.
···
0
..
.
−1
S .
eλn t
Do you think you understand this? If so, think again. What exactly do you mean by
an infinite sum? Actually there is no problem here. You can do this just fine and the
sums converge in the sense that the ij th entries converge in the partial sums. Think
about this. You know what you need from calculus to see this.
8. Show that if A is similar to B then AT is similar to B T .
¡
¢T
¡ ¢−1
This is easy. A = S −1 BS and so AT = S T B T S −1 = S T B T S T
.
9. Suppose Am = 0 for some m a positive integer. Show that if A is diagonalizable, then
A = 0.
Since Am = 0 suppose S −1 AS = D. Then raising to the mth power, Dm = S −1 Am S =
0. Therefore, D = 0. But then A = S0S −1 = 0.
1
1 −6
10. Find the complex eigenvalues and eigenvectors of the matrix 7 −5 −6 .
−1 7
2
Determine whether the matrix is defective.
After wading through much affliction you find the eigenvalues are −6, 2 + 6i, 2 − 6i.
Since these are distinct, the matrix cannot be defective. We must find the eigenvectors
11.4. EXERCISES WITH ANSWERS
199
and
−1
1
0
= −6. The augmented matrix to row
| 0
| 0
| 0
|
|
|
−1
1 .
−1
0
0 .
0
200
SPECTRAL THEORY
Some Special Matrices
12.0.1
Outcomes
A. Define symmetric matrix, skew-symmetric matrix, and orthogonal matrix. Prove identities involving these types of matrices.
B. Characterize and determine the eigenvalues and eigenvectors of symmetric, skewsymmetric, and orthogonal matrices. Derive basic facts concerning these matrices.
C. Define an orthonormal set of vectors. Determine whether a set of vectors is orthonormal.
D. Relate the orthogonality of a matrix to the orthonormality of its column (or row)
vectors.
E. Diagonalize a symmetric matrix. In particular, given a symmetric matrix, A, find an
orthogonal matrix, U and a diagonal matrix, D such that U T AU = D.
F. Understand and use the Gram Schmidt process.
G. Understand and use the technique of least square approximations.
12.1
Symmetric And Orthogonal Matrices
12.1.1
Orthogonal Matrices
Remember that to find the inverse of a matrix was often a long process. However, it was
very easy to take the transpose of a matrix. For some matrices, the transpose equals the
inverse and when the matrix has all real entries, and this is true, it is called an orthogonal
matrix.
Definition 12.1.1 A real n × n matrix, U is called an Orthogonal matrix if U U T =
U T U = I.
Example 12.1.2 Show the matrix,
U =
√1
2
√1
2
is orthogonal.
201
√1
2
− √12
202
SOME SPECIAL MATRICES
UUT =
√1
2
√1
2
√1
2
− √12
1
Example 12.1.3 Let U = 0
0
√1
2
√1
2
√1
2
− √12
µ
=
1 0
0 1
¶
.
0
0
0 −1 . Is U orthogonal?
−1 0
The answer is yes. This is because the columns form an orthonormal set of vectors as
well as the rows. As discussed above this is equivalent to U T U = I.
The answer is yes. This is because the columns (rows) form an orthonormal set of
vectors.
The importance of orthogonal matrices is that they change components of vectors relative to different Cartesian coordinate systems. Geometrically, the orthogonal matrices are
exactly those which preserve all distances in the sense that if x ∈ Rn and U is orthogonal,
then ||U x|| = ||x|| because
2
T
2
||U x|| = (U x) U x = xT U T U x = xT Ix = ||x|| .
Observation 12.1.7 Suppose U is an orthogonal matrix. Then det (U ) = ±1.
This is easy to see from the properties of determinants. Thus
¡ ¢
¡
¢
2
det (U ) = det U T det (U ) = det U T U = det (I) = 1.
Orthogonal matrices are divided into two classes, proper and improper. The proper
orthogonal matrices are those whose determinant equals 1 and the improper ones are those
whose determinants equal -1. The reason for the distinction is that the improper orthogonal matrices are sometimes considered to have no physical significance since they cause a
change in orientation which would correspond to material passing through itself in a non
physical manner. Thus in considering which coordinate systems must be considered in
certain applications, you only need to consider those which are related by a proper orthogonal transformation. Geometrically, the linear transformations determined by the proper
orthogonal matrices correspond to the composition of rotations.
12.1.2
Symmetric And Skew Symmetric Matrices
Definition 12.1.8 A real n × n matrix, A, is symmetric if AT = A. If A = −AT , then
A is called skew symmetric.
Theorem 12.1.9 The eigenvalues of a real symmetric matrix are real. The eigenvalues of
a real skew symmetric matrix are 0 or pure imaginary.
Proof: The proof of this theorem is in [9]. It is best understood as a special case of
more general considerations. However, here is a proof in this special case.
Recall that for a complex number, a + ib, the complex conjugate, denoted by a + ib is
given by the formula a + ib = a − ib. The notation, x will denote the vector which has every
entry replaced by its complex conjugate.
Suppose A is a real symmetric matrix and Ax = λx. Then
¡ ¢T
λxT x = Ax x = xT AT x = xT Ax = λxT x.
Dividing by xT x on both sides yields λ = λ which says λ is real. (Why?)
Next suppose A = −AT so A is skew symmetric and Ax = λx. Then
¡ ¢T
λxT x = Ax x = xT AT x = −xT Ax = −λxT x
and so, dividing by xT x as before, λ = −λ. Letting λ = a + ib, this means a − ib = −a − ib
and so a = 0. Thus λ is pure imaginary.
It remains to find an orthonormal basis. You can check that the dot product of any of
these vectors with another of them gives zero and so it suffices choose z in each case such
that the resulting vector has length 1. First consider the vectors for λ = 3. It is required to
choose z such that
1√
2 5
z − 34
1
is a unit vector. In other words, you need
1√ 1√
2 5
2 5
3
z − 4 · z − 34 = 1.
1
1
But the above dot product equals
45 2
16 z
and this equals 1 when z =
4
15
√
5. Therefore, the
eigenvector which is desired is
1
2
√
5
4√
5 − 34 =
15
1
2
3
√
− 51 5
√
4
15 5
.
2
Next find the eigenvector for λ = −1. The same process requires that 1 = 45
4 z which
√
2
happens when z = 15
5. Therefore, an eigenvector for λ = −1 which has unit length is
Therefore, the eigenvectors are of the form
√
√
3
5y − 25 5z
− 10
.
y
z
This is a two dimensional eigenspace.
Before going further, we want to point out that no matter how we choose y and z the
resulting vector will be orthogonal to the eigenvector for λ = 2. This is a special case of a
general result which states that eigenvectors for distinct eigenvalues of a symmetric matrix
are orthogonal. This is explained in Problem ??. For this case you need to show the
following dot product equals zero.
2
√
√
− 3 5y − 52 5z
3
1 √ 10
5 5 ·
(12.2)
y
√
4
z
15 5
This is left for you to do.
It remains to find the third vector in the orthonormal basis. This merely involves choosing
y and z in 12.2 in such a way that the resulting vector has dot product with the two given
vectors equal to zero. Thus you need
√
√
3
− 23
− 10
5y − 25 5z
3√
1√
0 =
5y +
5z = 0.
· √
y
5
5
1
5
3
z
The dot product with the eigenvector for λ = 2 is automatically equal to zero and so all
that you need is the above equation. This is satisfied when z = − 13 y. Therefore, the vector
we want is of the form
and so this is indeed an orthonormal basis.
Because of the repeated eigenvalue, there would have been many other orthonormal
bases which could have been obtained. It was pretty arbitrary for to take y = 0 in the above
argument. We could just as easily have taken z = 0 or even y = z = 1. Any such change
would have resulted in a different orthonormal basis. Geometrically, what is happening is
the eigenspace for λ = 1 was two dimensional. It can be visualized as a plane in three
dimensional space which passes through the origin. There are infinitely many different pairs
of perpendicular unit vectors in this plane.
12.1.3
Diagonalizing A Symmetric Matrix
Recall the following definition:
Definition 12.1.17 An n × n matrix, A = (aij ) is called a diagonal matrix if aij = 0
whenever i 6= j. For example, a diagonal matrix is of the form indicated below where ∗
denotes a number.
∗ 0 ··· 0
..
0 ∗
.
.
.
.. 0
..
0 ···
0 ∗
Definition 12.1.18 An n × n matrix, A is said to be non defective or diagonalizable
if there exists an invertible matrix, S such that S −1 AS = D where D is a diagonal matrix
as described above.
Some matrices are non defective and some are not. As indicated in Theorem 12.1.13 if
A is a real symmetric matrix, there exists an orthogonal matrix, U such that U T AU = D
a diagonal matrix. Therefore, every symmetric matrix is non defective because if U is an
orthogonal matrix, its inverse is U T . In the following example, this orthogonal matrix will
be found.
1 0 0
0 23 12
. Find an orthogonal matrix, U such that U T AU
Example 12.1.19 Let A =
1
3
0 2 2
is a diagonal matrix.
In this case, a tedious computation shows the eigenvalues are 2 and 1. First we will find
an eigenvector for the eigenvalue 2. This involves row reducing the following augmented
matrix.
1
0
0
| 0
0 2 − 32
− 21
| 0
3
1
2− 2 | 0
0 −2
12.1. SYMMETRIC AND ORTHOGONAL MATRICES
The row reduced echelon form is
Now what if instead of numbers, the entries, A, B, C, D, E, F, G are matrices of a size such
that the multiplications and additions needed in the above formula all make sense. Would
the formula be true in this case? I will show below that this is true.
Suppose A is a matrix of the form
A11 · · · A1m
..
..
A = ...
(12.3)
.
.
Ar1
···
Arm
where Aij is a si × pj matrix where si is constant for j = 1, · · · , m for each i = 1, · · · , r.
Such a matrix is called a block matrix, also a partitioned matrix. How do you get the
block Aij ? Here is how for A an m × n matrix:
z¡
si ×m
}|
0
Isi ×si
z
n×pj
}| {
0
{¢
0 A Ipj ×pj .
0
(12.4)
In the block column matrix on the right, you need to have cj − 1 rows of zeros above the
small pj × pj identity matrix where the columns of A involved in Aij are cj , · · · , cj + pj and
in the block row matrix on the left, you need to have ri − 1 columns of zeros to the left
of the si × si identity matrix where the rows of A involved in Aij are ri , · · · , ri + si . An
important observation to make is that the matrix on the right specifies columns to use in
the block and the one on the left specifies the rows used. Thus the block Aij in this case
is a matrix of size si × pj . There is no overlap between the blocks of A. Thus the identity
n × n identity matrix corresponding to multiplication on the right of A is of the form
Ip1 ×p1
0
..
.
0
12.2. FUNDAMENTAL THEORY AND GENERALIZATIONS*
and A is a block matrix of the form
A11
..
.
Ap1
···
..
.
···
213
A1m
..
.
Apm
(12.6)
and that for all i, j, it makes sense to multiply Bis Asj for all s ∈ {1, · · · , p}. (That is the
two matrices, Bis and Asj are conformable.) and that
P for fixed ij, it follows Bis Asj is the
same size for each s so that it makes sense to write s Bis Asj .
The following theorem says essentially that when you take the product of two matrices,
you can do it two ways. One way is to simply multiply them forming BA. The other way
is to partition both matrices, formally multiply the blocks to get another block matrix and
this one will be BA partitioned. Before presenting this theorem, here is a simple lemma
which is really a special case of the theorem.
Lemma 12.2.1 Consider the following product.
0
¡
I 0 I
0
0
¢
where the first is n × r and the second is r × n. The small identity matrix I is an r × r matrix
and there are l zero rows above I and l zero columns to the left of I in the right matrix.
Then the product of these matrices is a block matrix of the form
0 0 0
0 I 0
0 0 0
Proof: From the definition of the way you multiply matrices, the product is
0
0
0
0
0
0
I 0 · · · I 0 I e1 · · · I er I 0 · · · I 0
0
0
0
0
0
0
which yields the claimed result. In the formula ej refers to the column vector of length r
which has a 1 in the j th position. This proves the lemma.
Theorem 12.2.2 Let B be a q × p block matrix as in 12.5 and let A be a p × n block matrix
as in 12.6 such that Bis is conformable with Asj and each product, Bis Asj for s = 1, · · · , p
is of the same size so they can be added. Then BA can be obtained as a block matrix such
that the ij th block is of the form
X
Bis Asj .
(12.7)
s
Proof: From 12.4
Bis Asj =
¡
0
Iri ×ri
0
¢
0
B Ips ×ps
0
¡
0
Ips ×ps
0
¢
0
A Iqj ×qj
0
where here it is assumed Bis is ri × ps and Asj is ps × qj . The product involves the sth
block in the ith row of blocks for B and the sth block in the j th column of A. Thus there
The vectors, {uj }j=1 , generated in this way are therefore an orthonormal basis because
each vector has unit length.
The process by which these vectors were generated is called the Gram Schmidt process.
12.2.3
Schur's Theorem∗
Every matrix is related to an upper triangular matrix in a particularly significant way. This
is Shur's theorem and it is the most important theorem in the spectral theory of matrices.
The important result which makes this theorem possible is the Gram Schmidt procedure of
Lemma 12.2.7.
Definition 12.2.8 An n × n matrix, U, is unitary if U U ∗ = I = U ∗ U where U ∗ is defined
∗
. Note that every real orthogonal
to be the transpose of the conjugate of U. Thus Uij = Uji
∗
matrix is unitary. For A any matrix, A just defined as the conjugate of the transpose is
called the adjoint.
∗
This proves the lemma.
Theorem 12.2.10 Let A be an n × n matrix. Then there exists a unitary matrix, U such
that
U ∗ AU = T,
(12.9)
where T is an upper triangular matrix having the eigenvalues of A on the main diagonal
listed according to multiplicity as roots of the characteristic equation.
Proof: Let v1 be a unit eigenvector for A . Then there exists λ1 such that
Av1 = λ1 v1 , |v1 | = 1.
Extend {v1 } to a basis using Theorem 7.4.20 and then use the Gram Schmidt procedure to
obtain {v1 , · · · , vn }, an orthonormal basis in Fn . Let U0 be a matrix whose ith column is
vi . Then from the above, it follows U0 is unitary. Then U0∗ AU0 is of the form
λ1 ∗ · · · ∗
0
..
.
A
1
where A2 is an n − 2 × n − 2 matrix. Continuing in this way, there exists a unitary matrix,
U given as the product of the Ui in the above construction such that
U ∗ AU = T
where T is some upper triangular matrix similar to A which consequently has the same
eigenvalues with the same multiplicities as A.
QnSince the matrix is upper triangular, the
characteristic equation for both A and T is i=1 (λ − λi ) where the λi are the diagonal
entries of T. Therefore, the λi are the eigenvalues.
As a simple consequence of the above theorem, here is an interesting lemma.
Lemma 12.2.11 Let A be of the form
This proves the lemma.
Definition 12.2.12 An n × n matrix, A is called Hermitian if A = A∗ . Thus a real
symmetric matrix is Hermitian.
12.2. FUNDAMENTAL THEORY AND GENERALIZATIONS*
219
Theorem 12.2.13 If A is Hermitian, there exists a unitary matrix, U such that
U ∗ AU = D
(12.10)
where D is a diagonal matrix. That is, D has nonzero entries only on the main diagonal.
Furthermore, the columns of U are an orthonormal basis for Fn .
Proof: From Schur's theorem above, there exists U unitary such that
U ∗ AU = T
where T is an upper triangular matrix. Then from Lemma 12.2.9
∗
where the entries denote the columns of AU and U D respectively. Therefore, Aui = λi ui
and since the matrix is unitary, the ij th entry of U ∗ U equals δ ij and so
δ ij = uTi uj = uTi uj = ui · uj .
This proves the corollary because it shows the vectors {ui } form an orthonormal basis. This
proves the theorem.
Corollary 12.2.14 If A is Hermitian, then all the eigenvalues of A are real.
Proof: Since A is Hermitian, there exists unitary, U such that U ∗ AU = D, a diagonal
matrix whose diagonal entries are the eigenvalues of A. Therefore, D∗ = U ∗ A∗ U = U ∗ AU =
D showing D is real.
Corollary 12.2.15 If A is a real symmetric (A = AT )matrix, then A is Hermitian and
there exists a real unitary matrix, U such that U T AU = D where D is a diagonal matrix.
Proof: This follows from Corollary 12.2.14 which says the eigenvalues are all real. Then
if Ax = λx, the same is true of x. and so in the construction for Shur's theorem, you
can always deal exclusively with real eigenvectors as long as your matrices are real and
symmetric. When you construct the matrix which reduces the problem to a smaller one
having A1 in the lower right corner, use the Gram Schmidt process on Rn using the real dot
product to construct vectors, v2 , · · · , vn in Rn such that {v1 , · · · , vn } is an orthonormal
basis for Rn . The matrix A1 is symmetric also. This is because for j, k ≥ 2
¡
¢T
A1kj = vkT Avj = vkT Avj = vjT Avk = A1jk .
