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What's New in Version 1.2
We improved Quick Math CAS Calculator making it a faster and better Application than ever before. We added over 15 new features that YOU requested by your feedback and beside that, we improved the stability of the App and done some minor bug fixes. New features include:
- Matrix Calculus (Matrix multiplication, reverse, addition, substraction and raise to the power of...). New interfaces were uniquely created for the Matrix Calculus making it very easy to use.
- Relative position between spheres and planes (circle of intersection or boundary points)
- Relative position of two spheres (plane of intersection, circle of intersection or boundary point)
- Relative position of line and sphere (point of intersection)
- Improved curve discussion feature: It runs now faster, more stable and it can deal with complex functions. In addition, you can now specify the boundaries between you want to run the curve discussion.
- The integrated memory is faster and preloads when starting the App.
- And much more... Just have a look at where we present the update in detail.
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Learning algebra prepares you for pre- calculus and geometry. Its a great course to take if you want to major or even minor in a mathematics or science. With algebra I learned how to find volumes of shapes, the Pythagorean theorem, slope formula, point slope formula, and midpoint formula. I wish that I had the needed calculator required for the class. But over all the class was fun and easy to take.
Course highlights:
Throughout the course we learned the Pythagorean theorem, point slope formula. How to use a number line and graphing.
Hours per week:
6-8 hours
Advice for students:
Have a separate notebook and binder for the class. Find the right graphing calculator because you'll need for years to come. Also do not think you don't need to study or take notes because you receive a bunch of formulas and equations. | 677.169 | 1 |
Page 5
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SECTION A: SPECIFICATION SUMMARY COURSE OUTLINE This course is designed to meet the needs of students who wish to continue the study of Mathematics beyond GCSE but for whom AS units may not be immediately appropriate. The content consists of four areas of Pure Mathematics: · Algebra; · Co-ordinate Geometry; · Trigonometry; · Calculus. Each of these is used to support a topic from a recognised branch of Applied Mathematics. AIMS · To introduce students to the power and elegance of advanced mathematics.…read more
Page 6
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SECTION B: GENERAL INFORMATION 1 Introduction 1.1 RATIONALE The target students for this course are those who have taken Higher GCSE either at the end of Year 10 or in January of Year 11, or who will take GCSE at the same time as Additional Mathematics.…read more
Page 7
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PROGRESSION The content of this course is contained in the AS units of OCR and other Examining Bodies. Consequently a student who has taken this unit is, by design, well-prepared to continue to Mathematics at AS and A level. 1.7 OVERLAP WITH OTHER QUALIFICATIONS The content of this unit includes some of the content of the Higher Tier GCSE Mathematics: the remainder is a subset of the GCE Mathematics AS units. 1.…read more
Page 8
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Assessment Objectives The assessment will test the ability of candidates to: · recall and use manipulative techniques (AO1; weighting 25-35); · interpret and use mathematical data, symbols and terminology(AO2; weighting 25-35); · recognise the appropriate mathematical procedure for a given situation(AO3; weighting 10-20); · formulate problems into mathematical terms and select and apply appropriate techniques of solution(AO4; weighting 10-20); · pursue a mathematical argument rigorously and with a high level of algebraic skill(AO5; weighting 10-20). 4 Scheme of Assessment 4.…read more
Page 9
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GRADING The examination result is reported as a grade A, B, C, D, E or U. 4.5 CERTIFICATION Candidates will receive a certificate entitled: OCR Free Standing Mathematics Qualification (Advanced): Additional Mathematics 4.6 ASSESSMENT OF ICT Candidates are expected to use calculators effectively, know how to enter complex calculations and use an extended range of function keys, including trigonometrical and statistical functions relevant to the course and content. 4.…read more
Page 10
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Know that the equation of a circle, centre (0,0), radius r is x2 + y2 = r2. · Know that (x a)2 + (y b)2 = r2 is the equation of a circle with centre (a, b) and radius r. · Be able to illustrate linear inequalities in two variables. · Be able to use the definitions of sin, cos and tan for any angle (measured in degrees only). · Be able to apply trigonometry to right angled triangles.…read more | 677.169 | 1 |
What's the Relationship?
Twelfth graders explore data and conduct analysis based upon using logarithmic substitution. They can use a graphing calculator to help in the process or other technology can be substituted like a computer program | 677.169 | 1 |
Number System
LCM and HCF
Fractions
Surds and Indices
Square Root and Cube Root
Simplification
Number Series
Average
Ratio and Proportion
Problem Based on Age
Mixture and Alligation
Percentage
Profit, Loss and Discount
Simple and Compound Interest
Time and Work
Time, Speed and Distance
Partnership
Algebra
Graph of Linear Equations
Geometry
Coordinate Geometry
Area and Perimeter
Volume and Surface Area
Trigonometry
Height and Distance
Data Interpretation | 677.169 | 1 |
Differentiation and Functions in Mathematics Course
Important information
Course
Online
When: Flexible
Description
This intermediate math course continues our free online maths suite of courses. It covers rules and applications of differentiation, straight line graphs, graphing circular functions, logs and indices, the Binomial theorem, inverse functions, and factors of polynomials. This course is ideal for second-level students, anyone studying for an exam, and those interested in re-igniting their knowledge of mathematicsDifferentiation and Functions in Mathematics CourseAlison
Free
Do you see something that is not right in this course? Let us know if there are any mistakes and you will help users like yourself. | 677.169 | 1 |
...
Show More Student-Centered Mathematics eBook Series. Each of the three grade band eBook DVDs, K-3, 3-5 and 5-8,feature grade specific lessons in action, personal interviews with the author, instructional tips and strategies, and more. What makes the eBook so unique? From the Van de Walle Professional Mathematics Series. Hear legendary mathematician, John Van de Walle speak about the Big Ideas in each chapter through a series of personal interviews. See excerpts from Van de Walle's professional development workshops without leaving the comfort of your home or school. Observe lessons in action through video of classrooms. Explore tips and activities you can use in your classroom. The eBook is available for purchase in the following package configurations: Single License Package (e-Book DVD & Book): Users with a DVD computer drive can take advantage of the larger video windows available in this single-user, single-disc package. School Network License Package (e-Book DVD & Book): This version will give all teachers within a single school access to this rich professional-development tool. Once installed, the school network version allows for multiple access and progressive downloading across a Local Area Network (LAN).* District Network License Package (e-Book DVD & Book): This package is the most economical way for a district or school board to purchase for multiple schools. This network-installable version allows for multiple access and progressive downloading across a LAN or high-speed Wide Area Network. *For order information, including pricing, please contact your local sales representative | 677.169 | 1 |
Providing reference and review material on many math topics, this site also features practice exams, a message board, reference tables, and a list of recommended books. Over 2,500 pages give short lessons on algebra, trigonometry, calculus, and matrix algebra. The explanations are complete and extensive, flow well from one to the next, and include plenty of sample exercises. This resource is part of the Teaching Quantitative Skills in the Geosciences collection. | 677.169 | 1 |
Introduction to Tensor Analysis and the Calculus of Moving
Most students will find that some problems will require repeated and persistent effort to solve. It is very challenging, but it does give students a chance to get students up and moving. Parker, Cosmic Time Travel: A Scientific Odyssey (1991) Cambridge: Perseus Publishing. Likewise, the problem of computing a quantity on a manifold which is invariant under differentiable mappings is inherently global, since any local invariant will be trivial in the sense that it is already exhibited in the topology of Rn.
In higher dimensions, the Riemann curvature tensor is an important pointwise invariant associated to a Riemannian manifold that measures how close it is to being flat. An important class of Riemannian manifolds is the Riemannian symmetric spaces, whose curvature is not necessarily constant. These are the closest analogues to the "ordinary" plane and space considered in Euclidean and non-Euclidean geometry Projective Duality and Homogeneous Spaces (Encyclopaedia of Mathematical Sciences) projectsforpreschoolers.com. The range of topics covered is wide including Topology topics like Homotopy, Homology, Cohomology theory and others like Manifolds, Riemannian Geometry, Complex Manifolds, Fibre Bundles and Characteristics Classes. I believe this book gives you a solid base in the modern mathematics that are being used among the physicists and mathematicians that you certainly may need to know and from where you will be in a position to further extent (if you wish) into more technical advanced mathematical books on specific topics, also it is self contained but the only shortcoming is that it brings not many exercises but still my advice, get it is a superb book ref.: Surveys in Differential download for free The book has insight and makes many good remarks. However, chapter 15 on Differential Geometry is perhaps too brief considering the importance of understanding this material, which is applied in the chapters thereinafter. The book is suitable for second to third year student in theoretical physics Differential Geometry, Group read online terrific.cc. I would call this a presentation of classical differential geometry from a modern viewpoint, since do Carmo practically gives the abstract definitions of a manifold, but by a sleight of hand specialises them to curves and surfaces , source: Geometry of Vector Sheaves: An Axiomatic Approach to Differential Geometry Volume II: Geometry. Examples and Applications (Mathematics and Its Applications) (Vol 1) Cootie Catcher is an interactive version (requires Macromedia Shockwave Plug-in). Virtual Fingertip Fortune Teller requires Macromedia Flash Player. The companion Fingertip Fortune Teller can be printed and assembled Symplectic Geometry: An read online This is arguably one of the deepest and most beautiful results in modern geometry, and it is surely a must know for any geometer / topologist Foliations on Riemannian Manifolds and Submanifolds luxuryflatneemrana.com. How many lines are contained in a general surface of degree three in space Non-Riemannian Geometry (Colloquium Publications) Non-Riemannian Geometry (Colloquium?
This notion can also be defined locally, i.e. for small neighborhoods of points. Any two regular curves are locally isometric. However, the Theorema Egregium of Carl Friedrich Gauss showed that already for surfaces, the existence of a local isometry imposes strong compatibility conditions on their metrics: the Gaussian curvatures at the corresponding points must be the same download. Differential topology is the study of the (infinitesimal, local, and global) properties of structures on manifolds having no non-trivial local moduli, whereas differential geometry is the study of the (infinitesimal, local, and global) properties of structures on manifolds having non-trivial local moduli , cited: Mixed Hodge Structures (Ergebnisse der Mathematik und ihrer Grenzgebiete. 3. Folge / A Series of Modern Surveys in Mathematics) read for free. We will then talk about Khovanov homology of knots, which is a "categorification" of the Jones polynomial constructed by M ref.: Locally Toric Manifolds and Singular Bohr-Sommerfeld Leaves (Memoirs of the American Mathematical Society) download online. During the 18th century, Euler applied these ideas to establish what is still today the classical theory of most general curves and surfaces, largely used in engineering , cited: Advances in Lorentzian Geometry: Proceedings of the Lorentzian Geometry Conference in Berlin (Ams/Ip Studies in Advanced Mathematics)
The Mathematics of Minkowski Space-Time: With an Introduction to Commutative Hypercomplex Numbers (Frontiers in Mathematics)
I NON-RIEMANNIAN GEOMETRY. download online. It will be apparent to the reader that these constitute a powerful weapon for analysing the geometrical properties of surfaces, and of systems of curves on a surface online. A major feature of life sciences in the 21st century is their transformation from phenomenological and descriptive disciplines to quantitative and predictive ones epub. A diffeomorphism between two symplectic manifolds which preserves the symplectic form is called a symplectomorphism , source: Geography of Order and Chaos in Mechanics: Investigations of Quasi-Integrable Systems with Analytical, Numerical, and Graphical Tools (Progress in Mathematical Physics) This formula was discovered by Isaac Newton and Leibniz for plane curves in the 17th century and by the Swiss mathematician Leonhard Euler for curves in space in the 18th century. (Note that the derivative of the tangent to the curve is not the same as the second derivative studied in calculus, which is the rate of change of the tangent to the curve as one moves along the x-axis.) With these definitions in place, it is now possible to compute the ideal inner radius r of the annular strip that goes into making the strake shown in the figure Hermitian Analysis: From Fourier Series to Cauchy-Riemann Geometry (Cornerstones) nssiti.com. To see this implemented in Mathematica visit the code page. [Jul 6, 2010] This project started in spring 2009 Geometry and Nonlinear Partial Differential Equations: Dedicated to Professor Buqing Su in Honor of His 100th Birthday : Proceedings of the Conference ... (Ams/Ip Studies in Advanced Mathematics) read online. Classical instruments allowed in geometric constructions are those with compass and straightedge. However, some problems turned out to be difficult or impossible to solve by these means alone, and ingenious constructions using parabolas and other curves, as well as mechanical devices, were found download. More technical than a "popular" book, this text is a readable "semi-technical" work. Epstein, Relativity Visualized (1985) San Francisco: Insight Press. This is a popular book sort of in the "for Dummies" style. Faber, Differential Geometry and Relativity Theory, An Introduction, Pure and Applied Mathematics, A Program of Monographs, Textbooks, and Lecture Notes #76 (1983) NY: Marcel Dekker epub.
Foliations on Riemannian Manifolds (Universitext)
Mathematical Aspects of Evolving Interfaces: Lectures given at the C.I.M.-C.I.M.E. joint Euro-Summer School held in Madeira Funchal, Portugal, July 3-9, 2000 (Lecture Notes in Mathematics)
Boundary Element Topics: Proceedings of the Final Conference of the Priority Research Programme Boundary Element Methods 1989-1995 of the German Research Foundation October 2-4, 1995 in Stuttgart
Scalar and Asymptotic Scalar Derivatives: Theory and Applications (Springer Optimization and Its Applications)
The Two-Dimensional Riemann Problem in Gas Dynamics (Monographs and Surveys in Pure and Applied Mathematics)
The Mathematics of Knots: Theory and Application (Contributions in Mathematical and Computational Sciences)
The Statistical Theory of Shape (Springer Series in Statistics)
Denjoy Integration in Abstract Spaces (Memoirs of the American Mathematical Society)
Differential Geometry and Related Topics
Index Theorem. 1 (Translations of Mathematical Monographs)
Differential Geometry (06) by Graustein, William C [Paperback (2006)]
Topics in Physical Mathematics
A Quantum Kirwan Map: Bubbling and Fredholm Theory for Symplectic Vortices over the Plane (Memoirs of the American Mathematical Society)
There COMPLEX GEOMETRY; DIFFERENTIAL read pdf COMPLEX GEOMETRY; DIFFERENTIAL GEOMETRY;. The aim of the event is to invite researchers from different places for reporting recent progress and discussing further problems appearing at the intersection of analysis, geometry, and applications , cited: ElementaryDifferential Geometry 2nd Second edition byO'Neill download pdf. I pdf Catastrophe Theory: Second read epub read epub. Poincaré Duality Angles on Riemannian Manifolds With Boundary — Geometry/Topology Seminar, Duke University, Sept. 15, 2009. Linking Integrals in Hyperspheres — Bi-Co Math Colloquium, Bryn Mawr College, Apr. 13, 2009. Poincaré Duality Angles for Riemannian Manifolds With Boundary — Geometry–Topology Seminar, Temple University, Dec. 2, 2008 , source: Integral Geometry and Inverse read online read online. Differential geometry is a wide field that borrows techniques from analysis, topology, and algebra. It also has important connections to physics: Einstein's general theory of relativity is entirely built upon it, to name only one example. Algebraic geometry is a complement to differential geometry. It's hard to convey in just a few words what the subject is all about epub. I am interested in differential geometry and magnetic monopoles. I am a PhD student of Prof Michael Singer and Dr Jason Lotay, and work in the field of complex Kähler geometry , cited: The metric theory of Banach manifolds (Lecture notes in mathematics ; 662) projectsforpreschoolers.com. Note: Some of you may have studied point-set topology (metric and topological spaces, continuous maps, compactness, etc.). The content of this course is different: it is usually called algebraic and differential topology. This course is designed so that familiarity with point-set topology is unnecessary The Scalar-Tensor Theory of Gravitation (Cambridge Monographs on Mathematical Physics) The Scalar-Tensor Theory of Gravitation. Legendrian Fronts for Affine Varieties, Symplectic Techniques in Hamiltonian Dynamics, ICMAT (6/2016). Legendrians and Mirror Symmetry, Georgia Topology Conference, U. Contact Topology from the Legendrian viewpoint, Submanifolds in Contact Topology, U ref.: An Introduction to read pdf 99propertyguru.in. In fact, we do not have a classification of the possible fundamental groups. I will discuss some of what is known about this problem. Along the way, we will discuss a question of S.-S. Chern posed in the 1960s, important examples by R. Shankar in the 1990s, and more recent classification results in the presence of symmetry by X Clifford Algebras and their read epub projectsforpreschoolers.com. | 677.169 | 1 |
Mathematics for the Analysis of Algorithms
4.11 - 1251 ratings - Source
This monograph collects some fundamental mathematical techniques that are required for the analysis of algorithms. It builds on the fundamentals of combinatorial analysis and complex variable theory to present many of the major paradigms used in the precise analysis of algorithms, emphasizing the more difficult notions. The authors cover recurrence relations, operator methods, and asymptotic analysis in a format that is concise enough for easy reference yet detailed enough for those with little background with the material.The book is very well written. The style and the mathematical exposition make the book pleasant to read.
Title
:
Mathematics for the Analysis of Algorithms
Author
:
Daniel H. Greene, Donald E. Knuth
Publisher
:
Springer Science & Business Media - 1990-09-01
ISBN-13
:
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NCERT solutions for class 11 mathematics are providing to good lab and book of syllabus help you approach the exam in way every student to the basics of the subject. The NCERT questions and examples are usually the questions every student and teacher targets .The better knowledge to in the Physics subject high quality learning to improve your understanding to check on our scholarslearning site
This course will give you a great basic understanding of event planning. It will also explain some of the details of how the business works and what an event planner actually does. Lastly, it will give you the basics to help get you started in
Rishikul Yogshala is registerd 200 hour yoga teacher training school, located in the beautiful & spiritually-heightened Rishikesh valley in India, at the root valley of the exotic Himalayan mountain range, Rishikul Yogshala is shares the scenic beauty around. We focus on removing stress, tension, ego and unwanted thoughts by adapting a holistic & traditional philosophy, expanding one's own consciousness, and by using its method of practiceIIT Jee Main 2016 , as quote scholars learning is as good teacher that help in your study . which study material are available here it gives you best opportunity for preparation . which student extremely serious about his education they should do preparation with best educational study material of this site. .
Welcome to Charis Montessori As our Teacher-student ratio is 1:6, the teacher is able to focus on each child development and needs. Since 1999, Charis Montessori started off through sowing the seed of love to the young children. Charis in Greek means Grace, Kindness and favor. In a mix age group setup, learning is made relevant and fun to the young minds.
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Indian National Mathematics Olympiad is a comprehensive book which has everything that as a student you need. With brainstorming questions for practice, especially for students of class 9, 10 preparing for exams. Not only increasing their aptitude to solve different Olympiad questions, but also ensures a steady growth in the understanding of each and every problem and enhancing the ability to recognize your mistakes.
Indian National Mathematics Olympiad has plenty of examples that can be used as practice problems. Theory is a part of certain chapters that ensure better understanding of basic concepts. Sample examples are set after each topic, along with Additional Solved Examples for better comprehension. Each topic consists of Theorems with their proof. Latest Solved Question Papers, Latest Solved Question Papers, Equations, Inequalities, Combinatorics, Geometry, and Functional Equations along with RMO and INMO Solved Papers2012-2013 is a thorough guide for students.
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Rajeev Manocha has written other resourceful books for students appearing for competitive exams like Mathematics in 40 Days for JEE Main 2013, 4th Edition, Preparing for AIEEE Chemistry in 40 Days, AIEEE Mathematics In 40 Days.
Indian National Mathematics Olympiad enables you to prepare for competitive exams like RMO and INMO and is a must buy for those students that will guide them best for these exams. Order now! This paperback is listed online with the ISBN-10: 9350945339 and ISBN-13: 978-9350945339. | 677.169 | 1 |
Hardcover | January 9, 2017 with whole-number arithmetic, fractions, and rational numbers. Each chapter covers these topics from multiple perspectives, including genetic disorders, cognition, instruction, and neural networks. Covers innovative measures and recent methodological advances in mathematical thinking and learning Contains contributions that improve instruction and education in these domains Informs policy aimed at increasing the level of mathematical proficiency in the general public
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David C. Geary is a cognitive developmental and evolutionary psychologist at the University of Missouri. He has wide ranging interests but his primary areas of research and scholarly work are children's mathematical cognition and learning and Darwin's sexual selection as largely but not solely related to human sex differences.Professor... | 677.169 | 1 |
Elementary Algebra, Student Support Edition
4.11 - 1251 ratings - Source Important Notice: Media content referenced within the product description or the product text may not be available in the ebook version.Section 2.3 Algebra and Problem Solving 101 In Exercises 1a€"6, match the verbal
phrase with the cor- rect algebraic expression. (a) (b) 2.3 ... Developing Skills (c) (
d) (e) (f) 12x 3 11x 1 3 12 3x 3x 12 3x 12 11 1 3 x 1. Twelve ... Answers to all
Review: Concepts, Skills, and Problem Solving exercises are given in the back of
the textbook. ... Chemical Reaction The rate of change in a chem- ical reaction is
where is the original amount, is the new amount, and is a constant of
proportionality.
Title
:
Elementary Algebra, Student Support Edition
Author
:
Ron Larson, Robert P. Hostetler
Publisher
:
Cengage Learning - 2007-01-02
ISBN-13
:
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Practical Problems in Mathematics for Information Technology
4.11 - 1251 ratings - Source
Create a new approach to explaining the math and logic fundamentals required in the information technology industry. Practical Problems in Mathematics for Information Technology is an exciting new resource for building a solid foundation in the mathematical skills that are used in a number of areas, such as networking, systems administration, programming, database management, web programming, and computer repair. By presenting examples, problems, and exercises that are taken directly from these concentration areas, readers will not only build their mathematical know-how, but they will achieve the added benefit of being fully prepared for the types of challenges they are likely to encounter on the job. Real-world examples from the industry are included throughout this new book. Important Notice: Media content referenced within the product description or the product text may not be available in the ebook version.Instructors may choose to use this book as a standalone text or as a
supplemental workbook to a theory-based text. The Answers to Odd-Numbered
Problems are provided at the end of this book along with a Glossary of technical
terms and an Appendix ... ABOUT THE AUTHOR Todd Verge has a Bachelor of
Science Degree in Physics from Dalhousie University, a Masters Degree in
Philosophy fromanbsp;...
Title
:
Practical Problems in Mathematics for Information Technology
Author
:
Todd Verge
Publisher
:
Cengage Learning - 2008-03-13
ISBN-13
:
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Saxon Math - Level I
Jul 11 - Jul 15 (Monday, Tuesday, Wednesday, Thursday, Friday)
BASIS Independent Brooklyn
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Meeting Dates
From Jul 11, 2016 to Jul 15, 2016
About This Activity
This critical program will give students the opportunity learn and/or review essential topics to help set them up for success in math for the year to come. Incoming students as well as current students desiring review would benefit from the focused approach and further practice. The program is designed to allow flexibility by providing the choice between one of two one-week courses, concentrating on preliminary material, or a two-week course, introducing material from the target math | 677.169 | 1 |
Factor Around the Room - Higher Order Polynomials
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Getting tired of practicing factoring with worksheets? Use this activity in your classroom as a fun way to practice factoring higher order polynomials!
This activity will get students up out of their chairs and working to complete all 16 problems scattered around the room in four problem sets. These problems are a combination of higher order polynomials of the following types: greatest common factor, trinomials, and special binomials. (See preview | 677.169 | 1 |
IMPACT (Interweaving Mathematics Pedagogy and Content for Teaching) is an exciting new series of advanced textbooks for teacher education which aims to advance the teaching of mathematics by integrating mathematics content teaching with the broader research and theoretical base of mathematics education.
The Learning and Teaching of Algebra is a primer for teachers and researchers that provides a pedagogical framework for the teaching and learning of algebra grounded in theory and research.
Areas covered include:
• Algebra: Setting the Scene
• Some Lessons From History
• Seeing Algebra Through the Eyes of a Learner
• Emphases in Algebra Teaching
• Algebra Education in the Digital Era
This guide will be essential reading for trainee and qualified teachers of mathematics, graduate students, curriculum developers and all those who are interested in the "problématique" of teaching and learning algebra. It allows you to get involved in the wealth of knowledge that teachers can draw upon to assist learners, helping you gain the insights that mastering algebra pr | 677.169 | 1 |
PrecalculusChapter R:Review of Prerequisites
SectionR.1Sets and the Real Number LineSectionR.2Exponents and RadicalsSectionR.3Polynomials and Factoring Obj 8Factor Expressions Containing Negative and Rational ExponentsProblem Recognition Exercises Simplifying Algebraic ExpressionsSectionR.4Rational Expressions and More Operations on RadicalsSectionR.5EquationsSection R.6Complex Numbers and Equations with Complex SolutionsSectionR.7Linear, Compound, and Absolute Value InequalitiesSectionR.8Applications of Equations and Inequalities
Chapter 1:Functions and Relations
Section1.1The Rectangular Coordinate System and Graphing UtilitiesSection1.2CirclesSection1.3Functions and RelationsSection1.4Linear Equations in Two Variables and Linear FunctionsSection1.5Applications of Linear Equations and ModelingProblem Recognition Exercises Comparing Graphs of EquationsSection1.6Transformations of GraphsSection1.7Analyzing Graphs of Functions and Piecewise-Defined FunctionsSection1.8Algebra of Functions and Function Composition
Chapter 2:Polynomial and Rational Functions
Section2.1Quadratic Functions and ApplicationsSection2.2Introduction to Polynomial FunctionsSection2.3Division of Polynomials and the Remainder and Factor TheoremsSection2.4Zeros of PolynomialsSection2.5Rational FunctionsProblem Recognition Exercises Polynomial and Rational FunctionsSection2.6Polynomial and Rational InequalitiesProblem Recognition Exercises Solving Equations and InequalitiesSection2.7Variation
Chapter 4:Trigonometric Functions
Section4.1Angles and Their Measure and Special TrianglesSection4.2Trigonometric Functions Defined on the Unit CircleSection4.3Trigonometric Functions Defined on Right TrianglesSection4.4Graphs of the Sine and Cosine Functions Section4.5Graphs of Other Trigonometric FunctionsProblem Recognition Exercises Comparing Graphical Characteristics of Trigonometric FunctionsSection4.6Inverse Trigonometric Functions
Chapter 6:Applications of Trigonometric Functions
Section6.1Applications of Right TrianglesSection6.2The Law of SinesSection6.3The Law of CosinesProblem Recognition Exercises Solving Triangles Using a Variety of ToolsSection6.4Harmonic Motion and Combinations of Trigonometric Functions
Chapter 8:Systems of Equations and Inequalities
Section8.1Systems of Linear Equations in Two Variables and ApplicationsSection8.2Systems of Linear Equations in Three Variables and ApplicationsSection8.3Partial Fraction DecompositionSection8.4Systems of Nonlinear Equations in Two VariablesSection8.5Inequalities and Systems of Inequalities in Two VariablesProblem Recognition Exercises Equations and Inequalities in Two VariablesSection8.6Linear Programming
Online:Chapter 12:Preview of Calculus
Section12.1Introduction to Limits through Tables and GraphsSection12.2Algebraic Properties of LimitsProblem Recognition Exercises Using a Variety of Methods to Evaluate LimitsSection12.3Derivatives: The Tangent ProblemSection12.4Integrals: The Area Problem
Online:Section A-1Proof of the Binomial TheoremA-2Definition of Conics From a Fixed Point and Fixed Line | 677.169 | 1 |
ayn Martin-Gay firmly believes that every student can succeed, and her developmental math textbooks and video resources are motivated by this belief. Basic College Mathematics with Early Integers, Second Edition was written to help students effectively make the transition from arithmetic to algebra. The new edition offers new resources like the Student Organizer and now includes Student Resources in the back of the book to help students on their quest for success. | 677.169 | 1 |
Math Phonics - Pre-Algebra
4.11 - 1251 ratings - Source
Basic math skills to prepare them for algebra. Her fun methods and concrete examples will help younger students begin to grasp the principles of algebra before they actually have to deal with the complete course. Included are easy-to-understand explanations and instructions, wall charts, games, activity pages and worksheets. As in all her Math Phonics books, the author emphasizes three important principles: understanding, learning and mastery. Students will learn about integers, exponents and scientific notation, expressions, graphing, slope, binomials and trinomials. In addition to helpful math rules and facts, a complete answer key is provided. As students enjoy the quick tips and alternative techniques for math mastery, teachers will appreciate the easy-going approach to a difficult subject.Middle school and high school students in the 21st century are quite familiar with
graphing and all the types of ... Elm Peterson Buchanan F i r s t S t M a i n S t V
i n e S t -5 For graphing, we use two number lines which form four right ...
LESSON. PLAN. 8: GRAPHING. y (5, 7) +5 x -5 +5 -5 y II I x III IV Lesson Plan 8:
Graphing.
Title
:
Math Phonics - Pre-Algebra
Author
:
Marilyn B. Hein
Publisher
:
Lorenz Educational Press - 2004-03-01
ISBN-13
:
Continue
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Mathematica Navigator: Graphics and Methods of Applied Mathematics
Mathematica Navigator contains a great deal of material not easily found elsewhere in a well-organized form, with sufficient detail and illustrative examples. This book will serve excellently as a Mathematica handbook; it starts with the basics, goes carefully through the main material of Mathematica, and covers some advanced topics. Mathematica packages are integrated into the text, so that the reader gets a comprehensive overview of the features of Mathematica. This book allows a new user to begin working with Mathematica and proceed to quite a high level. Old users will find much new material, allowing them to raise their knowledge and skill to an even higher level. In addition, registered readers can send questions to the author concerning the use of Mathematica in areas treated in the book.
Part tutorial and part handbook, this title shows beginners how to use recent Mathematica versions
From the Publisher:
Praised as both a tutorial and a handbook, this new edition of Mathematica Navigator emphasizes applied mathematical and statistical methods, graphics, and programming. No previous experience with Mathematica is required, yet most chapters also integrate advanced techniques. The book and the included CD-ROM are valuable resources for both beginners and experienced Mathematica users.
Features: - Based on Mathematica 5.0, but can also be used with older versions of Mathematica - With the enclosed CD-ROM, the entire book can be installed into the help system of Mathematica - Comprehensive coverage from basic, introductory information through to more advanced topics - Studies several real data sets and many classical mathematical models | 677.169 | 1 |
Maths for Economics
4.11 - 1251 ratings - Source
Maths for Economics provides a solid foundation in mathematical principles and methods used in economics, beginning by revisiting basic skills in arithmetic, algebra and equation solving and slowly building to more advanced topics, using a carefully calculated learning gradient.Solving a pair of simultaneous equations means finding a pair of values for x and
ythat satisfies both equations. ... Graphical. solution. of. simultaneous.
linearequations. EXAMPLE 3.18 The first pair of simultaneous equations that we
anbsp;...
Title
:
Maths for Economics
Author
:
Geoff Renshaw
Publisher
:
Oxford University Press - 2011-12-01
ISBN-13
:
Continue
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Linear Systems: Using Matrix Operations
For this linear systems worksheet, students use matrix multiplication as an elimination technique to solve linear systems. They compare products and explore trends. This four-page worksheet contains explanations, examples and five | 677.169 | 1 |
A Book of Set Theory
4.11 - 1251 ratings - Source
qThis accessible approach to set theory for upper-level undergraduates poses rigorous but simple arguments. Each definition is accompanied by commentary that motivates and explains new concepts. A historical introduction is followed by discussions of classes and sets, functions, natural and cardinal numbers, the arithmetic of ordinal numbers, and related topics. 1971 edition with new material by the authorq--aquot;This accessible approach to set theory for upper-level undergraduates poses rigorous but simple arguments.
Title
:
A Book of Set Theory
Author
:
Charles C Pinter
Publisher
:
Courier Corporation - 2014-07-23
ISBN-13
:
Continue
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Algebra: Introductory and Intermediate: An Applied Approach
4.11 - 1251 ratings - Source
As in previous editions, the focus in ALGEBRA: INTRODUCTORY a INTERMEDIATE Eighth.EXAMPLE a#39; 4 I YOU TRY IT a#39; 4 I I 2 3 1 _ _ 2 1 4 Slmpllfy: a€" a€" a€"2 + a€" Slmplif :
a€" a€" a€" + a€" x x xy b a ab Solution Your ... 2_32+i22.2_12.2+i.5 x x xy x xy x y xy
x _2xy 3y x a€"Ea€"E E _2xya€"3y+x _T : Simplify: x a€" 4_x Simplify: Q+ a_9 2xa€"4anbsp;...