Therefore, continuing this way, the process of the proof delivers only real vectors and real
matrices.
220
SOME SPECIAL MATRICES
12.3
Least Square Approximation
A very important technique is that of the least square approximation.
Lemma 12.3.1 Let A be an m × n matrix and let A (Fn ) denote the set of vectors in Fm
which are of the form Ax for some x ∈ Fn . Then A (Fn ) is a subspace of Fm .
Proof: Let Ax and Ay be two points of A (Fn ) . It suffices to verify that if a, b are
scalars, then aAx + bAy is also in A (Fn ) . But aAx + bAy = A (ax + by) because A is
linear. This proves the lemma.
Theorem 12.3.2 Let y ∈ Fm and let A be an m × n matrix. Then there exists x ∈ Fn
2
minimizing the function, |y−Ax| . Furthermore, x minimizes this function if and only if
(y−Ax) · Aw = 0
for all w ∈ Fn .
Proof: Let {f1 , · · · , fr } be an orthonormal basis for A (Fn ). Since
A (Fn ) = span (f1 , · · · , fr ) ,
it follows that there exists y1 , · · · , yr that minimize
¯
¯2
r
¯
¯
X
¯
¯
y k fk ¯ ,
¯y−
¯
¯
k=1
This proves the theorem.
Recall the definition of the adjoint of a matrix.
Definition 12.3.3 Let A be an m × n matrix. Then
A∗ ≡ (AT ).
This means you take the transpose of A and then replace each entry by its conjugate. This
matrix is called the adjoint. Thus in the case of real matrices having only real entries, the
adjoint is just the transpose.
Lemma 12.3.4 Let A be an m × n matrix. Then
Ax · y = x·A∗ y
Proof: This follows from the definition.
Ax · y
=
X
Aij xj yi
i,j
=
X
xj A∗ji yi
i,j
= x·A∗ y.
This proves the lemma.
The next corollary gives the technique of least squares.
Corollary 12.3.5 A value of x which solves the problem of Theorem 12.3.2 is obtained by
solving the equation
A∗ Ax = A∗ y
and furthermore, there exists a solution to this system of equations.
Proof: For x the unique minimizer of Theorem 12.3.2, (y−Ax) · Aw = 0 for all w ∈ Fn
and from Lemma 12.3.4, this is the same as saying
A∗ (y−Ax) · w = 0
for all w ∈ Fn . This implies
A∗ y − A∗ Ax = 0.
Therefore, there is a unique solution to the equation of this corollary and it solves the
minimization problem of Theorem 12.3.2.
222
12.3.1
SOME SPECIAL MATRICES
The Least Squares Regression Line
For the situation of the least squares regression line discussed here I will specialize to the
case of Rn rather than Fn because it seems this case is by far the most interesting and
the extra details are not justified by an increase in utility. Thus, everywhere you see A∗ it
suffices to place AT .
An important application of Corollary 12.3.5 is the problem of finding the least squares
n
regression line in statistics. Suppose you are given points in the plane, {(xi , yi )}i=1 and you
would like to find constants m and b such that the line y = mx + b goes through all these
points. Of course this will be impossible in general. Therefore, try to find m, b to get as
close as possible. The desired system is
y1
x1 1
µ
¶
.. ..
.. m
. = .
.
b
yn
xn 1
which is of the form y = Ax and it is desired to choose m and b to make
¯
¯2
¯ µ
y1 ¯¯
¶
¯
m
¯
¯
− ... ¯
¯A
b
¯
¯
¯
yn ¯
as small as possible. According to Theorem 12.3.2 and Corollary 12.3.5, the best values for
m and b occur as the solution to
y1
µ
¶
m
AT A
= AT ...
b
yn
where
and one would use the same technique as above. Many other similar problems are important,
including many in higher dimensions and they are all solved the same way.
12.4. THE RIGHT POLAR FACTORIZATION∗
12.3.2
223
The Fredholm Alternative
The next major result is called the Fredholm alternative. It comes from Theorem 12.3.2
and Lemma 12.3.4.
Theorem 12.3.6 Let A be an m × n matrix. Then there exists x ∈ Fn such that Ax = y
if and only if whenever A∗ z = 0 it follows that z · y = 0.
Proof: First suppose that for some x ∈ Fn , Ax = y. Then letting A∗ z = 0 and using
Lemma 12.3.4
y · z = Ax · z = x · A∗ z = x · 0 = 0.
This proves half the theorem.
To do the other half, suppose that whenever, A∗ z = 0 it follows that z · y = 0. It is
necessary to show there exists x ∈ Fn such that y = Ax. From Theorem 12.3.2 there exists
2
x minimizing |y − Ax| which therefore satisfies
(y − Ax) · Aw = 0
(12.11)
for all w ∈ Fn . Therefore, for all w ∈ Fn ,
A∗ (y − Ax) · w = 0
which shows that A∗ (y − Ax) = 0. (Why?) Therefore, by assumption,
(y − Ax) · y = 0.
Now by 12.11 with w = x,
(y − Ax) · (y−Ax) = (y − Ax) · y− (y − Ax) · Ax = 0
showing that y = Ax. This proves the theorem.
The following corollary is also called the Fredholm alternative.
Corollary 12.3.7 Let A be an m × n matrix. Then A is onto if and only if A∗ is one to
one.
Proof: Suppose first A is onto. Then by Theorem 12.3.6, it follows that for all y ∈ Fm ,
y · z = 0 whenever A∗ z = 0. Therefore, let y = z where A∗ z = 0 and conclude that z · z = 0
whenever A∗ z = 0. If A∗ x = A∗ y, then A∗ (x − y) = 0 and so x − y = 0. Thus A∗ is one to
one.
Now let y ∈ Fm be given. y · z = 0 whenever A∗ z = 0 because, since A∗ is assumed to
be one to one, and 0 is a solution to this equation, it must be the only solution. Therefore,
by Theorem 12.3.6 there exists x such that Ax = y therefore, A is onto.
12.4
The Right Polar Factorization∗
The right polar factorization involves writing a matrix as a product of two other matrices,
one which preserves distances and the other which stretches and distorts. First here are
some lemmas which review and add to many of the topics discussed so far about adjoints
and orthonormal sets and such things.
Lemma 12.4.1 Let A be a Hermitian matrix such that all its eigenvalues are nonnegative.
Then there exists a Hermitian matrix, A1/2 such that A1/2 has all nonnegative eigenvalues
¡
¢2
and A1/2 = A.
224
SOME SPECIAL MATRICES
Proof: Since A is Hermitian, there exists a diagonal matrix D having all real nonnegative
entries and a unitary matrix U such that A = U ∗ DU. Then denote by D1/2 the matrix which
is obtained by replacing each diagonal entry of D with its square root. Thus D1/2 D1/2 = D.
Then define
A1/2 ≡ U ∗ D1/2 U.
Then
Proof: This follows from the definition. From the properties of the dot product and
using the fact that the given set of vectors is orthonormal,
¯
¯2
r
r
r
¯X
¯
X
X
¯
¯
ck xk ¯ =
ck xk ,
cj xj
¯
¯
¯
k=1
=
k=1
X
ck cj (xk , xj ) =
j=1
r
X
2
|ck | .
k=1
k,j
This proves the lemma.
Next it is helpful to recall the Gram Schmidt algorithm and observe a certain property
stated in the next lemma.
Lemma 12.4.3 Suppose {w1 , · · · , wr , vr+1 , · · · , vp } is a linearly independent set of vectors
such that {w1 , · · · , wr } is an orthonormal set of vectors. Then when the Gram Schmidt
process is applied to the vectors in the given order, it will not change any of the w1 , · · · , wr .
Proof: Let {u1 , · · · , up } be the orthonormal set delivered by the Gram Schmidt process.
Then u1 = w1 because by definition, u1 ≡ w1 / |w1 | = w1 . Now suppose uj = wj for all
j ≤ k ≤ r. Then if k < r, consider the definition of uk+1 .
wk+1 −
Pk+1
j=1 (wk+1 , uj ) uj
¯
uk+1 ≡ ¯¯
Pk+1
¯
¯wk+1 − j=1 (wk+1 , uj ) uj ¯
By induction, uj = wj and so this reduces to wk+1 / |wk+1 | = wk+1 . This proves the lemma.
This lemma immediately implies the following lemma.
12.4. THE RIGHT POLAR FACTORIZATION∗
225
Lemma 12.4.4 Let V be a subspace of dimension p and let {w1 , · · · , wr } be an orthonormal
set of vectors in V . Then this orthonormal set of vectors may be extended to an orthonormal
basis for V,
{w1 , · · · , wr , yr+1 , · · · , yp }
Proof: First extend the given linearly independent set {w1 , · · · , wr } to a basis for V
and then apply the Gram Schmidt theorem to the resulting basis. Since {w1 , · · · , wr } is
orthonormal it follows from Lemma 12.4.3 the result is of the desired form, an orthonormal
basis extending {w1 , · · · , wr }. This proves the lemma.
Here is another lemma about preserving distance.
Lemma 12.4.5 Suppose R is an m × n matrix with m > n and R preserves distances.
Then R∗ R = I.
Proof: Since R preserves distances, |Rx| = |x| for every x. Therefore from the axioms
of the dot product,
2
= |(R∗ Rx − x, y)|
Thus |(R∗ Rx − x, y)| = 0 for all x, y because the given x, y were arbitrary. Let y =
R∗ Rx − x to conclude that for all x,
R∗ Rx − x = 0
which says R∗ R = I since x is arbitrary. This proves the lemma.
With this preparation, here is the big theorem about the right polar factorization.
Theorem 12.4.6 Let F be an m × n matrix where m ≥ n. Then there exists a Hermitian
n × n matrix, U which has all nonnegative eigenvalues and an m × n matrix, R which
preserves distances and satisfies R∗ R = I such that
F = RU.
In this section, A will be an m × n matrix. To begin with, here is a simple lemma.
Lemma 12.5.1 Let A be an m × n matrix. Then A∗ A is self adjoint and all its eigenvalues
are nonnegative.
2
Proof: It is obvious that A∗ A is self adjoint. Suppose A∗ Ax = λx. Then λ |x| =
(λx, x) = (A∗ Ax, x) = (Ax,Ax) ≥ 0.
Definition 12.5.2 Let A be an m × n matrix. The singular values of A are the square roots
of the positive eigenvalues of A∗ A.
With this definition and lemma here is the main theorem on the singular value decomposition.
Theorem 12.5.3 Let A be an m × n matrix. Then there exist unitary matrices, U and V
of the appropriate size such that
µ
¶
σ 0
∗
U AV =
0 0
where σ is of the form
This proves the corollary.
The singular value decomposition also has a very interesting connection to the problem
of least squares solutions. Recall that it was desired to find x such that |Ax − y| is as small
as possible. Lemma 12.3.2 shows that there is a solution to this problem which can be found
by solving the system A∗ Ax = A∗ y. Each x which solves this system solves the minimization
problem as was shown in the lemma just mentioned. Now consider this equation for the
solutions of the minimization problem in terms of the singular value decomposition.
A∗
A. Apply Gauss-Seidel iteration to approximate a solution to a linear system of equations.
B. Apply Jacobi iteration to approximate a solution to a linear system of equations.
13.1
Iterative Methods For Linear Systems
Consider the problem of solving the equation
Ax = b
(13.1)
where A is an n × n matrix. In many applications, the matrix A is huge and composed
mainly of zeros. For such matrices, the method of Gauss elimination (row operations) is
not a good way to solve the system because the row operations can destroy the zeros and
storing all those zeros takes a lot of room in a computer. These systems are called sparse.
To solve them it is common to use an iterative technique. The idea is to obtain a sequence
of approximate solutions which get close to the true solution after a sufficient number of
iterations.
∞
The matrix on the left in 13.3 is obtained by retaining the main diagonal of A and
setting every other entry equal to zero. The matrix on the right in 13.3 is obtained from A
by setting every diagonal entry equal to zero and retaining all the other entries unchanged.
Example 13.1.4 Use the Jacobi method to solve the system
3 1 0 0
x1
1
1 4 1 0 x2 2
0 2 5 1 x3 = 3
0 0 2 4
x4
4
In terms of
3
0
0
0
The Gauss Seidel method differs from the Jacobi method in using xk+1
for all j < i in going
j
from xk to xk+1 . This is why it is called the method of successive corrections. The precise
description of this method is in the following definition.
Definition 13.1.5 The Gauss Seidel method, also called the method of successive
corrections is given as follows. For A = (aij ) , the iterates for the problem Ax = b are
obtained according to the formula
i
X
aij xr+1
=−
j
j=1
In terms of matrices, letting
n
X
aij xrj + bi .
j=i+1
a11
..
A= .
an1
···
..
.
···
a1n
..
.
ann
(13.4)
13.1. ITERATIVE METHODS FOR LINEAR SYSTEMS
235
The iterates are defined as
=
a11
a21
.
..
an1
0
0
−
.
..
0
0
···
..
.
..
.
a22
..
.
···
···
..
.
..
.
0
0
..
.
···
0
ann
ann−1
a12
0
..
.
a1n
..
.
an−1n
0
xr+1
1
xr+1
2
..
.
xr+1
n
xr1
xr2
.. +
.
xrn
b1
b2
..
.
(13.5)
bn
In words, you set every entry in the original matrix which is strictly above the main
diagonal equal to zero to obtain the matrix on the left. To get the matrix on the right,
you set every entry of A which is on or below the main diagonal equal to zero. Using the
iteration procedure of 13.4 directly, the Gauss Seidel method makes use of the very latest
information which is available at that stage of the computation.
The following example is the same as the example used to illustrate the Jacobi method.
Example 13.1.6 Use the Gauss Seidel method to solve the system
3
1
0
0
1
4
2
0
0
1
5
2
0
x1
x2
0
1 x3
4
x4
1
2
=
3
4
In terms of matrices, this procedure is
3
1
0
0
0
4
2
0
0
0
5
2
r+1
x1
0
xr+1
0
2
0 xr+1
3
4
xr+1
4
0
0
= −
0
0
1
0
0
0
r
0
x1
xr2
0
1 xr3
0
xr4
0
1
0
0
1
2
+ .
3
4
As before, let x1 be the zero vector. Thus the first iteration is to solve
The iterates seem to be getting farther from the actual solution. Why is the process which
worked so well in the other examples not working here? A better question might be: Why
does either process ever work at all?. A complete answer to this question is given in [9].
Both iterative procedures for solving
Ax = b
are of the form
(13.6)
Bxr+1 = −Cxr + b
where A = B + C. In the Jacobi procedure, the matrix C was obtained by setting the
diagonal of A equal to zero and leaving all other entries the same while the matrix, B
was obtained by making every entry of A equal to zero other than the diagonal entries
which are left unchanged. In the Gauss Seidel procedure, the matrix B was obtained from
A by making every entry strictly above the main diagonal equal to zero and leaving the
others unchanged and C was obtained from A by making every entry on or below the main
diagonal equal to zero and leaving the others unchanged. Thus in the Jacobi procedure,
B is a diagonal matrix while in the Gauss Seidel procedure, B is lower triangular. Using
matrices to explicitly solve for the iterates, yields
xr+1 = −B −1 Cxr + B −1 b.
(13.7)
This is what you would never have the computer do but this is what will allow the statement
of a theorem which gives the condition for convergence of these and all other similar methods.
Theorem 13.1.8 Let A = B + C and suppose all eigenvalues of B −1 C have absolute value
less than 1 where A = B + C. Then the iterates in 13.7 converge to the unique solution of
13.6.
A complete explanation of this important result is found in [9]. It depends on a theorem
of Gelfand which is completely proved in this reference. Theorem 13.1.8 is very remarkable
because it gives an algebraic condition for convergence which is essentially an analytical
question.
Numerical Methods For Solving
The Eigenvalue Problem
14.0.3
Outcomes
A. Apply the power method with scaling to approximate the dominant eigenvector corresponding to a dominant eigenvalue.
B. Use the shifted inverse power method to find the eigenvector and eigenvalue close to
some number.
C. Approximate an eigenvalue of a symmetric matrix by computing the Rayleigh quotient and finding the associated error bound. Illustrate why the Rayleigh quotient
approximates the dominant eigenvalue.
14.1
The Power Method For Eigenvalues
As indicated earlier, the eigenvalue eigenvector problem is extremely difficult. Consider for
example what happens if you cannot find the eigenvalues exactly. Then you can't find an
eigenvector because there isn't one due to the fact that A − λI is invertible whenever λ
is not exactly equal to an eigenvalue. Therefore the straightforward way of solving this
problem fails right away, even if you can approximate the eigenvalues. The power method
allows you to approximate the largest eigenvalue and also the eigenvector which goes with
it. By considering the inverse of the matrix, you can also find the smallest eigenvalue.
The method works in the situation of a nondefective matrix, A which has an eigenvalue of
algebraic multiplicity 1, λn which has the property that |λk | < |λn | for all k 6= n. Note that
for a real matrix this excludes the case that λn could be complex. Why? Such an eigenvalue
is called a dominant eigenvalue.
Let {x1 , · · · , xn } be a basis of eigenvectors for Fn such that Axn = λn xn . Now let u1 be
some nonzero vector. Since {x1 , · · · , xn } is a basis, there exists unique scalars, ci such that
u1 =
n
X
ck xk .
k=1
Assume you have not been so unlucky as to pick u1 in such a way that cn = 0. Then let
Auk = uk+1 so that
n−1
X
m
(14.1)
ck λm
um = Am u1 =
k xk + λn cn xn .
k=1
239
240
NUMERICAL METHODS FOR SOLVING THE EIGENVALUE PROBLEM
For large m the last term, λm
vector on the
n cn xn , determines quite well the direction of the
Pn−1
right. This is because |λn | is larger than |λk | and so for a large, m, the sum, k=1 ck λm
k xk ,
on the right is fairly insignificant. Therefore, for large m, um is essentially a multiple of the
eigenvector, xn , the one which goes with λn . The only problem is that there is no control
of the size of the vectors um . You can fix this by scaling. Let S2 denote the entry of Au1
which is largest in absolute value. We call this a scaling factor. Then u2 will not be just
Au1 but Au1 /S2 . Next let S3 denote the entry of Au2 which has largest absolute value and
define u3 ≡ Au2 /S3 . Continue this way. The scaling just described does not destroy the
relative insignificance of the term involving a sum in 14.1. Indeed it amounts to nothing
more than changing the units of length. Also note that from this scaling procedure, the
absolute value of the largest element of uk is always equal to 1. Therefore, for large m,
um =
λm
n cn xn
+ (relatively insignificant term) .
S2 S3 · · · Sm
Therefore, the entry of Aum which has the largest absolute value is essentially equal to the
entry having largest absolute value of
¶
µ m
λm+1 cn xn
λn cn xn
= n
≈ λn um
A
S2 S3 · · · Sm
S2 S3 · · · Sm
and so for large m, it must be the case that λn ≈ Sm+1 . This suggests the following
procedure.
Finding the largest eigenvalue with its eigenvector.
1. Start with a vector, u1 which you hope has a component in the direction of xn . The
T
vector, (1, · · · , 1) is usually a pretty good choice.
2. If uk is known,
uk+1 =
Auk
Sk+1
where Sk+1 is the entry of Auk which has largest absolute value.
3. When the scaling factors, Sk are not changing much, Sk+1 will be close to the eigenvalue and uk+1 will be close to an eigenvector.
4. Check your answer to see if it worked well.
At this point, you could stop because the scaling factors are not changing by much.
They went from 11. 895 to 12. 053. It looks like the eigenvalue is something like 12 which is
in fact the case. The eigenvector is approximately u11 . The true eigenvector for λ = 12 is
1
−.5
0
and so you see this is pretty close. If you didn't know this, observe
5 −14 11
1.0
11. 974
−4
=
4
−4
−. 499 26
−5. 991 2
3
6
−3
−1. 466 2 × 10−3
8. 838 6 × 10−3
and
This method can find various eigenvalues and eigenvectors. It is a significant generalization
of the above simple procedure and yields very good results. The situation is this: You have
a number, α which is close to λ, some eigenvalue of an n × n matrix, A. You don't know
λ but you know that α is closer to λ than to any other eigenvalue. Your problem is to find
both λ and an eigenvector which goes with λ. Another way to look at this is to start with
α and seek the eigenvalue, λ, which is closest to α along with an eigenvector associated
with λ. If α is an eigenvalue of A, then you have what you want. Therefore, we will always
−1
assume α is not an eigenvalue of A and so (A − αI) exists. The method is based on the
following lemma. When using this method it is nice to choose α fairly close to an eigenvalue.
14.2. THE SHIFTED INVERSE POWER METHOD
243
Otherwise, the method will converge slowly. In order to get some idea where to start, you
could use Gerschgorin's theorem but this theorem will only give a rough idea where to look.
There isn't a really good way to know how to choose α for general cases. As we mentioned
earlier, the eigenvalue problem is very difficult to solve in general.
n
Lemma 14.2.1 Let {λk }k=1 be the eigenvalues of A. If xk is an eigenvector of A for the
−1
eigenvalue λk , then xk is an eigenvector for (A − αI)
corresponding to the eigenvalue
1
λk −α .
Proof: Let λk and xk be as described in the statement of the lemma. Then
(A − αI) xk = (λk − α) xk
and so
1
−1
xk = (A − αI) xk .
λk − α
This proves the lemma.
In explaining why the method works, we will assume A is nondefective. This is not
necessary! One can use Gelfand's theorem on the spectral radius which is presented in
−1
[9] and invariance of (A − αI) on generalized eigenspaces to prove more general results.
It suffices to assume that the eigenspace for λk has dimension equal to the multiplicity of
the eigenvalue λk but even this is not necessary to obtain convergence of the method. This
method is better than might be supposed from the following explanation.
Pick u1 , an initial vector and let Axk = λk xk , where {x1 , · · · , xn } is a basis of eigenvectors which exists from the assumption that A is nondefective. Assume α is closer to λn
than to any other eigenvalue. Since A is nondefective, there exist constants, ak such that
u1 =
n
X
ak xk .
k=1
Possibly λn is a repeated eigenvalue. Then combining the terms in the sum which involve
eigenvectors for λn , a simpler description of u1 is
u1 =
m
X
aj xj + y
j=1
where y is an eigenvector for λn which is assumed not equal to 0. (If you are unlucky in your
choice for u1 , this might not happen and things won't work.) Now the iteration procedure
is defined as
−1
(A − αI) uk
uk+1 ≡
Sk
−1
where Sk is the element of (A − αI)
14.2.1,
Pm
uk+1
j=1
=
³
=
uk which has largest absolute value. From Lemma
³
aj
1
λn −α
1
λj −α
´k
S2 · · · Sk
´k
³
xj +
1
λn −α
´k
S2 · · · Sk
m
X
j=1
µ
aj
λn − α
λj − α
¶k
y
xj + y .
244
NUMERICAL METHODS FOR SOLVING THE EIGENVALUE PROBLEM
Now it is being assumed that λn is the eigenvalue which is closest to α and so for large k,
the term,
µ
¶k
m
X
λn − α
aj
xj ≡ Ek
λj − α
j=1
is very small while for every k ≥ 1, uk is a moderate sized vector because every entry has
absolute value less than or equal to 1. Thus
³
uk+1 =
Therefore, for large k, uk is approximately equal to an eigenvector of (A − αI)
−1
(A − αI)
uk ≈
−1
. Therefore,
1
uk
λn − α
and so you could take the dot product of both sides with uk and approximate λn by solving
the following for λn .
−1
(A − αI) uk · uk
1
=
2
λ
n−α
|uk |
T
How else can you find the eigenvalue from this? Suppose uk = (w1 , · · · , wn ) and from
the construction |wi | ≤ 1 and wk = 1 for some k. Then
−1
Sk uk+1 = (A − αI)
uk ≈ (A − αI)
−1
(Ck−1 y) =
−1
1
1
(Ck−1 y) ≈
uk .
λn − α
λn − α
Hence the entry of (A − αI) uk which has largest absolute value is approximately
and so it is likely that you can estimate λn using the formula
Sk =
−1
1
λn −α
1
.
λn − α
Of course this would fail if (A − αI) uk had more than one entry having equal absolute
value.
Here is how you use the shifted inverse power method to find the eigenvalue
and eigenvector closest to α.
−1
1. Find (A − αI)
.
14.2. THE SHIFTED INVERSE POWER METHOD
245
Pm
2. Pick u1 . It is important that u1 = j=1 aj xj +y where y is an eigenvector which goes
with the eigenvalue closest to α and the sum is in an "invariant subspace corresponding
to the other eigenvalues". Of course you have no way of knowing whether this is so
but it typically is so. If things don't work out, just start with a different u1 . You were
unlucky in your choice.
3. If uk has been obtained,
−1
uk+1 =
(A − αI)
Sk
uk
where Sk is the element of uk which has largest absolute value.
4. When the scaling factors, Sk are not changing much and the uk are not changing
much, find the approximation to the eigenvalue by solving
Sk =
1
λ−α
for λ. The eigenvector is approximated by uk+1 .
5. Check your work by multiplying by the original matrix
found works.
5 −14
4
Example 14.2.2 Find the eigenvalue of A = −4
3
6
Also find an eigenvector which goes with this eigenvalue.
to see how well what you have
11
−4 which is closest to −7.
−3
In this case the eigenvalues are −6, 0, and 12 so the correct answer is −6 for the eigenvalue. Then from the above procedure, we will start with an initial vector,
1
u1 ≡ 1 .
1
Then we must solve the following equation.
5 −14 11
1
−4
4
−4 + 7 0
3
6
−3
0
If you use your graphing calculator to graph this polynomial, you find there is an eigenvalue
somewhere between −.9 and −.8 and that this is the middle eigenvalue. Of course you could
zoom in and find it very accurately without much trouble but what about the eigenvector
which goes with it? If you try to solve
1 0 0
1 2 3
x
0
(−.8) 0 1 0 − 2 1 4 y = 0
0 0 1
3 4 2
z
0
there will be only the zero solution because the matrix on the left will be invertible and the
same will be true if you replace −.8 with a better approximation like −.86 or −.855. This is
because all these are only approximations to the eigenvalue and so the matrix in the above
is nonsingular for all of these. Therefore, you will only get the zero solution and
Eigenvectors are never equal to zero!
1. 0
−. 856 9
−. 856 9 −. 587 77 = . 503 7
−. 227 14
. 194 6
Thus the vector of 14.2 is very close to the desired eigenvector, just as −. 856 9 is very close
to the desired eigenvalue. For practical purposes, we have found both the eigenvector and
the eigenvalue.
2 1 3
Example 14.2.4 Find the eigenvalues and eigenvectors of the matrix, A = 2 1 1 .
3 2 1
248
NUMERICAL METHODS FOR SOLVING THE EIGENVALUE PROBLEM
This is only a 3×3 matrix and so it is not hard to estimate the eigenvalues. Just get
the characteristic equation, graph it using a calculator and zoom in to find the eigenvalues.
If you do this, you find there is an eigenvalue near −1.2, one near −.4, and one near 5.5.
(The characteristic equation is 2 + 8λ + 4λ2 − λ3 = 0.) Of course we have no idea what the
eigenvectors are.
Lets first try to find the eigenvector and a better approximation for the eigenvalue near
−1.2. In this case, let α = −1.2. Then
−25. 357 143 −33. 928 571 50.0
−1
12. 5
17. 5
−25.0 .
(A − αI) =
23. 214 286
30. 357 143 −45.0
T
You see at this point the scaling factors have definitely settled down and so it seems our
eigenvalue would be obtained by solving
1
= −54. 113 379
λ − (−1.2)
14.2. THE SHIFTED INVERSE POWER METHOD
249
and this yields λ = −1. 218 479 7 as an approximation to the eigenvalue and the eigenvector
would be obtained by dividing by −54. 113 379 which gives
1. 000 000
u5 = −. 491 830 9 .
−. 908 883 0
How well does it work?
What about complex eigenvalues? If your matrix is real, you won't see these by graphing
the characteristic equation on your calculator. Will the shifted inverse power method find
these eigenvalues and their associated eigenvectors? The answer is yes. However, for a real
matrix, you must pick α to be complex. This is because the eigenvalues occur in conjugate
pairs so if you don't pick it complex, it will be the same distance between any conjugate
pair of complex numbers and so nothing in the above argument for convergence implies you
will get convergence to a complex number. Also, the process of iteration will yield only real
vectors and scalars.
Example 14.2.5 Find the complex eigenvalues
trix,
5 −8
1 0
0 1
It took more iterations than before because α was not very close to 1 + i.
This illustrates an interesting topic which leads to many related topics. If you have a
polynomial, x4 + ax3 + bx2 + cx + d, you can consider it as the characteristic polynomial of
a certain matrix, called a companion matrix. In this case,
−a −b −c −d
1
0
0
0
.
0
1
0
0
0
0
1
0
The above example was just a companion matrix for λ3 − 5λ2 + 8λ − 6. You can see the
pattern which will enable you to obtain a companion matrix for any polynomial of the form
λn + a1 λn−1 + · · · + an−1 λ + an . This illustrates that one way to find the complex zeros
of a polynomial is to use the shifted inverse power method on a companion matrix for the
polynomial. Doubtless there are better ways but this does illustrate how impressive this
procedure is. Do you have a better way?
254
NUMERICAL METHODS FOR SOLVING THE EIGENVALUE PROBLEM
14.3
The Rayleigh Quotient
There are many specialized results concerning the eigenvalues and eigenvectors for Hermitian
matrices. A matrix, A is Hermitian if A = A∗ where A∗ means to take the transpose of the
conjugate of A. In the case of a real matrix, Hermitian reduces to symmetric. Recall also
that for x ∈ Fn ,
n
X
2
2
|xj | .
|x| = x∗ x =
j=1
The following corollary gives the theoretical foundation for the spectral theory of Hermitian matrices. This is a corollary of a theorem which is proved Corollary 12.2.14 and
Theorem 12.2.13 on Page 219.
Corollary 14.3.1 If A is Hermitian, then all the eigenvalues of A are real and there exists
an orthonormal basis of eigenvectors.
n
Thus for {xk }k=1 this orthonormal basis,
x∗i xj
½
= δ ij ≡
1 if i = j
0 if i 6= j
For x ∈ Fn , x 6= 0, the Rayleigh quotient is defined by
x∗ Ax
|x|
2
.
n
Now let the eigenvalues of A be λ1 ≤ λ2 ≤ · · · ≤ λn and Axk = λk xk where {xk }k=1 is
the above orthonormal basis of eigenvectors mentioned in the corollary. Then if x is an
arbitrary vector, there exist constants, ai such that
x=
In other words, the Rayleigh quotient is always between the largest and the smallest eigenvalues of A. When x = xn , the Rayleigh quotient equals the largest eigenvalue and when x = x1
the Rayleigh quotient equals the smallest eigenvalue. Suppose you calculate a Rayleigh quotient. How close is it to some eigenvalue?
14.3. THE RAYLEIGH QUOTIENT
255
Theorem 14.3.2 Let x 6= 0 and form the Rayleigh quotient,
x∗ Ax
|x|
≡ q.
2
Then there exists an eigenvalue of A, denoted here by λq such that
|λq − q| ≤
1 2 3
T
Example 14.3.3 Consider the symmetric matrix, A = 2 2 1 . Let x = (1, 1, 1) .
3 1 4
How close is the Rayleigh quotient to some eigenvalue of A? Find the eigenvector and eigenvalue to several decimal places.
Everything is real and so there is no need to worry about taking conjugates. Therefore,
the Rayleigh quotient is
¡
You see that for practical purposes, this has found the eigenvalue and an eigenvector.
258
NUMERICAL METHODS FOR SOLVING THE EIGENVALUE PROBLEM
Vector Spaces
It is time to consider the idea of a Vector space.
Definition 15.0.4 A vector space is an Abelian group of "vectors" satisfying the axioms of
an Abelian group,
v + w = w + v,
the commutative law of addition,
(v + w) + z = v+ (w + z) ,
the associative law for addition,
v + 0 = v,
the existence of an additive identity,
v+ (−v) = 0,
the existence of an additive inverse, along with a field of "scalars", F which are allowed to
multiply the vectors according to the following rules. (The Greek letters denote scalars.)
α (v + w) = αv+αw,
(15.1)
(α + β) v =αv+βv,
(15.2)
α (βv) = αβ (v) ,
(15.3)
1v = v.
(15.4)
The field of scalars is usually R or C and the vector space will be called real or complex
depending on whether the field is R or C. However, other fields are also possible. For
example, one could use the field of rational numbers or even the field of the integers mod p
for p a prime. A vector space is also called a linear space.
For example, Rn with the usual conventions is an example of a real vector space and Cn
is an example of a complex vector space. Up to now, the discussion has been for Rn or Cn
and all that is taking place is an increase in generality and abstraction.