Title
:
Algebra: Introductory and Intermediate: An Applied Approach
Author
:
Richard Aufmann, Joanne Lockwood
Publisher
:
Cengage Learning - 2010-02-16
ISBN-13
:
Continue
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Analysis and Probability
Probability theory is a rapidly expanding field and is used in many areas of science and technology. Beginning from a basis of abstract analysis, this mathematics book develops the knowledge needed for advanced students to develop a complex understanding of probability. The first part of the book systematically presents concepts and results from analysis before embarking on the study of probability theory. The initial section will also be useful for those interested in topology, measure theory, real analysis and functional analysis. The second part of the book presents the concepts, methodology and fundamental results of probability theory. Exercises are included throughout the text, not just at the end, to teach each concept fully as it is explained, including presentations of interesting extensions of the theory. The complete and detailed nature of the book makes it ideal as a reference book or for self-study in probability and related fields.
Covers a wide range of subjects including f-expansions, Fuk-Nagaev inequalities and Markov triples.Provides multiple clearly worked exercises with complete proofs.Guides readers through examples so they can understand and write research papers independently | 677.169 | 1 |
Vedic Mathematics: Elementary level
4.11 - 1251 ratings - Source
Vedic Mathematics was reconstructed from ancient vedic texts early last century by Sri Bharati Tirthaji (1884-1960). It is a complete systems of mathematics which has many surprising properties and applies at all levels and areas of mathematics, pure and applied. The system is based on sixteen word-formulae that relate to the way in which we use our mind.The Arya Samaj was one of several socio-religious movements which were founded in the nineteenth century.
Title
:
Vedic Mathematics: Elementary level
Author
:
Kenneth R. Williams
Publisher
:
Motilal Banarsidass Publishe - 2005-01-01
ISBN-13
:
Continue
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handyCalc Calculator
handyCalc is a powerful calculator with automatic suggestion and solving which makes it easier to learn and use.With almost all the features you can imagine on a calculator, waiting for you to explore.* currency convert, unit convert, graph, solve equations | 677.169 | 1 |
Essential Mathematics for Games and Interactive Applications: A Programmer's Guide authors' popular tutorials at the Game Developers Conference, Essential Mathematics for Games and Interactive Applications presents the core mathematics necessary for sophisticated 3D graphics and interactive physical simulations. The book begins with linear algebra and matrix manipulation and expands on this foundation to cover such topics as texture filtering, interpolation, animation, and basic game physics. Essential Mathematics focuses on the issues of 3D game development important to programmers and includes optimization guidance throughout.
*Covers concepts in sufficient detail for a programmer to understand the foundations of 3D without feeling overwhelmed by proofs and theory
*Companion CD-ROM with code examples built around a shared code base, including a math library covering all the topics presented in the book, a core vector/matrix math engine, and libraries to support basic 3D rendering and interaction
*Provides guidance for students trying to understand how games are actually developed, including optimization techniques | 677.169 | 1 |
A Boolean Look at Inequalities
Students explore the concept of boolean algebra. In this boolean algebra lesson plan, students use boolean algebra to determine when one curve is greater than another in a system of inequalities. Students use their graphing calculator to determine at what point the boolean test turns from false to true. | 677.169 | 1 |
Description
Graph and interact with your Math formulas! Cited among the 10 Best Graphing Apps for iPhone! (
* Please note: "Math Graphing" is also available for the iPad. Search on "Math Graphing XL" to learn more.
MATH GRAPHING is a unique 1D graphing calculator to graph mathematical expressions of arbitrary complexity. "Math Graphing" can help students improve their math skills by developing some visual intuition of mathematical expressions or advanced users who need some scientific capabilities only available with expensive desktop software.
MATH GRAPHING provides the following functionalities:
- Multiple expressions with quasi-unlimited number of variables can be combined to produce simple or complex formulas.
- Interactive sliders can be created to visually investigate the role of important parameters on the graphical representation of the formulas.
- Trace mode to display coordinates and derivative of marker on selected curve.
- Solver tool to solve y = f(x) where x or y are unknown, local/global minima/maxima/extrema (NEW!)
- Formula graphs can be saved to the device's Photos Album, or the formulas be exported through email with embedded graphs.
- A list of favorite formulas can be created for editing or archival purposes.
- Several formulas can be plotted simultaneously in different colors and styles.
- Customizable graph appearance: axes labels, title, curve color and style, ticks number, grid, etc.
- Single-precision calculator (NEW!)
Example of mathematical expression:
# Gabor function
x = 0:1:100
sigma = 1:50; freq = 0:0.1; phase = 0:180
u = cos(2*pi*(x-50)*freq+phase*pi/180)
v = e^(-((x-50)^2)/(2*sigma^2))
y = u*v
plot(u,'lr'); plot(v,'lg'); plot(y,'lw')
xlabel('time'); ylabel('amplitude')
title('Gabor')
Please visit our "Math Graphing" web page ( to learn more about "Math Graphing"!
About KyberVision: we provide consulting, research and development services in Vision Sciences. We also develop vision-related applications for the iPad/iPhone/iPod Touch platform, that are spin-off products of Psykinematix, our Mac OS X flagship product in the field of Visual Psychophysics. Feel free to subscribe to our newsletter | 677.169 | 1 |
Solve Math problems and plot functions.
Full featured scientific calculator . It can help you solve basic calculations to college calculus.
With "Maths Solver" You can solve complex Math problems or plot multiple functions with accuracy and speed.
No network access required! which means you don't need internet connected to use its features.
You can plot functions and zoom-in , zoom-out (Press back button to make graph full screen by hiding keyboard )
It covers following areas of Mathematics.
Features:-
Basic Algebra
Polynomial
Solve equation and system of linear equations
Roots
Trignometry
Derivative
Indefinite and Definite Integration
Limits
Set operations i.e Union, Intersection, Mean, Median, Max, Min
Matrix : Determinant, Eigenvalues, EnigenVectors, Transpose, Power
Vectors
Curl and Divergence
2D function plots
For full list of features see Catalog and Examples in | 677.169 | 1 |
high school Algebra offering from Mathalicious is a fantastic collection of real-world math lessons, each broken down into three sections; Plan, Teach, and Reflect. The Plan section has clear teacher instructions, while high-quality video clips hold students' attention. The heart of each lesson is in the Teach section, which includes a preview activity and a set of problems. In Reflect, users can post comments and share ideas. Not all of the Common Core Algebra standards are addressed, but three are covered well through these explorations of real-world math.
Have your students complete the preview activity with a partner. Ask them to write a "3-2-1 Summary" of the activity: 3 things they learned, 2 things they found most interesting, and 1 thing they still have a question about. Follow with a whole-class discussion for further exploration on the topic. Afterward, have students return to their partners to complete the lesson problems collaboratively.
Read MoreRead LessReasoning With Equations And Inequalities
HSA.REI: Solve Equations And Inequalities In One Variable
HSA.REI.4
Solve quadratic equations in one variable.
Seeing Structure In Expressions
HSA.SSE: Interpret The Structure Of Expressions
HSA.SSE.1
Interpret expressions that represent a quantity in terms of its context.
Write Expressions In Equivalent Forms To Solve Problems
HSA.SSE.4
Derive the formula for the sum of a finite geometric series (when the common ratio is not 1), and use the formula to solve problems. For example, calculate mortgage payments.★ | 677.169 | 1 |
GRE Math Questions
What's one great way to get better at GRE math? Well, do GRE math questions. Below is a overview of the GRE quant landscape, so you know which resources to use depending on your level.
Beginner
Just thinking about the formula for the area of a circle causes beads of sweat to form at your brow. Your experience with learning math has been typically to become frustrated because math books seem vague and tend to jump around. Prepping for the GRE, however, doesn't have to be that way, if you use the following resources.
Manhattan GRE guides
Ever open up a book to the math section and quickly feel overwhelmed? With the Manhattan GRE guides by your side, fear not. Everything is clearly explained, and practice questions at the end of each chapter build off what was talked about in that chapter (vs. other books that throw in information it assumes you know).
The Princeton Review: Cracking the new GRE
This book has great tips for those just starting out. You will definitely learn how to tackle GRE math questions(instead of feeling that you're the one getting tackled). Big picture strategies such as approximation, process of elimination, and backsolving/plugging in are covered. Most of the questions are fairly easy. The only draw back is the book contains very few questions.
Magoosh
Magoosh's GRE lesson videos are a great place to learn the fundamentals and then do practice questions related to the fundamentals covered in each video. The pause and playback feature make Magoosh a great way to learn, especially for those who prefer a human voice to a book.
Intermediate
You understand a fair amount of basics, but you still need a review of the fundamentals. At the same time, you'll need questions that pull you a little out of your comfort zone without overwhelming you completely.
Manhattan GRE 5lbs. book
This book contains oodles of practice problems for just about every concept covered on the GRE. The explanations are excellent so rarely will you feel flustered when you miss a question. Make sure you know the concepts first (you may want to go through Magoosh's lesson videos) before you tackle the question, as this book does not introduce concepts. It simply provides practice questions.
Official material
Don't forget to use the actual stuff. Though you can refresh/learn concepts, the Official Guide is best suited as a source of actual practice. Remember, there is nothing like do questions written by the GRE folks themselves. Speaking of which you might also want to try the official guide for the old test, which was called Practicing to Take the GRE.
Advanced
For the advanced student, you already know your basics. You simply want lots of tough practice questions. Basically copy the advice for the intermediate section, except for the old official guide (those questions are probably a little too easy). Here are a few other books to help you sharpen your "quant teeth."
Nova's GRE math prep book
This is a great book for practice. The introductions of concepts are severely limited, so I would never recommend this book to anyone but those with a solid grounding in math. The math questions though can be tough, and will prepare you for most of what you'll see test day in the tough math section.
The perfect 170
You are a quant guru, and your program requires a score in the top 5%. Anything below 165 is not an option. Your one goal is the perfect score—a 170. So you don't want to waste your time with mediocre prep material, or prep material that is too easy. You want only the tough stuff.
GRE Official Guide
Mostly to do all the practice tests, just to get you into test shape. After all the GRE will have many easy questions as well. You don't only want to do tough questions.
Manhattan 5 lbs. prep book
This recommendation is mostly because once you buy the book you get access to Manhattan GRE's online challenge problem archive.
Magoosh
We have some of the most difficult GRE questions out there. If you can handle these, you can definitely handle anything the GRE will throw at you test day.
GMAT material
Besides the Data Sufficiency section, the GMAT problem solving overlaps very well that on the GRE, except for the GMAT questions are even more difficult. Over on the GMAT blog, I've rounded up some of the best GMAT resources for those GRE overachievers seeking a perfect score on math. 🙂2 Responses to GRE Math Questions
This is Nag from India , I have done master in technology in 2015 (in computer network engineering ) along with that I have 9 years of software development in computer network product development. I am planning to take MS in any US universities , my question is universities in us will consider my masters or bachelor degree to get admitted .
The exact answer to your question will depend on the individual university or program you want to apply to. So, I recommend contacting those programs directly to discuss how they will take into consider your previous studies and experience | 677.169 | 1 |
Trying to understand geometry but feel like you're stuck in another dimension? Here's your solution. Geometry Demystified, Second Edition helps you grasp the essential concepts with ease.
Written in a step-by-step format, this practical guide begins with two dimensions, reviewing points, lines, angles, and distances, then covers triangles, quadrilaterals, polygons, and the Cartesian plane. The book goes on to discuss three dimensions, explaining surface area, volume, vectors, Cartesian three-space, alternative coordinates, hyperspace, and warped spacePlane geometry and solid geometry
Using a drafting compass and straight edge
Solving pairs of equations
Working with vectors in three-space
Polar coordinates
Cartesian n-space
Simple enough for a beginner, but challenging enough for an advanced student, Geometry Demystified, Second Edition helps you master this fundamental mathematics Provides a self-paced method for learning the general concepts and fundamentals of geometry, and includes multiple-choice questions at the end of each chapter and a final exam. Bookseller Inventory # ABE_book_new_0071756264 | 677.169 | 1 |
This document provides teaching guidelines and student material for a unit intended for use in 12th grade algebra classes. Time allotment is from four to six hours of classroom time. The objective of this capsule is to teach students how to solve compound interest problems using arithmetic, logorithms, and calculators. Prerequisites for the unit are second year algebra, a working knowledge of logorithms and exponential equations, and an available calculator. The student material is divided into three sections. Section I focuses on solving compound interest problems, deriving the compound interest formula, and analyzing several savings accounts to determine which pays the most. Section II emphasizes using the compound interest formula, while Section III concentrates on solving compound interest problems for the rate of interest for the time the money was in the account. Seven activity sheets provide word problems relating to compound interest and a comparative analysis chart. Three supplementary activities focus on computer program problem solving and further comparative analysis. The teacher's guide provides a list of needed materials, advance preparation information, keys to activities and tests, a bibliography, and copies of all worksheets and assessments. (CK) | 677.169 | 1 |
PreCalculus With Limits
Description: Engineers looking for an accessible approach to calculus will appreciate Precalculus, 2nd Edition. The book offers a clear writing style that helps reduce any math anxiety they may have while developing their problem-solving skills. It incorporatesEngineers looking for an accessible approach to calculus will appreciate Precalculus, 2nd Edition. The book offers a clear writing style that helps reduce any math anxiety they may have while developing their problem-solving skills. It incorporates Parallel Words and Math boxes that provide detailed annotations which follow a multi-modal approach. Your Turn exercises reinforce concepts by allowing them to see the connection between the exercises and examples. A five-step problem solving method is also used to help engineers gain a stronger understanding of word problems | 677.169 | 1 |
CRAN Differential Equations
Differential equations (DE) are mathematical equations that describe how a quantity changes as a function of one or several (independent) variables, often time or space. Differential equations play an important role in biology, chemistry, physics, engineering, economy and other disciplines. Differential equations can be separated into stochastic versus deterministic DEs. Problems can be split into initial value problems versus boundary value problems. One also distinguishes ordinary differential equations from partial differential equations, differential algebraic equations and delay differential equations. All these types of DEs can be solved in R. DE problems can be classified to be either stiff or nonstiff; the former type of problems are much more difficult to solve
Keywords for this software
Anything in here will be replaced on browsers that support the canvas element | 677.169 | 1 |
Mathematics Summer School in July
Starting Industrial Design in September 2016? Then we would like to inform you about our Brush up your Maths! Summer Course as part of the CuriousU festival which you can follow before you start your study at the UT.
Brush up your Maths!
Mathematics is a very important tool for science based engineering, so it is essential to understand basic mathematics. This course will help you achieve exactly that goal!
Our course will enable you to become a more accurate user of mathematics. You will develop or improve your mathematical skills needed to study in an engineering context. Our course is a perfect preparation for our own math and engineering lectures offered in your engineering programme.
Strengthen the base of Mathematics
During the summer school you'll practice your mathematic skills which will be useful for the rest of your life. Especially for a technical education it's very important to strengthen the base of mathematics. Besides that there will be fun as well: the math course is divided in an educational part and a math recreation part. The educational part covers the basics of math. The recreation part consist of workshops w focusing on e.g. "the history of maths", "the role of zero" or "math quizzes" etc.
Besides the math courses there will be other activities as part of the CuriousU festival like: academic courses, music, sports, theatre, inspirational speakers. Check for more information.
Participation in the course
The course will take place from 14 - 23 August, 2016 You can register for the course through the website
The costs are € 800
Waiver
We offer one free ticket for this summer school experience. If you are interested you can send a motivation letter to a.vandenboomgaard@utwente.nl before July 15 2016.
In this letter you should let us know (max. 100 words) why the course "Mathematics for Engineering" is important for you and why you should be the one to get the free ticket. The best motivation will win the free ticket! | 677.169 | 1 |
Algebra 1 - Two-Step Equations Partner Problems
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PRODUCT DESCRIPTION
Algebra 1 - Two-Step Equations Partner Problems
This activity includes two sets of twenty problems that partners can work on at the same time. The two sets have the same answers so students can verify they are doing it correctly as they go. This would also be a great supplement for a struggling student to check as they go.
It is provided in both pdf and word formats. All the problems are worked out step by step | 677.169 | 1 |
Mentored means that you have tutors assisting you with learning, that you can ask questions, and get help.
Adaptive means that your curriculum is customized to your learning needs. You can advance at your own pace as long as you advance. You also have opportunities to regain mastery and attempt higher scores through re-testing.
While your Math Zone course will only meet for 3 hours of seat time, it is crucial that you spend another 6 hours independently completing practice problems either in the Math Zone during tutoring hours or remotely from home or elsewhere.
The most important part in all of this is learning personal responsibility. Indeed, this is part of what we teach. Unlike high school, the Math Zone requires that each student become directly engaged in their own learning. Students learn to take charge of the time they spend studying. We are here to help. Your instructors and Math Zone Staff will follow up with you if you are not meeting your obligation. This is a united effort.
Each Math Zone class is designed to turn out students who have demonstrated proficiency in mathematics. We want to level the playing field for students so that they can fully succeed in any upper level math class they need to take. Acquiring math competency is an essential part of student success at UNH.
How it works
Students who are assigned to the Math Zone are enrolled in one of the aforementioned courses. You will be provided detailed instructions on how to access your Math Zone course and how to attend classes each week. Students are also shown how to access their UNH Blackboard account, and also how to access the Pearson MyLabs Plus web site to gain access to the instructional materials.
This is not an online class. Testing and some of your weekly studying must be done in the Math Zone.
Even though this is not an online class, you can access your course to study on your own any place that you have an Internet connection. This means:
If you need help with the math or the lesson, don't wait, ask;
If the computers or technology are causing a problem, let us know now, not tomorrow or the next time the problem comes up
We want students to interact with the faculty, we want you to be engaged in your own learning. We are committed to providing every students an individualized learning opportunity to get through the basic math they need. | 677.169 | 1 |
MTG Foundation Course For Class 9 - Mathematics
Product Code : 9788189255329
Quick Overview
MTG Foundation Course For Class 9 - Mathematics
• Exhaustive coverage of the topics that help building Foundation for IIT/AIEEE while studying class 9th syllabus • Each concept of class 9th is explained with solved examples, illustrations activities and concept maps. • Complete practice of a topic through variety of questions such as MCQs, True-False, Matrix Match, Assertion and Reason, Passage comprehension, Subjective problems of different weightage. • Solved NCERT questions. • In short, a complete study package to prepare you to ace in school exams and competitive exams as well.
Chapters included:
Number Systems
Polynomials
Coordinate Geometry
Linear Equations in Two Variables
Introduction to Euclid's Geometry
Lines and Angles
Triangles
Quadrilaterals
Areas of Parallelograms and Triangles
Circles
Constructions
Heron's Formula
Surface Areas and Volumes
Statistics
Probability
Questions from Science Olympiad and International Mathematics Olympiad | 677.169 | 1 |
Mathematics Explorations: Grades 6-9: Teacher Edition:
I asked him what he thought about that. When we would come to ask him about the problems we were stuck on, he would say, "Oh yes, that one got me for a while, too. During my first week I started by simply observing my cooperating teacher. When it is time to display the results teachers should be able to support this step through meaningful discussions regarding the diverse ways that the problem has been attempted and solved.
Of course, they do so in a discordant, chaotic way, but this can just as easily be said of the same students English essays or history papers. They simply have less practice with this particular kind of argument, and so they are expectantly less coherent The 32 Most Effective SAT Math read for free The 32 Most Effective SAT Math. This text identifies, isolates, and emphasizes the essential patterns, illustrating them in several subject areas of mathematics. There are a limited number of key vocabulary words from logic ("and", "or", "not", "if... then", "if and only if", "for all", and "there exists") which are frequently used in mathematics epub. Complete printable PDFs on a CD-ROM for ease of use and daily lesson planning. Designed to be used in conjunction with the print Transition Guide, this online intervention and transition resource makes it easy to locate and print previous grade-level Reteach and Extra Practice worksheets to address transition-related knowledge and skill gaps , e.g. Let's Prepare for the Grade 8 Math Assessment thewhyredcompany.com. Describing the Use of and Assessing Perceived Effectiveness of Technology in Calculus III, Gavin LaRose, University of Michigan In third-semester (multivariable) calculus at the University of Michigan, a Computer Algebra System (Maple) is used for short in-class demonstrations pdf. Obviously to be able to implement this type of strategy means teachers need to determine where to start instruction , cited: Learn Business Math by GoLearningBus Learn Business Math by GoLearningBus. A discipline (a organized, formal field of study) such as mathematics tends to be defined by the types of problems it addresses, the methods it uses to address these problems, and the results it has achieved. One way to organize this set of information is to divide it into the following three categories (of course, they overlap each other): Mathematics as a human endeavor Problem Solving: Reasoning and read online However, what I have seen so far is not encouraging. It is downright scary when scientists, mathematicians, and university professors all seem to agree that the new reform "integrated math" does not do justice. My post is probably just touching the tip of the iceberg. I'd urge you to read the comments from various university professors on the following blog- Geometry Out Loud: Learning Mathematics Through Reading and Writing Activities
While ultimately I have to dish out judgement on their suggestions (this name doesn't make sense, that idea doesn't pan out for this reason, etc.), I make an honest effort to explain why and reiterate our goals, showing the discrepancy, and then requesting another suggestion. Unfortunately that keeps it as a bona-fide lecture, but if I had a week, the students would ideally critique each other's work Measurement and Statistics for Teachers Measurement and Statistics for Teachers. For example, it is better, to ask "Why does an airplane normally require a longer takeoff run at Denver than at New Orleans?" instead of, "Would you expect an airplane to require a longer takeoff run at Denver or at New Orleans?" Mathematics is a broad discipline with many diverse applications in physical sciences, life sciences, and engineering as well as social and managerial sciences. The Department of Mathematics provides a variety of concentrations leading to Baccalaureate, Masters, and PhD degrees. Our faculty-student ratio is high compared to many universities , source: Hands on History: A Resource for Teaching Mathematics (Notes) (Mathematical Association of America Notes)
Like video or a textbook, CBT is an aid to the instructor. The instructor must be actively involved with the students when using instructional aids ref.: World History Vol 2 1900-1968 read online balancestudios.net. As a result, teachers need strategies that will help them reach all of their students and ensure that students learn what they need to know Harcourt School Publishers read pdf With Long Division you'd be subtracting 74 from 85; neither of which are real numbers in the problem Answers to Times Tables Tests read epub Once testing is finished, test administrators must collect the math reference sheets and shred any that may contain student writing Right-Brained Addition & download epub balancestudios.net. If you'd like to contribute to the library and help this list grow we'd love to hear from you. Assessment provides educators with a better understanding of what students are learning and engages students more deeply in the process of learning content How To Work with Data & Probability, Grade 3 (Math) Please email resume or curriculum vitae, plus cover letter stating the specific areas you are interested in teaching to: Aurora University is an Equal Opportunity Employer and is committed to becoming a model university of the twenty-first century. Toward this effort we are determined to have a faculty, staff, and student body reflecting the cultural demographics of the greater Chicago area Operation Order! read online read online. Math is quite important in each of these areas! Thus, our math education system faces the ongoing challenge of helping to prepare students for life in a world where math is both an important discipline in its own right and is also an important part of our changing world. This challenge is becoming greater as the pace of technological change in our world continues to increase Horizons Mathematics, Grade 6: Student Workbook, Book 2 This is the same company that designs standardized tests for most New York students in grades three through eight. In 2010, Pearson inked a 5-year contract with the state of New York for 32 million dollars (US). Can you see the conflict of interest here? The same company that is being paid to design standardized tests for New York students is also being paid to design curriculum for New York students download.
Still, Luna said the $4 million would put Idaho on the road to helping teachers be better math instructors and put instruction about math concepts in the early elementary grades like Whitaker does at Taft , source: Mathematics Course 1: Additional Examples on Transparencies English for the English: a read epub English for the English: a Chapter on. Near the top of this list I found the article: Kennedy, M. The role of preservice teacher education. Retrieved 2/18/2016 from Kennedy's article is chapter 3 from the book: Darling-Hammond, L., & Sykes, G. (1999) pdf. If you need to teach it, we have it covered. Start your free trial to gain instant access to thousands of expertly curated worksheets, activities, and lessons created by educational publishers and teachers. Help your students understand the process of our national elections, from the President down to local representatives, with our election activities , e.g. Assessment, Learning and Judgement in Higher Education read for free. pdf? The study involved 34 matched pairs of students in grades 6 and 7 who either had been diagnosed with learning disabilities or were categorized as at risk for learning problems , e.g. Building Real-Life Math Skills: 16 Lessons With Reproducible Activity Sheets That Teach Measurement, Estimation, Data Analysis, Time, Money, and Other Practical Math Skills Building Real-Life Math Skills: 16. I knew how to do the problems, but I always got my answers with wrong sign epub. It is weak in mathematics as a human endeavor and as a discipline of study , source: Houghton Mifflin Mathmatics California: Math Center Resources Level 5 read pdf. Richard Askey explained the relevance of this book in an article by the same name in the Fall 1999 edition of American Educator. You'll be introduced to typical scenarios in the classroom, such as teaching subtraction and regrouping, multi-digit multiplication, division by fractions, and the relationship between area and perimeter , source: Educational Encounters: Nordic read for free Use both intentionally planned experiences and everyday activities as natural vehicles for developing children's mathematical knowledge. Provide a mathematically rich environment, which includes manipulatives, blocks, puzzles, number books and board games, and incorporate the language of mathematics throughout the day The Building Blocks of Early read epub sgsgreen.com. Young students are especially inquisitive and curious. Every teacher has an opportunity to help their students develop habits of mind that will last a lifetime. Cultivate these habits of mind in yourself, and learn to cultivate them in your students Math Practice Pages for the download for free download for free. This includes giving ourselves the flexibility of moving from one classroom to another. The key for us is having pre-tests that are specific enough that the information they cover can be taught and measured within three weeks (or fewer). Beyond that, we also create shared assessments on a weekly basis to determine what has been retained or needs to be revisited. With our first round of flexible grouping, the average test scores went from 40% on the pre-test to 95% on the post-test in a three week period , e.g. Mathematical Knowledge in Teaching: 50 (Mathematics Education Library) | 677.169 | 1 |
The following is a list of links to useful textbooks
in mathematics, available for free on the Internet. All books are
legally safe to download, The books are in printable format - Postscript (PS) or Portable Document Format
(PDF). You are free to download, read and print them. This is a partial
list of "free textbooks" in math, the list is to be updated regularly.
By definition, a circle is the locus of all points equidistant from a fixed center. However, the circle of Apollonius is defined differently. In the following post I will define and construct an Apollonian circle. | 677.169 | 1 |
Math to Build on
4.11 - 1251 ratings - Source
Geared toward individuals involved in construction, this guide offers a refresher course in basic math, providing formulas and exercises for determining measures, angles, and volumeFinding Arc Length Using a Fraction of a Circle Leta#39;s take a closer look at finding
arc length when you know the fraction of ... Leta#39;s make the radius of this arc 1.125
aquot;. c = 27tr C = 2x3.1416x1.125aquot; c = 7.0686aquot; Second: Multiply the fraction by theanbsp;...
Title
:
Math to Build on
Author
:
Johnny E. Hamilton, Margaret S. Hamilton
Publisher
:
- 1993-01-01
ISBN-13
:
Continue
You Must CONTINUE and create a free account to access unlimited downloads & streaming | 677.169 | 1 |
Macmillan CAS Technology
4.11 - 1251 ratings - Source
This practical How to... guide is excellent introductory CAS resources for students of VCE Further Mathematics. It contains worked examples specifically using CAS technology (CASIO Classpad and TI-Nspire CAS). Written by an experienced author who is a leader in the field, the teaching material is broken into small chapters so that the content is manageable for students (and teachers) new to the technology. At the end of each section there are questions for the students to complete using CASThis practical How to... guide is excellent introductory CAS resources for students of VCE Further Mathematics.
Title
:
Macmillan CAS Technology
Author
:
Peter Flynn
Publisher
:
- 2009
ISBN-13
:
Continue
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Thinking Mathematically unfolds the processes which lie at the heart of mathematics. It demonstrates how to encourage, develop, and foster the processes which seem to come naturally to mathematicians. In this way, a deep seated awareness of the nature of mathematical thinking can grow. The book is increasingly used to provide students at a tertiary level with some experience of mathematical thinking processes. | 677.169 | 1 |
Note: Citations are based on reference standards. However, formatting rules can vary widely between applications and fields of interest or study. The specific requirements or preferences of your reviewing publisher, classroom teacher, institution or organization should be applied.
Discrete mathematics stands among the leading disciplines of mathematics and theoretical computer science. This volume offers a blend of research and survey papers reflecting the topics in contemporary discrete mathematics. It also discusses topics such as graph theory, partially ordered sets, computational complexity issues and applications.Read more... | 677.169 | 1 |
Complex Numbers Complex
Section 13.1 &13.2 Concept tests
Setting up your cards
Press the GO button Press 2 then 7 (that is, you chose the channel 27 for the frequency); after you pressed 7, you should get a green light
(Hint. Use both hands to press the
Math 322, Sample Exam # 2
Partial credit is possible, but you must show all work.
Name: I hereby testify that this is individual work. Signed: 1 2 3 2
1. (a) Find the eigenvalues and eigenvectors of the matrix A =
(b) By starting with the denition A = , | 677.169 | 1 |
This practical guide eases you into the subject of probability using familiar items such as coins, cards, and dice. As you progress, you will master concepts such as addition and multiplication rules, odds and expectation, probability distributions, and more. You'll learn the relationship between probability and normal distribution, as well as how to use the recently developed Monte Carlo method of simulation. Detailed examples make it easy to understand the material, and end-of-chapter quizzes and a final exam help reinforce key ideas.
It's a no-brainer! You'll learn about:
Classical probability
Game theory
Actuarial science
Addition rules
Bayes' theorem
Odds and expectation
Binomial distribution
Simple enough for a beginner, but challenging enough for an advanced student, Probability Demystified, Second Edition, helps you master this essential subject.
"synopsis" may belong to another edition of this title.
About the Author:
Allan G. Bluman is professor emeritus at the Community College of Allegheny County, South Campus. He taught mathematics and statistics in high school, college, and graduate school for more than 35 years. Mr. Bluman is the recipient of "An Apple for the Teacher Award" for bringing excellence to the learning environment and the "Most Successful Revision of a Textbook" award from McGraw-Hill. He is the author of three mathematics textbooks and several highly successful books in the DeMYSTiFieD series, including Pre-Algrebra DeMYSTiFieD, Math Word Problems DeMYSTiFieD and Business Math DeMYSTiFieD. Professional, 2012. Book Condition: New. Brand New, Unread Copy in Perfect Condition. A+ Customer Service! Summary: The second edition of one the top DeMYSTiFieD bestsellers is updated with all-new quizzes and test questions, clearer explanations of the material, and a completely refreshed interior and exterior design. Bookseller Inventory # ABE_book_new_007178097180971
Book Description McGraw-Hill Professional884121
Book Description McGraw-Hill Education - Europe. Paperback. Book Condition: new. BRAND NEW, Probability Demystified (2nd Revised edition), Allan G. Bluman, "Probability Sidelight," "math-phobic" Provides chapter-opening objectives that describe what you'll learn in each step Includes questions at the end of every chapter to reinforce learning and pinpoint weaknesses Offers "Still Struggling?" elements throughout to help you with difficult subtopics Concludes with a final exam for overall self-assessment. Bookseller Inventory # B9780071780971 | 677.169 | 1 |
Precalculus Advice
Precalculus Documents
Showing 1 to 11 of 11
PSYCHOMETRICS
TOPIC 1: THE PROBLEM OF MEASUREMENT IN PSYCHOLOGY
introduction
Psychometrics is the branch of psychology that deals with issues related to the measurement of
psychological aspects. The psychodiagnostic and psychological evaluation are taken
CONSULTATION ON SETS
What is a set?