Definition 15.0.5 If {v1 , · · · , vn } ⊆ V, a vector space, then
( n
)
X
span (v1 , · · · , vn ) ≡
αi vi : αi ∈ F .
i=1
A subset, W ⊆ V is said to be a subspace if it is also a vector space with the same field of
scalars. Thus W ⊆ V is a subspace if ax + by ∈ W whenever a, b ∈ F and x, y ∈ W. The
span of a set of vectors as just described is an example of a subspace.
259
implies
α1 = · · · = αn = 0
and {v1 , · · · , vn } is called a basis for V if
span (v1 , · · · , vn ) = V
and {v1 , · · · , vn } is linearly independent. The set of vectors is linearly dependent if it is not
linearly independent.
The next theorem is called the exchange theorem. It is very important that you understand this theorem. There are two kinds of people who go further in linear algebra, those
who understand this theorem and its corollary presented later and those who don't. Those
who do understand these theorems are able to proceed and learn more linear algebra while
those who don't are doomed to wander in the wilderness of confusion and sink into the
swamp of despair. Therefore, I am giving multiple proofs. Try to understand at least one
of them. Several amount to the same thing, just worded differently.
Theorem 15.0.7 Let {x1 , · · · , xr } be a linearly independent set of vectors such that each
xi is in the span{y1 , · · · , ys } . Then r ≤ s.
Proof 1: Let
xk =
s
X
ajk yj
j=1
If r > s, then the matrix A = (ajk ) has more columns than rows. By Corollary 7.2.8
one of these columns is a linear combination of the others. This implies there exist scalars
c1 , · · · , cr such that
r
X
ajk ck = 0, j = 1, · · · , r
k=1
and not all the di can equal zero. (If they were all equal to zero, it would follow that the set,
{x1 , · · · , xr } would be dependent since one of the vectors in it would be a linear combination
of the others.)
Now since{v1 , · ·Definition 15.0.10 A vector space V is of dimension n if it has a basis consisting of n
vectors. This is well defined thanks to Corollary 15.0.8. It is always assumed here that
n < ∞ in this case, such a vector space is said to be finite dimensional.
Theorem 15.0.11 If V = span (u1 , · · · , un ) then some subset of {u1 , · · · , un } is a basis for V. Also, if {u1 , · · · , uk } ⊆ V is linearly independent and the vector space is finite
dimensional, then the set, {u1 , · · · , uk }, can be enlarged to obtain a basis of V.
Proof: Let
S = {E ⊆ {u1 , · · · , un } such that span (E) = V }.
For E ∈ S, let |E| denote the number of elements of E. Let
m ≡ min{|E| such that E ∈ S}.
Thus there exist vectors
{v1 , · · · , vm } ⊆ {u1 , · · · , un }
such that
span (v1 , · · · , vm ) = V
and m is as small as possible for this to happen. If this set is linearly independent, it follows
it is a basis for V and the theorem is proved. On the other hand, if the set is not linearly
independent, then there exist scalars,
c1 , · · · , cm
such that
0=
m
X
c i vi
i=1
and not all the ci are equal to zero. Suppose ck 6= 0. Then the vector, vk may be solved for
in terms of the other vectors. Consequently,
V = span (v1 , · · · , vk−1 , vk+1 , · · · , vm )
264
VECTOR SPACES
contradicting the definition of m. This proves the first part of the theorem.
To obtain the second part, begin with {u1 , · · · , uk } and suppose a basis for V is {v1 , · · · , vn }.
If
span (u1 , · · · , uk ) = V,
then k = n. If not, there exists a vector,
uk+1 ∈
/ span (u1 , · · · , uk ) .
Then {u1 , · · · , uk , uk+1 } is also linearly independent. Continue adding vectors in this way
until n linearly independent vectors have been obtained. Then span (u1 , · · · , un ) = V
because if it did not do so, there would exist un+1 as just described and {u1 , · · · , un+1 }
would be a linearly independent set of vectors having n+1 elements even though {v1 , · · · , vn }
is a basis. This would contradict Theorem 15.0.7. Therefore, this list is a basis and this
proves the theorem.
It is useful to emphasize some of the ideas used in the above proof.
Lemma 15.0.12 Suppose v ∈
/ span (u1 , · · · , uk ) and {u1 , · · · , uk } is linearly independent.
Then {u1 , · · · , uk , v} is also linearly independent.
Pk
Proof: Suppose i=1 ci ui + dv = 0. It is required to verify that each ci = 0 and that
d = 0. But if d 6= 0, then you can solve for v as a linear combination of the vectors,
{u1 , · · · , uk },
k ³ ´
X
ci
v=−
ui
d
i=1
Pk
contrary to assumption. Therefore, d = 0. But then i=1 ci ui = 0 and the linear independence of {u1 , · · · , uk } implies each ci = 0 also. This proves the lemma.
Theorem 15.0.13 Let V be a nonzero subspace of a finite dimensional vector space, W of
dimension, n. Then V has a basis with no more than n vectors.
Proof: Let v1 ∈ V where v1 6= 0. If span {v1 } = V, stop. {v1 } is a basis for V .
Otherwise, there exists v2 ∈ V which is not in span {v1 } . By Lemma 15.0.12 {v1 , v2 } is a
linearly independent set of vectors. If span {v1 , v2 } = V stop, {v1 , v2 } is a basis for V. If
span {v1 , v2 } 6= V, then there exists v3 ∈
/ span {v1 , v2 } and {v1 , v2 , v3 } is a larger linearly
independent set of vectors. Continuing this way, the process must stop before n + 1 steps
because if not, it would be possible to obtain n + 1 linearly independent vectors contrary to
the exchange theorem, Theorems 15.0.7. This proves the theorem.
Linear Transformations
16.1
Matrix Multiplication As A Linear Transformation
Definition 16.1.1 Let V and W be two finite dimensional vector spaces. A function, L
which maps V to W is called a linear transformation and L ∈ L (V, W ) if for all scalars α
and β, and vectors v, w,
L (αv+βw) = αL (v) + βL (w) .
An example of a linear transformation is familiar matrix multiplication. Let A = (aij )
be an m × n matrix. Then an example of a linear transformation L : Fn → Fm is given by
(Lv)i ≡
and so, since {w1 , · · · , wm } is a basis, dil = 0 for each i = 1, · · · , m. Since l is arbitrary,
this shows dil = 0 for all i and l. Thus these linear transformations form a basis and this
shows the dimension of L (V, W ) is mn as claimed.
16.3
Eigenvalues And Eigenvectors Of Linear Transformations
Let V be a finite dimensional vector space. For example, it could be a subspace of Cn . Also
suppose A ∈ L (V, V ) . Does A have eigenvalues and eigenvectors just like the case where A
is a n × n matrix?
Theorem 16.3.1 Let V be a nonzero finite dimensional complex vector space of dimension
n. Suppose also the field of scalars equals C.1 Suppose A ∈ L (V, V ) . Then there exists
v 6= 0 and λ ∈ C such that
Av = λv.
1 All that is really needed is that the minimal polynomial can be completely factored in the given field.
The complex numbers have this property from the fundamental theorem of algebra.
16.3. EIGENVALUES AND EIGENVECTORS OF LINEAR TRANSFORMATIONS
267
2
Proof: Consider the linear transformations, I, A, A2 , · · · , An . There are n2 + 1 of these
transformations and so by Theorem 16.2.2 the set is linearly dependent. Thus there exist
constants, ci ∈ C such that
n2
X
ck Ak = 0.
c0 I +
k=1
This implies there exists a polynomial, q (λ) which has the property that q (A) = 0. In fact,
Pn2
q (λ) ≡ c0 + k=1 ck λk . Dividing by the leading term, it can be assumed this polynomial
is of the form λm + cm−1 λm−1 + · · · + c1 λ + c0 , a monic polynomial. Now consider all such
monic polynomials, q such that q (A) = 0 and pick one which has the smallest degree. This
is called the minimal polynomial and will be denoted here by p (λ) . By the fundamental
theorem of algebra, p (λ) is of the form
p (λ) =
This proves the theorem.
Corollary 16.3.2 In the above theorem, each of the scalars, λk has the property that there
exists a nonzero v such that (A − λi I) v = 0. Furthermore the λi are the only scalars with
this property.
Proof: For the first claim, just factor out (A − λi I) instead of (A − λp I) . Next suppose
(A − µI) v = 0 for some µ and v 6= 0. Then
0 =
p
Y
(A − λk I) v =
k=1
= (µ − λp )
Ãp−1
Y
p−1
Y
k=1
= (µ − λp )
!
(A − λk I) v
k=1
Ãp−2
Y
(A − λk I) (Av − λp v)
!
(A − λk I) (Av − λp−1 v)
k=1
= (µ − λp ) (µ − λp−1 )
Ãp−2
Y
k=1
!
(A − λk I)
268
LINEAR TRANSFORMATIONS
continuing this way yields
=
p
Y
(µ − λk ) v,
k=1
a contradiction unless µ = λk for some k.
Therefore, these are eigenvectors and eigenvalues with the usual meaning. This leads to
the following definition.
Definition 16.3.3 For A ∈ L (V, V ) where dim (V ) = n, the scalars, λk in the minimal
polynomial,
p
Y
p (λ) =
(λ − λk )
k=1
are called the eigenvalues of A. The collection of eigenvalues of A is denoted by σ (A). For
λ an eigenvalue of A ∈ L (V, V ) , the generalized eigenspace is defined as
m
showing that x ∈ ker (A − λI)
. Therefore, continuing this way, it follows that for all
k ∈ N,
m
m+k
ker (A − λI) = ker (A − λI)
.
Therefore, this shows 16.1.
The following theorem is of major importance and will be the basis for the very important
theorems concerning block diagonal matrices.
16.3. EIGENVALUES AND EIGENVECTORS OF LINEAR TRANSFORMATIONS
269
Theorem 16.3.5 Let V be a complex vector space of dimension n and suppose σ (A) =
{λ1 , · · · , λk } where the λi are the distinct eigenvalues of A. Denote by Vi the generalized
eigenspace for λi and let ri be the multiplicity of λi . By this is meant that
ri
Vi = ker (A − λi I)
(16.2)
and ri is the smallest integer with this property. Then
V =
k
X
Vi .
(16.3)
i=1
Proof: This is proved by induction on k. First suppose there is only one eigenvalue, λ1
of algebraic multiplicity m. Then by the definition of eigenvalues given in Definition 16.3.3,
A satisfies an equation of the form
r
(A − λ1 I) = 0
r
where r is as small as possible for this to take place. Thus ker (A − λ1 I) = V and the
theorem is proved in the case of one eigenvalue.
Now suppose the theorem is true for any i ≤ k − 1 where k ≥ 2 and suppose σ (A) =
{λ1 , · · · , λk } .
m
Claim 1: Let µ 6= λi , Then (A − µI) : Vi → Vi and is one to one and onto for every
m ∈ N.
m
k
Proof: It is clear that (A − µI) maps Vi to Vi because if v ∈ Vi then (A − λi I) v = 0
for some k ∈ N. Consequently,
k
In this section the vector space will be Cn and the linear transformations will be n × n
matrices.
Definition 16.4.1 Let A and B be two n × n matrices. Then A is similar to B, written as
A ∼ B when there exists an invertible matrix, S such that A = S −1 BS.
Theorem 16.4.2 Let A be an n×n matrix. Letting λ1 , λ2 , · · · , λr be the distinct eigenvalues
of A,arranged in any order, there exist square matrices, P1 , · · · , Pr such that A is similar
to the block diagonal matrix,
P1 · · · 0
P = ... . . . ...
0
···
Pr
272
LINEAR TRANSFORMATIONS
in which Pk has the single eigenvalue λk . Denoting by rk the size of Pk it follows that rk
equals the dimension of the generalized eigenspace for λk ,
m
and Crk ABrk is an rk × rk matrix.
What about the eigenvalues of Crk ABrk ? The only eigenvalue of A restricted to Vλk is
λk because if Ax = µx for some x ∈ Vλk and µ 6= λk , then as in Claim 1 of Theorem 16.3.5,
rk
Therefore, λ = λk because, as noted above, λk is the only eigenvalue of A restricted to Vλk .
Now letting Pk = Crk ABrk , this proves the theorem.
The above theorem contains a result which is of sufficient importance to state as a
corollary.
Corollary 16.4.3 Let A be an n × n matrix and let Dk denote a basis for the generalized
eigenspace for λk . Then {D1 , · · · , Dr } is a basis for Cn .
More can be said. Recall Theorem 12.2.10 on Page 217. From this theorem, there exist
unitary matrices, Uk such that Uk∗ Pk Uk = Tk where Tk is an upper triangular matrix of the
form
λk · · · ∗
..
.. ≡ T
..
.
k
.
.
0
where Tk is an upper triangular matrix having only λk on the main diagonal. The diagonal
blocks can be arranged in any order desired. If Tk is an mk × mk matrix, then
m
mk = dim {x : (A − λk I) x = 0 for some m ∈ N} .
Furthermore, mk is the multiplicity of λk as a zero of the characteristic polynomial of A.
Proof: The only thing which remains is the assertion that mk equals the multiplicity
of λk as a zero of the characteristic polynomial. However, this is clear from the observation
that since T is similar to A they have the same characteristic polynomial because
¡
¢
det (A − λI) = det S (T − λI) S −1
¢
¡
= det (S) det S −1 det (T − λI)
¡
¢
= det SS −1 det (T − λI)
= det (T − λI)
and the observation that since T is upper triangular, the characteristic polynomial of T is
of the form
r
Y
m
(λk − λ) k .
k=1
The above corollary has tremendous significance especially if it is pushed even further
resulting in the Jordan Canonical form. This form involves still more similarity transformations resulting in an especially revealing and simple form for each of the Tk , but the result
of the above corollary is sufficient for most applications.
It is significant because it enables one to obtain great understanding of powers of A by
using the matrix T. From Corollary 16.4.4 there exists an n × n matrix, S 2 such that
A = S −1 T S.
Therefore, A2 = S −1 T SS −1 T S = S −1 T 2 S and continuing this way, it follows
Ak = S −1 T k S.
where T is given in the above corollary. Consider T k . By block multiplication,
k
0
T1
..
Tk =
.
.
k
0
Tr
The matrix, Ts is an ms × ms matrix which is of the form
α ··· ∗
Ts = ... . . . ...
0
···
(16.5)
α
which can be written in the form
Ts = D + N
for D a multiple of the identity and N an upper triangular matrix with zeros down the main
diagonal. Therefore, by the Cayley Hamilton theorem, N ms = 0 because the characteristic
2 The
S here is written as S −1 in the corollary.
16.5. THE MATRIX OF A LINEAR TRANSFORMATION
275
equation for N is just λms = 0. Such a transformation is called nilpotent. You can see
N ms = 0 directly also, without having to use the Cayley Hamilton theorem. Now since D
is just a multiple of the identity, it follows that DN = N D. Therefore, the usual binomial
theorem may be applied and this yields the following equations for k ≥ ms .
Tsk
k
= (D + N ) =
k µ ¶
X
k
Dk−j N j
j
j=0
ms µ ¶
X
k
=
Dk−j N j ,
j
j=0
(16.6)
the third equation holding because N ms = 0. Thus Tsk is of the form
k
α ··· ∗
.. .
..
Tsk = ...
.
.
0 · · · αk
Lemma 16.4.5 Suppose T is of the form Ts described above in 16.5 where the constant, α,
on the main diagonal is less than one in absolute value. Then
¡ ¢
lim T k ij = 0.
k→∞
for ei the standard basis vectors for Fn consisting of
0
..
.
ei ≡
1
.
..
0
where the one is in the ith slot. It is clear that q defined in this way, is one to one, onto,
and linear. For v ∈ V, q −1 (v) is a list of scalars called the components of v with respect to
the basis {v1 , · · · , vn }.
Definition 16.5.1 Given a linear transformation L, mapping V to W, where {v1 , · · · , vn }
is a basis of V and {w1 , · · · , wm } is a basis for W, an m × n matrix A = (aij )is called the
matrix of the transformation L with respect to the given choice of bases for V and W , if
whenever v ∈ V, then multiplication of the components of v by (aij ) yields the components
of Lv.
The following diagram is descriptive of the definition. Here qV and qW are the maps
defined above with reference to the bases, {v1 , · · · , vn } and {w1 , · · · , wm } respectively.
{v1 , · · · , vn }
Example 16.5.2 Let
V ≡ { polynomials of degree 3 or less},
W ≡ { polynomials of degree 2 or less},
and L ≡ D where D is the differentiation operator. A basis for V is {1,x, x2 , x3 } and a
basis for W is {1, x, x2 }.
What is the matrix of this linear transformation with respect to this basis? Using 16.9,
¡
¢ ¡
¢
0 1 2x 3x2 = 1 x x2 C.
It follows from this that
0 1
C= 0 0
0 0
0
2
0
0
0 .