It is the group throughout a well differentiated in the mind or intuition objects, therefore, these
objects are clearly defined and differentiated.
It is the gathering, group or collection of distinct elements that have
FAMILY Bernoulli
It is a family of mathematicians from Antwerp late sixteenth century took up residence in
Switzerland. Effectively contributed to the spread of differential calculus and its influence lasted
until the eighteenth century ended.
Belonging t
Logic
Make an example of the following laws of involvement: Modus ponens, Modus tollens,
Hypothetical Syllogism.
2. List the steps to build the truth table.
1. calculate the number of rows that have the table. This will be equal to 2 raised to the number
GEORG FRIEDRICH BERNHARD RIEMANN
Historical evolution of Mathematics
Georg Riemann was born on September 17 in Breselenz (Germany) and died on July 20, 1866
in Selasca, Italy. He studied at the University of Gottingen, which was assistant professor (1854)
WOMEN OF SCIENCE HIGHLIGHTS OF S. XX
The American Fanya Montalvo, was once the only girl enrolled in the career of Physics at
Chicago. His specialty, Mathematical Psychology, which allowed him to study human behavior
from a mathematical basis: subsequentl
STATUS OF XVII CENTURY
The calculus was created to solve major scientific problems of the seventeenth century, for
example, to obtain lengths of curves, areas and volumes of geometric bodies, tangent to a
curve and maxima and minima of functions.
Many of
Simon Denis Poisson
Historical evolution of Mathematics
Simon Denis Poisson was born on June 21, 1781 in Pithiviers, France and died on April 25,
1840 in Sceaux (near Paris), France.
The most important work of Poisson was a series of writings of Definite
FUNCTION
A function is a correspondence between joint that occurs when each of the elements of the first
set is connected with one element of the second set. We are in the presence of a function when
each element of the first set only get a single arrow.
How we choose the items?
The choice of items is made by the author of the test, taking into account:
To be representative of the attribute that I want to measure, how it manifests that attribute?
direct observation, define what the demonstration, taking f | 677.169 | 1 |
Gems of Geometry
4.11 - 1251 ratings - Source
Based on a series of lectures for adult students, this lively and entertaining book proves that, far from being a dusty, dull subject, geometry is in fact full of beauty and fascination. The author's infectious enthusiasm is put to use in explaining many of the key concepts in the field, starting with the Golden Number and taking the reader on a geometrical journey via Shapes and Solids, through the Fourth Dimension, finishing up with Einstein's Theories of Relativity. Equally suitable as a gift for a youngster or as a nostalgic journey back into the world of mathematics for older readers, John Barnes' book is the perfect antidote for anyone whose maths lessons at school are a source of painful memories. Where once geometry was a source of confusion and frustration, Barnes brings enlightenment and entertainment. In this second edition, stimulated by recent lectures at Oxford, further material and extra illustrations have been added on many topics including Coloured Cubes, Chaos and Crystals.Equally ideal as an educational gift for a youngster or as a nostalgic journey back into the world of mathematics for older readers, John Barnesa#39; book brings enlightenment and entertainment. a€œ... unlike your average student textbook, this ...
Title
:
Gems of Geometry
Author
:
John Barnes
Publisher
:
Springer Science & Business Media - 2012-08-17
ISBN-13
:
Continue
You Must CONTINUE and create a free account to access unlimited downloads & streaming | 677.169 | 1 |
Math 4 Test 2 Study Guide
For the test on Friday, April 11, youre responsible for the sections we covered in chapters 2 and 3 of Lay,
plus section 4.1.
The key to preparing for the test is working lots of problems. Re-working homework problems would be
he
Math 4 Test 2 April 11, 2014
Name _
Put all your answers, together with your work, on separate paper.
You can use a calculator unless the problem says without using a calculator, in which case you need to
show the steps for working the problem manually. H
Math 4 Test 1 Study Guide
Updated 3/4/2014, 3/8/2014
Youre responsible for all the sections of Chapter 1 in Lay, except section1.6 and 1.10, which are on
applications of linear systems. The applications are important, of course, but since everyone chose w
Math 4 Syllabus Quiz
Name _
Short Answers
1.
What days does the class meet?
_
2.
What is the name of the textbook? _
3.
What is your instructors name?
4.
Where is your instructors office? _
5.
What day of the week will homework usually be collected? _
6.
Math 4 Final Exam Take-home Portion
Name _
Spring 2014 Jim Riley
This is due at the start of the timed final exam: Wednesday, May 28, 2014, at 11:30 a.m.
Exact answers only no decimal approximations. You can use technology for row-reducing matrices;
there
Math 4 Quiz 1 February 14, 2014
Name _
Due in class on Wednesday, February 19, 2014
This is a take-home quiz. You can use the textbook, your own notes, and any other materials (including
online ones) that are already published. Please dont consult other s | 677.169 | 1 |
Advanced Calculus for Applications
The text provides advanced undergraduates with the necessary background in advanced calculus topics, providing the foundation for partial differential equations and analysis. Readers of this text should be well-prepared to study from graduate-level texts and publications of similar level.
Book Description Paperback. Book Condition: New. 2nd. 151mm x 37mm x 233mm. Paperback. The text provides advanced undergraduates with the necessary background in advanced calculus topics, providing the foundation for partial differential equations and analysis. Readers of th.Shipping may be from multiple locations in the US or from the UK, depending on stock availability. 733 pages. 0.912. Bookseller Inventory # 9780130111890 | 677.169 | 1 |
Integrated content includes algebra, statistics, probability, trigonometry, discrete mathematics and data analysis. Integration, occurs within and across lessons and exercises at the point of instruction. Each chapter opens with a focus on the prerequisite skills that are needed for the chapter. Real-World Applications and Interdisciplinary Connections help to make the geometric concepts exciting and relevant. Author : Boyd ISBN : 0078228808 Language : English No of Pages : 910 Edition : 1 Publication Date : 5/26/2000 Format/Binding : Hardcover Book dimensions : 10.9x8.7x1.4 Book weight : 0.05 | 677.169 | 1 |
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First impressions of Key Maths 7 suggest a colourfully presented, yet relatively traditional textbook. Worked examples are displayed in yellow boxes and followed by exercises such as "copy and complete . . .", and "work through these questions". For teachers who prefer to work in a more formal manner, this may be a course worth looking at.
This first year of a new course for key stage 3 will be followed by Key Maths 8 later this year and Key Maths 9 in 1997. The pupils' books are aimed at different ability levels: Book 7 1 is for those entering key stage 3 at levels 2 or 3, Book 7 2 for levels 3 or 4. I am not convinced there is sufficient difference between them to cater for the full range of pupil attainment at this stage.
The content has been planned using a spiral curriculum. The material is definitely not presented by level, but careful thought has been given to construct a course in which ideas build on one another, and where difficult new ideas are introduced gradually. Although access to a scientific calculator is assumed throughout, students are instructed, for some exercises, not to use their calculators.
Each of the 16 chapters in the pupil's books is intended to be a coherent unit of work lasting about two weeks. Book 7 1 and Book 7 2 are designed to be used in parallel; they have the same chapter titles, and much of the material in common. The bulk of each chapter is core material, followed by "Questions" and "Extension" questions for homework. The Extensions in Book 7 1 are the same as the Questions in Book 7 2, providing a bridge between the two books.
Each chapter concludes with a clear summary page which provides a revision of the essential ideas and a "Test Yourself" page of questions, with answers at the back of the book. Each book ends with a Help Yourself section for pupils who find difficulties with number skills and need to practise them individually or at home with parents.
The Teacher's File is a very substantial ring-binder containing notes, photocopy masters for worksheets used in the chapters, and homework sheets. The Teacher's Notes are very full, with a page devoted to each section of each chapter. Homework is suggested for each section; the prepared homework sheets are simply photocopy masters of those questions, so that pupils do not have to take the text-books home. A helpful feature on each page of the Teacher's Notes, in view of the rate at which mathematics at key stage 3 has been changing over recent years, is a paragraph of notes for the non-specialist. Activities are suggested to supplement the textbooks.
The Question Bank contains more than 200 pages of assessment questions, with mark schemes and answers, designed as a resource from which teachers can devise their own tests. A bank of 20 mental tests is also planned, but not yet available.
Finally, at the back of the Teacher's File, I discovered a small set of worksheets and lesson ideas for introducing graphic calculators to Year 7 pupils. These are based on the new Texas TI-80 calculator, and they contain some good ideas, encouraging pupils to experiment and to work on open questions.
A BUYER'S GUIDE TO KEY STAGE 3 MATHEMATICS COURSES
Peter Wilder lists the important points to bear in mind:
* Is the pupil material attractive and engaging? Does it provide positive images for all pupils?
* Are the appropriate levels of the national curriculum covered, including Ma1 Using and Applying Mathematics?
* Does the course encourage variety? In the words of the Cockcroft Report: Are there opportunities for exposition, discussion, consolidation and practice, practical work, problem solving, investigation? Does it include individual, small group and whole class activities?
* What is the teacher's role? Is there adequate back-up for a variety of forms of assessment? Will the material enhance or restrict the role of the teacher?
* Is there provision for the least able? Consider the reading level required.
* Are there appropriate and challenging extensions for the most able? Does the material provide opportunities for work outside and beyond the national curriculum?
* Does the material encourage appropriate use of calculators and computers, as well as developing mental arithmetic and approximation strategies?
* Is there a range of applications and contexts giving a vision of the importance and power of mathematics for all pupils? | 677.169 | 1 |
The
'This book does a very good job of filling an important niche - giving mathematics students a thorough introduction to Mathematica without overwhelming them with the long manual.' Mathematics Teacher
' … a down-to-earth approach makes the book eminently readable and instructive...almost casually the reader learns how to make the program perform nearly any mathematical operation, from integration to finding eigenvalues or plotting intricate surfaces. A welcome book for anyone who needs to use Mathematica.' CHOICE
'Generally, the text is friendly and easy to follow...Overall, this is an approachable and helpful text that looks like it has been tested in the classroom.' Computing Reviews
'The book can be used as a supplementary text in calculus or algebra courses and also as a textbook for students who wish to learn Mathematica and use it to solve mathematical problems. The book is nicely written and is recommended to anyone studying mathematics.' European Mathematical Society
'The unique feature of this book is that it presents concepts in an order that closely follows a standard mathematics curriculum, rather than structure the book along features of the software … Mathematica commands are introduced as a means of solving problems and illuminating the underlying mathematical principles.' ZDM - The International Journal on Mathematics Education
'The book is written in a fashion that really every student should be able to follow it. The examples are neatly chosen, not too complicated, and Mathematica's capabilities are introduced deliberately … for the purpose and the level it is written for, the book is excellent.' Zentralblatt MATH
'… a book that most students can benefit from.' MAA Reviews
Book Description:
A new edition of this well loved book completely rewritten for Mathematica 6. This book can be used in a variety of courses, from precalculus to linear algebra. Used as a supplementary text it will aid in bridging the gap between the mathematics in the course and Mathematica. Cambridge University Press. Paperback. Condizione libro: new. BRAND NEW, The Student's Introduction to Mathematica: A Handbook for Precalculus, Calculus, and Linear Algebra (2nd Revised edition), Bruce F. Torrence, Eve A. Torrence, B9780521717892
Descrizione libro Paperback. Condizione libro: New. Not Signed; book. Codice libro della libreria ria9780521717892_rkm17892 | 677.169 | 1 |
Math in Sports (edX)
Come learn how you can use mathematics to get a deeper insight into both the sports you love and everyday life.
In this course you will learn to use some mathematical tools that can help predict and analyze sporting performances and outcomes. This course will help coaches, players, and enthusiasts to make educated decisions about strategy, training, and execution. We will discuss topics such as the myth of the Hot Hand and the curse of the Sports Illustrated cover; how understanding data can improve athletic performance; and how best to pick your Fantasy Football team. We will also see how elementary Calculus provides insight into the biomechanics of sports and how game theory can help improve an athlete's strategy on the field.
In this course you will learn:
- How a basic understanding of probability and statistics can be used to analyze sports and other real life situations.
- How to model physical systems, such as a golf swing or a high jump, using basic equations of motion.
- How to best pick your Fantasy Football, March Madness, and World Cup winners by using ranking theory to help you determine athletic and team performance.
By the end of the course, you will have a better understanding of math, how math is used in the sports we love, and in our everyday lives | 677.169 | 1 |
Archived Information
Mathematics Curriculum Framework - November 2000
Guiding Philosophy
This curriculum framework envisions all students in the Commonwealth achieving mathematical competence through a strong mathematics program that emphasizes problem solving, communicating, reasoning and proof, making connections, and using representations. Acquiring such competence depends in large part on a clear, comprehensive, coherent, and developmentally appropriate set of standards to guide curriculum expectations.
Problem Solving
Problem solving is both a means of developing students' knowledge of mathematics and a critical outcome of a good mathematics education. As such, it is an essential component of the curriculumvers, students need many opportunities to formulate questions, model problem situations in a variety of ways, generalize mathematical relationships, and solve problems in both mathematical and everyday contexts.
Communicating
The ability to express mathematical ideas coherently to different audiences is an important skill in a technological society. Students develop this skill and deepen their understanding of mathematics when they use accurate mathematical language to talk and write about what they are doing. They clarify mathematical ideas as they discuss them with peers, and reflect on strategies and solutions. By talking and writing about mathematics, students learn how to make convincing arguments and to represent mathematical ideas verbally, pictorially, and symbolically.
Reasoning and Proof
From the early grades on, students develop their reasoning skills by making and testing mathematical conjectures, drawing logical conclusions, and justifying their thinking in developmentally appropriate ways. As they advance through the grades, students' arguments become more sophisticated and they are able to construct formal proofs. By doing so, students learn what mathematical reasoning entails.
Making Connections
Mathematics is not a collection of separate strands or standards. Rather, it is an integrated field of study. Students develop a perspective of the mathematics field as an integrated whole by understanding connections within and outside of the discipline. It is important for teachers to demonstrate the significance and relevance of the subject by encouraging students to explore the connections that exist within mathematics, with other disciplines, and between mathematics and students' own experiences.
Representations
Mathematics involves using various types of representations for mathematical objects and actions, including numbers, shapes, operations, and relations. These representations can be numerals or diagrams, algebraic expressions or graphs, or matrices that model a method for solving a system of equations. Students must learn to use a repertoire of mathematical representations. When they can do so, they have a set of tools that significantly expands their capacity to think mathematically. | 677.169 | 1 |
A First Course in Mathematical Analysis
4.11 - 1251 ratings - Source
Mathematical Analysis (often called Advanced Calculus) is generally found by students to be one of their hardest courses in Mathematics. This text uses the so-called sequential approach to continuity, differentiability and integration to make it easier to understand the subject.Topics that are generally glossed over in the standard Calculus courses are given careful study here. For example, what exactly is a 'continuous' function? And how exactly can one give a careful definition of 'integral'? The latter question is often one of the mysterious points in a Calculus course - and it is quite difficult to give a rigorous treatment of integration! The text has a large number of diagrams and helpful margin notes; and uses many graded examples and exercises, often with complete solutions, to guide students through the tricky points. It is suitable for self-study or use in parallel with a standard university course on the subject.This text uses the so-called sequential approach to continuity, differentiability and integration to make it easier to understand the subject.Topics that are generally glossed over in the standard Calculus courses are given careful study ...
Title
:
A First Course in Mathematical Analysis
Author
:
David Alexander Brannan
Publisher
:
Cambridge University Press - 2006-08-17
ISBN-13
:
Continue
You Must CONTINUE and create a free account to access unlimited downloads & streaming | 677.169 | 1 |
Matching Trigonometric Functions, Graphs and Descriptions
PDF (Acrobat) Document File
Be sure that you have an application to open this file type before downloading and/or purchasing.
0.43 MB | 28 pages
PRODUCT DESCRIPTION
The cards are designed to be run off on card stock and cut into 3.5 inches by 5 inches cards. The cards have been randomly numbered and an answer key has been provided to make it easy to check student matchings.
Each set of four cards match a function, a graph, a description of the period, amplitude, and two points on the graph and some description of the function. You can make up a lesson that includes any number of four matched cards. It is not necessary to use all 25 functions.
After you decide which cards you want to use in the lesson, randomly handout the cards to the students. One student will have a function, another student will have a graph of the function, another will have the period, amplitude, and y-intercept of the function, and a fourth student will have a description of the function. The challenge for students is to find their matching cards.
The matching of the cards can be checked by using the following chart using the symbol on the bottom of the card | 677.169 | 1 |
Algebra Essentials Resource Bundle
PDF (Acrobat) Document File
Be sure that you have an application to open this file type before downloading and/or purchasing.
34.96 MB | 109 pages
PRODUCT DESCRIPTION
Do you find that your students struggle with the most basic Algebra concepts? Are you often re-teaching integer operations or exponent properties? Then this lesson bundle is for you!
This collection of resources is intended to supplement any early Algebra curriculum (whether Pre-Algebra or Algebra 1). Several important concepts are covered, providing lessons to teach in a way that makes sense so that students retain more.
This huge collection includes inquiry and problem based lessons to guide students to a conceptual understanding.
In addition, it includes fun and unique skills practice pages, such as challenge mazes and correcting the mistake.
And finally, it includes an extensive problem based project to continue to make sense of linear equations in a real world context.
Your students will love getting creative as they design their own amusement park! (If you're interested in the project only, it is available separately! Just search for "Build Your Own Amusement Park" in my shop!).
In addition, this resource is full of teaching tips so you can teach with confidence, as well as answer keys and a rubric to make it easy and low-prep to use!
*Please note: this is intended for personal use in the home or classroom only. To use in multiple classrooms, please purchase multiple licenses | 677.169 | 1 |
ECON 432 Documents
Showing 1 to 30 of 138
Microeconomics II
CIDE, MsC Economics
List of Problems
1. There are three people, Amy (A), Bart (B) and Chris (C): A and B have hats. These three
people are arranged in a room so that B can see everything that A does and C can see everything
that B does,
Some Model Answers for HW on Chapter 2
#12: For any integer n>2, show that there are at least two elements in U(n)
that satisfy x2=1.
Clearly 1 is one such element. The other is n-1. Every U(n)
contains n-1, since n-1 is always relatively prime to n. Furt
Notes on Cyclic Groups
09/13/06 Radford (revision of same dated 10/07/03) Z denotes the group of integers under addition.
Let G be a group and a G. We dene the power an for non-negative integers n inductively as follows: a0 = e and an = aan1 for n > 0. If
1
Groups: denition and examples
Algebra is based on arithmetics, and below we abstract from arithmetic its core ideas.
Section 1.2 abstracts from elementary algebra those properties that enable us to solve a
single linear equation in one variable, for exa | 677.169 | 1 |
Principles of Mathematical Modeling
4.11 - 1251 ratings - Source
The and social decision making. Prospective students should have already completed courses in elementary algebra, trigonometry, and first-year calculus and have some familiarity with differential equations and basic physics.-This book provides a readable and informative introduction to the development and application of mathematical models in science and engineering.
Title
:
Principles of Mathematical Modeling
Author
:
Clive L. Dym
Publisher
:
Academic Press - 2004
ISBN-13
:
Continue
You Must CONTINUE and create a free account to access unlimited downloads & streaming | 677.169 | 1 |
Analyze this:
Our Intro to Psych Course is only $329.
*Based on an average of 32 semester credits per year per student. Source
Tutorial
Introduction
An important topic in College Algebra and Precalculus is rate of change. We have lots of language for various ways of thinking about rate of change, including:
average rate of change
slope
difference quotient
Graphing calculators are well set up for analyzing values of functions, but they are not so well set up for analyzing rates of change. In particular, an important tool for this is the idea of finite differences. This packet shows how to use a different tool-a computer spreadsheet-to analyze finite differences (in particular first and second differences).
The set up: Making a table for a function
Whether you use Microsoft Excel (and you probably do) or some other spreadsheet (such as Apple's Numbers or OpenOffice or Google Docs, etc.), the basic functions are all the same. Except for the precise appearance of the tools, I doubt there is any meaningful difference between what you'll see in these videos and what your computer will look like. | 677.169 | 1 |
Chapter 10 Equations and Facts
Equation of a Sphere
r^2=(x-a)^2+(y-b)^2+(z-c)^2
Vectors have both _ and _
magnitude and direction
A+B (both vectors) =
(A1+B1)i + (A2 +B2)j + (A3+B3)k
A Unit Vector has a magnitude of _
one
The three standard unit vectors a
Your analysis, based on the concepts covered in this course, will address each of the following:
1. Business Strategy Analysis: Develop an understanding of the business and
competitive strategies of the company. Which of the three generic competitive
stra: REAL ANALYSIS I
MIDTERM, OCTOBER 9, 2013
I. Theory
1. Let (X, d) be a metric space.
(a) What is the Borel -algebra, BX ?
The Borel -algebra is the smallest -algebra that contains all the open sets in X .
(b) Give three dierent families of
MATHEMATICS 580
HOMEWORK 1 SOLUTIONS
DUE SEPTEMBER 4
1. We will show that the sequence cfw_xn is Cauchy, hence convergent since X is
complete. Since the series converges, given > 0 we can nd N such that m
n N implies m n d(xk , xk+1 ) < . By the triangl
HOMEWORK 5
SOLUTIONS
1. The point of this exercise is to show that the measurability assumption of the
function f (x, y ) on the product space is important. Consider the measure spaces
(X, M, ) = (Y, N , ) = ([0, 1], L, m).
Notice both of
Math 238 Project
Due by April 26, 2013
Note: You will get credit for the project provided it appears that you made a good faith eort to
answer the questions below. You should fully work one of the two problems, but try to do both!
You can document your wo
Review for Final
Note: The only topic which might be on the exam, but is not on this review, is direction elds.
In particular, I decided not to put any questions about systems on the exam. The review over
represents section 7.6 compared to the nal, but th
Homework #9 Solution
3.71 A warehouse contains ten printing machines, four of which are defective. A company randomly selects
ve of the machines for purchase. What is the probability that all ve of the machines are not defective?
ans: There are four machi | 677.169 | 1 |
The journal covers a wide spectrum of subjects, including statistics and various topics in mathematical sciences such as mathematical analyses, number theory, probability, algebra, geometry and topology, function and number theory, differential equations and linear operators. | 677.169 | 1 |
Differential Equations with the TI-86
Students explore how to graph differential equations using a TI calculator. In this math activity, students solve systems of equations using the calculator. Students explore graphing and use it to interpret experimental | 677.169 | 1 |
Junior Cycle - Mathematics
Subject Group: Science
These subjects demonstrate how to explore nature using carefully planned methods, and teach the basic methods and findings of scientific investigation.
Brief Description:
In Mathematics, you will develop your problem-solving skills and your ability to present logical arguments. You will be better able to use what you learned in Mathematics in real life situations in everyday life and work.
How will Mathematics be useful to me?
Studying mathematics prepares you for business calculations, for handling your money sensibly and for courses in sciences, engineering and technology. You should see mathematics as an opportunity to strengthen your thinking skills.
Note: The Revised Junior Certificate Mathematics Syllabus for Examination in 2015 and in 2016 are available through the 'Course Outline' link below. | 677.169 | 1 |
The use of softwares as teaching tools has been gradually introduced in schools. The objective of this study was to investigate wheter high school students identify, out of Cartesian plane, graphics of equations and functions in the geometric shapes of know trademark logos, and understand the meaning and the graphical representation of an inequality, with one proposed activity using the GrafEq software. Notes and photographs were used to record the activity, in addition to the application of a questionnaire to the students to also assess the activity. The responses to the questionnaire as well as the development of the activity by the students, were used to proposed a reflection on what is a problem, and how to create learning environments through Mathematical Modeling and Problem Solving | 677.169 | 1 |
Discrete Mathematics in the Schools
4.11 - 1251 ratings - Source
This book provides teachers of all levels with a great deal of valuable material to help them introduce discrete mathematics into their classrooms.For example, the problem of finding the shortest route through key cities between
Boston and Miami can be posed as early as ... For example, Mrs. Smith wants to
determine the optimal time to trade in her Ford Taurus. She has consulted with
the Ford dealer and determined cost and trade-in projections as indicated in the
following table: Taurus LE year price new trade-in price 1996 1997 1998 1999
2000anbsp;...
Title
:
Discrete Mathematics in the Schools
Author
:
Joseph G. Rosenstein
Publisher
:
American Mathematical Soc. -
ISBN-13
:
Continue
You Must CONTINUE and create a free account to access unlimited downloads & streaming | 677.169 | 1 |
Plane Trigonometry and Complex Numbers
Consider a right-angled triangle ABC as shown in the figure below. You will need to use your graphing calculator. University of Arizona Published in 2002, 62 pages Published in 1919, 356 pages Michael Barr, Charles Wells That reduces the number of factors of sines and cosines by 1. The first four programs in this mathematics education software series are a continuation of the Algebra by Chapter Series and is appropriate for Intermediate Algebra and Pre Calculus students.
Pages: 402
Publisher: vector small press (November 30, 2010)
ISBN: 0982831404
Practical Geodesy: Comprising Chain Surveying, and the Use of Surveying Instruments Levelling, and Tracing of Contours Together With Trigonometrical Colonial Mining and Maritime Surveying
Spherical trigonometry: for the use of colleges and schools. With numerous examples.
Use of Trigonometric Table (#1985, Edition 1)
Elements of Plane and Spherical Trigonometry with Logarithmic and Other Mathematical Tables and Examples of Their Use and Hints On the Art of Computation
Note that atan2(0, 0) is not well-defined. The cofunctions of the hyperbolic sine, cosine, and tangent (cosech/csch and cotanh/coth are aliases) The area (also known as the inverse) functions of the hyperbolic sine, cosine, and tangent The area cofunctions of the hyperbolic sine, cosine, and tangent (acsch/acosech and acoth/acotanh are aliases) The trigonometric constant pi and some of handy multiples of it are also defined. cannot be computed for all arguments because that would mean dividing by zero or taking logarithm of zero By Karl J. Smith - Student Solutions Manual for Smith's Essentials of Trigonometry, 4th: 4th (fourth) Edition. We can compare this to the fact that half way around the circle is 180° and get π radians = 180° Peason Custom Mathematics MA 105-Intor to Trigonometry (Custom Edition for Montgomery College Germantown). Jeremy Rouse teaches his Multivariate Calculus class outdoors on Manchester Plaza. Jason Parsley teaches his differential geometry class. Raynor talks with students at the MAA meeting. Jason Parsely teaches his freshman seminar on the mathematics of elections, Professor Jason Parsley spent the summer working with a student to model voting patterns Functions, Statistics, & Trigonometry 2nd EDITION. The student applies the process standards in mathematics to generate new understandings by extending existing knowledge. The student generates new mathematical understandings through problems involving numerical data that arise in everyday life, society, and the workplace Plane trigonometry, (Prentice-Hall mathematics series; A. A. Bennett, editor). The WV Department of Education claims no affiliation or association with, nor endorsement or sponsorship by, any of these sites or their operators. What is the Millennium Mathematics Project? The Millennium Mathematics Project (MMP) is a maths education and outreach initiative for ages 3 to 19 and the general public download Plane Trigonometry and Complex Numbers pdf. These trigonometric or Fourier series have found numerous applications in almost every branch of science, from optics and acoustics to radio transmission and earthquake analysis. Their extension to nonperiodic functions played a key role in the development of quantum mechanics in the early years of the 20th century EBENE UND SPHARISCHE TRIGONOMETRIE. The student uses mathematical processes to acquire and demonstrate mathematical understanding Plane Trigonometry and Complex Numbers online.
These Logarithmic, Trigonometric, and Other Mathematical Tables. Students will learn to work with various types of functions in symbolic, graphical, numerical and verbal form." The University of California at Irvine awards the certificate of completion. 10-12 (graph above: absolute value of the gamma function in the complex plane, source: Wikimedia Commons) Here is an unordered list of online mathematics books, textbooks, monographs, lecture notes, and other mathematics related documents freely available on the web A Graphical Approach to Algebra and Trigonometry (6th Edition). Round to the nearest hundredth of a degree. Recall that the tangent of an angle is equal to the ratio of the opposite side to the adjacent side of the triangle Elements of Geometry: Containing the First six Books of Euclid, With a Supplement on the Quadrature of the Circle, and the Geometry of Solids ; to ... Elements of Plane and Spherical Trigonometry. Ultimate Triangle Calculator -- Solve for angles and sides of a triangle given three pieces of information (e.g., 3 sides, side-angle-side, etc.) We will assume knowledge of the following well-known, basic indefinite integral formulas: Most of the following problems are average Practical Mathematics.
In addition, we may use Responses for the purposes set forth under the "Use of Responses: License Grant" section below Tb Alg Trig Analy Geo 12e. If cos beta / cos beta = n and sin alpha / sin beta = m, then show that (m2-n2) sin2 beta = 1-n2 i want mcq questions regarding ch.trignometric ratios and identities of class10th. a round balloon of radius r subtends an angle alpha at the eye of the observer while the angle of elevation of its centre is beta. prove that the height of the centre of the balloon is alpha sin beta cosec alpha/2 Introduction to geometry (Modules in technical mathematics). Technology Masters (Merrill Algebra 2 with Trigonometry). To them the scientific method is a myth that is used by scientists as they actually proceed through other means to achieve any consensus. Their works invariably show that scientific results were the result of politics and personalities and not based upon higher fundaments. However, it is no great trick to prove a proposition when that proposition happens to be your primary assumption! Trigonometry Flipper! For these angles we've to make fractions for which we've to write 0, 1, 2, 3 and 4 in the numerators and write 4 in the denominator of each fraction. After that take the square root of each of these fractions and there you are. Refer to the table below for better understanding Catalogue of stars for the epoch Jan. 1, 1892 from observations by the Great trigonometrical survey. Spherical Trigonometry for Dummies. We are a partner in the NCETM consortium, with responsibility for the secondary mathematics strand and also manage the government-funded Further Mathematics Support Programme Logarithmic tables of numbers and trigonometrical functions. In practice this process takes longer and is more likely to lead to inaccuracies than finding the cosine direct from a table. Consequently separate tables for cosines are included at the end of this book. There is one difference between the sine and cosine tables which you need to remember when you are using them. We saw in section 50, that as angles in the first quadrant increase, sines increase but cosines decrease Plane Trigonometry, 1940, 336 pages.. | 677.169 | 1 |
Showing 1 to 1 of 1
I recommend this course because it sort of connects Elementary Linear Algebra to previous calculus courses.
Course highlights:
I loved seeing how this course connected with other things I was learning in my physics class at the time. I learned how nice it is to work with vectors to find out so much information about a certain graph. It can be applied to a wide range of disciplines including engineering.
Hours per week:
6-8 hours
Advice for students:
Spend more time studying the material, and do more practice problems. Practice is key. If you are having trouble with understanding the material, don't hesitate to go visit the professor during his office hours. | 677.169 | 1 |
NAVIGATION
Algebra
Announcements
Welcome to the world of exponents and polynomials!
Since September we have reviewed the basics of algebra: equations, factoring, distributing, combining like terms, reviewing the different mathematical properties, writing expressions, additive inverses, integers and the number line, formulas, percents, and proportions, etc! We have mastered all these skills and then put these skills to work as we dove into the world of inequalities. Here we learned that instead of having one exact solution to a problem; we could have a solution set that would make a mathematical statement true. That unique sign allowed us to narrow down the possible solutions to our mathematical sentences and most importantly figure out what the solutions could not be! Then using that knowledge, we explored conjunctions, disjunctions, and absolute value inequality sentences.
Now that we have mastered those terms, we are ready to expand our knowledge about exponents and explore the world of polynomials! Here we will use the advanced skills on how to deal with exponents, use scientific notation, etc. Then we will be able to identify monomials, binomials, and polynomials which will lead us into using them in complex algebraic expressions and equations. So ready or not, here we come! :) | 677.169 | 1 |
Description
The Calculus I Flashcard has seven chapters containing 110 flashcards. It covers the first-semester of calculus including limits, differentiation, application of differentiation, continuity, and integration. Below is the chapter summary.----Calculus…..