3
Now consider the important case where V = Fn , W = Fm , and the basis chosen is the
standard basis of vectors ei described above. Let L be a linear transformation from Fn to
Fm and let A be the matrix of the transformation with respect to these bases. In this case
the coordinate maps qV and qW are simply the identity map and the requirement that A is
the matrix of the transformation amounts to
π i (Lb) = π i (Ab)
where π i denotes the map which takes a vector in Fm and returns the ith entry in the vector,
the ith component of the vector with respect to the standard basis vectors. Thus, if the
components of the vector in Fn with respect to the standard basis are (b1 , · · · , bn ) ,
¡
¢T X
b = b1 · · · bn
=
bi ei ,
i
then
π i (Lb) ≡ (Lb)i =
X
aij bj .
j
What about the situation where different pairs of bases are chosen for V and W ? How
are the two matrices with respect to these choices related? Consider the following diagram
which illustrates the situation.
Fn A2 Fm
−
→
q2 ↓
◦ p2 ↓
V −
L
W
→
q1 ↑
◦ p1 ↑
Fn A1 Fm
−
→
In this diagram qi and pi are coordinate maps as described above. From the diagram,
−1
p−1
1 p2 A2 q2 q1 = A1 ,
where q2−1 q1 and p−1
1 p2 are one to one, onto, and linear maps.
Definition 16.5.3 In the special case where V = W and only one basis is used for V = W,
this becomes
q1−1 q2 A2 q2−1 q1 = A1 .
278
LINEAR TRANSFORMATIONS
Letting S be the matrix of the linear transformation q2−1 q1 with respect to the standard basis
vectors in Fn ,
S −1 A2 S = A1 .
(16.10)
When this occurs, A1 is said to be similar to A2 and A → S −1 AS is called a similarity
transformation.
Here is some terminology.
Definition 16.5.4 Let S be a set. The symbol, ∼ is called an equivalence relation on S if
it satisfies the following axioms.
1. x ∼ x
for all x ∈ S. (Reflexive)
2. If x ∼ y then y ∼ x. (Symmetric)
3. If x ∼ y and y ∼ z, then x ∼ z. (Transitive)
Definition 16.5.5 [x] denotes the set of all elements of S which are equivalent to x and
[x] is called the equivalence class determined by x or just the equivalence class of x.
With the above definition one can prove the following simple theorem which you should
do if you have not seen it.
Theorem 16.5.6 Let ∼ be an equivalence class defined on a set, S and let H denote the
set of equivalence classes. Then if [x] and [y] are two of these equivalence classes, either
x ∼ y and [x] = [y] or it is not true that x ∼ y and [x] ∩ [y] = ∅.
Theorem 16.5.7 In the vector space of n × n matrices, define
A∼B
if there exists an invertible matrix S such that
A = S −1 BS.
Then ∼ is an equivalence relation and A ∼ B if and only if whenever V is an n dimensional
vector space, there exists L ∈ L (V, V ) and bases {v1 , · · · , vn } and {w1 , · · · , wn } such that
A is the matrix of L with respect to {v1 , · · · , vn } and B is the matrix of L with respect to
{w1 , · · · , wn }.
Proof: A ∼ A because S = I works in the definition. If A ∼ B , then B ∼ A, because
A = S −1 BS
implies
If A ∼ B and B ∼ C, then
and so
B = SAS −1 .
A = S −1 BS, B = T −1 CT
−1
A = S −1 T −1 CT S = (T S)
CT S
which implies A ∼ C. This verifies the first part of the conclusion.
Now let V be an n dimensional vector space, A ∼ B and pick a basis for V,
{v1 , · · · , vn }.
16.5. THE MATRIX OF A LINEAR TRANSFORMATION
Define L ∈ L (V, V ) by
X
Lvi ≡
279
aji vj
j
where A = (aij ) . Then if B = (bij ) , and S = (sij ) is the matrix which provides the similarity
transformation,
A = S −1 BS,
between A and B, it follows that
Lvi =
X
¡
¢
sir brs s−1 sj vj .
(16.11)
r,s,j
Now define
X¡
wi ≡
s−1
¢
ij
vj .
j
Then from 16.11,
X¡
s−1
¢
ki
Lvi =
i
X ¡
¢
¡
¢
s−1 ki sir brs s−1 sj vj
i,j,r,s
and so
X
Lwk =
bks ws .
s
This proves the theorem because the if part of the conclusion was established earlier.
What if the linear transformation consists of multiplication by a matrix A and you want
to find the matrix of this linear transformation with respect to another basis? Is there an
easy way to do it? The answer is yes.
Proposition 16.5.8 Let A be an m×n matrix and let L be the linear transformation which
is defined by
à n
!
n
m X
n
X
X
X
L
x k ek ≡
(Aek ) xk ≡
Aik xk ei
k=1
i=1 k=1
k=1
In simple language, to find Lx, you multiply on the left of x by A. Then the matrix M of
this linear transformation with respect to the bases {u1 , · · · , un } for Fn and {w1 , · · · , wm }
for Fm is given by
M=
where
¡
w1
···
wm
¢
¡
w1
···
wm
¢−1
A
¡
u1
···
un
¢
is the m × m matrix which has wj as its j th column.
Proof: Consider the following diagram.
{u1 , · · · , un }
Fn
qV ↑
Fn
L
→
◦
→
M
Fm
↑ qW
Fm
{w1 , · · · , wm }
Here the coordinate maps are defined in the usual way. Thus
qV
¡
x1
···
xn
¢T
≡
n
X
i=1
xi ui .
280
LINEAR TRANSFORMATIONS
Therefore, qV can be considered the same as multiplication of a vector in Fn on the left by
the matrix
¡
¢
u1 · · · un .
Similar considerations apply to qW . Thus it is desired to have the following for an arbitrary
x ∈ Fn .
¡
¢
¡
¢
A u1 · · · un x = w1 · · · wn M x
Therefore, the conclusion of the proposition follows. This proves the proposition.
Definition 16.5.9 An n × n matrix, A, is diagonalizable if there exists an invertible n × n
matrix, S such that S −1 AS = D, where D is a diagonal matrix. Thus D has zero entries
everywhere except on the main diagonal. Write diag (λ1 · · · , λn ) to denote the diagonal
matrix having the λi down the main diagonal.
Which matrices are diagonalizable?
Theorem 16.5.10 Let A be an n × n matrix. Then A is diagonalizable if and only if Fn
has a basis of eigenvectors of A. In this case, S of Definition 16.5.9 consists of the n × n
matrix whose columns are the eigenvectors of A and D = diag (λ1 , · · · , λn ) .
Proof: Suppose first that Fn has a basis of eigenvectors,
{v1 , · · ·, vn } where Avi = λi vi .
uT1
..
−1
Then let S denote the matrix (v1 · · · vn ) and let S
≡ . where uTi vj = δ ij ≡
½
Next suppose A is diagonalizable so S −1 AS = D ≡ diag (λ1 , · · · , λn ) . Then the columns
of S form a basis because S −1 is given to exist. It only remains to verify that these columns
of A are eigenvectors. But letting S = (v1 · · · vn ) , AS = SD and so (Av1 · · · Avn ) =
(λ1 v1 · · · λn vn ) which shows that Avi = λi vi . This proves the theorem.
It makes sense to speak of the determinant of a linear transformation as described in the
following corollary.
Corollary 16.5.11 Let L ∈ L (V, V ) where V is an n dimensional vector space and let A
be the matrix of this linear transformation with respect to a basis on V. Then it is possible
to define
det (L) ≡ det (A) .
16.5. THE MATRIX OF A LINEAR TRANSFORMATION
281
Proof: Each choice of basis for V determines a matrix for L with respect to the basis.
If A and B are two such matrices, it follows from Theorem 16.5.7 that
A = S −1 BS
and so
But
and so
det (A) = det (B)
which proves the corollary.
Definition 16.5.12 Let A ∈ L (X, Y ) where X and Y are finite dimensional vector spaces.
Define rank (A) to equal the dimension of A (X) .
The following theorem explains how the rank of A is related to the rank of the matrix
of A.
Theorem 16.5.13 Let A ∈ L (X, Y ). Then rank (A) = rank (M ) where M is the matrix
of A taken with respect to a pair of bases for the vector spaces X, and Y.
Proof: Recall the diagram which describes what is meant by the matrix of A. Here the
two bases are as indicated.
{v1 , · · · , vn }
4. det (L) 6= 0
5. If Lv = 0 then v = 0.
Pn
n
Proof: Suppose
first L is one to one and let {vi }i=1 be a basis. Then if i=1 ci Lvi = 0
Pn
it follows L ( i=1P
ci vi ) = 0 which means that since L (0) = 0, and L is one to one, it must
n
be the case that i=1 ci vi = 0. Since {vi } is a basis, each ci = 0 which shows {Lvi } is a
linearly independent set. Since there are n of these, it must be that this is a basis.
Now suppose 2.). Then letting {vi } P
be a basis, and yP
∈ V, it follows from part 2.) that
n
n
there are constants, {ci } such that y = i=1 ci Lvi = L ( i=1 ci vi ) . Thus L is onto. It has
been shown that 2.) implies 3.).
Now suppose 3.). Then the operation consisting of multiplication by the matrix of L, ML ,
must be onto. However, the vectors in Fn so obtained, consist of linear combinations of the
columns of ML . Therefore, the column rank of ML is n. By Theorem 7.5.7 this equals the
determinant rank and so det (ML ) ≡ det (L) 6= 0.
Now assume 4.) If Lv = 0 for some v 6= 0, it follows that ML x = 0 for some x 6= 0.
Therefore, the columns of ML are linearly dependent and so by Theorem 7.5.7, det (ML ) =
det (L) = 0 contrary to 4.). Therefore, 4.) implies 5.).
Now suppose 5.) and suppose Lv = Lw. Then L (v − w) = 0 and so by 5.), v − w = 0
showing that L is one to one. This proves the theorem.
Also it is important to note that composition of linear transformation corresponds to
multiplication of the matrices. Consider the following diagram.
X
A
Y
−
→
qX ↑
◦
↑ qY
Fn MA Fm
−−→
B
Z
−
→
◦
↑ qZ
M B Fp
−−→
where A and B are two linear transformations, A ∈ L (X, Y ) and B ∈ L (Y, Z) . Then
B ◦ A ∈ L (X, Z) and so it has a matrix with respect to bases given on X and Z, the
coordinate maps for these bases being qX and qZ respectively. Then
−1
−1
B ◦ A = qZ MB qY qY−1 MA qX
= qZ MB MA qX
.
But this shows that MB MA plays the role of MB◦A , the matrix of B ◦ A. Hence the matrix
of B ◦ A equals the product of the two matrices MA and MB . Of course it is interesting
to note that although MB◦A must be unique, the matrices, MB and MA are not unique,
depending on the basis chosen for Y .
Theorem 16.5.15 The matrix of the composition of linear transformations equals the product of the matrices of these linear transformations.
16.5.1
Some Geometrically Defined Linear Transformations
If T is any linear transformation which maps Fn to Fm , there is always an m × n matrix, A
with the property that
Ax = T x
(16.12)
for all x ∈ Fn . You simply take the matrix of the linear transformation with respect to the
standard basis. What is the form of A? Suppose T : Fn → Fm is a linear transformation
and you want to find the matrix defined by this linear transformation as described in 16.12.
Then if x ∈ Fn it follows
n
X
x=
x i ei
i=1
16.5. THE MATRIX OF A LINEAR TRANSFORMATION
283This proves the following theorem.
Theorem 16.5.16 Let T be a linear transformation from Fn to Fm . Then the matrix, A
satisfying 16.12Example 16.5.17 Determine the matrix for the transformation mapping R2 to R2 which
consists of rotating every vector counter clockwise through an angle of θ.
µ ¶
µ ¶
1
0
Let e1 ≡
and e2 ≡
. These identify the geometric vectors which point
0
1
along the positive x axis and positive y axis as shown.
e2 6
e1
From Theorem 16.5.16, you only need to find T e1 and T e2 , the first being the first
column of the desired matrix, A and the second being the second column. From drawing a
picture and doing a little geometry, you see that
µ
¶
µ
¶
cos θ
− sin θ
T e1 =
, T e2 =
.
sin θ
cos θ
Therefore, from Theorem 16.5.16,
µ
A=
cos θ
sin θ
− sin θ
cos θ
¶
284
LINEAR TRANSFORMATIONS
Example 16.5.18As an application, I will consider the problem of rotating counter clockwise about a given
unit vector which is possibly not one of the unit vectors in coordinate directions. First
consider a pair of perpendicular unit vectors, u1 and u2 and the problem of rotating in the
counterclockwise direction about u3 where u3 = u1 × u2 so that u1 , u2 , u3 forms a right
handed orthogonal coordinate system. Thus the vector u3 is coming out of the page.
I want to write this transformation in terms of the usual basis vectors, {e1 , e2 , e3 }. From
Proposition 16.5.8, if A is this matrix,
cos θ − sin θ 0
sin θ
cos θ 0
0
0
1
¡
¢−1 ¡
¢
u1 u2 u3
=
A u1 u2 u3
and so you can solve for A if you know the ui .
Suppose the unit vector about which the counterclockwise rotation takes place is (a, b, c).
Then I obtain vectors, u1 and u2 such that {u1 , u2 , u3 } is a right handed orthogonal system
with u3 = (a, b, c) and then use the above result. It is of course somewhat arbitrary how
this is accomplished. I will assume, however that |c| 6= 1 since otherwise you are looking at
either clockwise or counter clockwise rotation about the positive z axis and this is a problem
which has been dealt with earlier. (If c = −1, it amounts to clockwise rotation about the
positive z axis while if c = 1, it is counterclockwise rotation about the positive z axis.) Then
let u3 = (a, b, c) and u2 ≡ √a21+b2 (b, −a, 0) . This one is perpendicular to u3 . If {u1 , u2 , u3 }
is to be a right hand system it is necessary to have
¡
An important application of the above theory is to the Euler angles, important in the
mechanics of rotating bodies. Lagrange studied these things back in the 1700's. To describe
the Euler angles consider the following picture in which x1 , x2 and x3 are the usual coordinate
axes fixed in space and the axes labeled with a superscript denote other coordinate axes.
Here is the picture.
x3 = x13
x23
» x12
»
»
»
φ
»
x2
¡
φ¤¤
¡
x1 ¡ ¤
x11
x13
C
Cθ
C
2
C
»» x2
»
»
θ
C»
x12
¡
¡¡
x11 = x21
x23 = x33
3
»» x2
»
»
ψ
»
x22
¡¤
ψ
¡ ¤
x21 ¡ ¤
x31
We obtain φ by rotating counter clockwise about the fixed x3 axis. Thus this rotation
and it is desired to rotate counter clockwise through an angle of ψ about this vector. Thus,
in this case,
a = cos2 φ + sin2 φ cos θ, b = cos φ sin φ (1 − cos θ) , c = − sin φ sin θ.
and you could substitute in to the formula of Theorem 16.14 and obtain a matrix which
represents the linear transformation obtained by rotating counter clockwise about the positive x23 axis, M3 (φ, θ, ψ) . Then what would be the matrix with respect to the usual basis
for the linear transformation which is obtained as a composition of the three just described?
By Theorem 16.5.15, this matrix equals the product of these three,
M3 (φ, θ, ψ) M2 (φ, θ) M1 (φ) .
I leave the details to you. There are procedures due to Lagrange which will allow you to
write differential equations for the Euler angles in a rotating body. To give an idea how
these angles apply, consider the following picture.
16.5. THE MATRIX OF A LINEAR TRANSFORMATION
289
x3
¡
¡
R ¡ψ
x3 (t)
θ
x2
¡
¡
¡
¡
¡
x1 ¡
¡
φ
line of nodes
This is as far as I will go on this topic. The point is, it is possible to give a systematic
description in terms of matrix multiplication of a very elaborate geometrical description of
a composition of linear transformations. You see from the picture it is possible to describe
the motion of the spinning top shown in terms of these Euler angles. I think you can also
see that the end result would be pretty horrendous but this is because it involves using the
basis corresponding to a fixed in space coordinate system. You wouldn't do this for the
application to a spinning top.
Not surprisingly, this also has applications to computer graphics.
In words, there is an unbroken string of ones down the super diagonal and the number, α
filling every space on the main diagonal with zeros everywhere else. A matrix is strictly
upper triangular if it is of the form
0 ∗ ∗
.. . .
.
. ∗ ,
0
···
291
0
292
THE JORDAN CANONICAL FORM*
where there are zeroes on the main diagonal and below the main diagonal.
The Jordan canonical form involves each of the upper triangular matrices in the conclusion of Corollary 16.4.4 being a block diagonal matrix with the blocks being Jordan blocks
in which the size of the blocks decreases from the upper left to the lower right. The idea
is to show that every square matrix is similar to a unique such matrix which is in Jordan
canonical form.