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Does calculus have you dazed and confused? Calculus FTW Free has detailed step-by-step solutions to example problems from first year calculus! NOTE: This is the FREE version of Calculus FTW. It*** IMPORTANT NOTICE ***This is NOT an advanced mathematical solver or anything like that. Only a little mental arithmetics game called Calculus! Please read the description and look at the icon… [more..] | 677.169 | 1 |
Reviews for Modeling of Quantitative Basis 1 functions polynomial functions exponential functions logarithmic functions trigonometric functions review numerical exercises 2 Derivatives Derivatives of simple functions X 1 Knan1 sinx cosx cosx sinx ex 3 eX ax axlna lnx lx Derivative of compound functions dy 0 du dx Example ysinx2 ux2 then dy dy du 2 7ioi s1nu 39 u 39cos x 2x dx du dx 3 Integration the inverse of derivative Here are a few examples I kdx kx C n1 wax x C n l Ie dx 6x C Icosxdx sinx C Isinxdx cosx C Definitive Integration NewtonLeibniz Formula Ifxdx Fb Fa where fxF x 4 Regression j a bx Sum ofsquares SST SSR SSE R2 Slope intercept Positive correlationnegative correlation II Concepts 1 Ecosystems 2 Components of System Models a reservoirs stocks b processes ows c converters parameters 111 N1 3095 H000 0 d interrelationships 3 feedback loops a positive feedbacks b negative feedbacks 4 System Behavior Patterns Linear Exponential Logistic overshotcrash a Graphic view b System Diagrams c Difference Equations d Impact of model parameters on model behaviors Model calibration Validation and Sensitivity Analysis Three radiation laws a Planck s Equation b StefenBoltzmann s Law c Wien s Law 9939 gt1 Leaf Area Index Ecosystem Productions and Respirations a 9 NPP NEP Ra Rh 9519 0 D ID ID ID ID ID ID I ONUIAUJNt O39 Evapotranspiration Soil water potential Permanent Wilting Point Field Capacity and Soil Porosity Mineralization and immobilization Nitrogen cycle carbon cycle water cycle Greenhouse gases Missing carbon sink Short Questions How do you understand that Earth is a dynamic and evolving system What is the smallest ecosystem you observed in daily life and how life is sustained within the ecosystem What is the solar elevation angle at Chapel Hill latitude36 at noon time on March 21 How does stoma control the loss of water from plants How is transpiration different from evaporation How are soil porosity field capacity and volumetric soil moisture related How relatively humidity vapor pressure saturated vapor pressure vapor pressure deficit are related What are the sources of water for precipitation on land Draw the nitrogen cycling loop for the terrestrial ecosystem Please draw the global watercarbon cycle loop no numbers are needed Essay Questions 1 2 9399 Discuss all possible feedbacks both positive and negative of deforestation on global warming If you would implement the Beer s law as we discussed in class for the entire global what do you need What are the challenges you will face What kind of errors do you expect Discuss how water move in the soil to roots and through plants into the atmosphere How the water potential gradient is created to maintain the ow of water Discuss the factors and the mechanisms that in uence plant photosynthesis Discuss how deforestation can in uence the aquatic animals in the streams that ow out from the area Based on Henry s law can you predict D0 will increase or decrease in a lake for different seasons If yes explain the prediction How does global water cycle in uences global climate Derivatives and Integration 1 Derivatives Example 1 Traveling from Chapel Hill to Raleigh DistanceD30 miles leaving Chapel Hill at 800am arriving Raleigh City Hall at 845am Therefore the speed of your travel is vDt 30miles075hr40 mileshr This actually the average speed you may travel at a different speed at any particular moment If one asks what your speed is at 815am we may have to figure that out in the following way We know Dvt if we know the distance traveled from Chapel Hill at 815am D8 15am and the distance your traveled two minute later ie D815am Then the distance you traveled from 8 15am to 817am is D815amD815am2min then the speed at 815am can be estimated as D815amD817ammile 2 min The actual speed at 8 15 may still be different from the above estimation but it is a better estimation than 40 mileshr In fact the shorter the time you allow your car to travel after 815am the more accurate the speed you calculate Let t stand for time and At for the time allowed for travel the speed at 815am can be written as V N Dt At Dt t78153m At Example 2 How many of you have watched the last launch of space shuttle Endeavour on August 8 2007 How fast the shuttle is traveling at the time it is just off the launch pat How fast the shuttle is traveling at the just before it reached orbit 18000 mileshour In order to get rid of the gravitation of Earth an object has to travel at an accelerating speed of 79kms2 If you do a plot of time and distance the shuttle is traveling it would look like miles seconds The last Endeavour launch took place at 636pm on August 8 If I ask how fast the shuttle is traveling at 637pm how would to gure it out the speed The shorter the At is the more accurate the speed Mathematically V 1 DtAt Dt 2815 A1213 At In general If a function yfX exists at X0 when X increased AX at X0 ie X X0AX the function has a corresponding increase AyfX0AXfxo if the limit of the ratio of Ay to AX eXists when AX9 0 the limit is called the derivative of yfX at XX0 y39 f 39x 1 gzl A13le A1301 Ax Examples yfXC Cconstant This means regardless of what X value is y is always X Thus fXC fXAXC 1 Q1 C C0 AEEIBAx A1301 Ax y Thus the derivative of any constant is zero yxx fXX fXAXXAX y 1 xAx x1 g1 A1301 Ax A1301 Ax A1301 Ax Hm fXX2 fXAXxAX2X22XAXAX2 y 1 Q1 x2 2xAxAx2 x2 1 2xAxAx2 2x A1301 Ax A1301 Ax A1301 Ax In general Kn anl For convenience we can gure out the derivatives for the commonly used functions and put them in a table for later use so that we don t have to do this again and again Here they are C 0 X l Xquot nX 391 sinX c0sX tagc lc0s2X ctagX lsinzx e e ax axlna lnX UK The Geometric Meaning of the Derivatives y x A f 90 Ay 7 ta Ax gap When Ax O the angle p9 0L therefore the derivatives of yfx at x f c is the slope of the tangent line passing x y The functions we provided with derivative are very simple functions We often work with 2 2x more complex functions that are made from the Simple ones for example s1nx e etc we call these functions compound functions as they are functions containing fuctions Where sinxz can be written as sinu where ux2 Similarly e2x can be written as eu where u2x Here are the rules for taking derivatives for compound functions If yfgx is the compound function of yfu and ugx if the derivatives for ugx exists at x and yfu exists at ugx then the derivative of the compound function yfgx with respective to x is dy dy du f f gtxlt a dx du dx or X 11 g X Examples 1 ysinX2 and ysinu uX2 y sinu Xu sinxz XX2 2xcosxz 2 ysin2x Let usinX ysin2Xuza wan dx du dx u2 39 2u2sinx d sinx39 cosx dx 0 2sinx cosx dx du dx 3 yeZXsinx Let u a z du z dx 4 ye2XCOSZX Integration The inverse of derivatives Let me ask the inverse question in Example 1 of derivatives if I travel at 40 mph on I40 east where am I in 45 minutes how far away I am from Chapel Hill We know we traveled 30 miles in 45 min at that speed is that suf cient to know where we are What else do we need to know In derivative we can write dSdt40 The inverse of that is intergration ie j dS 40 j dt S40tC Where C is a constant determined by the initial condition eg X2 2X in fact XZC 2X where C is a constant IZxdx x2 C Similarly we can create a table of integration 1 jkdx kx C n1 2 Ixquotdx x C where nil n l 3 Ie dx e C 4 Icosxdx sinx C 5 Isinxdx cosx C De nite Integration Given a function fX which is bounded on a b Randomly insert n points within a b so that aX0ltX1lt ltXnb separate the interval ab into n smaller intervals X0X1 X1 X2 Kn1 Xn The lengths of the invervals are respectively AX1X1X0 AX2XzX1 AX XnXn1 Take any number s from any interval above calculate the product f8iAXi and sum the product S i f8 Ax Let 7 be the maximum length of the n intervals if X90 regardless of how a b is separated and how si is taken from the interval Xi1 Xi S is always approach a nite limit The limit is the de nite integration of fX in the interval a b Newton Leibniz Formula fxdx Fb Fa where fXF X 17 I f xdx is the area under the curve from a to b Examples 7r2 1 Isinxdx 0 1 2 Iaquot dx 0 3 szdx | 677.169 | 1 |
Students will explore the concept of functions as relationships, as machines, as equations, and as graphs.
Materials:
At minimum students should have worksheets, graph paper, and pencil. This lesson can be enriched with a computer spreadsheet or graphing calculator.
Introduction:
Have students come up to the front of the class. Choose males and females to pair up. Have one male choose to be with two females and watch the reaction of the class.
Development:
By using students, you are modeling ordered pairs. To find out if the ordered pairs are a function, you must link a male to a female. If the male is "cheating" on his woman...this situation is not functioning properly in this relationship. Hence the pair cannot function. Not a function.
Practice:
Guided Practice: Using an example 1: Determine whether each relation is a function. Explain. This mapping diagram represents a function since, for each element of the domain; there is only one corresponding element in the range. It does not matter if two elements of the domain are paired with same element in the range. Next this table represents a relation that is not a function. The element 2 in the domain is paired with both 5 and 4 in the range. If you are given that x is 2, you cannot determine the value of y.
Using Example 2: You can use the vertical line test to see if a graph represents a function. If no vertical line can be drawn so that it intersects the graph more than once, then the graph is a function. If a vertical line can be drawn so that it intersects the graph at two or more points, the relation is not a function.
Accommodations:
Have students who are struggling with both concepts to focus on the concept they can do.
Checking For Understanding:
Using an "Exit Ticket", have students take a 1-3 question quiz to exit the class.
Closure:
Review several questions from the independent practice assignment.
Evaluation:
Lesson was a success. Students claimed that the lesson was easy to follow.
Teacher Reflections:
The kids loved the Focusing event. The conversations about fidelity are still going on! | 677.169 | 1 |
A unified approach to numerical modelling which integrates aspects of continuum mechanics, differential equations, and numerical analysis, is presented in this book. The text is an explanation how to formulate a mathematical description of the phenomena under consideration, devise techniques for solving the governing equations, then refine the model and interpret the results. Emphasis is put on physical applications, and the three major classes of partial differential equations - elliptic, parabolic and hyperbolic - are related to steady-state systems, dissipative systems, and nondissipative systems, respectively. Some higher-order equations, nonlinear equations, and coupled systems of equations are also | 677.169 | 1 |
Dad wrote the book
In Riverview, Michigan an engineer father started out helping his sons with their math, then rewrote their textbook and finally wrote a series of math textbook that are being used by schools and parents. The Detroit News reports Nicholas Aggor's books are catching on:
Riverview Community School District teachers liked the books so much they started using them in classes last year, and district officials in June made it part of the curriculum for elementary and middle school starting in fall.
. . . Interest in the series has multiplied exponentially among Metro Detroit districts. Administrators in Wyandotte, Taylor and other districts asked teachers to review the series. In nearby River Rouge, curriculum director Paula Daniels said she wants copies by September for parents as a tutoring guide.
The "MathMaster Series" hasn't yet been published: Aggor copies the books himself and hands them out for free.
Teachers and parents like the books' step-by-step instructions, the blend of basics and concepts, easy-to-understand examples and the close match with the state's content standards.
Aggor, an immigrant from Ghana, wored as an automotive engineer before devoting himself to the books. His wife as a school principal in Ghana.
Comments
I'm a HS math teacher and would very much like to get a copy of his textbook. I'm always looking for new ways to teach Alegbra and Geometry and hope this will help me get those other students who struggle with their math.
People are still designing new bicyles too,(a 100 year old design). It is a way to make money and a lot of people think the texts can be better than they are. On the other hand: Translations of Euclid's Elements were still in use early in the 20th century.
Bicycles are different: there are always new materials and new technology to include. Grammar school math, on the other hand, is basically unchanged since Euclid. All we need is for one team to write the definitive book and we're done. Forever.
My intent was to sarcastically point out that there is obviously a reason that we're still writing those textbooks. A) There is good money to be made in textbooks and B) There are people who spend their careers coming up with new and different ways to teach things – not necessarily better ways, but new and different.
I don't mind people trying to come up with new and different ways to teach things, but I'd like a rule in place that you can't switch to a new plan until it has been proven better than the old one in a peer-reviewed study. | 677.169 | 1 |
Use this savings calculator to see how a consistent approach to investing can make your money grow. Whether saving for a house, a car, or other special purchase, the savings calculator will help you d... More: lessons, discussions, ratings, reviews,...
The rule says that to find the number of years required to double your money at a given interest rate, you just divide the interest rate into 72. For example, if you want to know how long it will take... More: lessons, discussions, ratings, reviews,...
The "Exponential Decay" Java applet was used on the discussion board with an online Algebra II course. Students were asked to use the applet, and then go to the discussion board and discuss the questi... More: lessons, discussions, ratings, reviews,...
Explore examples of exponential growth and decay, bounded growth, and carrying capacity in this simulation of the effects of birth and death rates and food supply on a deer population. Includes questi... More: lessons, discussions, ratings, reviews,...
Americans today owe more money than ever before. The fact that 'interest never sleeps' means that the situation will continue to worsen unless steps are taken at the individual level to reduce or elim... More: lessons, discussions, ratings, reviews,...
This set of problems consists of a handful of randomly selected graphs for which the user is to find a Hamiltonian circuit or trail (i.e., a walk in the graph that uses every vertex exactly once) if iThis has a full explanation of how to graph a hyperbola from a given equation. If you scroll down to the middle of the page, there is a flash movie that takes you through the graphing process step by... More: lessons, discussions, ratings, reviews,...
Flash introduction to finding the equation of an hyperbola centered on (0,0) and with its major axis on the x-axis. With step-by-step instructions and an illustrated glossary, students can learn how t... More: lessons, discussions, ratings, reviews,...
Flash introduction to finding the equation of an hyperbola centered on (0,0) and with its major axis on the y-axis. With step-by-step instruction and an illustrated glossary, students can learn how to | 677.169 | 1 |
Volume by Cylindrical Shells - Classwork
( x, y)
There is another way to determine the volume of a curve that is rotated about an axis - the method of cylindrical
shells. Let us take some function y = f ( x) and rotate it about the y-axis as shown above.
AP* Calculus Free-response Question Type Analysis and Notes
Revised to include the 2013 Exam
By Lin McMullin
General note: AP Questions often test several diverse ideas or concepts in the same
question. The type names are meant only as a guide and may ref
AP Calculus BC Exam
2013
SECTION I: Multiple Choice
DO NOT OPEN THIS BOOKLET UNTIL YOU ARE TOLD TO DO SO.
At a Glance
Total Time
1 hour, 45 minutes
Number of Questions
45
Percent of Total Score
50%
Writing Instrument
Pencil required
Part A
Instructions
S
&#'()*+,-.,(/#,0"%.#,1,2"%+3)*4
In geometry and physics, concepts such as temperature, mass, time, length, area, and volume can be >uantified with
a single real number. These are called +'%"%*,56%.(-(-#+ and the real number associated with it is called a
WHAT TO EXPECT ON THE CALCULUS BC FREE RESPONSE QUESTIONS FOR MAY 8TH, 2013
1 Infinite Series (Power, Taylor, Maclaurin)
Taylor: f a f ' a x a
f ' a
f ' a
f iv a
2
3
4
x a
x a
x a .
2!
3!
4!
When a=0, it is called a Maclaurin series.
Memorize t
YOUR ECONOMETRICS PAPER
BASIC TIPS
There are a couple of websites that you can browse to give you some ideas for topics and
data. Think about what you want to do with this paper. Econometrics is a great tool to
market when looking for jobs. A well-written
EC3313: Industrial Economics
Maris Goldmanis
Assessed Autumn Coursework
Due: Noon on Tuesday, January 19, 2015
Instructions to candidates:
This is a take-home test. It is due in the Department Office at noon on Tuesday,
January 19, 2015.
This test is as
Section 4.3
Difference Equations
to
Differential Equations
The Fundamental Theorem
of Calculus
We are now ready to make the long-promised connection between dierentiation and integration, between areas and tangent lines. We will look at two closely relate
Sections 4.9: Antiderivatives
For the last few sections, we have been developing the notion of the
derivative and determining rules which allow us to dierentiate all different types of functions. In this section, and for the remainder of the
semester, we
Sections 5.1: Areas and Distances
In this section we shall consider problems closely related to the problems we considered at the beginning of the semester (the tangent and
velocity problems). Specically, we shall consider the problem of nding the area of
Sections 5.2: The Denite Integral
In this section we shall formalize the ideas from the last section to
functions in general.
1. The Definite Integral
We start with a formal denition.
Denition 1.1. Suppose f (x) is a continuous function on the interval
[a | 677.169 | 1 |
The first assignment 8
. There are also a set of ten online assignments which are usually collected weekly. Often-times when a problem can be reduced to one of linear algebra it is "solved". rather than abstract mathematics. The main idea of the course is to emphasize the concepts of vector spaces and linear transformations as mathematical structures that can be used to model the world around us.Preface
These linear algebra lecture notes are designed to be presented as twenty five. fifty minute lectures suitable for sophomores likely to use the material for applications but still requiring a solid foundation in this fundamental branch of mathematics. Watch an introductory video below:
Introductory Video
The notes are designed to be used in conjunction with a set of online homework exercises which help the students read the lecture notes and learn basic linear algebra skills. In practical terms. Instead. students learn explicit skills such as Gaussian elimination and diagonalization in order that vectors and linear transformations become calculational tools. the course aims to produce students who can perform computations with large linear systems while at the same time understand the concepts behind these techniques. There are relatively few worked examples or illustrations in these notes. "algebra" and "applications"). Once "persuaded" of this truth. this material is instead covered by a series of "linear algebra how-to videos". Interspersed among the lecture notes are links to simple online problems that test whether students are actively reading the notes. The "scripts" They can be viewed by clicking on the take one icon for these movies are found at the end of the notes if students prefer to read this material in a traditional format and can be easily reached via the script icon . These notes do not devote much space to applications (there are already a plethora of textbooks with titles involving some permutation of the words "linear". In addition there are two sets of sample midterm problems with solutions as well as a sample final exam. . they attempt to explain the fundamental concepts carefully enough that students will realize for their own selves when the particular application they encounter in future studies is ripe for a solution via linear algebra.
They range from simple tests of understanding of the material in the lectures to more difficult problems. It can efficiently check whether a student has answered an explicit. We have found that there tend to be relatively few questions from students in office hours about the Webwork assignments.ucdavis. and ensure that basic computational skills are mastered.edu/index. online homework system which originated at the University of Rochester. logical quantifiers and basic methods of proof).html and the notes are also hyperlinked to Wikipedia where students can rapidly access further details and background material for many of the concepts. Most students rapidly realize that it is best to print out the Webwork assignments and solve them on paper before entering the answers online.edu/webwork2/MAT22A-WaldronWinter-2012/ Webwork is an open source. problem correctly.org/?video#Linear Algebra
9
.math. all of them require thinking. Videos of linear algebra lectures are available online from at least two sources:
ˆ The Khan Academy. rather than blind application of mathematical "recipes". Instead.khanacademy. functions. Those who do not tend to fare poorly on midterm examinations. These exercises are all available at The remaining nine assignments are devoted to the usual matrix and vector gymnastics expected from any sophomore linear algebra class. The problem sets chosen to accompany these notes could contribute roughly 20% of a student's grade. the student's focus was primarily on understanding ideas. Office hour questions reflected this and offered an excellent chance to give students tips how to present written answers in a way that would convince the person grading their work that they deserved full credit! Each lecture concludes with references to the comprehensive online textbooks of Jim Hefferon and Rob Beezer: designed to ensure familiarity with some basic mathematic notions (sets. typically computation-based. by assigning 20% of the grade to written assignments drawn from problems chosen randomly from the review exercises at the end of each lecture.smcvt.
Pearson 2001. David C. a useful technique for instructors is to project these using a computer. Rohit Thomas has spent a great deal of time editing these notes and the accompanying webworks and has improved them immeasurably. Lipson. as well as awkwardly explained concepts. There are many "cartoon" type images for the important theorems and formalæ . An army of 400 students. Lloyd. B. 1978. ˆ "Introduction to Linear Algebra". Lipschutz and M. ˆ "Algebra and Geometry". McGraw-Hill 2008. Gilbert Strang. Click here to see some of their work. ˆ "Linear Algebra and Its Applications". Springer 1997.edu/courses/mathematics/18-06-linear-algebra-spring -2010/video-lectures/
There are also an array of useful commercially available texts. D.mit. Kolman and D. exhaustive list includes
A non-
ˆ "Introductory Linear Algebra. S. An Applied First Course". Wellesley Cambridge Press 2009. Axler. Professor Gilbert Strang. many useful online math resources. Lay. Katrina Glaeser and Travis Scrimshaw have spent many 10
. ˆ "Schaum's Outline of Linear Algebra". In a classroom with a projector. These can be downloaded at: Lecture Materials There are still many errors in the notes.
A good strategy is to find your favorite among these in the University Library. ˆ "Linear Algebra Done Right". Hill. Fu Liu. S. There are many. They provide a colorful relief for students from (often illegible) scribbles on a blackboard. CBRC. Students have also started contributing to these notes. Addison–Weseley 2011.ˆ MIT OpenCourseWare. Stephen Pon and Gerry Puckett have already found many of them. Holten and J. A partial list is given in Appendix I.
1
What is Linear Algebra?
Video Overview
Three bears go into a cave. For example. . subtraction.. two come out. how much does Jill like Andrew. We could arrange these in a square array 9 4 ··· 10 6 .. . how much does Andrew like Andrew. Would it make sense to replace such an array by an abstract symbol M ? In the case of numbers. how much does Andrew like Jill. there exists a myriad of mathematical operations (addition. . n − 1 come out. suppose I asked everybody in this room to rate the likeability of everybody else on a scale from 1 to 10.) that can be performed with the symbol n that could provide useful information about the real world system at hand. etcetera). so much so that we often introduce abstract pronumerals to represent them: n bears go into a cave. . Would you go in? A single number alone is not sufficient to model more complicated real world situations.. multiplication.. The array M is often called a matrix and is an example of a 12
. Would you go in? Brian Butterworth
Numbers are highly useful tools for surviving in the modern world. In a room full of n people (or bears sic) there would be n2 ratings to keep track of (how much Jill likes Jill. the pronumeral n was more than a placeholder for a particular piece of information.
Along the way we'll learn about matrices and how to manipulate them.
This vector in the example is just a list of two numbers. we'll learn about three main topics: Linear Systems. and write it like this: (x. apples and oranges don't mix.otherwise if I see the vector (1. so if I want to keep track of the number of apples and oranges I have. Let x be the number of apples I have. I won't know whether I have two apples or two oranges. and y be the number of oranges I have. (To understand why having an abstract theory of linear transformations might be incredibly useful and even lucrative. Everything will carefully defined later.more general abstract structure called a linear transformation on which many mathematical operations can also be defined. like so:
13
. we can represent it with a point in the plane with the corresponding coordinates.
Example Suppose I have a bunch of apples and oranges. The order here matters! I should remember to always write the number of apples first and then the number of oranges . We'll call this list a vector. 2). we just want to start with some simple examples to get an idea of the things we'll be working with. As everyone knows. so if we want to. For now. try replacing "likeability ratings" with the number of times internet websites link to one another!) In this course. we'll illustrate some of the basic ideas of the course in the case of two by two matrices. I should put them in a list. Vector Spaces. y). and Linear Transformations.
for the points that don't have integer coordinates). and twice as many oranges as apples. First. and no places where x and y are multiplied together. Then 14
. no fractional or negative powers of x or y. you get on the left side x + y − (−2x + y) = 3x. The collection of all such vectors—all the points in our apple-orange plane—is an example of a vector space. we can imagine each point as some combination of apples and oranges (or parts thereof.
Example There are 27 pieces of fruit in a barrel. Then if you subtract the second equation from the first. and no variables are multiplied together: There are no powers of x or y greater than one.Oranges (x. Then each point corresponds to some vector. Reading homework: problem 1. and on the left side you get 27 − 0 = 27. How many apples and oranges are in the barrel? How to solve this conundrum? We can re-write the question mathematically as follows: x + y = 27 y = 2x
This is an example of a Linear System. It's a collection of equations in which variables are multiplied by constants and summed.1 Notice that we can solve the system by manipulating the equations involved. y)
Apples
In the plane. notice that the second equation is the same as −2x + y = 0.
Systems of Linear Equations
15
. We can multiply the matrix by the vector to get back the linear system using the following rule for multiplying matrices by vectors: a b c d x y = ax + by cx + dy (1)
Reading homework: problem 1.
References
Hefferon. Sections WILA and SSLE Wikipedia. because it takes one vector and turns it into another in a "linear" way. we then see that y = 18. so we learn that x = 9. by working with the list of equations as an object in itself. Chapter SLE. First we rewrite the equations tidily:
x + y = 27 2x − y = 0 We can express this set of equations with a matrix as follows: 1 1 2 −1 x y = 27 0
The square list of numbers is an example of a matrix.2
A 3 × 3 matrix example
The matrix is an example of a Linear Transformation. Let's do it again. We'll learn a general method called Gaussian Elimination. Section 1 Beezer. Our next task is to solve linear systems. Then there are 9 apples and 18 oranges. Using the second equation. Chapter One.3x = 27.
i. Next recall that multiplication of ordinary numbers is associative. Let M be a matrix and u and v vectors: a b x w M= . Matrix Multiplication: Let M and N be matrices M= and v a vector v= x y . compute the vector M (N v).u = . in fact it is the matrix obtained by multiplying the matrices M and N . We need to be careful reading this equation because N v is a vector and so is M (N v). namely the order of brackets does not matter: (xy)z = x(yz). (b) Check that your definition obeys M v + M u = M (u + v). 2. Use your result for M (N v) to find the matrix M N . This means that M N must be a matrix. Pablo is a nutritionist who knows that oranges always have twice as much sugar as apples. (M N )v should also be a vector. 3.e. c d y z (a) Propose a definition for u + v. that is M (N v) = (M N )v .
Compute the vector N v using the rule given above..Review Problems
1. he represents the barrel like so:
16
. When considering the sugar intake of schoolchildren eating a barrel of fruit. Now multiply this vector by the matrix M . Let us try to demand the same property for matrices and vectors. a b c d and N = e f g h . Therefore the right hand side.v = .
From that. That step is repeated until there is an equation with only one variable. To illustrate how it can lead to wrong conclusions. Find this y. Find the resulting y. One often taught in high school is to solve one of the equations for a variable. then substitute the resulting expression into other equations. and is also more likely to lead to errors.
Hint
4. Write your answer as a matrix. and then back-substitution can be done. f )
sugar
Find a linear transformation relating Pablo's representation to the one in the lecture. the first number in the solution is derived. Hint: Let λ represent the amount of sugar in each apple. since it involves more arithmetic operations. we will use the system x + 3y = 1 2x + y = −3 2x + 2y = 0 (a) Solve the first equation for x and substitute that expression into the second equation.fruit
(s. (b) Again solve the first equation for x. This method takes longer than Gauss' method. but this time substitute that expression into the third equation. There are methods for solving linear systems other than Gauss' method. 17
.
What extra step must a user of this method take to avoid erroneously concluding a system has a solution?
18
.
You can check that if 0 1 v is any vector. since it will show up in any y linear system we deal with. we can write the solution as: 1 0 0 1 The matrix I = x y = 9 18
1 0 is called the Identity Matrix .1
Gaussian Elimination
Notation for Linear Systems
In Lecture 1 we studied the linear system
x + y = 27 2x − y = 0 and found that x = 9 y = 18
We learned to write the linear system using a matrix and two vectors like so: 1 1 2 −1 x y = 27 0
Likewise. which looks like this for the linear system we've been dealing with: 1 1 27 2 −1 0 x . then Iv = v. A useful shorthand for a linear system is an Augmented Matrix . The solution to the linear system looks like this: We don't bother writing the vector 1 0 9 0 1 18
19
.2
2.
Equivalence Relations for Linear Systems
Equivalence Example
It often happens that two mathematical objects will appear to be different but in fact are exactly the same.Augmented Matrix Notation
Here's another example of an augmented matrix. The number of equations in the linear system is the number of rows r in the augmented matrix. here's the general case. . . . . . the fractions 1 and 12 describe the same number. and the number of columns k in the matrix left of the vertical line is the number of unknowns. 2 We could certainly call the two fractions equivalent. . . ar ar · · · ar b r 2 1 k Reading homework: problem 2.1 Here's the idea: Gaussian Elimination is a set of rules for taking a general augmented matrix and turning it into a very simple augmented matrix consisting of the identity matrix on the left and a bunch of numbers (the solution) on the right. In our running example. . we've noticed that the two augmented matrices 20
. The best-known example of this are 6 fractions. . . . a1 a1 · · · a1 b 1 2 1 k a2 a2 · · · a2 b 2 k 1 2 . For example. for a linear system with three equations and four unknowns: 1 3 2 0 9 6 2 0 −2 0 −1 0 1 1 3 And finally.
we write: 1 1 27 2 −1 0 ∼ 1 0 9 0 1 18
The symbol ∼ is read "is equivalent to". every rational number has a reduced fraction. and so on. To denote this. 2 −1 0
1 0 9 0 1 18
both contain the same information: x = 9. We usually call this a reduced fraction. There's a theorem telling us that every rational number can be specified by a unique fraction whose numerator and denominator are relatively prime. 100 . such as 2 .
A 3 × 3 example
2. that reduced fraction is unique. 27 . but slower.1 1 27 . This is an example of a canonical form.2 Reduced Row Echelon Form
Since there are many different augmented matrices that have the same set of solutions. This canonical form is called Reduced Row Echelon Form. RREF looks like this in general:
21
. A small excursion into the philosophy of mathematical notation: Suppose I have a large pile of equivalent fractions. y = 18. or RREF for short. Two augmented matrices corresponding to linear systems that actually have solutions are said to be (row) equivalent if they have the same solutions. Most 4 54 200 people will agree that their favorite way to write the number represented by 1 all these different factors is 2 . which is an extremely impressive way of saying "favorite way of writing it down". we should find a canonical form for writing our augmented matrices. To say that again. in which the numerator and denominator are relatively prime. and furthermore.
)
Reading homework: problem 2. Then y = 1 − 3λ and x = 4 − 7λ. (There are always as many of these as there are columns in the matrix before the vertical line. x4 . so lets just call them x1 . we see that w = 2 immediately. we get the following: x + 7z y + 3z =4 =1 w=2 Solving from the bottom variables up. Set z = λ. z is not a pivot. on the other hand is the number of linear equations.
(Note that you get to be creative here.
Example 1 0 0 0 1 0 0 0 0 1 0 0 1 2 0 0 0 0 1 0 1 2 . More concisely: −7 4 x −3 y 1 = + λ 1 z 0 0 2 w So we can read off the solution set directly from the RREF. x2 . we could have used λ and µ or any other names we like for a pair of unknowns.When we write this augmented matrix as a system of linear equations. the number of rows.) To begin with we immediately notice that there are no pivots in the second and fourth columns so x2 and x4 are undetermined and we set them to x2 = λ1 . so it is still undetermined. x3 .)
23
. x4 = λ2 . x5 .2 You need to become very adept at reading off solutions of linear systems from the RREF of their augmented matrix. The general method is to work from the bottom up and set any non-pivot variables to unknowns. but one for every choice of λ. (Notice that we use the word "set" because there is not just one solution. 3 0
Here we were not told the names of the variables. Here is another example.
Perhaps unsurprisingly in light of the previous discussions of RREF. 0 0 0 1 2 0 24
. Chapter One. we have a theorem: Theorem 2. Next lecture. 0 0 1 3 0 0 Observe. we will prove it.1. Section RREF Wikipedia. Next we see from the second last row that x5 = 3.Working from the bottom up we see that the last row just says 0 = 0. 1 0 0 0 3 1 0 1 0 0 1 2 0 0 1 0 1 3 . As a challenge. State whether the following augmented matrices are in RREF and compute their solution sets. Section 1 Beezer. The second row says x3 = 2 − 2x4 = 2 − 2λ2 . The top row then gives x1 = 1 − x2 − x4 = 1 − λ1 − λ2 . look carefully at this solution and make sure you can see how every part of it comes from the original augmented matrix without every having to reintroduce variables and equations. that since no variables were given at the beginning. Again we can write this solution as a vector 1 −1 −1 0 1 0 2 + λ1 0 + λ2 −2 . Every augmented matrix is row-equivalent to a unique augmented matrix in reduced row echelon form.