Note that in the conclusion of Corollary 16.4.4 each of the triangular matrices is of the
form αI + N where N is a strictly upper triangular matrix. The existence of the Jordan
canonical form follows quickly from the following lemma.
Lemma A.0.25 Let N be an n × n matrix which is strictly upper triangular. Then there
exists an invertible matrix, S such that
Jr1 (0)
0
Jr2 (0)
S −1 N S =
..
.
0
Jrs (0)
where r1 ≥ r2 ≥ · · · ≥ rs ≥ 1 and
where the wq come from {B1 , · · · , Bk−1 , Bk } . By induction, some of these wq must come
from Bk . Let vik be the one for which i is as large as possible. Then do N i−1 to both sides
to obtain v1k , the eigenvector upon which the chain Bk is based, is a linear combination of
{B1 , · · · , Bk−1 } contrary to the construction. Since {B1 , · · · , Bk } is linearly independent,
the process terminates. This proves the claim.
Claim 3: Suppose N w = 0. (w is an eigenvector) Then there exists scalars, ci such
that
s
X
w=
ci v1i .
i=1
Recall that v1i is the eigenvector in the ith chain on which this chain is based.
Proof of Claim 3: From the construction, w ∈ span (B1 , · · · , Bs ) since otherwise, it
could serve as a base for another chain. Therefore,
w=
ri
s X
X
cki vki .
i=1 k=1
Now apply N to both sides.
0=
ri
s X
X
i
cki vk−1
i=1 k=2
and so by Claim 2, cki = 0 if k ≥ 2. Therefore,
w=
s
X
c1i v1i
i=1
and this proves the claim.
It remains to verify that span (B1 , · · · , Bs ) = Fn . Suppose w ∈
/ span (B1 , · · · , Bs ) . By
Claim 3 this implies w is not an eigenvector since all the eigenvectors are in span (B1 , · · · , Bs ) .
Since N n = 0, there exists a smallest integer, k ≥ 2 such that N k w = 0 but N k−1 w 6= 0.
Then k ≤ min (r1 , · · · , rs ) because there exists a chain of length k based on the eigenvector,
N k−1 w, namely
N k−1 w,N k−2 w,N k−3 w, · · · , w
and this chain must be no longer than the preceding chains because of the construction in
which a longest possible chain was chosen at each step. Since N k−1 w is an eigenvector, it
follows from Claim 3 that
N k−1 w =
as claimed. This proves the lemma.
Now let the upper triangular matrices, Tk be given in the conclusion of Corollary 16.4.4.
Thus, as noted earlier,
Tk = λk Irk ×rk + Nk
where Nk is a strictly upper triangular matrix of the sort just discussed in Lemma A.0.25.
Therefore, there exists Sk such that Sk−1 Nk Sk is of the form given in Lemma A.0.25. Now
Sk−1 λk Irk ×rk Sk = λk Irk ×rk and so Sk−1 Tk Sk is of the form
Ji1 (λk )
0
Ji2 (λk )
..
.
0
where i1 ≥ i2 ≥ · · · ≥ is and
where Tk is an upper triangular mk × mk matrix having only λk on the main diagonal.
By Corollary A.0.26 There exist matrices, Sk such that Sk−1 Tk Sk = J (λk ) where J (λk ) is
described in 1.3. Now let M be the block diagonal matrix given by
S1
0
..
M =
.
.
0
Sr
It follows that M −1 S −1 ASM = M −1 T M and this is of the desired form. This proves the
theorem.
What about the uniqueness of the Jordan canonical form? Obviously if you change the
order of the eigenvalues, you get a different Jordan canonical form but it turns out that if
the order of the eigenvalues is the same, then the Jordan canonical form is unique. In fact,
it is the same for any two similar matrices.
Theorem A.0.28 Let A and B be two similar matrices. Let JA and JB be Jordan forms of
A and B respectively, made up of the blocks JA (λi ) and JB (λi ) respectively. Then JA and
JB are identical except possibly for the order of the J (λi ) where the λi are defined above.
Proof: First note that for λi an eigenvalue, the matrices JA (λi ) and JB (λi ) are both
of size mi × mi because the two matrices A and B, being similar, have exactly the same
characteristic equation and the size of a block equals the algebraic multiplicity of the eigenvalue as a zero of the characteristic equation. It is only necessary to worry about the
number and size of the Jordan blocks making up JA (λi ) and JB (λi ) . Let the eigenvalues
m
of A and B be {λ1 , · · · , λr } . Consider the two sequences of numbers {rank (A − λI) } and
m
{rank (B − λI) }. Since A and B are similar, these two sequences coincide. (Why?) Also,
m
m
for the same reason, {rank (JA − λI) } coincides with {rank (JB − λI) } . Now pick λk an
m
m
eigenvalue and consider {rank (JA − λk I) } and {rank (JB − λk I) } . Then
JA − λk I =
and in going to m + 1 a discrepancy must occur because the sum on the right will contribute
less to the decrease in rank than the sum on the left. This proves the theorem.
298
THE JORDAN CANONICAL FORM*
An Assortment Of Worked
Exercises And Examples
B.1
Worked Exercises Page ??
1. Here is an augmented matrix in which ∗ denotes an arbitrary number and ¥ denotes
a nonzero number. Determine whether the given augmented matrix is consistent. If
consistent, is the solution unique?
¥ ∗ ∗ ∗ ∗ | ∗
0 ¥ ∗ ∗ 0 | ∗
0 0 ¥ ∗ ∗ | ¥
0 0 0 0 ¥ | ∗
In this case the system is consistent and there is an infinite set of solutions. To see
it is consistent, the bottom equation would yield a unique solution for x5 . Then
letting x4 = t, and substituting in to the other equations, beginning with the equation
determined by the third row and then proceeding up to the next row followed by the
first row, you get a solution for each value of t. There is a free variable which comes
from the fourth column which is why you can say x4 = t. Therefore, the solution is
infinite.
2 0 ¥ | ¥
0 0 ∗ | 0
In this case there is no solution because you could use a row operation to place a 0 in
the third row and third column position, like this:
¥ ∗ ∗ | ∗
0 0 ¥ | ¥
0 0 0 | ¥
This would give a row of zeros equal to something nonzero.
3. Find h such that
Therefore, there is a unique solution. In particular the system is consistent.
5. Find the point, (x1 , y1 ) which lies on both lines, 5x + 3y = 1 and 4x − y = 3.
You solve the system of equations whose augmented matrix is
µ
¶
5 3 | 1
4 −1 | 3
A reduced echelon form is
µ
17
0
0
17
10
−11
¶
and so the solution is x = 17/10 and y = −11/17.
6. Do the three lines, 3xThis is asking for the solution to the three
is
3 2
2 −1
4 3
A reduced echelon form is
1
0
0
0
1
0
equations shown. The augmented matrix
| 1
| 1
| 3
|
|
|
0
0
1
and this would require 0x + 0y = 1 which is impossible so there is no solution to this
system of equations and hence no point on each of the three lines.
A reduced echelon form for the matrix is
1 0 8 2
0 1 −4 0 .
0 0 0 0
Therefore, y = 4z and x = 2 − 8z. Apparently z can equal anything so we let z = t
and then the solution is
x = 2 − 8t, y = 4t, z = t.
8. Find the point, (x1 , y1 ) which lies on both lines, x + 2y = 1 and 3x − y = 3.
The solution is y = 0 and x = 1.
9. Find the point of intersection of the two lines x + y = 3 and x + 2y = 1.
The solution is (5, −2) .
10. Do the three lines, xTo solve this set up the augmented matrix and go to work on it. The augmented
matrix is
1 2 | 1
2 −1 | 1
4 3 | 3
A reduced echelon matrix for this is
1 0 |
0 1 |
0 0 |
3
5
1
5
0
Therefore, there is a point in the intersection of these and it is y = 1/5 and x = 3/5.
Thus the point is (3/5, 1/5) .
11. Do the three planes, x + 2y − 3z = 2, x + y + z = 1, and 3x + 2y + 2z = 0 have
a common point of intersection? If so, find one and if not, tell why there is no such
point.
You need to find (x, y, z) which solves each equation. The augmented matrix is
1 2 −3 | 2
1 1 1 | 1
3 2 2 | 0
A reduced echelon form for the matrix is
1 0 0
0 1 0
0 0 1
It follows there exists a solution but the solution is not unique because x4 is a free
variable. You can pick it to be anything you like and the system will yield values for
the other variables.
13 ¥ ∗ | ∗
0 0 ¥ | ∗
In this case there is a unique solution to the
row operations and reduce this to something
¥ 0 0
0 ¥ 0
0 0 ¥
is the augmented matrix of an inconsistent matrix.
Take −3 times the top row and add to 2 times the bottom. This yields
µ
¶
2
h
| 4
0 12 − 3h | 2
Now
¡ if h = 4 the¢system is inconsistent because it would have the bottom row equal
to 0 0 | 2 .
16. Choose h and k such that the augmented matrix shown has one solution. Then choose
h and k such that the system has no solutions. Finally, choose h and k such that the
system has infinitely many solutions.
µ
¶
1 h | 2
.
2 4 | k
If h 6= 2 then k can be anything and the system represented by the augmented matrix
will have a unique solution. Suppose then that h = 2. Then taking −2 times the top
row and adding to the bottom row gives
¶
µ
1 2 |
2
0 0 | k−4
If k 6= 4 there is no solution. However, if k = 4 you are left with the single equation,
x + 2y = 2 and there are infinitely many solutions to this. In fact anything of the form
(2 − 2y, y) will work just fine.
17. Determine if the system is consistent.
x + 2y + z − w = 2
x−y+z+w =1
2x + y − z = 1
4x + 2y + z = 5
This system is inconsistent. To see this, write the augmented matrix and do row
operations. The augmented matrix is
1 2
1 −1 | 2
1 −1 1
1 | 1
2 1 −1 0 | 1
4 2
1
0 | 5
A reduced echelon form for this matrix is
1
1 0 0
3
0 1 0 −2
3
0 0 1 0
0 0 0 0
and the bottom row shows there is no solution.
Does there exist a value of t for which this matrix fails to have an inverse? Explain.
t
e
e−t cos t
e−t sin t
det et −e−t cos t − e−t sin t −e−t sin t + e−t cos t
et
2e−t sin t
−2e−t cos t
= 5et e2(−t) cos2 t + 5et e2(−t) sin2 t = 5e−t which is never equal to zero for any value of
t and so there is no value of t for which the matrix has no inverse.
8. Use the formula for the inverse in terms
inverse of the matrix
1
0
4
1. Find the rank of the following matrices. If the rank is r, identify r columns in the
original matrix which have the property that every other column may be written
as a linear combination of these. Also find a basis for the row and column spaces of
the matrices.
9 2 0
3 7 1
(a)
6 1 0
0 2 1
From using row operations we obtain the row reduced echelon form which is
1 0 0
0 1 0
0 0 1
0 0 0
Therefore, a basis for the column space of the original matrix is the first three
columns of the original matrix. A basis for the row space is just
¡
¢ ¡
¢
¡
¢
1 0 0 , 0 1 0 , and 0 0 1 .
1. Find the matrix for the linear transformation which rotates every vector in R2 through
an angle of 5π/12.
You note that 5π/12 = 2π/3 − π/4. Therefore, you can first rotate through −π/4
and then rotate through 2π/3 to get the rotation through 5π/12. The matrix of the
transformation with respect to the usual coordinates which rotates through −π/4 is
µ √
√2/2
− 2/2
and this is the matrix of the desired transformation. Note this shows that
cos (5π/12) = −
sin (5π/12) =
1√
1√ √
2+
3 2 ≈ . 258 819 05
4
4
1√ √
1√
3 2+
2 ≈ . 965 925 83.
4
4
314
AN ASSORTMENT OF WORKED EXERCISES AND EXAMPLES
2. Find the matrix for the linear transformation which rotates every vector in R2 through
an angle of 2π/3 and then reflects across the x axis.
What does it do to e1 ? First you rotate e1 through the given angle to obtain
µ
¶
−1/2
√
3/2
and then this becomes
µ
−1/2
√
− 3/2
¶
.
This is the first column of the desired matrix. Next e2 first is rotated through the
given angle to give
µ √
¶
− 3/2
−1/2
and then it is reflected across the x axis to give
µ √
¶
− 3/2
1/2
and this gives the second column of the desired matrix. Thus the matrix is
√
µ
¶
−1/2
−
3/2
√
.
− 3/2
1/2
T
5. If A, B, and C are each n × n matrices and ABC is invertible, why are each of A, B,
and C invertible.
0 6= det (ABC) = det (A) det (B) det (C) and so none of det (A) , det (B) , or det (C)
can equal zero. Therefore, each is invertible. You should do this another way, showing
that each of A, B, and C is one to one and then using a theorem presented earlier.
6. Give an example of a 3 × 1 matrix with the property that the linear transformation
determined by this matrix is one to one but not onto.
0
1
1
Here is one. 0 . If 0 x = 0 , then x = 0 but this is certainly not onto
0
0
0
1
as a map from R1 to R3 because it does not ever yield 1 .
0
7. Find the matrix of the linear transformation from R3 to R3 which first rotates every
vector through an angle of π/4 about the z axis when viewed from the positive z axis
and then rotates every vector through an angle of π/6 about the x axis when viewed
from the positive x axis.
The matrix of the linear transformation which accomplishes the first rotation is
√
√
√2/2 −√ 2/2 0
2/2
2/2 0
0
0
1
and the matrix which accomplishes the second rotation is
1 √0
0
0
3/2 √
−1/2
3/2
0 1/2
Therefore, the matrix of the desired linear transformation is
√
√
1 √0
0
√2/2 −√ 2/2 0
0
3/2 √
−1/2 2/2
2/2 0
0 1/2
3/2
0
0
1
√
1√
2
−√12 √2
0
2 √
√
= 14 √
3 2 41 √
3 2 −√12
1
1
1
4 2
4 2
2 3
This might not be the first thing you would think of.
At this point we stop because the matrix on the right is in upper triangular form.
3. Find the LU factorization of the coefficient matrix using Dolittle's method and use it
to solve the system of equations.
x + 2y + 3z = 5
2x + 3y + 3z = 6
3x + 5y + 4z = 11
The coefficient matrix
1
2
3
Then we first solve
which yields
is
2
3
5
3
1 0
3 = 2 1
4
3 1
1 0
2 1
3 1
0
1
0 0
1
0
2
−1
0
0
u
5
0 v = 6
1
w
11
u
5
v = −4
w
0
3
−3 .
−2
B.7. WORKED EXERCISES PAGE ??
Next we solve
z
317
=(u,v,w)T
}|
{
3
x
5
−3 y = −4
−2
z
0
1 2
0 −1
0 0
which yields
B.7
−3
x
y = 4
0
z
Worked Exercises Page ??
12
and so λ = λ showing that λ must be real.
23
318
AN ASSORTMENT OF WORKED EXERCISES AND EXAMPLES
−1 0 |
0 1 |
0 0 |
0
0
0
Therefore, the eigenspace is of the form
t
t
0
This is only one dimensional and so the matrix is defective.
1
4. Find the complex eigenvalues and eigenvectors of the matrix 7
−1
Determine whether the matrix is defective.
1
−5
7
−6
−6 .
2
After wading through much affliction you find the eigenvalues are −6, 2 + 6i, 2 − 6i.
Since these are distinct, the matrix cannot be defective. We must find the eigenvectorsB.7. WORKED EXERCISES PAGE ??
and
319−1
1
0
= −6. The augmented matrix to row
| 0
| 0
| 0
|
|
|
0
0 .
0
−1
1 .
−1
5. You own a trailer rental company in a large city and you have four locations, one
in the South East, one in the North East, one in the North West, and one in the
South West. Denote these locations by SE,NE,NW, and SW respectively. Suppose
you observe that in a typical day, .7 of the trailers starting in SE stay in SE, .1 of the
trailers in NE go to SE, .1 of the trailers in NW end up in SE, .2 of the trailers in SW
end up in SE, .1 of the trailers in SE end up in NE,.7 of the trailers in NE end up in
NE,.2 of the trailers in NW end up in NE,.1 of the trailers in SW end up in NE, .2
of the trailers in SE end up in NW, .1 of the trailers in NE end up in NW, .6 of the
trailers in NW end up in NW, .2 of the trailers in SW end up in NW, 0 of the trailers
in SE end up in SW, .1 of the trailers in NE end up in SW, .1 of the trailers in NW
end up in SW, .5 of the trailers in SW end up in SW. You begin with 20 trailers in
each location. Approximately how many will you have in each location after a long
time? Will any location ever run out of trailers?
It sometimes helps to write down a table summarizing the given information.
SE
NE
NW
SW
SE
.7
.1
.2
0
NE
.1
.7
.1
.1
NW
.1
.2
.6
.1
SW
.2
.1
.2
.5
320
AN ASSORTMENT OF WORKED EXERCISES AND EXAMPLES
Then the migration matrix is
7/10
1/10
1/5
0
1/10 1/10 1/5
7/10 1/5 1/10
1/10 3/5 1/5
1/10 1/10 1/2
All we have to do is find the eigenvector (In this case the eigenspace will be one
dimensional because some power of the matrix has all positive entries.) corresponding
to λ = 1 which has all the entries add to 20. This will be the long time population.
Remember, these processes conserve the sum of the entries. We must row reduce
−3/10 1/10
1/10
1/5 | 0
1/10 −3/10 1/5
1/10 | 0
1/5
1/10 −2/5 1/5 | 0
0
1/10
1/10 −1/2 | 0
The row reduced echelon form is
Thus there will be about 5.6 trailers in SE, 6.4 in NE, 5.6 in NW, and 2.4 in SW. In
particular, it appears no location will run out of trailers.
The Fundamental Theorem Of
Algebra
The fundamental theorem of algebra states that every non constant polynomial having
coefficients in C has a zero in C. If C is replaced by R, this is not true because of the
example, x2 + 1 = 0. This theorem is a very remarkable result and notwithstanding its title,
all the best proofs of it depend on either analysis or topology. It was first proved by Gauss
in 1797. The proof given here follows Rudin [11]. See also Hardy [6] for a similar proof,
more discussion and references. The best proof is found in the theory of complex analysis.
Recall De Moivre's theorem from trigonometry which is listed here for convenience.
Theorem C.0.1 Let r > 0 be given. Then if n is a positive integer,
n
It follows from the above inequality that for |z − w| < δ, |az n − awn | < ε. The function of
the lemma is just the sum of functions of this sort and so it follows that it is also continuous.
Theorem C.0.4 (Fundamental theorem of Algebra) Let p (z) be a nonconstant polynomial.
Then there exists z ∈ C such that p (z) = 0.
321 | 677.169 | 1 |
Mathematics of Games (And as a Game!)
Welcome to Students
What can I get out of this course?
In this course you'll learn the basics of game theory and strategic
thinking, as a refreshing change of scenery from the traditional math
sequence and its emphasis on algebraic computation. We'll learn to recognize
that mathematical thinking arises as a refinement of the basic cognitive
heuristics that we all use when making decisions in everyday life's
interactive contexts.
How is the course structured?
Three strands of investigation set the framework within which various games
can be analyzed. We'll talk about (i) individual puzzles pitting the
``player'' against the impartial laws of logic, (ii) competitive games
pitting the player against one rational opponent, and (iii) probabilistic
games pitting players against ``controlled randomness'', where informed
strategy decisions can help turn the odds to their favor.
What books do I need to purchase?
The first two books listed below will be discussed as a class, so you have
to obtain a personal copy. Other texts are listed here in case you want
further references or ideas for individual research projects.
Fisher, Len. Rock, Paper, Scissors: Game Theory in
Everyday Life
Lewis, Michael. Moneyball
Davis, Morton. Game Theory: A Nontechnical
Introduction
Dixit, Avinash. Thinking Strategically: The
Competitive Edge in Business, Politics, and Everyday Life | 677.169 | 1 |
Completing the Square Notes
Be sure that you have an application to open
this file type before downloading and/or purchasing.
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Product Description
This is a one-sided notes page intended for the beginner over completing the square.
The notes begin by explaining the purpose of the completing the square technique. Then a schematic shows the equation of a quadratic from standard form to vertex form.
Next, a worked out example is shown including steps. The second example includes the steps and has the student fill in parts of the process. Finally, the third example repeats the same steps, but has the student do all of the work.
The three examples also include identifying the vertex once the completing the square process is completed. All examples have "a" as 1 and "b" as an even number. Key included. | 677.169 | 1 |
Tuesday, March 14, 2017
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DPLS Scientific Calculator
Description
This is a highly functional and easy to use Scientific Calculator. It incorporates extensive science tools including a triangle calculator, vector calculator, shape calculator, kinematic calculator, half-life calculator, maths calculator, gas laws calculator, statistical calculator, molar mass calculator, pH calculator and one of the most extensive measurement converters available. Over 300 commonly used compounds can be quickly called to list their name, formula, molecular mass and CAS number. Over 100 constants can be called to list their numerical value and uncertainty value. Over 300 formulas and equations are listed for science, maths, trigonometry and statistics. It lists over 200 science symbols, Greek symbols, maths and statistics symbols. Numerous science data reference systems can be called including SI units, derived quantities, maths laws, atomic structures, organic compounds, ions, homologous series, frequencies, shapes, angle types, conjugate pairs and acids and bases. The system contains extensive listings of electrical circuit symbols, piping and process symbols and material properties. It contains 8 colour coded periodic tables that can be used to find 30 types of property data for each element. It contains full electron configuration of the elements, radiation types and quantum numbers. Interactive flowcharts can be called for mechanical and electrical units. A glossary system contains the phonetic alphabet as well as extensive lists of science and medicine fields, and acronyms and abbreviations used in science and computing. It lists the activity series and describes chemical reaction types. It contains an event timer, yearly calendar and world times system. A help system explains operation, and system contents can be quickly located with a search system.
Scientific Calculator Precision 54 for scientists, engineers, teachers, and students. The calculator was designed with purpose to fit Netbooks and Notebooks with small display. Of course, the calculator can be used on laptop and desktop computers..
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College Scientific Calculator 36 for scientists, engineers, teachers, and students. Calculates mathematical formulas of any length and complexity. Calculation history can be stored into text file or printed. | 677.169 | 1 |
What is MathStudio and why I created it
For more info, or if you think something is missing/wrong,
please contact me.
What is MathStudio ?
MathStudio is an open-source project conceived to make
typing and resolution of mathematical expressions easier and more comfortable.
The target of this program is not advanced\engineering calculus (maybe these
features will be added in future, but, by now, they are missing).
There are other programs devoted to the same purpose, but very very few are free and
even fewer are easy to use; many of these force the user to write
input data using one row only (everything is typed at the same level,
exponents and bases are on the same row) and a lot of brackets to
make the operation order explicit. This math is really different from
the math you can do on a piece of paper !
The aim of this project is exactly to reduce this gap providing
both cross-platform open source c++ libraries (probably wrappers
for other languages, python, basic, java, will be added in future)
which can be easily embedded in other open source programs (these
two libraries are MathCore and MathGUI
and they can be downloaded
using developer links) and a program, MathStudio which demonstrates
their usage.
Example:
If I want to solve the following equation:
in a lot of programs, I should type
((36*x^2+12*x+1) / (81*a^2) = SQRT(1296*x^4)
Typing expressions like the one in the example above, is much
easier in MathStudio; you can see some screenshots which demonstrate this in the
Screenshots page.
MathStudio not only allows the user to write expressions and
equations in a simple way but it also allows to solve them with few, simple operations, generating a
series of intermediate steps, obtained simplifying the initial expression:
The image on the left is just an example costructed using a paint program (I hope
to be able to replace it with some screenshots in the next month) but it should give
you an idea of one of the key-feature of MathStudio: the intermediate steps...
Also notice that the styles which compose the output (the fonts, the colors) can be
configured by the user. MathStudio, in fact, can be used not only to solve expressions,
equations and inequalities but also to create "snapshots" of mathematical data
to insert as bitmaps in the programs which do not support the input of such expressions
(many word processor, for example).
MathStudio is a program aimed to be an easy and
powerful symbolic calculator. It is different from other programs of this type both
because of the way the user inputs the expression and because of its data
processing system: flexible and highly customizable, able to show to the user
the operations and the intermediate steps done toward the resolution.
This picture sums up the two basic features
of the program: a
WYSIWYG (What You See Is What You Get, i.e. a visual tool) equation
editor and a powerful math engine.
For a detailed list of MathStudio features, please visit the
Features page.
Why did you create it ?
My first answer is: I never found a program with the features of MathStudio and, in
general, the open-source mathematics programs are very difficult to use; when I try to use
Maxima, MatLab or Mathematica I always have to spend a lot of time trying to correct
bugs and problems in the input data: to give you an indea these are some errors and their
reasons...
This error is generated because I missed the '*' operator between the '3' and the 'x'
Command Solve[12*x = 0, x]; on Mathematica:
Set:: write : "Tag Times
in 12 x is Protected."
Solve:: eqf : "0 is not
a well-formed equation."
I did not manage to understand
the trigger of this error before the demo of the
program expired...
I can continue reporting errors such these for other 3-4 pages... but I don't want to
annoy you. I imagine that these problems are very easy to solve for experienced users of
Maxima and Mathematica, but I want a program where the user can intuitively input any
expression and solve without such problems...
In the end, the main reason is that I want an easy program to do math... a program where
if I want to solve an equation, I do not have to learn the syntax of program-specific commands and
to spend hours trying to understand what I missed when I get incomprehensible errors.
The second answer is: price. Professional programs cost something like 1800$. MathStudio is free.
I think that, for non-professional uses, MathStudio is really competitive !!!
Last answer: I like mathematics and that's why I chose to code in my free time such a program :-) | 677.169 | 1 |
Free With This Purchase
Description
The catalyst to boost and sharpen logical thinking and reasoning skills.
All the topics in this book are grouped as under:
VERBAL REASONING & NON-VERBAL REASONING
For each question type, worked out examples shows exactly what the question demands, and explains how to tackle it in three or four clear steps. The approach is systematic and highly effective, lending itself to active learning.
Each topic is dealt with in a separate unit.
Each unit begins with a brief explanation of the question types of the procedure to solve them.
Sufficient Number of examples with clear step-by-step explanation to arrive the solution.
Ample number of practice questions.
This book contains several interesting and investigative problems with worked out examples and explanations.
World's First Reasoning Curriculum @ Schools Reasoning Quotient is a Must. Prepare Now! Reasoning Trainer Plus
The catalyst to boost and sharpen logical thinking and reasoning skills.
All the topics in this book are grouped as under:
VERBAL REASONING & NON-VERBAL REASONING
For each question type, worked out examples shows exactly what the question demands, and explains how to tackle it in three or four clear steps. The approach is systematic and highly effective, lending itself to active learning.
Each topic is dealt with in a separate unit.
Each unit begins with a brief explanation of the question types of the procedure to solve them.
Sufficient Number of examples with clear step-by-step explanation to arrive the solution.
Ample number of practice questions.
This book contains several interesting and investigative problems with worked out examples and explanations.
The Following Topics are covered. VERBAL REASONING 1. Change one word into another 2. Three missing letters make a word 3. Insert a letter 4. Create word 5. Dot representation 6. Puzzle Test 7. Odd one out 8. Matching Pairs 9. Series 10. Direction Sense 11. Mathematical Reasoning 12. Blood Relations NON-VERBAL REASONING 13. Which one is different? 14. Analogy 15. What comes next ? 16. Mirror and water images 17. Analytical Reasoning 18. Cubes and Dice & Key | 677.169 | 1 |
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Need More Info?
Condition
GOOD: This textbook is in good condition, showing a few signs of use and wear: some of the pages are creased, some of the cover's corners are slightly bent. There is no writing on the pages.
Description
The Bittinger/Ellenbogen/Johnson Algebra textbooks come with a great reputation for thoroughly presenting concepts and applications in Algebra. This particular book is third in a series of three texts in an algebra series.
The hefty volume includes color photos and illustrations. The answers for odd-numbered questions are included in the back. A "Feature Walkthrough" section at the front of the book emphasizes these new fourth edition features, such as Chapter Openers that contain testimonials of people who have applied the algebra covered in the chapter, Student Notes, Study Skills, Technology Connections that aid the student in using graphing calculators, Algebraic-Graphical Connection for concept visualization, Real-Data Applications, Connecting the Concepts, Concept Reinforcement Exercises, Exercises, Study Summary, and Cumulative Review.
The textbook can be used by advanced high schoolers and college students. Additional online support is available from the publisher.
Chapters
Introduction to Algebraic Expressions
Equations, Inequalities, and Problem Solving
Introduction to Graphing
Polynomials
Polynomials and Factoring
Rational Expressions and Equations
Functions and Graphs
Systems of Linear Equations and Problem Solving
Inequalities and Problem Solving
Exponents and Radicals
Quadratic Functions and Equations
Exponential and logarithmic Functions
Conic Sections
Sequences, Series, and the Binomial Theorem
Elementary Algebra Review
Details
SKU:
1573
Subject:
Algebra
Author:
Bittinger, Marvin L. and David J. Ellenbogen and Barbara L Johnson
Publisher:
Pearson Addison Wesley
Publish Date:
2006
Format:
Hardback
Weight (pounds):
5.15
Dimensions (W"xL"xH"):
8.8″x10.3″x1.6″
ISBN-13:
9780321233837
ISBN-10:
0321233832
Grade Level:
Grades 10-12, Grade 10, Grade 11, Grade 12, College
Related Products book was written "to provide a solid foundation in algebra for students who might have had no previous experience in algebra." This is a college textbook, which may also be used by an advanced high school student. traces the history of algebraic topology beginning with its creation by Henri Poincaré in 1900, and describing in detail the important ideas introduced in the theory before 1960." — From the back cover
"An hour of a programmer's time often costs more than the price of a book. By this measure, you hold a volume potentially worth thousands of dollars. That it can be purchased for a fraction of this cost I consider a modern miracle. The amount of information crammed into this book is incredible." — Eric Haines, co-author of Real-Time Rendering
"Matrix Groups for Undergraduates is concrete and example-driven, with geometric motivation and rigorous proofs. The story begins and ends with the rotation of a globe. In between, the author combines rigor and intuition to describe basic objects of Lie theory: Lie algebras, matrix exponentiation, Lie brackets, and maximal tori | 677.169 | 1 |
Variables in Algebra
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Sheryl Kish has created this powerpoint for Algebra 1 classes. The textbook used in this class is McDougal Littell's Algebra 1: Concepts and Skills. This powerpoint covers Chapter 1, section 1, called "Variables in Algebra." This section covers definitions of variables, values, variable expressions, discusses how to evaluate expressions, shows examples of evaluating real-life expressions, and ends with evaluating geometric expressions (basic area and perimeter formulas). This powerpoint starts with a "Do Now" or Warm-up, goes through definitions, examples and follows with practice problems for the students to complete. It ends with a sample homework assignment from the textbook used in class. | 677.169 | 1 |
Coursework in Mathematics: MEI discussion paper page 3 The current situation All GCSE Mathematics specifications contain two pieces of coursework, each worth. Investigations for GCSE Mathematics For teaching from September 2010 The coursework element was removed from GCSE Mathematics assessments in September 2007. Maths Coursework Help,GCSE Maths Statistics Help Coursework Online from our helpers and get ready to solve your Maths coursework problems on time. Maths Coursework Help - The Best Custom Essay, Thesis Papers, Term Papers, Research Papers Writing Service. Professional High-quality Reports, Reviews, Dissertations.
Maths gcse coursework
Support information for GCSE Maths students. Coursework : Home Lessons Checklist Class dates Exams Coursework Links Contact Other info. Tough GCSE topics broken down and explained by out team of expert teachers In this coursework a pass in GCSE Maths will open a lot of doors for you;. GCSE Coursework Writing your coursework (now often called controlled assessment). Get yourself a copy of GCSE Mathematics Coursework from Amazon! Use Computers.
Mathematics; Administration; Coursework Tasks;. for centres choosing to undertake coursework for the AQA A-level Mathematics and Statistics. GCSE scores vs 'AS. In the coursework file for Gcse maths there is a piece called Table Patterns which I think should be 7 pages long but shows only 3. How do I get to the oth. Has anyone got a link to the old booklet that describes the GCSE coursework tasks? I feel embarrassed to ask because coursework was so pointless and so.
It is time to start thinking as an adult and get the best GCSE coursework writing help on the market! Our team of experts is always ready to help you. Has anyone got a link to the old booklet that describes the GCSE coursework tasks? I feel embarrassed to ask because coursework was so pointless and so. Homework Expert Singapore Provides Maths course work, maths statistics coursework, mathematics coursework, maths gcse coursework, data handling coursework and maths. GCSE. GCSE resources with teacher and student feedback Maths Coursework. For this maths coursework, I will be investigating the volume of different sized open boxes. Maths Coursework Help,GCSE Maths Statistics Help Coursework Online from our helpers and get ready to solve your Maths coursework problems on time.
Coursework for GCSE mathematics is to be axed in England, the Education Secretary, Alan Johnson, has announced. All other GCSE coursework would have to be.
Coursework for GCSE mathematics is to be axed in England, the Education Secretary, Alan Johnson, has announced. All other GCSE coursework would have to be.