References
Hefferon. Chapter SLE. Row Echelon Form
Review Problems
1. we do not really need to state them in our solution. a well known fact! Note that a row of zeros save for a non-zero entry after the vertical line would be mathematically inconsistent and indicates that the system has NO solutions at all.
(For a fuller discussion of equivalence relations. y and z ∈ U .
Hints for Questions 4 and 5
26
. Problem 4) Show that row equivalence of augmented matrices is an equivalence relation. if x ∼ y and y ∼ z then x ∼ z. see Homework 0.ˆ Transitive: For any x.
Multiply the first row by λ so that this pivot entry is 1.
Example
Why do these preserve the linear system in question? Swapping rows is just changing the order of the equations begin considered. which certainly should not alter the solutions. The three elementary row operations are:
ˆ (Row Swap) Exchange any two rows. Swap rows so that the first entry in the first column is non-zero. This process is called Gauss–Jordan elimination or simply Gaussian elimination. 1. Scalar multiplication is just multiplying the equation by the same number on both sides. Add multiples of the first row to each other row so that the first entry of every other row is zero. if two equations share a common solution. then the associated variable is undetermined. 3. ˆ (Row Sum) Add a multiple of one row to another row.3
Elementary Row Operations
Our goal is to begin with an arbitrary matrix and apply operations that respect row equivalence until we have a matrix in Reduced Row Echelon Form (RREF). 2. If all entries in a given column are zero. 4. working column by column. ˆ (Scalar Multiplication) Multiply any row by a non-zero constant. Therefore we can define augmented matrices to be row equivalent if the are related by a sequence of elementary row operations. make a note of the undetermined variable(s) and then ignore all such columns. This definition can also be applied to augmented matrices corresponding to linear systems with no solutions at all! There is a very simple process for row-reducing a matrix. adding one to the other preserves the solution.
27
. which does not change the solution(s) of the equation. Likewise.
6. Before moving on to step 6. Reading homework: problem 3. add multiples of the first row any rows above that you have ignored to ensure there are zeros in the column above the current pivot entry. Now ignore the first row and first column and repeat steps 2-5 until the matrix is in RREF.1
Example 3x3 = 9 x1 1 3 x1 +5x2 −2x3 = 2 +2x2 =3
First we write the system as an augmented matrix:
28
.5.
2 Beezer. matrices A and B. Elementary Matrix Operations
Review Problems
1. smaller. Section 1. But this is a contradiction! Then A = B. Call ˆ ˆ the new. Row Echelon Form Wikipedia. A and B. labeled by the row operations you performed. Suppose Alice and Bob compute the RREF for a linear system but get different results. Chapter SLE. removing columns does not affect row equivalence. (Row Equivalence) (a) Solve the following linear system using Gauss-Jordan elimination: 2x1 + 6x1 + 3x1 + 3x1 + 6x1 + 5x2 − 8x3 + 2x2 −10x3 + 6x2 + 2x3 + 1x2 − 5x3 + 7x2 − 3x3 + 2x4 + 6x4 + 3x4 + 3x4 + 6x4 + 2x5 = 0 8x5 = 6 5x5 = 6 4x5 = 3 9x5 = 9
Be sure to set your work out carefully with equivalence signs ∼ between each step.Proof. 31
.
Explanation of the proof
References
Hefferon. The new matrices should look this: ˆ A= IN a 0 0 ˆ and B = IN b 0 0 . Chapter One. Working from the left. ˆ ˆ Now if A and B have the same solution. discard all columns except for the pivots and the first column in which A and B differ.1 and 1. By Review Problem 1b.
where IN is an N × N identity matrix and a and b are vectors. then we must have a = b. Section RREF Wikipedia.
" Write the following row echelon matrix as a system of equations. and show that the resulting two matrices. 2. Provide a counterexample to show that row echelon form is not unique. does the new matrix have any solutions? In general. If you 4 17 20 remove one of the rows of this matrix. In row echelon form.(b) Check that the following two matrices are row-equivalent: 1 4 7 10 2 9 6 0 and 0 −1 8 20 4 18 12 0
Now remove the third column from each matrix. viewed as augmented matrices (shown below) are row-equivalent: 1 4 7 2 9 6 and 0 −1 8 4 18 12
Explain why row-equivalence is never affected by removing columns. then solve the system using back-substitution. 1 4 10 (c) Check that the matrix 3 13 9 has no solutions. Once a system is in row echelon form. (Gaussian Elimination) Another method for solving linear systems is to use row operations to bring the augmented matrix to row echelon form. it can be solved by "back substitution. and we only require that all entries left of the pivots are zero. 2 3 1 6 0 1 1 2 0 0 3 3 32
. and show that the resulting two matrices (shown below) are row-equivalent: 1 4 10 2 9 0 and 0 −1 20 4 18 0
Now remove the fourth column from each of the original two matrices. the pivots are not necessarily set to one. not necessarily entries above a pivot. can row equivalence be affected by removing rows? Explain why or why not.
then any point in R3 is a solution. These types of "solution sets" are hard to visualize. In general. Line. plus the possibility of no solutions. No solutions. or else all of the equations coincide geometrically. there are k + 2 possible outcomes. If you start with no information. 3.1 34
. 2. for systems of equations with k unknowns. we look for the common intersection of the planes (if an intersection exists). any point on that line then gives a solution to the system of equations. Plane. Here we have five different possibilities: 1. Perhaps you only had one equation to begin with. Some of the equations are contradictory. 4. with two free parameters. There are three free parameters. but luckily "hyperplanes" behave like planes in R3 in many ways. consider the case of r equations in three variables. The planes have a unique point of intersection. corresponding to the number of free parameters in the solutions set. Unique Solution. you have a plane of solutions. To find solutions to the system of equations. The planes intersect in a common line. so no solutions exist. one can have a number of different outcomes. Each of these equations is the equation of a plane in threedimensional space.
Pictures and Explanation
Reading homework: problem 4. For example.4
Solution Sets for Systems of Linear Equations
For a system of equations with r equations and k unknowns. All of R3 . In this case.
Planes
5.
Overview
We begin by looking at the space Rn . We might sometimes interpret a point and a vector as the same object. we can draw a vector v from O to P . pn − o n to put the origin anywhere that seems most convenient in Rn . a favorite point in Rn . a vector has a magnitude and a direction. not just at the point with zero coordinates:
Remark A quick note on points versus vectors. . . . you encountered three-dimensional vectors. n-Vectors
In vector calculus classes.5
Vectors in Space. . Here v 2 denotes the second component of a vector v. which we can think of as the space of points with n coordinates. Now given any other point P . For more details. then and p1 − o 1 p2 − o 2 the components of the vector v are . . . . see Appendix D
Do not be confused by our use of a superscript to label components of a vector. rather than a number v squared! 43
. Just as in R3 . If O has coordinates (o1 . but they are slightly different concepts and should be treated as such. . pn ). This construction allows us . on ) p has coordinates (p1 . We then specify an origin O. . . Now we will develop the notion of n-vectors and learn some of their properties.
5 5 10 5
Notice that these are the same rules we saw in Lecture 4! In Lectures 1-4. n a + bn
Given a scalar λ. it is the only vector with zero magnitude. Notice that with respect to the usual notions of Euclidean geometry. . A special vector is the zero vector connecting the origin to itself. All of its components are zero. a = . and b = . . a+b= . . for example −5 5 5 0 a + b = and 3a − 2b = . . and the only one which points in no particular direction. . Thus. λa = . Now we are thinking of a vector as a magnitude and a direction in Rn . . and luckily the same rules apply. the scalar multiple
λa1 . λan
Example Let 4 1 3 2 a = and b = . . 2 3 1 4 Then. any single vector 44
.Most importantly. we can add vectors and multiply vectors by a scalar: Definition Given two vectors a and b whose components are given by a1 b1 . n a bn their sum is a1 + b 1 . we thought of a vector as being a list of numbers which captured information about a linear system.
46
. one usually assumes that k = n − 1 for a hyperplane inside Rn . .describes a plane in 6-dimensional space parallel to the xy-plane. vk in Rn with k ≤ n determines a k-dimensional hyperplane. then any point in the hyperplane can be written as:
k
{P +
i=1
λi vi |λi ∈ R}
When the dimension k is not specified. If the vectors do determine a k-dimensional hyperplane. unless any of the vectors vi lives in the same hyperplane determined by the other vectors.1
Directions and Magnitudes
n
Consider the Euclidean length of a vector: v = (v 1 )2 + (v 2 )2 + · · · (v n )2 =
i=1
(v i )2 .
5. we can then connect the ends of v and u with the vector v − u. .
Parametric Notation
We can generalize the notion of a plane: Definition A set of k vectors v1 .
Using the Law of Cosines. . Given two vectors v and u that span a plane in Rn . . we can then figure out the angle between two vectors.
v = u1 v 1 + u2 v 2 + u3 v 3 − u4 v 4 . the "squared-length" of a vector with coordinates x. Notice in the definition above that we first defined the dot product. For non-zero vectors u and v with an inner-product .1
Example Consider a four-dimensional space.1 (Cauchy-Schwarz Inequality). Thus u (cv + dw) = c u v + d u w . 4. with a special direction which we will call "time". z and t is 2 = x2 + y 2 + z 2 − t2 .3. There are. The Lorentzian inner product on R4 is given by u. and u u = 0 only when u itself is the 0-vector. . Bilinear. The easiest proof would use the definition of the angle between two vectors and the fact that cos θ ≤ 1. but note that it is not positive definite. we change our notion of length and angle as well. Reading homework: problem 5. strictly speaking speaking we did not check our assumption that we could apply the Law of Cosines to the 48
. So if we change our idea of the dot product. Notice that it is possible for v 2 ≤ 0 for non-vanishing v! v
Theorem 5. linear in both u and v. which is to say. | u. which have all of the properties listed above (except that in some contexts the positive definite requirement is relaxed). and (cu + dw) v = c u v + d w v . Other definitions of length and angle arise from inner products. The dot product determines the Euclidean length and angle between two vectors. many different useful ways to define lengths of vectors. in fact. This is of central importance in Einstein's theory of special relativity. we usually write u. Instead of writing for other inner products. Positive Definite: u u ≥ 0. v to avoid confusion. As a result. y. However. v | ≤1 u v Proof. and then defined everything else in terms of the dot product.
for the year 1988 that ordered list was f = (20. Captain Conundrum mowed lawns on weekends to help pay his college tuition bills. y
(a) Sketch X and M X in R2 for several values of X and θ. He also listed the number of times he mowed each lawn in a given year. (b) Compute
||M X|| ||X||
for arbitrary values of X and θ. 6) . 2. When he was young. 10. 1. 50. 2. 1000. 500. 100. 100) . He charged his customers according to the size of their lawns at a rate of 5¢ per square foot and meticulously kept a record of the areas of their lawns in an ordered list: A = (200. 1. 4. (n) Find the angle between the diagonal of the unit (hyper)-cube in Rn and one of the coordinate axes. (2) Find the angle between the diagonal of the unit square in R2 and one of the coordinate axes. (3) Find the angle between the diagonal of the unit cube in R3 and one of the coordinate axes. 1. 51
. (∞) What is the limit as n → ∞ of the angle between the diagonal of the unit (hyper)-cube in Rn and one of the coordinate axes? 3. 100. 200. Consider the matrix M = cos θ sin θ − sin θ cos θ and the vector X = x . (a) Pretend that A and f are vectors and compute A f . 300. 5. 50. (d) Suppose Captain Conundrum charged different customers different rates. How could you modify the expression in part 1c to compute the Captain's earnings? 2.Review Problems
1. (b) What quantity does the dot product A f measure? (c) How much did Captain Conundrum earn from mowing lawns in 1988? Write an expression for this amount in terms of the vectors A and f .
(Lorentzian Strangeness). 4. (b) Find and sketch the collection of all vectors in two-dimensional Lorentzian space-time with zero length. Approximately what is the real-world angle between the vectors 0 1 and ? What is the angle between these two vectors according 1 1 to the dot-product? Give a definition for an inner product so that the angles produced by the inner product are the actual angles between vectors. consider Rn with the Lorentzian inner product and metric defined above. Suppose in R2 I measure the x direction in inches and the y direction in miles. (c) Find and sketch the collection of all vectors in three-dimensional Lorentzian space-time with zero length.(c) Explain your result for (b) and describe the action of M geometrically. For this problem. 5. (a) Find a non-zero vector in two-dimensional Lorentzian space-time with zero length.
The Story of Your Life
52
.
) (+ii) (Additive Commutativity) u + v = v + u. (Scalar times a vector is a vector. (Scalar multiplication distributes over addition of scalars. d ∈ R: (+i) (Additive Closure) u + v ∈ V . (· i) (Multiplicative Closure) c · v ∈ V . Once complete. (Scalar multiplication distributes over addition of vectors. (+v) (Additive Inverse) For every u ∈ V there exists w ∈ V such that u + w = 0V . As it turns out. Definition A vector space (over R) is a set V with two operations + and · satisfying the following properties for all u. (Order of addition doesn't matter.) (+iv) (Zero) There is a special vector 0V ∈ V such that u + 0V = u for all u in V . The two key properties of vectors are that they can be added together and multiplied by scalars.
53
.) (· ii) (Distributivity) (c+d)·v = c·v +d·v. (Adding two vectors gives a vector. The main idea is to define vectors based on their most important properties.) (+iii) (Additive Associativity) (u + v) + w = u + (v + w) (Order of adding many vectors doesn't matter.) (· iii) (Distributivity) c·(u+v) = c·u+c·v. the notion of a vector applies to a much more general class of structures than this.6
Vector Spaces
Thus far we have thought of vectors as lists of numbers in Rn .) (· iv) (Associativity) (cd) · v = c · (d · v). v ∈ V and c. So we make the following definition. (· v) (Unity) 1 · v = v for all v ∈ V . but will also cover many other extremely useful notions of vectors. our new definition of vectors will include vectors in Rn . We do this in the hope of creating a mathematical structure applicable to a wide range of real-world problems.
we'll just write scalar multiplication with juxtaposition cv = c · v. . n3 . (In notation. . . We can think of these functions as infinite sequences: f (0) is the first term. 1. Then for example the function f (n) = n3 would look like this: f = {0. Let's check some axioms. (In notation: : V × V → R. (+i) (Additive Closure) f1 (n) + f2 (n) is indeed a function N → R. The scalar product is a function that takes a vector and a number and returns a vector. Remark It isn't hard to devise strange rules for addition or scalar multiplication that break some or all of the rules listed above.
Example of a vector space
One can also find many interesting vector spaces.) Once the properties of a vector space have been verified.) On the other hand. to avoid confusing the notation. V is the space of all infinite sequences. the dot product takes two vectors and returns a number. .
54
. . and so on. since the sum of two real numbers is a real number. f (1) is the second term. such as the following.}. The constant zero function g(n) = 0 works because then f (n) + g(n) = f (n) + 0 = f (n). . Thinking this way. this can be written · : R × V → V . Example V = {f | f : N → R} Here the vector space is the set of functions that take in a natural number n and return a real number. 27. The other axioms that should be checked come down to properties of the real numbers. though. Scalar multiplication is just as simple: c · f (n) = cf (n). The addition is just addition of functions: (f1 +f2 )(n) = f1 (n)+f2 (n). 8.Examples of each rule
Remark Don't confuse the scalar product · with the dot product . (+iv) (Zero) We need to propose a zero vector. .
The zero function is just the function such that 0(x) = 0 for every x. since the derivative commutes with scalar multiplication d d ( dx (cf ) = c dx f ). λ V = | λ. x. we defined vector spaces over the real numbers. y ∈ R . A field is a collection of "numbers" satisfying a number of properties. dx The addition is point-wise (f + g)(x) = f (x) + g(x) . Complex numbers are extremely useful because of a special property that they enjoy: every polynomial over the complex numbers factors into a product of linear
55
. since the base field is the complex
numbers.1 Example Another very important example of a vector space is the space of all differentiable functions: d f | f : R → R. we know that the sum of any two differentiable functions is differentiable. µ ∈ C µ describes states of an electron. For example. f exists . One other example of a field is the complex numbers. as is scalar multiplication c · f (x) = cf (x) . One can actually define vector spaces over any field. From calculus. vector spaces over C describe all possible states a system of particles can have. Other states. since the derivative distributes over addition.Reading homework: problem 6. In fact. as is the space of functions with infinitely many derivatives. A scalar multiple of a function is also differentiable. The rest of the vector space properties are inherited from addition and scalar multiplication in R. the set of functions with at least k derivatives is always a vector space. like i −i 1 0 describes spin "up" and describes spin 0 1
are permissible. Vector Spaces Over Other Fields Above. C = x + iy | i2 = −1. In quantum physics. where "down".
One example of this phenomenon occurs when diagonalizing matrices. the polynomial x2 + 1 doesn't factor over the real numbers. but over the complex numbers it factors into (x+i)(x−i). Check that the complex numbers C = {x + iy | x. Section VS Wikipedia:
ˆ Vector Space ˆ Field ˆ Spin
1 2
ˆ Galois Theory
Review Problems
1.1 Beezer.3. Since the rationals are a field. This property ends up having very far-reaching consequences: often in mathematics problems that are very difficult when working over the real numbers become relatively simple when working over the complex numbers. Make sure you state carefully what your rules for 56
. and occasionally work over Z2 = {0. Section I. 1} where 1 + 1 = 0. see Appendix E. the mathematics of vector spaces still apply to this special case. Chapter VS. y ∈ R = R2 with the usual addition and y scalar multiplication is a vector space.
2. For example. which we will learn about later in the course. For more on fields in general. we will work mainly over the real numbers and the complex numbers. Check that V = x | x. In this class. So instead rational approximations are used. Chapter One. however the full story of fields is typically covered in a class on abstract algebra or Galois theory. y ∈ R} form a vector space over C.
References
Hefferon. This is field is important in computer algebra: a real number given by an infinite string of numbers after the decimal point can't be stored by a computer.polynomials. Another useful field is the rational numbers Q.
lim f does not exist or is ± ∞
n→∞
Is this a vector space? Explain why or why not.
R 5. Consider the set of 2 × 4 matrices: V = a b c d e f g h | a. ˆ Identify the zero vector. b. Also. (b) Now consider the set of divergent sequences. Identify the zero vector in V .
Problem 5 hint
57
. with the same addition and scalar multiplication that we defined for the space of sequences: V = f | f : N → R. h ∈ C
Propose definitions for addition and scalar multiplication in V . with the same addition and scalar multiplication as before: V = f | f : N → R. c. e. Let P3 be the set of polynomials with real coefficients of degree three or less.
ˆ Propose a definition of addition and scalar multiplication to make R P3 a vector space. d. lim f ∈ R
n→∞
Is this still a vector space? Explain why or why not. and check that every matrix has an additive inverse.vector addition and scalar multiplication are. Propose a small change R to the definition of P3 to make it a vector space over C. explain what would happen if you used R as the base field (try comparing to problem 1). 4. f. (a) Consider the set of convergent sequences. and find the additive inverse for the vector −3 − 2x + x2 . 3. g.
R ˆ Show that P3 is not a vector space over C.
while on the right the operations occur in W . v ∈ V and c ∈ R: L(u + v) = L(u) + L(v) L(cv) = cL(v) Combining these two requirements into one equation. Definition A function L : V → W is linear if for all u. this means that for any u.1
Example Take L : R3 → R3 defined by: x x+y L y = y + z z 0 x a Call u = y . if adding vectors and then applying L were the same as applying L to two vectors and then adding them. v ∈ V and r. This is often called the linearity property of a linear transformation. It would be ideal if the map L preserved these operations. In formulas. Now check linearity. v = b . when multiplying by a scalar. s ∈ R we have L(ru + sv) = rL(u) + sL(v) Notice that on the left the addition and scalar multiplication occur in V . it didn't matter whether we multiplied before or after applying L. Reading homework: problem 7. Now suppose we have two vector spaces V and W and a map L between them: L: V → W Now.7
Linear Transformations
Recall that the key properties of vector spaces are vector addition and scalar multiplication. In other words. both V and W have notions of vector addition and scalar multiplication. it would be nice if. Likewise. z c
58
. we get the definition of a linear function or linear transformation.
dimension is the number of independent directions available. then I choose one of them as my next direction. linear transformations can always be represented by matrices. For finite dimensional vector spaces. 'Homomorphism' is a very general term which in mathematics means 'Structure-preserving map. we will start studying matrices intensively in the next few lectures. If there are any vectors in my vector space not in the plane determined by the first two directions. Section II.
References
Hefferon.) Beezer. and MLT. Some of the examples of vector spaces we have worked with have been finite dimensional. Section LT. and pick a direction.) The polynomial example above is an example of an infinite dimensional vector space. Rn will turn out to have dimension n. Chapter Three. LTC. (For example. (Note that Hefferon uses the term homomorphism for a linear map. Roughly speaking. The size of a minimal set of independent vectors is the dimension of the vector space. Chapter LT. Wikipedia:
ˆ Linear Transformation ˆ Dimension
Review Problems
1. Subsections LT.Foreshadowing Dimension. and is thus a type of homomorphism.' A linear map preserves the linear structure of a vector space. then I choose another direction that isn't in the line determined by the direction I chose. If there are any vectors in my vector space that aren't in that direction. In other words. though we haven't actually defined the idea of dimension mathematically yet. I choose a collection of independent vectors in the vector space. Show that the pair of conditions: (i) L(u + v) = L(u) + L(v) (ii) L(cv) = cL(v) is equivalent to the single condition: 61
. For that reason. You probably have some intuitive notion of what dimension means. To figure out dimension. I stand at the origin.
62
. scalars in R). Let z ∈ C. Recall that we can express z = a+bi where a. L(t) = t3 .ii)⇒(iii). and we can form the complex conjugate of z by taking z = a−bi (note that this is unique since z = z). Let Pn be the space of polynomials of degree n or less in the variable t.
Hint
3.ii). Suppose L is a linear transformation from P2 → P3 such that L(1) = 4. What would a matrix for integration look like? Be sure to think about what to do with the constant of integration. b) → (a. c such that L(a + bt + ct2 ) = 1 + 3t + 2t3 .
Finite degree example
4. (b) Show that z is not linear over C. Show that integration is a linear transformation on the vector space of polynomials.e. and L(t2 ) = t − 1.
ˆ Find L(1 + t + 2t2 ). and it is clear that c agrees with complex conjugation. 2. b ∈ R. ˆ Find L(a + bt + ct2 ). −b). and then show that (iii)⇒(i. (a) Show that c is a linear map over R (i. Your answer should have two parts.(iii) L(ru + sv) = rL(u) + sL(v) . Show that (i. So we can define a function from c : R2 → R2 which sends (a. b. ˆ Find all values a.
. . . k is a j rectangular array of real (or complex) numbers: m1 m1 · · · m1 1 2 k m2 m2 · · · m2 2 k 1 M = . you may have encountered image files with a .
2
This notation was first introduced by Albert Einstein.
63
. vr
1 A 1 × k matrix v = (vk ) = (vk ) is called a row vector . . r An r × 1 matrix v = (v1 ) = (v r ) is called a column vector . . .. r. . Finally. written
v = v1 v2 · · · vk . Matrices are a very useful and efficient way to store information:
Example In computer graphics. . . . The resulting matrix then has its rows shuffled a bit: by listing. say. . .8
Matrices
Definition An r × k matrix M = (mi ) for i = 1. . and then each entry of the matrix is a number indicating the color of a particular pixel in the image.gif extension. It is often useful to consider matrices whose entries are more general than the real numbers. . . j = 1. a compression algorithm is applied to the matrix to reduce the size of the file. then a web browser downloading the file can start displaying an incomplete version of the picture before the download is complete. These files are actually just matrices: at the start of the file the size of the matrix is given. every eighth row. r r m1 m2 · · · mr k The numbers mi are called entries. so we allow that possibility. written v1 v 2 v=. . The superscript indexes the row of j the matrix and the subscript indexes the column of the matrix in which mi j appears2 . . .
a graph is just a collection of vertices and some edges connecting vertices. where mi j indicates the number of edges between the vertices labeled i and j: 1 2 1 1 2 0 1 0 M = 1 1 0 1 1 0 1 3 This is an example of a symmetric matrix. the graph pictured above would have the following matrix. Notice that M1 = Rn is just the vector space of column vectors. and n scalar multiplication multiplies every entry. ranging from telephone networks to airline routes.Adjacency Matrix Example
Example Graphs occur in many applications.
For example. addition just adds corresponding entries in two matrices. A matrix can be used to indicate how many edges attach one vertex to another. j i
r The space of r × k matrices Mk is a vector space with the addition and scalar multiplication defined as follows:
M + N = (mi ) + (ni ) = (mi + ni ) j j j j rM = r(mi ) = (rmi ) j j In other words. In the subject of graph theory .
64
. since mi = mj .
A square matrix that is zero for all non-diagonal entries is called a diagonal matrix.} is called the diagonal of the matrix . Matrix Terminology The entries mi are called diagonal. then to make the product M N we must have k = s. For an r × k matrix M and an s × m matrix N . the columns and rows must match.1 Recall that r × k matrices can be used to represent linear transformations R → Rr via
k k
MV =
j=1
mi v j . we can use a matrix N = (ni ) to represent a linear transformaj tion s N r L : Mk −→ Mk via L(M )i l =
j=1 s
ni mj . Likewise. A common shorthand for keeping track of the sizes of the matrices involved in a given product is: r×k × k×m = r×m
Example Multiplying a (3 × 1) matrix and a (1 × 2) matrix yields a (3 × 2) matrix. . it is required that m = r. Likewise. . L(M ) = N M is a linear transformation. 2 Any r × r matrix is called a square matrix . 1 1·2 1·3 2 3 3 2 3 = 3 · 2 3 · 3 = 6 9 2 2·2 2·3 4 6
Reading homework: problem 8. for the product N M . i 1 m2 . j
which is the same rule we use when we multiply an r × k matrix by a k × 1 vector to produce an r × 1 vector.
66
. In other words. . j l
This is the same as the rule we use to multiply matrices.Notice that in order for the multiplication to make sense. and the set {m1 .
yn
Hint
5. 4. In terms of the entries of M and the entries of N . Let M be any m × n matrix. In terms of the entries of M and the entries of N . Write out a few of the entries of j j each matrix in the form given at the beginning of this chapter. .(a) Let M = (mi ) and let N = (ni ). and y = . In terms of the entries of M and the entries of N . what is the entry in row i and column j of M N ? (c) Take the transpose (M N )T and write out a few of its entries in the same form as in part a. Let x = . Show that M T M and M M T are symmetric. show that right matrix multiplication obeys linearity. Show that the dot . . what is the entry in row i and column j of (M N )T ? (d) Take the transposes N T and M T and write out a few of their entries in the same form as in part a. Show that right multiplication s R s by a k × m matrix R is a linear transformation Mk −→ Mm . 3.
Problem hint
70
. (b) Multiply out M N and write out a few of its entries in the same form as in part a. In other words. what is the entry in row i and column j of N T M T ? (f) Show that the answers you got in parts c and e are the same.) What are their sizes? x1 y1 . (e) Multiply out N T M T and write out a few of its entries in the same form as in part a. we showed that left multiplication by an r × s matrix N was s N r a linear transformation Mk −→ Mk . (Hint: use the result of the previous problem. xn product x y = xT 1 y. Above. be column vectors.
Give a few simple examples before you start explaining. Explain what happens to a matrix when: (a) You multiply it on the left by a diagonal matrix. (b) You multiply it on the right by a diagonal matrix.
71
.6.
9
9. ˆ Matrix operations on block matrices can be carried out by treating the blocks as matrix entries. D = (0).
M2 = =
A B C D
A B C D
A2 + BC AB + BD CA + DC CB + D2
72
. Here A = 7 8 9 1
ˆ The blocks of a block matrix must fit together to form a rectangle. For example. if there are large blocks of zeros in a matrix.1
ˆ There are many ways to cut up an n × n matrix into blocks.1
Properties of Matrices
Block Matrices
It is often convenient to partition a matrix M into smaller matrices called blocks. D C D A
Reading homework: problem 9. Often context or the entries of the matrix will suggest a useful way to divide the matrix into blocks. but does not. like so: 1 2 3 1 4 5 6 0 A B M = 7 8 9 1 = C D 0 1 2 0 1 2 3 1 4 5 6. So C B B A makes sense. In the example above. or blocks that look like an identity matrix. B = 0. C = 0 1 2 . it can be useful to partition the matrix accordingly.
respectively.
Notice how the end product of M N and N M are different. to three dimensions. We can generalize this. The special case of θ = 90◦ is shown. compared with the case of N followed by M .
rotates vectors in the plane by an angle θ. Because.
Trace
Matrices contain a great deal of information. In fact the following matrices built from a 2 × 2 rotation matrix.
Example You learned in a Review Problem that the matrix M= cos θ sin θ − sin θ cos θ . using block matrices. 75
. you can also use them to rotate objects built from a collection of vectors like pretty colored blocks! Here is a picture of M and then N acting on such a block. a 1 × 1 identity matrix and zeroes everywhere else cos θ sin θ 0 1 0 0 and N = 0 cos θ sin θ .Here is an example of matrices acting on objects in three dimensions that also shows matrices not commuting. Here we need to assume that n < ∞ otherwise there are subtleties with convergence that we'd have to address. they rotate single vectors. so finding ways to extract essential information is useful. M = − sin θ cos θ 0 0 0 1 0 − sin θ cos θ perform rotations by an angle θ in the xy and yz planes. so M N = N M here.
More on the trace function
Linear Systems Redux Recall that we can view a linear system as a matrix equation M X = V. It is called the inverse of M . if the linear system has no solution.Another useful property of the trace is that: tr M = tr M T This is true because the trace only uses the diagonal entries. then we would multiply both 1 1 sides of the equation M X = V by M to obtain the solution immediately: X = M V . and V an r × 1 matrix of constants. Clearly. since if it existed we could find a solution. 1 Therefore M only sometimes exists. since X = M V yields only a single solution. written M
References
Beezer: Part T. This is easy to check. where M M = I. so we have hope of finding a single solution. If we could compute M . trace is a linear transformation from matrices to the real numbers. On the other hand. For example: tr = 4 = tr = tr 2 3 1 3 1 3 Finally. with M an r × k matrix of coefficients. if the system has 1 1 more than one solution. then the number of equations r is the same as the number of unknowns k. it also seems unlikely that M would exist. and is usually −1 . X a k × 1 matrix of unknowns. Above we discussed functions of matrices. which are fixed T 1 1 1 2 1 2 by the transpose. Section T Wikipedia:
ˆ Trace (Linear Algebra) ˆ Block Matrix
77
. then there can be no hope of finding 1 M . If M is a square matrix. An extremely useful function would be 1 1 1 f (M ) = M .
then (A−1 A)T = AT (A−1 )T = I. Notice that B −1 A−1 AB = B −1 IB = I = ABB −1 A−1 . Similarly. Then: (A−1 )T = (AT )−1 80
. then A is the inverse of B. If A is a square matrix and B is the inverse of A. Then: (AB)−1 = B −1 A−1 Then much like the transpose. since AB = I = BA. Since I T = I.10. Finally. Then we have the identity: (A−1 )−1 = A 2.
3. (AA−1 )T = (A−1 )T AT = I. taking the inverse of a product reverses the order of the product.1
Three Properties of the Inverse
1. recall that (AB)T = B T AT .
81
. putting the ek 's together into an identity matrix. the n × n identity matrix can be viewed as a bunch of column vectors In = (e1 e2 · · · en ). we get: M I ∼ I M −1 I = I M −1
−1 Example Find 2 4 apply row reduction to −1 2 −3 1 0 . and then apply Gaussian row reduction to the left side of the matrix. we could even write A−T for the inverse of the transpose of A (or equivalently the transpose of the inverse).. rather than M −1 V to appear on the right side of our augmented matrix.e.As such. I. Start by writing the augmented matrix. To compute M −1 .
Example
10. To solve many linear systems at once.2 Finding Inverses
Suppose M is a square matrix and M X = V is a linear system with unique solution X0 . Once the identity matrix is on the left side of the augmented matrix. we would like M −1 . we can consider augmented matrices with a matrix on the right side instead of a column vector. Since there is a unique solution. then the reduced row echelon form of the linear system has an identity matrix on the left: M V ∼ I M −1 V Solving the linear system M X = V then tells us what M −1 V is. So. M −1 V . where ek is the column vector of zeroes with a 1 in the kth entry. then −2 5 the left side. This is achieved by solving the collection of systems M X = ek . then the solution of each of the individual linear systems is on the right.
Applying the algorithm again decodes the message. For example.
1 0 1 Example 0 1 1 is an invertible matrix over Z2 : 1 1 1 −1 1 0 1 0 1 1 0 1 1 = 1 0 1 1 1 1 1 1 1 This can be easily verified 1 0 0 1 1 1 by multiplying: 1 0 1 1 1 0 0 1 1 0 1 = 0 1 0 1 1 1 1 0 0 1
Application: Cryptography A very simple way to hide information is to use a substitution cipher. Computers can add and multiply individual bits very quickly. 1} with addition and multiplication given by the following tables: + 0 1 · 0 1 0 0 1 0 0 0 1 1 0 1 0 1 Notice that −1 = 1. For example. keeping track of a single one or zero.10. so we can apply all of the linear algebra we have learned thus far to matrices with Z2 entries. A matrix with entries in Z2 is sometimes called a bit matrix 3 . It turns out that Z2 is almost as good as the real or complex numbers (they are all fields). the ROT-13 cypher just exchanges a letter with the letter thirteen places before or after it in the alphabet.
3
84
. For example. in which the alphabet is permuted and each letter in a message is systematically exchanged for another.5
Bit Matrices
In computer science. HELLO becomes URYYB. information is recorded using binary strings of data. turning
Note that bits in a bit arithmetic shorthand do not "add" and "multiply" as elements in Z2 does since these operators corresponding to "bitwise or" and "bitwise and" respectively. since 1 + 1 = 0. Consider the set Z2 = {0. the following string contains an English word: 011011000110100101101110011001010110000101110010 A bit is the basic unit of information.
each string of eight characters would be multiplied by M −1 .URYYB back into HELLO. where the entries are zeros and ones). see "The Code Book. a single character is represented by a string of eight bits. when they are not singular:
85
. English characters are often stored in computers in the ASCII format. but the basic idea can be extended to create cryptographic systems that are practically uncrackable. is to choose an 8 × 8 invertible bit matrix M . and multiply each letter of the message by M .
References
Hefferon: Chapter Three. Section IV." by Simon Singh (1999. To make the message a bit tougher to decode. Substitution ciphers are easy to break. In ASCII. and 2 then use an appropriately-sized invertible matrix. Doubleday).
You are now ready to attempt the first sample midterm. which we can consider as a vector in Z8 (which is like vectors in R8 . Then to decode the message. For example. So long as a particular set of substitutions is not used on more than one message. then. For more on cryptography. a one-time pad is a system that uses a different substitution for each letter in the message. Find formulas for the inverses of the following matrices. one could consider pairs (or longer sequences) of letters as a single vector in Z16 (or a higher-dimensional space). One 2 way to create a substitution cipher.2 Beezer: Chapter M. the one-time pad is unbreakable. Section MISLE Wikipedia: Invertible Matrix
Review Problems
1.
1 0 (a) 0 a (b) 0 0
a b 1 c 0 1 b c d e 0 f
When are these matrices singular? 2.)
Hints for Problem 3
4. there is a unique solution to M X = V. Show that M must be singular. Show that M must be singular. Let A= (a) Compute: 86 0 1 1 1 1 0
. Finally. For those which are not singular. suppose that M is non-singular. Explain why the following statements are equivalent: (a) M X = V has a unique solution for every column vector V . think about the key words: First. Left and Right Inverses: So far we have only talked about inverses of square matrices. 3. pair them with their inverse. suppose that there is some column vector V such that the equation M X = V has no solutions. Let M be a square matrix. that is. Write down all 2×2 bit matrices and decide which of them are singular. Next. Show that no matter what the column vector V is. (b) M is non-singular. suppose that there is some column vector V such that the equation M X = V has two distinct solutions. This problem will explore the notion of a left and right inverse for a matrix that is not square. show that M can have no inverse. (In general for problems like this.
−1
iii.. (c) Does BA make sense? (Why not?) (d) Let A be an n × m matrix with n > m. i.e. ii. AAT . B := AT AAT
(b) Show that the matrix B above is a right inverse for A.
(f) True or false: Left and right inverses are unique. (e) Test your proposal for a left inverse for the simple example A= 1 2 . Suggest a formula for a left inverse C such that CA = I Hint: you may assume that AT A has an inverse. AAT
−1
. verify that AB = I . If false give a counterexample.
Left and Right Inverses
87
.i.
we would just continue the process. W such that X −1 exists. Again. the process is exactly the same as for a square matrix.
−2 1 3 . 2 1 U1 = −2 1 3 0 2 −5
Since U1 is upper triangular. then.
Block LDU Explanation
94
. From here. Then: Example Let's find the LU decomposition of M = U0 = L1 = 1 0 . and it can be removed by subtracting 2 times the first row of M to the second row of M . With a larger matrix. and U a rectangular matrix.
11. Then L will be an m × m matrix. our decomposition will consist of a 2 × 2 matrix and a 2 × 3 matrix.For matrices that are not square. where D is block diagonal. we're done. Then M can be decomposed as a block LDU decomposition. Given an m × n matrix M . Since M is −4 4 1 a 2 × 3 matrix. 0 1 The next step is to zero-out the first column of M below the diagonal. we start with L0 = I and U0 = M . and U will be an m × n matrix (of the same shape as M ). for example we could write M = LU with L a square lower unit triangular matrix.3
Block LDU Decomposition
Let M be a square block matrix with square blocks X. We create a sequence of matrices Li and Ui that is eventually the LU decomposition. Y. X Y Z W
This can be checked explicitly simply by block-multiplying these three matrices. LU decomposition still makes sense. as follows: M= Then: M= I 0 −1 ZX I X 0 0 W − ZX −1 Y I X −1 Y 0 I . Z. 1 0 Then we start with L0 = I2 = . There is only one row to cancel.
but since the top line of any permutation is always the same. . For example. 2. and then swap 1 and 3.12. 5. (It's simplest to use the minimum number of swaps.
ˆ There are n! permutations of n distinct objects. but you don't have to: it turns out that any way of building up the permutation from swaps will have have the same parity of swaps. n} to [n]. and write ρ(3) = 5 from the above example. The mathematics of permutations is extensive and interesting. n − 1 choices for the second once the first has been chosen. to build up the permutation 3 1 2 from the trivial permutation 1 2 3 .) If this number happens to be even. you can first swap 2 and 3. there is some number of swaps it takes to build up the permutation. if this number is odd.1
Permutations
Consider n objects labeled 1 through n and shuffle them. n! is even for all n ≥ 2. . In general we can write 1 2 3 4 5 σ(1) σ(2) σ(3) σ(4) σ(5). 3]. then σ is called an even permutation. either even or odd. In fact. . There is also one more notation called cycle notation. 2. Each possible shuffle is called a permutation σ. For example. then σ is an odd permutation. here is an example of a permutation of 5: 1 2 3 4 5 ρ= 4 2 5 1 3 We can consider a permutation σ as a function from the set of numbers [n] := {1. 1. but we do not discuss it here. and so on. ˆ For any given permutation σ. . we can omit the top line and just write: σ = σ(1) σ(2) σ(3) σ(4) σ(5) so we can just write ρ = [4. and exactly half of the
97
. since there are n choices for the first object. ˆ Every permutation can be built up by successively swapping pairs of objects. there are a few properties of permutations that we'll need.
Definition For an n × n matrix M . since it uses zero swaps.
Example Because there are many permutations of n.1. there are 24 = 4! permutations. If M has a row consisting entirely of zeros. Definition The sign function is a function sgn(σ) that sends permutations to the set {−1. writing the determinant this way for a general matrix gives a very long sum. and for n = 5. then mi = 0 σ(i) for every σ. −1 if σ is odd. there are already 120 = 5! permutations.permutations are even and the other half are odd. σ(1) σ(2) σ(n)
The sum is over all permutations of n. Then det M = 0. the determinant of M (sometimes written |M |) is given by: det M =
σ
sgn(σ) m1 m2 · · · mn . The last statement about the summands yields a nice property of the determinant: Theorem 12.
98
. Each summand is a product of a single entry from each row. see Problem 4. defined by: sgn(σ) = 1 if σ is even. It's worth noting that the trivial permutation (which sends i → i for every i) is an even permutation.
Permutation Example
Reading homework: problem 12.
For more on the swaps (also known as inversions) and the sign function.1 We can use permutations to give a definition of the determinant. 1}. but with the column numbers shuffled by the permutation σ. For n = 4.
then det M = 0. for any matrix M .e.Thus we see that swapping rows changes the sign of the determinant. In other words. If two rows of the matrix are identical. Reading homework: problem 12. This implies another nice property of the determinant. M = − det M .2
Elementary Matrices
Our next goal is to find matrices that emulate the Gaussian row operations on a matrix. and a matrix M equal to M after a row operation. i where the matrix Ej is the identity matrix with rows i and j swapped. Then we see the following:
Theorem 12. but leaves the matrix unchanged. If M has two identical rows. It is our first example of an elementary matrix and we will meet it again soon.
12. we wish to find a matrix R such that M = RM . then swapping the rows changes the sign of the matrix. rows i and j of M are swapped. I.
See Some Ideas Explained
We will first find a matrix that. when it multiplies a matrix M .
100
.2 Applying this result to M = I (the identity matrix) yields
i det Ej = −1 .2.
1 and I. This result hints at an even better one for determinants of products of determinants.3 Beezer. Section PDM Wikipedia:
ˆ Determinant ˆ Permutation ˆ Elementary Matrix
102
. Chapter Four. Chapter D. Section DM. Stare at it again before reading the next Lecture:
References
Hefferon.Moreover. i i det Ej = −1 and det Ej M is the matrix M with rows i and j swapped we have that i i det Ej M = det Ej det M . Chapter D. Section I. Subsection EM Beezer. since swapping a pair of rows flips the sign of the determinant.
1 (c) Find a matrix S2 (λ) that adds a multiple λ of row 2 to row 1 under left multiplication. 4. we can make a new permutation τi. This problem is a "hands-on" look at why the property describing the parity of permutations is true. and 2 and 3 both come before 1. For simplicity. for the permutation ρ = [4.j σ by exchanging the ith and jth entries of σ. 1]. it's the number of "numbers that appear left of smaller numbers" in the permutation. and 1. 2. i 3. 3. the inversion number is 5. For example. 3. 1 1 2 1 2 Prove that M is non-singular if and only if: m1 m2 m3 − m1 m2 m3 + m1 m2 m3 − m1 m2 m3 + m1 m2 m3 − m1 m2 m3 = 0 1 2 3 1 3 2 2 3 1 2 1 3 3 1 2 3 2 1
1 2. assume that m1 = 0 = m1 m2 − m2 m1 . Let M be a matrix and Sj M the same matrix with rows i and j switched. (a) What is the inversion number of the permutation µ = [1. Let M = m2 m2 m2 . The number 4 comes before 2.Review Problems
m1 m1 m1 3 2 1 1. 2.
4. Use row operations to put M into row 1 2 3 m3 m3 m3 1 2 3 echelon form. (a) What does the matrix E2 =
0 1 a b do to M = 1 0 d c left multiplication? What about right multiplication?
under
(b) Find elementary matrices R1 (λ) and R2 (λ) that respectively multiply rows 1 and 2 of M by λ but otherwise leave M the same under left multiplication. Given a permutation σ. 3] that exchanges 4 and 3 and leaves everything else alone? Is it an even or an odd permutation?
103
. The inversion number of a permutation σ is the number of pairs i < j such that σ(i) > σ(j). Explain every line of the series of equations proving that i det M = − det(Sj M ).
5
104
. 1] that exchanges 1 and 4 and leaves everything else alone? Is it an even or an odd permutation? (c) What is the inversion number of the permutation τ1. 3].4 ρ. However
The parity of an integer refers to whether the integer is even or odd. Thus permutations of [n] under composition are an example of a group.i+1 ).(b) What is the inversion number of the permutation ρ = [4.3 µ. and these are associative. In particular.4 ρ? Compare the parity of ρ to the parity of τ2. and to sort µ we have s3 µ = [1. 2. 2. Here the parity of a permutation µ refers to the parity of its inversion number. let µ = [1. 2. 4]. 4. (d) What is the inversion number of the permutation τ2. 3. For this problem. (b) We can compose simple transpositions together to represent a permutation (note that the sequence of compositions is not unique). (Extra credit) Here we will examine a (very) small set of the general properties about permutations and their applications. we will show that one way to compute the sign of a permutation is by finding the inversion number N of σ and we have sgn(σ) = (−1)N . and we have an inverse since it is clear that si si σ = σ. (e) What is the inversion number of the permutation τ3. we have an identity (the trivial permutation where the list is in order or we do nothing on our list).3 µ? Compare the parity5 of µ to the parity of τ1. 4. (a) Show that every permutation σ can be sorted by only taking simple (adjacent) transpositions si where si interchanges the numbers in position i and i + 1 of a permutation σ (in our other notation si = τi.4 ρ? Compare the parity of ρ to the parity of τ3. 3.
Problem 4 hints
5. For example s2 µ = [1. 3].4 ρ. 2.
n]. 2.j in terms of simple transpositions? Is τi. 2] s2 s1 [1. 3] = s1 [1.j an even or an odd permutation? What is det(Mτi. and we have det(Msi ) = det(Ei+1 ) = −1. N Thus prove that det(Mσ ) = (−1) where N is a number of simple transpositions needed to represent σ as a permutation. Consider using diagrams of these paths to help. .note that not all simple transpositions commute with each other since s1 s2 [1. . You can assume that Msi sj = Msi Msj (it is not hard to prove) and that det(AB) = det(A) det(B) from Chapter 13. (c) Show that every way of expressing σ can be obtained from using the relations proved in part 5b. we do not write it. When we consider our initial permutation to be the trivial permutation e = [1. 3. 1] (you will prove here when simple transpositions commute). 3] = s2 [2. 3. In other words. follow the path of the (n + 1)-th strand by looking at sn sn−1 · · · sk sk±1 · · · sn and argue why you can write this as a subexpression for any expression of e. (e) Show that si+1 si si+1 = τi. Show that si si = e (in shorthand s2 = e).i+2 . using the proved relations. 1. This is analogous to not writing 1 when multiplying.j ? (f) The minimal number of simple transpositions needed to express σ is called the length of σ. and si and sj commute for all |i − j| ≥ 2. for example si ≡ si e and µ = s3 ≡ s3 e. 2. . Hint: You to make sure det(Mσ ) is well-defined since there are infinite ways to represent σ as simple transpositions. 3] = [2. 1. 2. Therefore we can just represent a permutation i σ as the matrix Mσ 6 . and so give one way of writing τi. 2] = [3. Hint: Use induction on n. for example the length of µ is 1 since
6
Often people will just use σ for the matrix when the context is clear. . For the induction step. si+1 si si+1 = si si+1 si for i all i. show that for any expression w of simple transpositions representing the trivial permutation e.
105
. (d) The simple transpositions acts on an n-dimensional vector space i i V by si v = Ei+1 v (where Ej is an elementary matrix) for all vectors v ∈ V .j )? What is the inversion number of τi.
µ = s3 . Hint: Find an procedure which gives you a new permutation σ where σ = si σ for some i and the inversion number for σ is 1 less than the inversion number for σ.
106
. Note that this immediately implies that sgn(σρ) = sgn(σ) sgn(ρ) for any permutations σ and ρ. where σ is a permutation with N inversions. (g) Show that (−1)N = sgn(σ) = det(Mσ ). Show that the length of σ is equal to the inversion number of σ.
Suppose that RREF(M ) is the reduced row echelon form of M . then det(EM ) = det E det M . and we have seen that any row operation can be emulated with left matrix multiplication by an elementary matrix. Then RREF(M ) = E1 E2 · · · Ek M where each Ei is an elementary matrix.1.
Reading homework: problem 13. If E is any of the elementary matrices Ej .We have also proved the following theorem along the way:
i i Theorem 13. What is the determinant of a square matrix in reduced row echelon form? 110
.1
We have seen that any matrix M can be put into reduced row echelon form via a sequence of row operations. Sj (λ). Ri (λ).
this scales the determinant of M by λ. ˆ Otherwise. then some row of RREF(M ) contains only zeros. ˆ Additionally. det M = 0 if and only if M is invertible. Thus. since M is square. det RREF(M ) = λ det RREF(M ). the inverse of an elementary matrix is another elementary matrix. Ri (λ). Since we know the determinants of the elementary matrices. notice that det RREF(M ) = det(E1 E2 · · · Ek M ). every row of RREF(M ) has a pivot on the diagonal. 111
. and so det RREF(M ) = 0. if M is not invertible. then det RREF(M ) = 0 if and only if det M = 0. In fact.3.2. M = E1 E2 · · · Ek RREF(M ) .
Then we have shown: Theorem 13. For any square matrix M . det RREF(M ) = 1. for elementary matrices Ei and Fi . with reduced row echelon forms such that. Sj (λ) is invertible. this means that RREF(M ) is the identity matrix. suppose that M and N are square n × n matrices. Any elementary matrix Ej .ˆ If M is not invertible. Since each Ei has non-zero determinant. det RREF(M ) = det(E1 ) · · · det(Ek ) det M .
To obtain one last important result. by our previous observation. Then by the theorem above. we can immediately obtain the following:
Determinants and Inverses
i i Corollary 13. Then if M is invertible. Then we can multiply the zero row by any constant λ without changing M . except i for R (0).
Suppose M =
a b is invertible. Ej . Find the inverses of each of the elementary matrices. 5. (Extra Credit: (Directional) Derivative of the Determinant) Notice that det : Mn → R where Mn is the vector space of all n × n matrices. (Extra Credit) Let ei denote the matrix with a 1 in the i-th row and j-th j column and 0's everywhere else. Ri (λ). Make sure to show that the elementary matrix times its inverse is actually the identity.2. and what is first order term (the coefficient of t)? Can you express your results in terms of tr(A)? What about the first order term in det(A + tIn ) for any arbitrary n × n matrix A in terms of tr(A)? We note that the result of det(A + tI2 ) is a polynomial in the variable t and by taking t = −λ is what is known as the characteristic polynomial from Chapter 18. and so we can take directional derivatives of det.
4.
i i 3. Sj (λ). Let A be an arbitrary n × n matrix. and for all i and j compute the following: (a) lim (b) lim (c) lim (d) lim det(In + At) − det(In ) t→0 t det(In + tei ) − det(In ) j t→0 t det(I3 + tei ) − det(I3 ) j t→0 t det(I2 + tei ) − det(I2 ) j t→0 t
114
. Write M as a product of elemenc d tary row matrices times RREF(M ). Compute det(A + tI2 ). and let A be an arbitrary 2 × 2 matrix.
(Recall that what you are calculating is the directional derivative in the ei and A directions.) Can you express your results in terms of the j trace function? Hint: Use the results from Problem 4 and what you know about the derivatives of polynomials evaluated at 0 (i.e. what is p (0)?).
115
14
Properties of the Determinant
In Lecture 13 we showed that the determinant of a matrix is non-zero if and only if that matrix is invertible. We also showed that the determinant is a multiplicative function, in the sense that det(M N ) = det M det N . Now we will devise some methods for calculating the determinant. Recall that: det M =
σ
sgn(σ)m1 m2 · · · mn . σ(1) σ(2) σ(n)
A minor of an n × n matrix M is the determinant of any square matrix obtained from M by deleting rows and columns. In particular, any entry mi j of a square matrix M is associated to a minor obtained by deleting the ith row and jth column of M . It is possible to write the determinant of a matrix in terms of its minors as follows: det M =
σ
Here the symbols σ refer to permutations of n − 1 objects. What we're doing ˆ here is collecting up all of the terms of the original sum that contain the first row entry m1 for each column number j. Each term in that collection j is associated to a permutation sending 1 → j. The remainder of any such permutation maps the set {2, . . . , n} → {1, . . . , j − 1, j + 1, . . . , n}. We call this partial permutation σ = σ(2) · · · σ(n) . ˆ The last issue is that the permutation σ may not have the same sign as σ. ˆ From previous homework, we know that a permutation has the same parity as its inversion number. Removing 1 → j from a permutation reduces the inversion number by the number of elements right of j that are less than j. Since j comes first in the permutation j σ(2) · · · σ(n) , the inversion 116
Example
A fun exercise is to compute the determinant of a 4 × 4 matrix filled in order, from left to right, with the numbers 1, 2, 3, . . . 16. What do you observe? Try the same for a 5 × 5 matrix with 1, 2, 3 . . . 25. Is there a pattern? Can you explain it?
7
For any permutation σ, there is a unique inverse permutation σ −1 that undoes σ. If σ sends i → j, then σ −1 sends j → i. In the two-line notation for a permutation, this corresponds to just flipping the permutation over. For 1 2 3 example, if σ = , then we can find σ −1 by flipping the permutation 2 3 1 and then putting the columns in order: σ −1 = 2 3 1 1 2 3 = 1 2 3 3 1 2
Since any permutation can be built up by transpositions, one can also find the inverse of a permutation σ by undoing each of the transpositions used to build up σ; this shows that one can use the same number of transpositions to build σ and σ −1 . In particular, sgn σ = sgn σ −1 . Reading homework: problem 14.1 Then we can write out the above in formulas as follows: det M =
σ
sgn(σ)m1 m2 · · · mn σ(1) σ(2) σ(n) sgn(σ)m1
σ σ −1 (1)
= =
σ
m2
σ −1 (2)
σ · · · mn
−1 (n)
sgn(σ −1 )m1 sgn(σ)m1
σ
σ −1 (1)
m2
σ −1 (2)
· · · mσ n
−1 (n)
=
σ(1)
m2
σ(2)
σ(n) · · · mn
= det M T . The second-to-last equality is due to the existence of a unique inverse permutation: summing over permutations is the same as summing over all inverses of permutations. The final equality is by the definition of the transpose. 118
Let's multiply M adj M . For any matrix N , the i, j entry of M N is given by taking the dot product of the ith row of M and the jth column of N . Notice that the dot product of the ith row of M and the ith column of adj M is just the expansion by minors of det M in the ith row. Further, notice that the dot product of the ith row of M and the jth column of adj M with j = i is the same as expanding M by minors, but with the jth row replaced by the ith row. Since the determinant of any matrix with a row repeated is zero, then these dot products are zero as well. We know that the i, j entry of the product of two matrices is the dot product of the ith row of the first by the jth column of the second. Then: M adj M = (det M )I Thus, when det M = 0, the adjoint gives an explicit formula for M −1 . Theorem 14.3. For M a square matrix with det M = 0 (equivalently, if M is invertible), then 1 adj M M −1 = det M
Figure 1: A parallelepiped. Chapter D. Section DM.
This process for finding the inverse matrix is sometimes called Cramer's Rule . we know that the volume of this object is |u (v × w)|. Section DM. v.3
Application: Volume of a Parallelepiped
Given three vectors u.3 Beezer. the parallelepiped determined by the three vectors is the "squished" box whose edges are parallel to u.
14.1 and I. Then: Volume = det u v w
References
Hefferon. This is the same as expansion by minors of the matrix whose columns are u. Chapter Four. Subsection CD Wikipedia: 122
. From calculus. w in R3 . Chapter D. v. Subsection DD Beezer. and w as depicted in Figure 1. w. Section I. v.
Find and verify a similar formula for det M in terms of tr(M 3 ). Then. Suppose M = LU is an LU decomposition.ˆ Determinant ˆ Elementary Matrix ˆ Cramer's Rule
Review Problems
1. Let M = a b . (a) How many additions and multiplications does it take to compute the determinant of a general 2 × 2 matrix? (b) Write a formula for the number of additions and multiplications it takes to compute the determinant of a general n × n matrix using the definition of the determinant. (c) How many additions and multiplications does it take to compute the determinant of a general 3 × 3 matrix using expansion by minors? Assuming m = 2a. tr(M 2 ). computing 2 · 6 − 5 would take a + m seconds. is this faster than computing the determinant from the definition?
Problem 3 hint
123
. Assume that finding and multiplying by the sign of a permutation is free. Explain how you would efficiently compute det M in this case. In computer science. 3. and multiplying two numbers takes m seconds. Show: c d 1 1 det M = (tr M )2 − tr(M 2 ) 2 2 Suppose M is a 3 × 3 matrix. the complexity of an algorithm is (roughly) computed by counting the number of times a given operation is performed. Suppose adding or subtracting any two numbers takes a seconds. and tr M . for example. 2.
1
Subspaces
Definition We say that a subset U of a vector space V is a subspace of V if U is a vector space under the inherited addition and scalar multiplication operations of V . Basis: How do we label vectors? Can we write any vector as a sum of some basic set of vectors? How do we change our point of view from vectors labeled one way to vectors labeled in another way? Let's start at the top!
15.15
Subspaces and Spanning Sets
It is time to study vector spaces more carefully and answer some fundamental questions. or if one is a "linear combination" of the others? 3.
124
.
Example Consider a plane P in R3 through the origin: ax + by + cz = 0. Linear Independence: Given a collection of vectors. Dimension: Is there a consistent definition of how "big" a vector space is? 4. is there a way to tell whether they are independent.) 2. 1. Subspaces: When is a subset of a vector space itself a vector space? (This is the notion of a subspace.
125
. x This equation can be expressed as the homogeneous system a b c y = 0. All other vector space requirements hold for P because they hold for all vectors in R3 . then. P contains the origin (which can be derived from the above by setting µ = ν = 0). so is µX1 + νX2 : M (µX1 + νX2 ) = µM X1 + νM X2 = 0. by linearity of matrix multiplication. u2 in U . z or M X = 0 with M the matrix a b c . Proof. We know that the additive closure and multiplicative closure properties are satisfied. Additionally. Each of the other eight properties is true in U because it is true in V . Note that the requirements of the subspace theorem are often referred to as "closure". u2 in U and all constants µ. That is. then U is a vector space. Then U is a subspace if and only if µu1 + νu2 ∈ U for arbitrary u1 . ν. then it is a vector space. The other direction is almost as easy: we need to show that if µu1 +νu2 ∈ U for all u1 . If X1 and X2 are both solutions to M X = 0. and arbitrary constants µ. The details of this are left as an exercise. ν. One direction of this proof is easy: if U is a subspace. and so by the additive closure and multiplicative closure properties of vector spaces. Let U be a non-empty subset of a vector space V . we need to show that the ten properties of vector spaces are satisfied. u2 in U and all constants constants µ. So P is closed under addition and scalar multiplication. it has to be true that µu1 + νu2 ∈ U for all u1 .1 (Subspace Theorem). ν.
Theorem 15.
2
Building Subspaces
0 1 U = 0 . nor is U closed under vector addition. it clear that U is not a vector space. since any constant multiple of these vectors should also be in U . U is not a vector space. That is. 0 0
Consider the set
Because U consists of only two vectors. the 0-vector is not in U . do we end up with a vector in U ? And. Reading homework: problem 15. For example.From now on. to check if U itself forms a smaller vector space we need check only two things: if we add any two vectors in U . But we know that any two vectors define a plane:
In this case. 1 ⊂ R3 . then U is a vector space. the vectors in U define the xy-plane in R3 . Call this set of all linear combinations the span of U : 1 0 span(U ) = x 0 + y 1 x. We can consider the xy-plane as the set of all vectors that arise as a linear combination of the two vectors in U . If not. if we multiply any vector in U by any constant. if we have some set U of vectors that come from some bigger vector space V . y ∈ R . we can use this theorem to check if a set is a vector space.1
15. 0 0 126
. do we end up with a vector in U ? If the answer to both of these questions is yes.
.
8
127
. Let P = 1. (Try drawing a picture to verify this!)
Usually our vector spaces are defined over R. It can be any finite number. The elements of span(S) are linear combinations of vectors in the x-axis and the vector P. Any finite sum of the form (a constant times s1 plus a constant times s2 plus a constant times s3 and so on) is in the span of S.5 is in span(S). On the other hand.
0 3 and X ⊂ V be the x-axis. 0 0 0 Definition Let V be a vector space and S = {s1 . because 17. That is. . but it must be finite. s2 . which is a vector space. N must be a finite number. Similarly. 0 2 2 2 The vector 3 is in span(S). vector 0 0 0 0 any vector of the form x 0 x 0 + y 1 = y 0 0 0 is in span(S). (Why?) So span(S) is the xy-plane. any vector in span(S) must have a zero in the z-coordinate. Similarly. N ∈ N}.5 = 0 + 17. The coefficients ri should come from whatever our base field is (usually R). Then the span of S is the set: span(S) = {r1 s1 + r2 s2 + · · · + rN sN |ri ∈ R.Notice that any vector in the xy-plane is of the form x 1 0 y = x 0 + y 1 ∈ span(U ).} ⊂ V a subset of V . but in general we can have vector spaces defined over different base fields such as C or Z2 .5 1 . . and set Example Let V = R 0 S =X ∪P . the 0 0 0 0 −12 −12 −12 0 17. because 3 = 0 + 3 1 . the span of S is the set of all finite linear combinations8 of elements of S. In the definition above. It is important that we only allow finite linear combinations.
. . Definition We say that the vectors v1 . . Then there exist constants d1 . vn are linearly independent. .16
Linear Independence
Consider a plane P that includes the origin in R3 and a collection {u.
Usually our vector spaces are defined over R. More generally. . Otherwise. then P = span{u. v. v. d2 (not both zero) such that w = d1 u + d2 v. so we should be able to span the plane using only two of the vectors u. But any two vectors determines a plane. . . v2 . w}. v. The coefficients ci should come from whatever our base field is (usually R). and express the other as a linear combination of those two. Since w can be expressed in terms of u and v we say that it is not independent. v2 . but in general we can have vector spaces defined over different base fields such as C or Z2 . the relationship c1 u + c2 v + c3 w = 0 ci ∈ R. the vectors v1 . . w} whose span is P . Suppose u and v span P .
9
131
. some ci = 0
expresses the fact that u. v. . Then we could choose two of the vectors in {u. cn not all zero such that c1 v1 + c2 v2 + · · · + cn vn = 0. . w} of non-zero vectors in P :
If no two of u. . vn are linearly dependent if there exist constants9 c1 . w. w are not all independent. v. c2 . v and w are parallel. .
. we must check that every linear combination of our vectors with non-vanishing coefficients gives something other than the zero vector. .
Worked proof
Example Consider the vector space P2 (t) of polynomials of degree less than or equal to 2. . . or we can express one of the vectors as a linear combination of the other vectors. we must show that the equation c1 v1 + c2 v2 + · · · + cn vn = 0 has no solutions other than c1 = c2 = · · · = cn = 0. v5 } is linearly dependent. On the other hand. and we are done. because v4 = v1 + v2 .
Example Consider the following vectors in R3 : 0 2 v1 = 0 . Equivalently. 3
134
. . v2 . Set: v1 = 1 + t v2 = 1 + t 2 v3 = t + t 2 v4 = 2 + t + t2 v5 = 1 + t + t2 . v2 = 2 . . . .
We have seen two different ways to show a set of vectors is linearly dependent: we can either find a linear combination of the vectors which is equal to zero.
The set {v1 . 2 1 Are they linearly independent? We need to see whether the system c1 v1 + c2 v2 + c3 v3 = 0
1 v3 = 4 .Therefore we have expressed vk as a linear combination of the previous vectors. vn is linearly independent. to show that the set v1 . to check that a set of vectors is linearly independent.
Let B n be the space of n × 1 bit-valued matrices (i. find a basis of B 3 . . . (a) How many different vectors are there in B n ? (b) Find a collection S of vectors that span B 3 and are linearly independent. . column vectors) over the field Z2 := Z/2Z. (b) Demonstrate that v =
n i=1 (v
ei )ei . en } is linearly independent.e. (a) Show that the collection {e1 . (d) Would it be possible to span B 3 with only two vectors?