Investigations for GCSE Mathematics For teaching from September 2010 The coursework element was removed from GCSE Mathematics assessments in September 2007.
GCSE Coursework Writing your coursework (now often called controlled assessment). Get yourself a copy of GCSE Mathematics Coursework from Amazon! Use Computers. GCSE. GCSE resources with teacher and student feedback Maths Coursework. For this maths coursework, I will be investigating the volume of different sized open boxes. It is time to start thinking as an adult and get the best GCSE coursework writing help on the market! Our team of experts is always ready to help you. Maths Resources. teaching maths london; GCSE Guide - GCSE revision, exam papers, coursework help; AQA Maths Resource Zone – Maths Resource Zone ; Welcome to the. | 677.169 | 1 |
Courses Seeking advice while preparing for a course.
In the suggested background section it suggests "A familiarity with the rudiments of high-school algebra " . Unfortunately my schooling did not involve algebra at all. Im trying to remedy this in the 3 months i have before the course starts by using Khan Academy to bring my math skills up to speed. Im already starting to do decimal division and multiplication and am just about to start exponents, which is the first example of math needed. Im finding this stuff is mostly review so far thankfully. But the formula example in the second part looks more advanced and unfortunately i have no memories to pull from.
If you could please look at the attachment image and look at example 3 which starts with the words " A familiarity with the rudiments of high-school algebra, the ability to solve an equation like...." and let me know how deep this is i would be grateful. I am worried that 3 months might not be enough to get up to that level. I am wondering what the path is from finding composite and primes, multiplication and division of decimal numbers, basics on exponents, positive and negative numbers, which is all arithmetic and pre-algebra according to Khan and the math needed for this course. How far do i have to go?
I have also signed up for a Introduction to Mathematical Thinking course at coursea as prep for the Astronomy Class. This course starts sep 17th and i figure that even if i cant pass it, learning how to think about math in a more advanced way, will be nothing but a benefit.
hmm... i would probably focus more on getting your algebra down than getting super good at arithmetic. assuming you can add/subtract/multiply/divide numbers just fine, next you should gain some familiarity with exponents (more about understanding how they work than about actually calculating a lot of them). and then jump right into algebra and get as much practice as you can.
edit: you should also get some familiarity with "scientific notation" since its useful for dealing with really big and really small numbers. this can be done before, after, or during your algebra practice.
as far as algebra goes, i think getting used to using variables and performing various manipulations will prove more useful than trying to perfect specific techniques or focusing too much on specific types of problems, so once you can get a little comfortable with using variables and solving the most simple equations (linear equations) you might want to jump around a bit trying little bits of everything or even making up random problems to try and solve instead of watching 35 videos about quadratics or something like that (quadratics are useful and good to do after linear equations but watching all 35 vids they have at kahn academy is overkill since you probably wont need them much for this course).
more specifically about this course (which i'm considering taking myself) the example equation looks more complicated than it actually is since you can solve it (for D) by just multiplication/division and taking a cube root at the end (though i imagine it looks like complete nonsense if you're not used to equations with more than 1 variable)
also, when solving algebraic equations, you're doing anything you can to get the variable you're solving for to be all by it's self on 1 side of the equals sign, and anything else on the other side. this will be the goal no matter how complicated the problem looks. so if you want to solve something like "2x+4=6" for x you need to somehow get rid of the 2 and 4 so x will be by its self. if you subtract 4 that will get rid of the 4, but since you need to keep both sides of the equals sign equal, you need to subtract from both sides giving 2x+4-4=6-4 which simplifies to 2x=2, now to get rid of the 2 you can divide by 2, and again, to keep both sides equal you have to do it to both sides so (2x)/2=2/2 which simplifies to x=1. more concisely thats:
2x+4=6
subtract 4 from both sides
2x=2
divide both sides by 2
x=1
a seemingly more complicated example: solving something like "ax+by=c" for x is really the same thing even though it has all these other variables there
ax+by=c
subtract "by" from both sides
ax=c-by
divide both sides by "a"
x=(c-by)/a
being able to do stuff like that is pretty much the bare minimum before you can say you've learned some algebra but its a good starting point i think. if what i did above makes any sense to you (i didnt explain in much detail, and i jumped the difficulty up with no warning, so dont worry if it doesnt) then getting to a sufficient skill level shouldnt take too much longer.
finally, when you're stuck for a long time something like can be your best friend. it can solve just about anything if you ask it correctly and will usually have a "show steps" button to show hot it was done. | 677.169 | 1 |
Find a Harbor CityAs part of my major, I studied linear algebra. In addition, I've taken many in-depth courses on mathematics and proof. Linear algebra is often used as a gateway to the more proof-based aspects of math, and I like how a good course teaches both concrete concepts while also scaffolding an understanding of proof in general | 677.169 | 1 |
Calculator Deals!
All math students are required to have a graphing calculator for class - see the registration guide p.47 for more information. We recommend that you check this page or your teacher's bulletin board during the first few weeks of school for where to find the best deals on calculators each week!
Students may find apps for graphing calculators on their phones, but will NOT be allowed to use them on any assessments. Use of phones in the classrooms in other situations is up to the discretion of the teacher.
Also, please bring in your calculator Proof of Purchase/Point Value so the math department can obtain free equipment for the classroom. If you would need your proof of purchase later, you can obtain one from Mrs. Dawson. Thanks!
Finally, you may be able to find a more inexpensive calculator online or from a former student. | 677.169 | 1 |
In Laymen's Terms: a segue on why we do, in fact, need algebra. In a recent guest column [update: error fixed] in the New York Times, a professor asks "do we need algebra?" The short answer is "hell ... had to work without the benefit of mathematical notation. Most of it had yet to be developed, and in fact al.
Math and more: 5 apps students must check out to improve their performance in class With over 10,000 free videos organised like mini lessons and 150,000 common core aligned practice questions for students ... Photomath: Guidance to isolate x in algebra homework is now easy with math buddy – PhotoMath. Step by step breakdown of the.
If School Is Mostly a Waste of Time, Why Spend More Time There? Although I agree that current school schedules and calendars are not necessarily optimal, increasing the amount of time that kids spend in classrooms strikes me as a bad idea, at least until schools can give a plausible answer to this question.
How to Remove a Second Operating System from a PC You can remove the second operating system without affecting the functionality of the original operating system; however, removing the original operating system could cause your computer to stop working. Back up any files you want to keep that are.
What You Should Know About Common Core Math As a lead writer of the Mathematics Common Core standards ... whose knowledge of algebra is based on a million meaningless mnemonics often crumble when faced with problems posed in an unconventional way or using unconventional notation, as they inevitably.
If You Answer This Math Problem Correctly, You May Be an Atheist Go ahead. Do the problem. The wrong answer — the one you come up with when you don't put any thought into it or simply go with your gut — would be $10. The right answer — which requires a bit of analysis — would be $5. (The bat costs $105.
CBSE Class 12 Computer Science Syllabus 2017-2018 Converting expression from infix to postfix notation and evaluation ... Text file and Binary file; Common to both the options. Refer to unit 3 DATABASE AND SQL mentioned in case of Python for further details. Unit 4: Boolean Algebra Common to both the.
At Cal State, algebra is a civil rights issue - EdSource The culprit is Intermediate Algebra , a high-school level course of technical procedures that most college students will never use, either in college or in life. Many students pass a course on this content in high school ( Algebra II), but when they.
Teaching Math Under Common Core: Fact and Fiction, Part V Question: Is the Common Core Dumbing Math Down? Answer: No. Most of the backlash against some of the new approaches to arithmetic are due to their novelty, not their merit. When a sample addition problem began circulating on the internet, some parents. | 677.169 | 1 |
Description
The life sciences deal with a vast array of problems at different spatial, temporal, and organizational scales. The mathematics necessary to describe, model, and analyze these problems is similarly diverse, incorporating quantitative techniques that are rarely taught in standard undergraduate courses. This textbook provides an accessible introduction to these critical mathematical concepts, linking them to biological observation and theory while also presenting the computational tools needed to address problems not readily investigated using mathematics alone.
Proven in the classroom and requiring only a background in high school math, Mathematics for the Life Sciences doesn't just focus on calculus as do most other textbooks on the subject. It covers deterministic methods and those that incorporate uncertainty, problems in discrete and continuous time, probability, graphing and data analysis, matrix modeling, difference equations, differential equations, and much more. The book uses MATLAB throughout, explaining how to use it, write code, and connect models to data in examples chosen from across the life sciences.
Provides undergraduate life science students with a succinct overview of major mathematical concepts that are essential for modern biology
Covers all the major quantitative concepts that national reports have identified as the ideal components of an entry-level course for life science students
Provides good background for the MCAT, which now includes data-based and statistical reasoning
Uses MATLAB throughout, and MATLAB m-files with an R supplement are available online
Prepares students to read with comprehension the growing quantitative literature across the life sciences
A solutions manual for professors and an illustration package is available
About the authors
Erin N. Bodine is assistant professor of mathematics at Rhodes College. Suzanne Lenhart is Chancellor's Professor of Mathematics at the University of Tennessee. Louis J. Gross is Distinguished Professor of Ecology and Evolutionary Biology and Mathematics at the University of Tennessee.
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Mathematics for the Life Sciences provides present and future biologists with the mathematical concepts and tools needed to understand and use mathematical models and read advanced mathematical biology books. It presents mathematics in biological contexts, focusing on the central mathematical ideas, and providing detailed explanations. The author assumes no mathematics background beyond algebra and precalculus. Calculus is presented as a one-chapter primer that is suitable for readers who have not studied the subject before, as well as readers who have taken a calculus course and need a review. This primer is followed by a novel chapter on mathematical modeling that begins with discussions of biological data and the basic principles of modeling. The remainder of the chapter introduces the reader to topics in mechanistic modeling (deriving models from biological assumptions) and empirical modeling (using data to parameterize and select models). The modeling chapter contains a thorough treatment of key ideas and techniques that are often neglected in mathematics books. It also provides the reader with a sophisticated viewpoint and the essential background needed to make full use of the remainder of the book, which includes two chapters on probability and its applications to inferential statistics and three chapters on discrete and continuous dynamical systems.
The biological content of the book is self-contained and includes many basic biology topics such as the genetic code, Mendelian genetics, population dynamics, predator-prey relationships, epidemiology, and immunology. The large number of problem sets include some drill problems along with a large number of case studies. The latter are divided into step-by-step problems and sorted into the appropriate section, allowing readers to gradually develop complete investigations from understanding the biological assumptions to a complete analysis facilitatingLet's face it: most students don't take calculus because they find it intellectually stimulating. It's not ... at least for those who come up on the wrong side of the bell curve! There they are, minding their own business, working toward some non-science related degree, when ... BLAM! They get next semester's course schedule in the mail, and first on the list is the mother of all loathed college courses ... CALCULUS!
Not to fear--Idiot's Guides: Calculus IA Modern Introduction to Differential Equations, Second Edition, provides an introduction to the basic concepts of differential equations.
The book begins by introducing the basic concepts of differential equations, focusing on the analytical, graphical, and numerical aspects of first-order equations, including slope fields and phase lines. The discussions then cover methods of solving second-order homogeneous and nonhomogeneous linear equations with constant coefficients; systems of linear differential equations; the Laplace transform and its applications to the solution of differential equations and systems of differential equations; and systems of nonlinear equations. Each chapter concludes with a summary of the important concepts in the chapter. Figures and tables are provided within sections to help students visualize or summarize concepts. The book also includes examples and exercises drawn from biology, chemistry, and economics, as well as from traditional pure mathematics, physics, and engineering.
This book is designed for undergraduate students majoring in mathematics, the natural sciences, and engineering. However, students in economics, business, and the social sciences with the necessary background will also find the text useful.
Student friendly readability- assessible to the average studentEarly introduction of qualitative and numerical methodsLarge number of exercises taken from biology, chemistry, economics, physics and engineering Exercises are labeled depending on difficulty/sophisticationEnd of chapter summariesGroup projects
This book will quickly, efficiently--and above all, effectively--help you prepare to succeed on the AP Calculus exam. Right from the start, the text helps you identify the course topics you most need practice on, allowing you to focus your study efforts o
This introductory textbook on mathematical biology focuses on discrete models across a variety of biological subdisciplines. Biological topics treated include linear and non-linear models of populations, Markov models of molecular evolution, phylogenetic tree construction, genetics, and infectious disease models. The coverage of models of molecular evolution and phylogenetic tree construction from DNA sequence data is unique among books at this level. Computer investigations with MATLAB are incorporated throughout, in both exercises and more extensive projects, to give readers hands-on experience with the mathematical models developed. MATLAB programs accompany the text. Mathematical tools, such as matrix algebra, eigenvector analysis, and basic probability, are motivated by biological models and given self-contained developments, so that mathematical prerequisites are minimal.
This book offers a self-study program on how mathematics, computer science and science can be profitably and seamlessly intertwined. This book focuses on two variable ODE models, both linear and nonlinear, and highlights theoretical and computational tools using MATLAB to explain their solutions. It also shows how to solve cable models using separation of variables and the Fourier SeriesFrom economics and business to the biological sciences to physics and engineering, professionals successfully use the powerful mathematical tool of optimal control to make management and strategy decisions. Optimal Control Applied to Biological Models thoroughly develops the mathematical aspects of optimal control theory and provides insight into the application of this theory to biological models.
Focusing on mathematical concepts, the book first examines the most basic problem for continuous time ordinary differential equations (ODEs) before discussing more complicated problems, such as variations of the initial conditions, imposed bounds on the control, multiple states and controls, linear dependence on the control, and free terminal time. In addition, the authors introduce the optimal control of discrete systems and of partial differential equations (PDEs).
Featuring a user-friendly interface, the book contains fourteen interactive sections of various applications, including immunology and epidemic disease models, management decisions in harvesting, and resource allocation models. It also develops the underlying numerical methods of the applications and includes the MATLAB® codes on which the applications are based.
Requiring only basic knowledge of multivariable calculus, simple ODEs, and mathematical models, this text shows how to adjust controls in biological systems in order to achieve proper outcomes facilitatingThis volume stems from two DIMACS activities, the U.S.-Africa Advanced Study Institute and the DIMACS Workshop, both on Mathematical Modeling of Infectious Diseases in Africa, held in South Africa in the summer of 2007. It contains both tutorial papers and research papers. Students and researchers should find the papers on modeling and analyzing certain diseases currently affecting Africa very informative. In particular, they can learn basic principles of disease modeling and stability from the tutorial papers where continuous and discrete time models, optimal control, and stochastic features are introduced | 677.169 | 1 |
Unit 6: Expressions, Equations, and Inequalities
▼
Overview
Unit 6 is based on the driving question: How are equations and inequalities used to represent situations in the real world? In this unit, you will be learning content standards and objectives 6.EE.5-9. | 677.169 | 1 |
Curriculum - Mathematics
Mathematics at St. Brendan's college is designed to cater for all. Ever since Pythagoras mulled over whether or not he should call the longest side of a triangle the 'Hypotenuse', students have had an aversion to this subject. This school aims to change that unhealthy state of mind. We recognise that students achieve at different levels in mathematics so for all junior classes we allocate three separate journey groups: Developmental, Intermediate and Extension. This allows for more flexibility in the students learning and gives them the opportunity to learn about real life applications suitable for their level of ability. In the senior years, students can study either Maths A, Maths B or Pre-Vocational Maths. The latter is a non-OP subject. As an elective, students may also choose Maths C. This subject is very popular and is a pre-requisite to many university courses. Students seeking a trade based course tend to choose Pre-Vocational Maths or Maths A while Maths B is universally considered to be the generic pre-requisite subject for many tertiary courses. As you may be aware, Maths has a bad reputation Australiawide. Students seem to be leaving this subject in droves (the statistics are alarming). However this is not the case at SBC! The emphasis on technology usage (graphic calculators, computers and interactive technologies), journey groups and meaningful life related learning experiences have stemmed the flow. Numbers are high in the academic Maths subjects especially since the introduction of the elective Math/Science junior elective subject Quantum. This subject aims to specifically prepare students for an academic rigour in Maths B/C, Physics, Chemistry and IPT. The hands-on approach of Maths at SBC features many strange events. It's not unusual after school to see students playing a Theremin as they investigate the mathematical nature of sound waves. The sound of a mini rollercoaster and accelerometer combination hurtling down a track while connected to TI84 or TI nSpire graphic calculator may seem strange to many – but not in a Maths classroom. Other students can be seen wandering around the oval armed with surveying equipment or letting air rockets off while a game of football is being played nearby. None of these sights are unusual. To be accosted by a Lego Robot while walking into a Quantum class is a common experience and this may be closely followed by a remote helicopter buzzing past while being monitored by a CBR and graphic calculator. These rewarding and rich Math tasks are a feature of the Maths Department at SBC. | 677.169 | 1 |
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