Hint for Problem 1
2.Review Problems
1. (c) Write each other vector in B 3 as a linear combination of the vectors in the set S that you chose. . In other words. Let v be an arbitrary vector in Rn . with rules for adding and multiplying coefficients given here. Remember that this means that the coefficients in any linear combination can be only 0 or 1.. . . . en } is the same as what vector space?
138
. Let ei be the vector in Rn with a 1 in the ith position and 0's in every other position.
(c) The span{e1 . .
. Proof. This means that the dimension of a vector space does not depend on the basis. then c0 = c1 = · · · = cn = 0. . . In fact. and of a set of vectors that span V . so p(t) is the zero polynomial. and so there exist constants ci such that w = c1 v1 + · · · + cn vn . t. dimension is a very important way to characterize of any vector space V . Then Pn (t) is finite dimensional. such a set is called a basis of the subspace V . Suppose V is a finite-dimensional vector space.
Theorem 17.
139
. tn }. . we established the notion of a linearly independent set of vectors in a vector space V . then span S = V . Definition Let V be a vector space. . If S is a basis of V and S has only finitely many elements. and dim Pn (t) = n + 1. then we say that V is finite-dimensional. Let S = {v1 . . ai ∈ R
so Pn (t) = span{1. vn } be a basis for a vector space V . .
Example Pn (t) has a basis {1.1. . This set of vectors is linearly independent: If the polynomial p(t) = c0 1 + c1 t + · · · + cn tn = 0.17
Basis and Dimension
In Lecture 16. We saw that any set of vectors that span V can be reduced to some minimal collection of linearly independent vectors. Later in this section. Luckily this isn't what happens. . tn }. we will show that S and T must have the same number of vectors. . . t. . The number of vectors in S is the dimension of V . Then a set S is a basis for V if S is linearly independent and V = span S. and S and T are two different bases for V . One might worry that S and T have a different number of vectors. since every polynomial of degree less than or equal to n is a sum a0 1 + a1 t + · · · + an tn . then we would have to talk about the dimension of V in terms of the basis S or in terms of the basis T . Then every vector w ∈ V can be written uniquely as a linear combination of vectors in the basis S: w = c1 v1 + · · · + cn vn . Since S is a basis for V .
1 Proof. which is a contradiction since the vectors vi are assumed to be non-zero. vn } is linearly dependent. which contradicts the assumption that S is linearly independent. . . using that equation. ci = di . . . The idea of the proof is to start with the set S and replace vectors in S one at a time with vectors from T . such that after each replacement we still have a basis for V .
Proof of Theorem
Next. then m ≤ n. we can use this last equation to write one of the vectors in S as a linear combination of other vectors in S. Then we can write w1 as a linear combination of the vi . . we would like to establish a method for determining whether a collection of vectors forms a basis for Rn .Suppose there exists a second set of constants di such that w = d1 v1 + · · · + dn vn .
If it occurs exactly once that ci = di .2. If we have more than one i for which ci = di . . Since S spans V . . Then: 0V = w−w = c1 v1 + · · · + cn vn − d1 v1 + · · · + dn vn = (c1 − d1 )v1 + · · · + (cn − dn )vn . . we need to show that any two bases for a finite-dimensional vector space has the same number of vectors. then the set {w1 . . vn } is a basis for a vector space V and T = {w1 . wm } is a linearly independent set of vectors in V . . Lemma 17. But first. v1 . If S = {v1 . . . Reading homework: problem 17. we can express one of the vi in terms of w1 and the remaining vj with j = 140
. Then for every i. then the equation reduces to 0 = (ci − di )vi .
. . . But we still have some vector wn+1 in T that is not in Sn . . vin−m }. there was a unique way to express w1 in terms of the set S. and the set Sn = {w1 . we have m > n. Now we need to prove that S1 is a basis. vn } is linearly independent: By the previous theorem. wn } is a basis for V . . Then no vector in S1 can be expressed as a combination of other vectors in S1 . and had a non-zero coefficient on the vector vi . to obtain a contradiction. v1 . this process ends with the set Sm = {w1 . . The set S1 spans V : For any u ∈ V . . . Now. . wm .i. which is fine. . . vi1 . we can express u as a linear combination of vectors in S. we can write wn+1 as a combination of the vectors in Sn . . . Then S1 is a basis of V with n vectors. Then replacing w1 with its expression in terms of the collection S gives a way to express the vector vk as a linear combination of the vectors in S. we cannot express w1 as a linear combination of the vectors in {vj |j = i}. The set S1 = {w1 . . vi−1 . . . rewriting vi as such allows us to express u as a linear combination of the vectors in S1 . since the expression of w1 in terms of S was unique. which contradicts the linear independence of S. we need to show that S1 is linearly independent and that S1 spans V . On the other hand. But we can express vi as a linear combination of vectors in the collection S1 . . . Since Sn is a basis. suppose there is some k and constants ci such that vk = c0 w1 + c1 v1 + · · · + ci−1 vi−1 + ci+1 vi+1 + · · · + cn vn . which 141
. which demonstrates that S1 is linearly independent. . Otherwise. replacing one of the vi in S1 with w2 . vi+1 . If m ≤ n. We can now iterate this process. . and so on. Then we can discard one of the vi from this set to obtain a linearly independent set that still spans V .
. . Then it must be the case that m ≤ n. and has unit length.1
Bases in Rn . . Let S and T be two bases for V . . so rather than writing a set of vectors. any two bases for V have the same number of vectors. Reading homework: problem 17. . j. . this basis is often written {i. But (exchanging the roles of S and T in application of the lemma) we also see that n ≤ m. . the canonical ordered basis for Rn is (e1 . .
Worked Example
Corollary 17. Suppose S has n vectors and T has m vectors. . as desired. The vector with a one in the ith position and zeros everywhere else is written ei . .
From one of the review questions. we have that m ≤ n. Proof. . Then both are linearly independent sets that span V . k} for R3 . . and dim Rn = n. For example. Then m = n. . we know that 1 0 0 0 0 1 Rn = span . This is called an ordered basis. . For a finite-dimensional vector space V . .contradicts the linear independence of the set T . we would write a list.3. 0 0 1 and that this set of vectors is linearly independent. en ). Then by the previous lemma. Note that it is often convenient to order basis elements.2
17. e2 . This basis is often called the standard or canonical basis for Rn . It points in the direction of the ith coordinate axis. . . . as desired. In multivariable calculus classes. . The possibility to reorder basis vectors is not the only way in which bases are non-unique:
142
. So this set of vectors is a basis for Rn .
Bases are not unique. Let M be a matrix whose columns are the vectors vi . While there exists a unique way to express a vector in terms of any particular basis. (See Review Question 3. . Rescaling any vector in one of these sets is already enough to show that R2 has infinitely many bases. it turns out that there are still infinitely many bases for R2 . Then the above equation is equivalent to requiring that there is a unique solution to MX = 0 . But even if we require that all of the basis vectors have unit length. From the previous discussion.)
To see whether a collection of vectors S = {v1 . −1 1 1 0 are bases for R2 . we take an arbitrary vector w and solve the linear system w = x1 v1 + · · · + xn vn 143
. vm } is a basis for Rn . we have to check that they are linearly independent and that they span Rn . then there is no non-trivial solution of the equation 0 = x1 v1 + · · · + xn vn . bases themselves are far from unique. To see if S spans Rn . we also know that m must equal n. and . . so assume S has n vectors. . If S is linearly independent. both of the sets: 1 1 0 1 . For example. .
. there is exactly one set of constants c1 . cn so that c1 v1 + · · · + cn vn = w. For which x 0 is Sx a basis of R2 ?
2. then S is a basis of V . In other words: suppose that for every vector w in V . (Hint: Suppose that you have a set of n vectors which span V but do not form a basis. wm } be a collection of n linearly independent vectors in V . How many vectors are in the span of any one vector? Any two vectors? How many vectors are in the span of any k vectors. . and let {v1 . . . Show that this means that the set S is linearly independent and spans V . . . .(b) Let Sx =
1 . Suppose that V is an n-dimensional vector space. where x is a unit vector in R2 . (This is the converse to the theorem in the lecture. What must be true about them? How could you get a basis from this set? Use Corollary 17. . Show that if every vector w in V can be expressed uniquely as a linear combination of vectors in S. . . such that the vector you choose is not in the span of the previous vectors you've chosen.) (b) Show that any set of n vectors in V which span V forms a basis for V . Apply the method of Lemma 17. for k ≤ n?)
Hint for Problem 2
3. . . 1. . Let S be a collection of vectors {v1 . vn } be a basis for V .) 4. . vn } in a vector space V . Let B n be the vector space of column vectors with bit entries 0. .3 to derive a contradiction. . How many bases are there for B 3 ? B 4 ? Can you make a conjecture for the number of bases for B n ? (Hint: You can build up a basis for B n by choosing one vector at a time.)
145
. (a) Show that any n linearly independent vectors in V form a basis. x .2 to these two sets of vectors. (Hint: Let {w1 . Write down every basis for B 1 and B 2 .
146
. Do you think your proof could be modified to work for linear transformations Rn → R? Hint: Represent R3 as column vectors. show that the set of all linear transformations mapping R3 → R is itself a vector space. If you are stuck or just curious. Vectors are objects that you can add together. and argue that a linear transformation T : R3 → R is just a row vector. Find a basis for this vector space. see dual space.5.
e2 } it is the ordered pair (x. In the basis {e1 . This can be confusing. with bases S = {e1 . things like coordinate axes and "components of a vector" (x. On the other hand.18
Eigenvalues and Eigenvectors
Before discussing eigenvalues and eigenvectors. en } and T = {f1 . fm } respectively. the idea to keep firm in your mind is that the vector space and its elements—vectors—are what really "exist". t) = (2. 2) while in the basis {f1 . y) are just mathematical tools used to label vectors. f2 } is corresponds to (s. Since these are bases.1
Matrix of a Linear Transformation
Let V and W be vector spaces. . we need to have a better understanding of the relationship between linear transformations and matrices. there exist constants v i and wj such that any vectors v ∈ V and w ∈ W can be written as: v = v 1 e1 + v 2 e2 + · · · + v n en w = w 1 f1 + w 2 f2 + · · · + w m fm
147
. .
18. y) = (2. 1). . . . . as an example the plane R2
The information of the vector v can be transmitted in many ways. Typically they will correspond to configurations of the real world system you are trying to describe. Consider. . .
e. . and zeros everywhere else. 1 0 .
This is the matrix of L in the basis
Example Any vector in Rn can be written as a linear combination of the standard basis vectors {ei |i ∈ {1. . . . . . . . . . Then if the ith column of M equals L(ei ) for every i. . I. it suffices to know what L(ei ) is for every i. . y
acts by matrix multiplication in the same way that L does.
18. n}}. observe that M ei is equal to the ith column of M . Then the matrix representing L in the standard basis is just the matrix whose ith column is L(ei ). . 0 0 1 Then to find the matrix of any linear transformation L : Rn → Rn . we have y x y = ax + by cx + dy =L x . . For any matrix M . . .Now notice that for any vector a b c d Then the matrix a b c d
x . then M v = L(v) for every v ∈ Rn . 0 1 .2
Invariant Directions
Have a look at the linear transformation L depicted below:
151
. e2 = . en = . The vector ei has a one in the ith position.
Recall that a vector is a direction and a magnitude. L applied to both the direction and the magnitude of the vectors given to it. Now look at the second picture on that line. 1 0 or 0 1 changes 1 0 = −4 −10 and L 0 1 = 3 7 . There. Notice how the unit square with a corner at the origin get mapped to a parallelogram. Now lets try an explicit example to see if we can achieve the last picture:
Example Consider the linear transformation L such that L so that the matrix of L is −4 3 −10 7 . Clearly this is a very special situation that should correspond to a interesting properties of L. 5 −10 · 3 + 7 · 5 5
152
.It was picked at random by choosing a pair of vectors L(e1 ) and L(e2 ) as the outputs of L acting on the canonical basis vectors. two vectors f1 and f2 have been carefully chosen such that if the inputs into L are in the parallelogram spanned by f1 and f2 . The second line of the picture shows these superimposed on one another. Notice that 3 −4 · 3 + 3 · 5 3 L = = . the outputs also form a parallelogram with edges lying along the same two directions.
Then L fixes the direction of the vector v2 =
In short.Then L fixes the direction (and actually also the magnitude) of the vector v1 = In fact also the vector v2 = 1 2 has its direction fixed by M . notice that any vector with the same direction as v1 can be written as cv1 for some constant c.
3 . given a linear transformation L it is sometimes possible to find a vector v = 0 and constant λ = 0 such that L(v) = λv. Then L(cv1 ) = cL(v1 ) = cv1 .
1 but stretches v2 by a factor of 2. L(cv2 ) = cL(v2 ) = 2cv2 . 5
Reading homework: problem 18. Then L stretches every vector pointing in the same direction as v2 by a factor of 2. so L fixes every vector pointing in the same direction as v1 . Also notice that L 1 2 = −4 · 1 + 3 · 2 −10 · 1 + 7 · 2 = 2 4 =2 1 2 .1 Now. 2 Now notice that for any constant c.
153
.
We call the direction of the vector v an invariant direction. In fact, any vector pointing in the same direction also satisfies the equation: L(cv) = cL(v) = λcv. The vector v is called an eigenvector of L, and λ is an eigenvalue. Since the direction is all we really care about here, then any other vector cv (so long as c = 0) is an equally good choice of eigenvector. Notice that the relation "u and v point in the same direction" is an equivalence relation. In our example of the linear transformation L with matrix −4 3 −10 7 ,
we have seen that L enjoys the property of having two invariant directions, represented by eigenvectors v1 and v2 with eigenvalues 1 and 2, respectively. It would be very convenient if we could write any vector w as a linear combination of v1 and v2 . Suppose w = rv1 + sv2 for some constants r and s. Then: L(w) = L(rv1 + sv2 ) = rL(v1 ) + sL(v2 ) = rv1 + 2sv2 . Now L just multiplies the number r by 1 and the number s by 2. If we could write this as a matrix, it would look like: 1 0 0 2 s t
Here, s and t give the coordinates of w in terms of the vectors v1 and v2 . In the previous example, we multiplied the vector by the matrix L and came up with a complicated expression. In these coordinates, we can see that L is a very simple diagonal matrix, whose diagonal entries are exactly the eigenvalues of L. This process is called diagonalization. It makes complicated linear systems much easier to analyze. Reading homework: problem 18.2 Now that we've seen what eigenvalues and eigenvectors are, there are a number of questions that need to be answered. 154
ˆ How do we find eigenvectors and their eigenvalues? ˆ How many eigenvalues and (independent) eigenvectors does a given linear transformation have? ˆ When can a linear transformation be diagonalized?
We'll start by trying to find the eigenvectors for a linear transformation.
This is a homogeneous system, so it only has solutions when the matrix is singular. In other words, det 2−λ 2 16 6−λ = 0
⇔ (2 − λ)(6 − λ) − 32 = 0 ⇔ λ2 − 8λ − 20 = 0 ⇔ (λ − 10)(λ + 2) = 0
155
For any square n × n matrix M , the polynomial in λ given by PM (λ) = det(λI − M ) = (−1)n det(M − λI) is called the characteristic polynomial of M , and its roots are the eigenvalues of M . In this case, we see that L has two eigenvalues, λ1 = 10 and λ2 = −2. To find the eigenvectors, we need to deal with these two cases separately. To do so, we solve the 2−λ 2 x 0 linear system = with the particular eigenvalue λ plugged 16 6−λ y 0 in to the matrix. λ = 10: We solve the linear system −8 2 16 −4 x y = 0 . 0
x will do. Since we only 4x need the direction of the eigenvector, we can pick a value for x. Setting x = 1 1 is convenient, and gives the eigenvector v1 = . 4 Both equations say that y = 4x, so any vector λ = −2: We solve the linear system 4 2 16 8 x y = 0 . 0
Here again both equations agree, because we chose λ to make the system 1 . singular. We see that y = −2x works, so we can choose v2 = −2 In short, our process was the following:
ˆ Find the characteristic polynomial of the matrix M for L, given by11 det(λI − M ). ˆ Find the roots of the characteristic polynomial; these are the eigenvalues of L. ˆ For each eigenvalue λi , solve the linear system (M − λi I)v = 0 to obtain an eigenvector v associated to λi .
Jordan block example
11
It is often easier (and equivalent if you only need the roots) to compute det(M − λI).
(b) When θ = 0, explain how L acts on the plane. Draw a picture. (c) Do you expect L to have invariant directions? (d) Try to find real eigenvalues for L by solving the equation L(v) = λv. (e) Are there complex eigenvalues for L, assuming that i = exists? √ −1
3. Let L be the linear transformation L : R3 → R3 given by L(x, y, z) = (x + y, x + z, y + z). Let ei be the vector with a one in the ith position and zeros in all other positions. (a) Find Lei for each i.
12
Independence of vectors is explained here.
157
m1 m1 m1 1 2 3 (b) Given a matrix M = m2 m2 m2 , what can you say about 1 2 3 m3 m3 m3 1 2 3 M ei for each i? (c) Find a 3 × 3 matrix M representing L. Choose three nonzero vectors pointing in different directions and show that M v = Lv for each of your choices. (d) Find the eigenvectors and eigenvalues of M. 4. Let A be a matrix with eigenvector v with eigenvalue λ. Show that v is also an eigenvector for A2 and what is its eigenvalue? How about for An where n ∈ N? Suppose that A is invertible, show that v is also an eigenvector for A−1 . 5. A projection is a linear operator P such that P 2 = P . Let v be an eigenvector with eigenvalue λ for a projection P , what are all possible values of λ? Show that every projection P has at least one eigenvector. Note that every complex matrix has at least 1 eigenvector, but you need to prove the above for any field.
158
19
Eigenvalues and Eigenvectors II
In Lecture 18, we developed the idea of eigenvalues and eigenvectors in the case of linear transformations R2 → R2 . In this section, we will develop the idea more generally.
Eigenvalues
Definition For a linear transformation L : V → V , then λ is an eigenvalue of L with eigenvector v = 0V if Lv = λv. This equation says that the direction of v is invariant (unchanged) under L. Let's try to understand this equation better in terms of matrices. Let V be a finite-dimensional vector space and let L : V → V . Since we can represent L by a square matrix M , we find eigenvalues λ and associated eigenvectors v by solving the homogeneous system (M − λI)v = 0. This system has non-zero solutions if and only if the matrix M − λI is singular, and so we require that det(λI − M ) = 0. The left hand side of this equation is a polynomial in the variable λ called the characteristic polynomial PM (λ) of M . For an n × n matrix, the characteristic polynomial has degree n. Then PM (λ) = λn + c1 λn−1 + · · · + cn . Notice that PM (0) = det(−M ) = (−1)n det M . The fundamental theorem of algebra states that any polynomial can be factored into a product of linear terms over C. Then there exists a collection of n complex numbers λi (possibly with repetition) such that PM (λ) = (λ − λ1 )(λ − λ2 ) · · · (λ − λn ), 159 PM (λi ) = 0
More generally. . Notice that 1 + 0 = 1 is also an eigenvector 0 1 1 −1 1 of L with eigenvalue 1. a vector space contained within the larger vector space V : It contains 0V .}. A linear combination of the vi can be written c1 v1 + c2 v2 + · · · for some constants {c1 .1
Eigenspaces
−1 1 1 and 0 for L In the previous example. and then using the Taylor series to show that if two d functions are eigenvectors of dx with eigenvalues λ. . v2 . we found two eigenvectors 1 0 −1 1 0 with eigenvalue 1. . .
19. Then: L(c1 v1 + c2 v2 + · · · ) = c1 Lv1 + c2 Lv2 + · · · by linearity of L = c1 λv1 + c2 λv2 + · · · since Lvi = λvi = λ(c1 v1 + c2 v2 + · · · ).
More on eigenspaces
162
. and is closed under addition and scalar multiplication by the above calculation. then they are scalar multiples of each other. It is. let {v1 .} be eigenvectors of some linear transformation L with the same eigenvalue λ. since L0V = 0V = λ0V . In fact.d This is actually the whole collection of eigenvectors for dx . c2 . The space of all vectors with eigenvalue λ is called an eigenspace. . . In simple terms. a notion explored in Lecture 15. in fact. So every linear combination of the vi is an eigenvector of L with the same eigenvalue λ. any sum of eigenvectors is again an eigenvector if they share the same eigenvalue. All other vector space properties are inherited from the fact that V itself is a vector space. this can be proved using the fact that every infinitely differentiable function has a Taylor series with infinite radius of convergence. An eigenspace is an example of a subspace of V . any linear combination r 1 + s 0 of 0 1 these two eigenvectors will be another eigenvector with the same eigenvalue.
References
Hefferon.1: Representing Linear Maps with Matrices Hefferon. What do you find from this computation? Does something similar hold for 3 × 3 matrices? What about n × n matrices? 163
. c d we can plug M into its characteristic polynomial and find the matrix PM (M ). and then use properties of the determinant to give a general explanation. Now. Section II.1
You are now ready to attempt the second sample midterm. Chapter Three. Chapter E.3: Eigenvalues and Eigenvectors Beezer. Compute the characteristic polynomial PM (λ) of the matrix M = a b . since we can evaluate polynomials on square matrices. Chapter Five.Reading homework: problem 19. 2. Section EE Wikipedia:
ˆ Eigen* ˆ Characteristic Polynomial ˆ Linear Transformations (and matrices thereof)
Review Problems
1. Explain why the characteristic polynomial of an n×n matrix has degree n. Section III. Make your explanation easy to read by starting with some simple examples.
)
Hint
164
. v(3). v(2). Discrete dynamical system.) (c) Find all vectors v(0) such that v(0). (This is known as a discrete dynamical system whose initial condition is v(0). Let M be the matrix given by M= 3 2 . (b) Find all vectors v(0) such that v(0) = v(1) = v(2) = v(3) = · · · (Such a vector is known as a fixed point of the dynamical system. and so on using the rule Given any vector v(0) = v(t + 1) = M v(t) for all natural numbers t. . v(2). . all point in the same direction. v(3). . 2 3
x(0) . v(1). we can create an infinite sequence of y(0) vectors v(1).) (a) Find all eigenvectors and eigenvalues of M. (Any such vector describes an invariant curve of the dynamical system.3.
. then the standard basis vectors ei must already be a set of n linearly independent eigenvectors. λn .. .1
Diagonalization
Now suppose we are lucky. . . We are especially interested in the case that the matrix is written with respect to a basis of eigenvectors. . We have shown:
165
. the matrix of L in the basis of eigenvectors is diagonal: λ1 λ2 . In a basis of eigenvectors. λn where all entries off of the diagonal are zero.. L(vn ) = λn vn
As a result. . we are interested in how to write it as a matrix. the matrix of a linear transformation is diagonal. Suppose that V is any n-dimensional vector space. . On the other hand. and we have L : V → V . . In other words.
20. vn } is a set of linearly independent eigenvectors for L. . in which case it is a particularly nice matrix. .20
Diagonalization
Given a linear transformation. Then:
L(v1 ) = λ1 v1 L(v2 ) = λ2 v2 . L is diagonalizable if there exists a basis for V of eigenvectors for L. with eigenvalues λ1 . and the basis {v1 . . We call a linear transformation L : V → V diagonalizable if there exists a collection of n linearly independent eigenvectors for L. if an n × n matrix is diagonal.
j
Then we can write: vj =
k i
vk Qk Pji . j entry of the product of the matrices QP . (Here vi and ui are vectors. then the matrix for L in the basis S is diagonal if and only if S is a basis of eigenvectors for L. . each vj is an eigenvector for QP with eigenvalue 1. 166
. .Theorem 20. not components of vectors in a basis!) Then we may write each vi uniquely as a linear combination of the uj : vj =
i
ui Pji . . un } for a vector space V . j
But i Qk Pji is the k. since we can also write each ui uniquely as a linear combination of the vj : uj =
k
vk Qk . . .2
Change of Basis
Suppose we have two bases S = {v1 .
Non-diagonalizable example
Reading homework: problem 20. . The matrix P must have an inverse. Given a basis S for a vector space V and a linear transformation L : V → V . .2
20. then QP fixes each vj . so QP is the identity. which we can regard as entries of a square matrix P = (Pji ).1. the Pji are constants. . .
or in a matrix notation 1 1 1 P1 P2 · · · P n P 2 P 2 2 1 . . . As a result. . . vn } and T = {u1 . . n n P1 · · · Pn v1 v2 · · · vn = u1 u2 · · · un
Here. Since the j only expression for vj in the basis S is vj itself.
as a map between vector spaces. u2 to the new basis v1 . un . u2 have. .
Worked Change of Basis Example
167
. .
Changing basis changes the matrix of a linear transformation. so is just P =
1 √ 2 1 √ 2 1 √ 3 1 − √3
. . n P1 This says that the first column of the change of basis matrix P is really just the components of the vector v1 in the basis u1 . .
Example Suppose the vectors v1 and v2 form a basis for a vector space V and with respect to some other basis u1 . . components
1 √ 2 1 √ 2
and
1 √ 3 1 − √3
. . v2 . v2 ? Before answering note that the above statements mean v1 = u1 u2
1 √ 2 1 √ 2
=
u1 + u2 √ 2
and v2 = u1 u2
1 √ 3 1 − √3
=
u1 − u2 √ . u2 . There is a quick and dirty trick to obtain it: Look at the formula above relating the new basis vectors v1 . In particular focus on v1 for which 1 P1 P 2 1 v1 = u1 u2 · · · un . vn to the old ones u1 .
What is the change of basis matrix P from the old basis u1 . u2 . . . . Linear transformations are the actual objects of study of this course. .The matrix P is called a change of basis matrix. the linear transformation is the same no matter which basis we use. un . 3
The change of basis matrix has as its columns just the components of v1 and v2 . . . . . matrices are merely a convenient way of doing computations. However. not matrices. respectively.
the matrix P of eigenvectors is a change of basis matrix which diagonalizes M : −1 0 0 P −1 M P = 0 0 0 . In order for M to be diagonalizable. we conjugate the matrix of L by the change of basis matrix: M → P −1 M P . A square matrix M is diagonalizable if and only if there exists a basis of eigenvectors for M . Moreover. the eigenvectors of M form a basis of R.ˆ To get the matrix of a linear transformation in the new basis. Corollary 20. v3 to be linearly independent.2. then we say that M and N are similar . are −8 −2 −1 v1 = −1 . 3 0 1 we need the vectors v1 . and v3 = −1. 9 18 29 The eigenvalues of M are determined by det(M − λ) = −λ3 + λ2 + 2λ = 0. these eigenvectors are the columns of the change of basis matrix P which diagonalizes M . Reading homework: problem 20. Then the above discussion shows that diagonalizable matrices are similar to diagonal matrices. 0 0 2
169
. 0. v2 . and so M is diagonalizable. Moreover. Therefore. v2 = 1 . Notice that the matrix −8 −2 −1 P = v1 v2 v3 = −1 1 −1 3 0 1 is invertible because its determinant is −1.3
Example Let's try to diagonalize the matrix −14 −28 −44 M = −7 −14 −23 .
If for two matrices N and M there exists an invertible matrix P such that M = P −1 N P . So the eigenvalues of M −1. and and associated eigenvectors turn out to be 2.
t } for Pn (t) and {1. and d d : Pn (t) → Pn−1 (t) be the derivative operator. Chapter R. t. Let Pn (t) be the vector space of polynomials of degree n or less. 2. . even the order of the basis matters! 170
. . . When writing a matrix for a linear transformation. . Section SD Beezer. here is the key result of this Lecture
References
Hefferon. In fact. . we have seen that the choice of basis matters. Sections MR-CB Wikipedia:
ˆ Change of Basis ˆ Diagonalizable Matrix ˆ Similar Matrix
Review Problems
1. Recall that the derivative operator is linear from Chapter 7. t } for Pn−1 (t). . . t.2 × 2 Example
As a reminder. Chapter Three. Chapter E. Section V: Change of Basis Beezer. . Find the matrix of dt dt n n−1 in the bases {1.
ˆ Write each change of basis matrix between the standard basis {e1 . . λ 1 0 ˆ Can the matrix 0 λ 1 be diagonalized? Either diagonalize 0 0 λ it or explain why this is impossible. 0 0 0 · · · λ 1 0 0 0 ··· 0 λ Either diagonalize it or explain why this is impossible.ˆ Write all possible reorderings of the standard basis {e1 . . . . How are these matrices related?
3. Show that similarity of matrices is an equivalence relation. e3 } for R3 . . Note: It turns out that every matrix is similar to a block matrix whose diagonal blocks look like diagonal matrices or the ones above and whose off-diagonal blocks are all zero. .) ˆ Given the linear transformation L(x.) 5. (The definition of an equivalence relation is given in Homework 0. . 3x. Jordan form
ˆ Can the matrix
λ 1 be diagonalized? Either diagonalize it or 0 λ explain why this is impossible. Make as many observations as you can about these matrices: what are their entries? Do you notice anything about how many of each type of entry appears in each row and column? What are their determinants? (Note: These matrices are known as permutation matrices. . . 2z+x+y). . y. .
a b c d
diagonalizable? Include examples in
4. write the matrix M for L in the standard basis. . z) = (2y−z. When is the 2 × 2 matrix your answer. e2 . . This is called 171
. λ 1 0 ··· 0 0 0 λ 1 · · · 0 0 0 0 λ · · · 0 0 ˆ Can the n × n matrix . e2 . . be diagonalized? . e3 } and each of its reorderings. and two other reorderings of the standard basis. . .
. . Let A and B be commuting matrices (i. .e. Additionally suppose that A is diagonalizable with distinct eigenvalues.
172
. Show that Bv also has an eigenvalue of λ. they have the same eigenvalues and eigenvectors). and thus showing A and B can be simultaneously diagonalized (i. . 6. λ
is called a Jordan n-cell or a Jordan block where n is the size of the block. . .the Jordan form of the matrix and a like λ 1 0 0 λ 1 . .e. . AB = BA) and suppose that A has an eigenvector v with eigenvalue λ. Show that v is also an eigenvector of B. . 0 0 0
(maximal) block that look 0 0 . .
. . . en = . e2 = . Notice that the Kronecker delta gives the entries of the identity matrix.21
Orthonormal Bases
You may have noticed that we have only rarely used the dot product. xn ) ∈ Rn is (x1 )2 + (x2 )2 + · · · (xn )2 .
173
. The canonical/standard basis in Rn
ˆ Each of the standard basis vectors has unit length:
ei =
√ ei ei =
eT ei = 1. i=j
where δij is the Kronecker delta. we have seen that the dot product v w is the same as the matrix multiplication v T w. at right angles or perpendicular). 0 0 1 has many useful properties. . This is the inner product on Rn . . Now let us consider the case of Rn where the length of a vector (x1 . i
ˆ The standard basis vectors are orthogonal (in other words. . . . . . . . .
ei ej = eT ej = 0 when i = j i This is summarized by eT ej = δij = i 1 0 i=j . which gives a square matrix. . . Given column vectors v and w. . That is because many of the results we have obtained do not require a preferred notion of lengths of vectors. x2 . We can also form the outer product vwT .
vn } is an orthogonal (not orthonormal) basis i for Rn .
Hint for
2
3. Find a formula for the constants ci in terms of v and the vectors in S. Then we can write any vector v as v = i c vi for some constants ci . (e) Test your abstract formulae starting with u = 1 2 0 and v = 0 1 1 . and P = span{u. Suppose S = {v1 . how can you find a third vector perpendicular to both u and v ⊥ ? (d) Construct an orthonormal basis for R3 from u and v.2. . . v be independent vectors in R3 . (a) Is the vector v ⊥ = v −
u·v u u·u
in the plane P ?
(b) What is the angle between v ⊥ and u? (c) Given your solution to the above. . .
Hint for
3
180
. v} be the plane spanned by u and v. Let u.
v are linearly independent vectors in R3 . When v is not paru v⊥ allel to u. It is significant that we wrote this decomposition with u in mind. This set could then be normalized by dividing each vector by its length to obtain an orthonormal basis. u × v ⊥ } would be an orthogonal basis for R3 . u·u This is called an orthogonal decomposition because we have decomposed v into a sum of orthogonal vectors. u·u
u v
u·v u·u
u=v
v⊥ This new vector v ⊥ is orthogonal to u because u v⊥ = u v − u·v u u = 0. If u. |v⊥ | . {u. v is parallel to u. v ⊥ } is an orthogonal basis for span{u. then the set {u. Sometimes we write v = v ⊥ + v where: u·v v⊥ = v − u u·u u·v v = u. we can construct a new vector: u·v v⊥ = v − u. and normalizing these vectors we obtain |u| . u·u
Hence. 181
. v ⊥ = 0.22
Gram-Schmidt and Orthogonal Complements
Given a vector u and some other vector v not in the span of u. an orthonormal basis. v ⊥ . v}.
whose effect upon multiplying the two matrices precisely undoes what we we did to the second column of the first matrix. You can easily get the idea behind QR decomposition by working through a simple example. To begin with we write 7 2 −5 1 1 1 0 5 M = 1 14 −2 0 1 0 . save the + 1 in the second entry 5 of the first row. 5 1 1 0 The matrix on the right is almost the identity matrix. The Gram–Schmidt procedure suggests another matrix decomposition. eigenvalue problems and least squares approximations.
Example Find the QR decomposition of 2 −1 1 M = 1 3 −2 . 5 0 1 −2 0 0 1 In the first matrix the first two columns are mutually orthogonal because we simpy replaced the second column of M by the vector that the Gram–Schmidt procedure produces from the first two columns of M . We will use the matrix R to record the steps of the Gram– Schmidt procedure in such a way that the product QR equals M .
184
.In Lecture 11 we learned how to solve linear systems by decomposing a matrix M into a product of lower and upper triangular matrices M = LU . 0 1 −2 What we will do is to think of the columns of M as three vectors and use Gram– Schmidt to build an orthonormal basis from these that will become the columns of the orthogonal matrix Q. Socalled QR-decompositions are useful for solving linear systems. M = QR where Q is an orthogonal matrix and R is an upper triangular matrix. namely 7 −5 −1 2 14 1 5 = 3 − 1 .
U + V = U ⊕ V .Definition Given two subspaces U and V of a space W such that U ∩ V = {0W }. Then (u − u ) = −(v − v ). the direct sum of U and V is defined as: U ⊕ V = span(U ∪ V ) = {u + v|u ∈ U. Reading homework: problem 22. Then we could express 0 = (u − u ) + (v − v ). there is a natural candidate U ⊥ for this second subspace as shown here:
186
. v ∈ V }. Similarly. Since U and V are subspaces. That is.1. then (u − u ) = 0. Theorem 22.1 Given a subspace U in W . proving the theorem. v ∈ V . using the inner product. Then the expression w = u + v is unique. (v − v ) = 0. Since U ∩ V = {0}. The direct sum has a very nice property. we also have (u − u ) ∈ V . Proof. with u. there is only one way to write w as the sum of a vector in U and a vector in V . we would like to write W as the direct sum of U and something. Notice that when U ∩ V = {0W }. we have (u − u ) ∈ U and −(v − v ) ∈ V . Suppose that u + v = u + v . There is not a unique answer to this question as can be seen from this picture of subspaces in W = R3 :
However. Let w = u + v ∈ U ⊕ V . and v. u ∈ U . But since these elements are equal.
define: U ⊥ = {w ∈ W |w u = 0 for all u ∈ U }. it may help to watch this quick overview video:
Overview
Example Consider any plane P through the origin in R3 . y. Let U be a subspace of a finite-dimensional vector space W .The general definition is as follows: Definition Given a subspace U of a vector space W . This is also often called the orthogonal complement of U .
Theorem 22. Then P is a subspace. we have: u u = 0 ⇔ u = 0. then R3 = P ⊕ P ⊥ = {(x. y ∈ R} ⊕ {(0. since if u ∈ U and u ∈ U ⊥ .
187
. Proof. if P is the xy-plane. and P ⊥ is the line through the origin orthogonal to P . 0.2. 0)|x. Then the set U ⊥ is a subspace of W . We have U ∩ U ⊥ = {0}. For example. which requires a simple check. The set U ⊥ (pronounced "U -perp") is the set of all vectors in W orthogonal to every vector in U . To see that U ⊥ is a subspace. we only need to check closure. z)|z ∈ R}. Probably by now you may be feeling overwhelmed. and W = U ⊕ U ⊥ .
Find the QR factorization of 1 0 2 M = −1 2 0 . v}. Show that u and v ⊥ are also linearly independent. 3. but with three independent vectors u. 4.References
Hefferon. Chapter Three. Section VI. 6. what is the probability that it lies in the span of some other vectors?" 189
. w. −1 −2 2
Hint
2. Repeat the previous problem. Given any three vectors u. v ⊥ } are a basis for span{u. w. Subsection GSP Wikipedia:
ˆ Gram-Schmidt Process ˆ QR Decomposition ˆ Orthonormal Basis ˆ Direct Sum
Review Problems
1. Chapter V. Section O. and v ⊥ and w⊥ as defined in the lecture. v. when do v ⊥ or w⊥ vanish? 5. "If I choose a bit vector at random. This question will answer the question. Suppose u and v are linearly independent. For U a subspace of W . v.2: Gram-Schmidt Orthogonalization Beezer. Explain why {u. use the subspace theorem to check that U ⊥ is a subspace of W .
what must the degree of P be? Given that each coefficient must be either 0 or 1. once we write down a reasonable way of choosing a random real vector. what is the probability that it lies in the span of some other vectors? In fact. If P is the characteristic polynomial of a 3 × 3 bit matrix. what is the probability that it has 0 as an eigenvalue? (Assume that you are choosing a random matrix M in such a way as to make each characteristic polynomial equally likely. iii. Give some method for choosing a random bit vector v in B 3 . consider the bit matrix M whose columns are the vectors in S. the probability that it lies in the span of n − 1 other real vectors is 0!
190
. Given a collection S of k bit vectors in B 3 . Show that S is linearly independent if and only if the kernel of M is trivial. How can we tell whether S ∪ {v} is linearly independent? Do you think it is likely or unlikely that S ∪ {v} is linearly independent? Explain your reasoning.) What is the probability that the columns of M form a basis for B 3 ? (Hint: what is the relationship between the kernel of M and its eigenvalues?) Note: We could ask the same question for real vectors: If I choose a real vector at random. Suppose S is a collection of 2 linearly independent bit vectors in B 3 . if I choose a real vector in Rn at random. how many possibilities are there for P ? How many of these possible characteristic polynomials have 0 as a root? If M is a 3×3 bit matrix chosen at random. ii.i.
and that λ = λ. (c) Let x = . is it real. (h) Show that λ = λ. (On Reality of Eigenvalues) √ (a) Suppose z = x + iy where x. imaginary. Let λ be an eigenvalue of M with eigenvector x. What kind of numbers are zz and zz? (The complex number z is called the complex conjugate of z). x3 } is an orthonormal basis for R3 . what does this say about λ?
Problem 1 hint
a 2. Let x1 = b . Compute: x† M x x† x (e) Suppose Λ is a 1 × 1 matrix. ∈ Cn . Compute x† x. etc. Using the result of a previous part of this problem.Review Problems
1. we can compute N by applying complex conjugation to each entry of N . what can you say about the number x† x? (E. Compute (x† )T . i = −1. positive. and z = x − iy.) (d) Suppose M = M T is an n × n symmetric matrix with real entries. Using the result of part 1a. Compute zz and zz in terms of x and y. negative. zn complex matrix or a row vector). 195
. (b) Suppose that λ = x + iy is a complex number with x. Let x† = z 1 · · · z n ∈ Cn (a 1 × n . x2 . Does this determine the value of x or y? What kind of number must λ be? z1 . Find vectors x2 and x3 such c that {x1 . Then compute (x† M x)T . y ∈ R. so M x = λx. y ∈ R.. where a2 + b2 + c2 = 1.g. Note that for matrices AB + C = AB + C. What is ΛT ? (f) What is the size of the matrix x† M x? (g) For any matrix (or vector) N .
What is the sum of the dimensions of the eigenspaces of A? (c) Based on your answer to the previous part. 0 −2 2 (b) Find a basis for each eigenspace of A.
196
. Explain why your formula must work for any real n × n symmetric matrix. (Dimensions of Eigenspaces) 4 0 0 (a) Let A = 0 2 −2 .3. Find all eigenvalues of A. guess a formula for the sum of the dimensions of the eigenspaces of a real n×n symmetric matrix.
1. That is. Notice that surjectivity is a condition on the image of f . Notice that injectivity is a condition on the pre-image of f . Rank
Given a linear transformation L : V → W . suppose that f has an inverse function g. The image of f is the set of elements of T to which the function f maps. every t has no more than one pre-image. we would like to know whether it has an inverse. Theorem 24. let us first discuss inverses of arbitrary functions. every element t ∈ T has at least one pre-image. and for any vector w ∈ W . f is one-to-one if for any elements x = y ∈ S. That is. If f is both injective and surjective. we have that f (x) = f (y). Conversely. Therefore. Recall that S is called the domain of f .e. we simply define g(t) to be the unique pre-image f −1 (t) of t. and since f is injective. we would like to know whether there exists a linear transformation M : W → V such that for any vector v ∈ V . 197
. Proof. That is. we have M (L(v)) = v. i. Let f : S → T be a function from a set S to a set T . and the set ran(f ) = im(f ) = f (S) = {f (s)|s ∈ S} ⊂ T is called the range or image of f . The function f is onto if every element of T is mapped to by some element of S. The pre-image of a set U is the set of all elements of S which map to U . A function f : S → T has an inverse function g : T → S if and only if it is bijective. Onto functions are also called surjective functions. One-to-one functions are also called injective functions. So before we discuss which linear transformations have inverses.. The function f is one-to-one if different elements in S always map to different elements in T . When we later specialize to linear transformations. Nullity. A linear transformation is just a special kind of function from one vector space to another. there exists some s ∈ S such that f (s) = t. T is called the codomain of f . f is onto if for any t ∈ T . We can also talk about the pre-image of any subset U ⊂ T : f −1 (U ) = {s ∈ S|f (s) ∈ U } ⊂ S. to construct an inverse function g.24
Kernel. it is bijective. Since f is surjective. Suppose that f is bijective. Range. we'll also find some nice ways of creating subspaces. the things in T which you can get to by starting in S and applying f . we have L(M (w)) = w.
ˆ The function f is surjective:
Let t be any element of T . g(t) is an element of S which maps to t. we can figure out whether f is one-to-one. so in particular g(f (x)) = x and g(f (y)) = y. Then v1 − v2 = 0. Then we can find v1 = v2 such that Lv1 = Lv2 . You will show that a linear transformation is one-to-one if and only if 0V is the only vector that is sent to 0W : by looking at just one (very special) vector. we always know that a linear function sends 0V to 0W . The set of all vectors v such that Lv = 0W is called the kernel of L: ker L = {v ∈ V |Lv = 0W }.ˆ The function f is injective:
Suppose that we have x. For arbitrary functions between arbitrary sets. y ∈ S such that f (x) = f (y). Suppose L is not injective. Everything we said above for arbitrary functions is exactly the same for linear transformations. but a linear transformation between two vector spaces. Thus. Definition Let L : V → W be a linear transformation. For example.
Now let us restrict to the case that our function f is not just an arbitrary function. then finding the kernel of L is equivalent to solving the homogeneous system M X = 0. We must have that f (g(t)) = t. but L(v1 − v2 ) = 0. Theorem 24. we have g(f (x)) = g(f (y)) so x = y. Therefore. things aren't nearly so convenient! Let L : V → W be a linear transformation. However. Notice that if L has matrix M in some basis. Proof. So f is surjective. A linear transformation L is injective if and only if ker L = {0V } . the linear structure of vector spaces lets us say much more about one-to-one and onto functions than we can say about functions on general sets. But since f (x) = f (y). We must have that g(f (s)) = s for any s ∈ S.2. 198
. The proof of this theorem is an exercise. f is injective.
vp . We also know that linear transformations can be represented by matrices. u1 . Reading homework: problem 24. . But then d1 u1 + · · · + dq uq must be in the span of {v1 .Now we show that {L(u1 ). . . . .e. Here is a list of them. This contradicts the assumption that {v1 . .. . then d1 u1 + · · · + dq uq = 0. . .
201
. . . so we are done. We argue by contradiction: Suppose there exist constants dj (not all zero) such that 0 = d1 L(u1 ) + · · · + dq L(uq ) = L(d1 u1 + · · · + dq uq ). L(uq )} is linearly independent. uq } was a basis for V . . and we have seen many ways to tell whether a matrix is invertible. and so d1 u1 + · · · + dq uq is in the kernel of L. vp }.1
Summary
We have seen that a linear transformation has an inverse if and only if it is bijective (i. .
But since the uj are linearly independent. . since this was a basis for the kernel. one-to-one and onto).2
24. . .
The columns (or rows) of M are a basis for Rn . 14. The linear transformation L is injective. If v is any vector in Rn . Proof. The determinant of M is not equal to 0. 13. 11. The matrix M does not have 0 as an eigenvalue. 10. 2. If V is any vector in Rn . The matrix M is row-equivalent to the identity matrix. 7. The rest are left as exercises for the reader. The characteristic polynomial det(λI − M ) does not have 0 as a root. 16. 8. and let L : Rn → Rn be the linear transformation defined by L(v) = M v. 202
. The linear transformation L is surjective. 3. 9. then it can't be invertible. then L(x) = v has exactly one solution. and many of the statements above are no longer equivalent to each other.6 (Invertibility). The linear transformation L does not have 0 as an eigenvalue. The linear transformation L is bijective. Note: it is important that M be an n × n matrix! If M is not square. The matrix M is invertible. The columns (or rows) of M span Rn . 5. Let M be an n × n matrix. 6. The columns (or rows) of M are linearly independent. 4. The transpose matrix M T is invertible.Theorem 24. 12. 15. Then the following statements are equivalent: 1. Some were left as review questions or sample final questions. The homogeneous system M X = 0 has no non-zero solutions. Many of these equivalences were proved earlier in these notes. then the system M X = V has exactly one solution.
Show that ker L = {0V }. . . show that 0V is in ker L. suppose that L is one-to-one. or a direct proof seem most appropriate? (b) Now.6. . Chapter Three. and then show that there are no other vectors in ker L. . Let L : V → W be a linear transformation. Sections ILT-IVLT Wikipedia:
ˆ Rank ˆ Dimension Theorem ˆ Kernel of a Linear Operator
Review Problems
1. suppose that ker L = {0V }.
Hint for
1
2. L(vn )}. Show that ker L = {0V } if and only if L is one-to-one: (a) First.Discussion on Theorem 24. That is.
203
. Section II. Think about methods of proof–does a proof by contradiction. . Explain why L(V ) = span{L(v1 ).
References
Hefferon. vn } be a basis for V . Show that L is one-to-one. Let {v1 . .2: Rangespace and Nullspace (Recall that "homomorphism" is is used instead of "linear transformation" in Hefferon. . a proof by induction. . Chapter LT.) Beezer.
6. Choose a simple yet non-trivial linear transformation with a non-trivial kernel and verify the above claim for the transformation you choose. and consider the derivative operator ∂x . Suppose L : R4 → R3 whose matrix equivalent to the following matrix: 1 0 0 0 1 0 0 0 1
M in the standard basis is row −1 1 . 5. Find and describe and algorithm (i. consider P2 (x. vn } is a basis for ker L. (Extra credit) We will show some ways the dimension formula can break down with the vector space is infinite dimensional. 204
. . Claim: If {v1 . y is degree one and x2 y is ∂ ∂ degree three. a general procedure) for finding a basis for L(Rn ) when L : Rn → Rm . (a) Let R[x] be the vector space of all polynomials with coefficients in d R in the variable x. . the space of polynomials of degree two or less in x and y. Find the dimension of ∂ the kernel and image of ∂x . where L : V → W . (Recall that xy is degree two.3. then it is always possible to extend this set to a basis for V . What is ker D? Hint: Use the basis {xn | n ∈ N}. .e.) Let L = ∂x + ∂y . Now. y).) Find a basis for the kernel of L. 4 1 1 6 4. . Let D = dx be the usual derivative operator. use your algorithm to find a basis for L(R4 ) when L : R4 → R3 is the linear transformation whose matrix M in the standard basis is 2 1 1 4 0 1 0 5 . L(xy) = ∂ ∂ (xy) + ∂y (xy) = y + x. for example. Verify the ∂x dimension formula in this case. 1
Explain why the first three columns of the original matrix M form a basis for L(R4 ). Let Pn (x) be the space of polynomials in x of degree less than or equal ∂ to n. Finally. (For example. Show that the range of D is R[x].
. What is the kernel and range of M ? (c) Let 1 be the vector space of all absolutely convergent sequences s = {si }.). 0. P 2 = P ). is a projection (i. 0. . Suppose that dim ker L < ∞. . and has infinite range. . Define the map P : 1 → 1 by si → and for example P ((1.e.)) = (0. 1. Hint: Define ei as the sequence with a 1 in the i-th position and 0 everywhere else which you can think of as a standard basis vector.
205
. 0.). has infinite kernel. . . 1. . Also show when dim L(V ) < ∞ that dim ker L is infinite. . 0. .)) = (0. 0. P ((0. . . . 0.(b) Consider R[x] and let M : R[x] → R[x] be the linear map M (p(x)) = xp(x) which multiplies by x. What is P (ei ) when i is even? When i is odd? (d) Let V be an infinite dimensional vector space and L : V → V be a linear operator. si 0 if i is even if i is odd.
Show that P is linear. show that dim L(V ) is infinite.
But if v is not in the range of L then there will never be any solutions for L(x) = v. Then we can uniquely write v = v + v ⊥ . we can write W = L(U ) ⊕ L(U )⊥ . and v and x are given by column vectors V and X in these bases. with v ∈ L(U ) and v ⊥ ∈ L(U )⊥ . recall that L(V ) is spanned by the columns of M . "My work always tried to unite the Truth with the Beautiful. but so long as the observations are right on average. or M T M X = M T V. v ⊥ is just V − M X. I usually chose the Beautiful. but when I had to choose one or the other. For example.. M T (V − M X) = 0. Then we need to approximate MX − V ≈ 0 . In components.) Then v ⊥ must be perpendicular to the columns of M . such as when trying to fit a (perhaps linear) function to a "noisy" set of observations. we try to find the best approximation possible. suppose we measured the position of a bicycle on a racetrack once every five seconds. In terms of M . To do this. we can figure out a best-possible linear function of position of the bicycle in terms of time. respectively.e. or a space of solutions. Then we should solve L(u) = v . (In the natural basis. .. we try to find x that minimizes ||L(x) − v||. instead. 206
. a unique solution. . Thus. This method has many applications. M en . the columns of M are M e1 . this system may have no solutions. . As we have seen. Suppose M is the matrix for L in some bases for U and W . Our observations won't be exact. Solutions X to M T M X = M T V are called least squares solutions to MX = V . where L : U −→ W .25
Least Squares
linear
Consider the linear system L(x) = v. Note that if dim U = n and dim W = m then M can be represented by an m × n matrix and x and v as vectors in Rn and Rm . However. and v ∈ W is given." – Hermann Weyl. for many applications we do not need a exact solution of the system. and is the part we will eventually wish to minimize. i.
so could only be the zero vector. This is a very nice number.) However the square matrix M T M is invertible. the converse is often false.Notice that any solution X to M X = V is a least squares solution. but still have least squares solutions to M T M X = M T V . then M T is an n × m matrix. Then M T M is an n × n matrix. However.1 Now suppose that ker L = {0}. since (M T M )T = M T M .
ˆ Compute M T M and M T V . this is the least squares method. In a nutshell. suppose there was a vector X such that M T M X = 0. we can evaluate X T M T M X to obtain a number. and is equal to X = (M T M )−1 M T V. so that the only solution to M X = 0 is X = 0. Then it would follow that X T M T M X = |M X|2 = 0. for any vector X. To see this. ts 1 2 3 v m/s 11 19 31
207
. But we are assuming that ker L = {0} so M X = 0 implies X = 0. In other words the vector M X would have zero length. Thus the kernel of M T M is {0} so this matrix is invertible. so not necessarily square. Then. the equation M X = V may have no solutions at all. Observe that since M is an m × n matrix. though! It is just the length |M X|2 = (M X)T (M X) = X T M T M X. the least squares solution (the X that solves M T M X = M V ) is unique. Reading homework: problem 25. In fact. in this case.
Example Captain Conundrum falls off of the leaning tower of Pisa and makes three (rather shaky) measurements of his velocity at three different times. and is symmetric. (This need not mean that M is invertible because M is an n × m matrix. So. ˆ Solve (M T M )X = M T V by Gaussian elimination.
8. we say that U is a subspace of V when the set U is also a vector space. using the vector addition and scalar multiplication rules of the vector space V . (c) V = {vectors in R3 with at least one entry containing a 1}.(d) Use one of the above examples to show why the following statement is FALSE. usual matrix addition. Subspaces: If V is a vector space. c d kc d (b) V = {polynomials with complex coefficients of degree ≤ 3}. v ∈ U ) and multiplicative closure (r. (a) V = { 2 × 2 matrices with entries in R}. What are the four main things we need to define for a vector space? Which of the following is a vector space over R? For those that are not vector spaces. with usual addition and scalar multiplication of polynomials. where V = {f : R → R | f is integrable} is a vector space over R with usual addition and scalar multiplication. y ∈ R 0 213
. u ∈ V ) ensure that (i) the zero vector 0 ∈ U and (ii) every u ∈ U has an additive inverse. 9. with usual addition and scalar multiplication.) Explain why additive closure (u + w ∈ U ∀ u. i.u ∈ U ∀ r ∈ R. modify one part of the definition to make it into a vector space. a b ka b and k · = for k ∈ R.. all elements of U are also elements of V . What does it mean for a function to be linear? Check that integration is a linear function from V to V . Check whether the following choices of U are vector spaces: x (a) U = y : x. Square matrices with the same determinant are always row equivalent.e. The symbol ∀ means "for all" and ∈ means "is an element of". 10. In fact it suffices to check closure under addition and scalar multiplication to verify that U is a vector space. (Remember that U ⊂ V says that "U is a subset of V ".
In other words, the angle is θ OR −θ. You should draw two pictures, one where the angle between X and M X is θ, the other where it is −θ.
|X For Cauchy–Schwartz, ||X||(M X)| = | cos θ| = 1 when θ = 0, π. For the ||M X|| triangle equality M X = X achieves ||X + M X|| = ||X|| + ||M X||, which requires θ = 0.
6. This is a block matrix problem. Notice the that matrix M is really I I just M = , where I and 0 are the 3 × 3 identity zero matrices, 0 I respectively. But M2 = and M3 = so, M k = I I 0 I I 2I 0 I = I 3I 0 I I I 0 I I I 0 I = I 2I 0 I
As a check, count that the total number of 2 × 2 bit matrices is 2(number of entries) = 24 = 16. (d) To disprove this statement, we just need to find a single counterexample. All the unit determinant examples above are actually row equivalent to the identity matrix, so focus on the bit matrices with vanishing determinant. Then notice (for example), that 1 1 0 0 ∼ / 0 0 0 0 .
So we have found a pair of matrices that are not row equivalent but do have the same determinant. It follows that the statement is false. 8. We can call a function f : V −→ W linear if the sets V and W are vector spaces and f obeys f (αu + βv) = αf (u) + βf (v) , for all u, v ∈ V and α, β ∈ R. Now, integration is a linear transformation from the space V of all integrable functions (don't be confused between the definition of a linear function above, and integrable functions f (x) which here are the vec∞ tors in V ) to the real numbers R, because −∞ (αf (x) + βg(x))dx = ∞ ∞ α −∞ f (x)dx + β −∞ g(x)dx. 218
9. The four main ingredients are (i) a set V of vectors, (ii) a number field K (usually K = R), (iii) a rule for adding vectors (vector addition) and (iv) a way to multiply vectors by a number to produce a new vector (scalar multiplication). There are, of course, ten rules that these four ingredients must obey. (a) This is not a vector space. Notice that distributivity of scalar multiplication requires 2u = (1 + 1)u = u + u for any vector u but 2· which does not equal a b a b + c d c d This could be repaired by taking k· a b c d = ka kb kc kd . = 2a 2b 2c 2d . a b c d = 2a b 2c d
(b) This is a vector space. Although, the question does not ask you to, it is a useful exercise to verify that all ten vector space rules are satisfied. (c) This is not a vector space for many reasons. An easy one is that (1, −1, 0) and (−1, 1, 0) are both in the space, but their sum (0, 0, 0) is not (i.e., additive closure fails). The easiest way to repair this would be to drop the requirement that there be at least one entry equaling 1. 10. (i) Thanks to multiplicative closure, if u ∈ U , so is (−1) · u. But (−1) · u + u = (−1) · u + 1 · u = (−1 + 1) · u = 0.u = 0 (at each step in this chain of equalities we have used the fact that V is a vector space and therefore can use its vector space rules). In particular, this means that the zero vector of V is in U and is its zero vector also. (ii) Also, in V , for each u there is an element −u such that u + (−u) = 0. But by additive close, (−u) must also be in U , thus every u ∈ U has an additive inverse. 219
x y (a) This is a vector space. First we check additive closure: let 0 z x z and w be arbitrary vectors in U . But since y + w = 0 0 0 x+z y + w, so is their sum (because vectors in U are those whose 0 third component Multiplicative closure is similar: for vanishes). x αx y = αy , which also has no third component, any α ∈ R, α 0 0 so is in U . (b) This is not a vector space for various reasons. A simple one is 1 2 0 is in U but the vector u + u = 0 is not in U that u = z 2z (it has a 2 in the first component, but vectors in U always have a 1 there).
What do you think would happen to the permanent of an n × n matrix M if (include a brief explanation with each answer): (a) You multiplied M by a number λ. (b) You multiplied a row of M by a number λ. (c) You took the transpose of M . (d) You swapped two rows of M . 5. Let X be an n × 1 matrix subject to X T X = (1) , and define H = I − 2XX T , (where I is the n × n identity matrix). Show H = H T = H −1 . 6. Suppose λ is an eigenvalue of the matrix M with associated eigenvector v. Is v an eigenvector of M k (where k is any positive integer)? If so, what would the associated eigenvalue be? Now suppose that the matrix N is nilpotent, i.e. Nk = 0 for some integer k ≥ 2. Show that 0 is the only eigenvalue of N . 7. Let M = 3 −5 1 −3 . Compute M 12 . (Hint: 212 = 4096.)
except that j i we need only reshuffle two matrix elements Mσ(i) and Mσ(j) (in the case where rows i and j were swapped). since HH = I.X T = I . . because in the formula for the permanent this amounts to summing over permutations of rows rather than σ(1) σ(2) σ(n) columns. = λk v . The argument is almost the same as in the previous part. (d) Swapping two rows also leaves the permanent unchanged. Therefore the permanent is multiplied by an overall factor λ.1. . We know M v = λv. Hence M 2 v = M M v = M λv = λM v = λ2 v . 5.i (a) Multiplying M by λ replaces every matrix element Mσ(j) in the i formula for the permanent by λMσ(j) . Mn back into its original order using the inverse permutation σ −1 . Now we compute H 2 = (I − 2XX T )(I − 2XX T ) = I − 4XX T + 4XX T XX T = I − 4XX T + 4X(X T X)X T = I − 4XX T + 4X. lets call (1) = 1 (the 1 × 1 identity matrix). 226
. So. and therefore the permanent is unchanged. Firstly. which demonstrates the first equality. Then we calculate H T = (I−2XX T )T = I T −2(XX T )T = I−2(X T )T X T = I−2XX T = H . But summing over permutations is equivalent to summing over inverse permutations. . and similarly M k v = λM k−1 v = . But we could then sort the product M1 M2 . Then we use the fact that summing over all permutations σ or over all permutations σ obtained by swapping a pair in σ are equivalent operations. we have H −1 = H. i (b) Multiplying the ith row by λ replaces Mσ(j) in the formula for the i permanent by λMσ(j) .
(c) The permanent of a matrix transposed equals the permanent of the original matrix. 6. . and therefore produces an overall factor λn . So v is an eigenvector of M k with eigenvalue λk .
The kernel of a linear transformation. An eigenvector.
2. the sum of currents entering any junction vanishes. The general solution to a linear system of equations. Define the following terms: (a) (b) (c) (d) (e) (f) (g) (h) (i) (j) (k) (l) (m) (n) (o) (p) (q) An orthogonal matrix. The direct sum of a pair of subspaces of a vector space. A homogeneous solution to a linear system of equations. The span of a set of vectors. A basis for a vector space. A subspace of a vector space. The change in voltage (measured in Volts) around any loop due to batteries | and resistors /\/\/\/\ (given by the product of the current measured in Amps and resistance measured in Ohms) equals zero. The dimension of a vector space. A particular solution to a linear system of equations. Consider the circuit
1 Ohm I Amps 60 Volts 3 Ohms
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2 Ohms 13 Amps 80 Volts 3 Ohms J Amps V Volts
. The orthogonal complement to a subspace of a vector space.C
Sample Final Problems and Solutions
1. An equivalence relation. The nullity of a linear transformation. The rank of a linear transformation. The characteristic polynomial of a square matrix. Also. The image of a linear transformation. Kirchoff 's laws: Electrical circuits are easy to analyze using systems of equations.
1 with 1+1 = 0). (h) Do you think the set of all linear transformations from B 3 to B is a vector space using the addition rule above? If you answer yes. . (b) Your answer to part (a) should be a list of vectors v1 . (a) Find a basis for B 3 . (g) Suppose L1 : B 3 → B and L2 : B 3 → B are linear transformations. L(vn ) completely determines L. (f) Use the notation of part (e) to list all linear transformations L : B3 → B . . In this problem the scalars in the vector spaces are bits (0.(b) Determine all possible eigenvectors associated with each eigenvalue. (d) What is the dimension of the vector space B 3 . 1} is a linear transformation. . Let us assume that the rule found by the engineers applies to all subsequent orbits. . what will happen to the satellite if the initial mistake in its location is in a direction given by an eigenvector. Define a new map (αL1 + βL2 ) : B 3 → B by (αL1 + βL2 )(v) = αL1 (v) + βL2 (v). L(v2 ). How many different linear transformations did you find? Compare your answer to part (c). 10.
235
. and α and β are bits. . k-component column vectors. What number did you find for n? (c) How many elements are there in the set B 3 . vn . . give a basis for this vector space and state its dimension. . The space B k is the vector space of bit-valued. v2 . Is this map a linear transformation? Explain. Explain why specifying L(v1 ). Discuss case by case. (e) Suppose L : B 3 → B = {0.
(b) Find the eigenvalues and eigenvectors of the matrix M . a. c c Here M is a 3 × 3 matrix. 20 Here. (c) The number |ω| is often called a characteristic frequency. −y − z Moreover. They notice that the force on x y the center of the bridge when it is displaced by an amount X = z is given by −x − y F = −x − 2y − z .11. A team of distinguished. dt2
Since the engineers are worried the bridge might start swaying in the heavy channel winds. b. Which 3 × 3 matrix M did the engineers find? Justify your answer. post-doctoral engineers analyzes the design for a bridge across the English channel. c and ω are constants which we aim to calculate. having read Newton's Principiæ.
19
236
. so the exact units. X = cos(ωt) c (a) By plugging their proposed solution in the above equations the engineers find an eigenvalue problem a a M b = −ω 2 b . they search for an oscillatory solution to this equation of the form20 a b . coordinate system and constant of proportionality are state secrets. What characteristic frequencies do you find for the proposed bridge?
The bridge is intended for French and English military vehicles. they know that force is proportional to acceleration so that19 F = d2 X .
You need not multiply out any of the matrix expressions you find. You may call your answers λ and µ for the rest of the problem to save writing. C and f . Conic Sections: The equation for the most general conic section is given by ax2 + 2bxy + dy 2 + 2cx + 2ey + f = 0 . Make sure you give formulas for the new unknown column vector Y and constant g in terms of X. Be sure to also state your result for D. 237
. the quadratic equation mx2 + 2cx + f = 0 can be rewritten by completing the square m x+ c m
2
=
c2 −f. (e) Is there a direction in which displacing the bridge yields no force? If so give a vector in that direction. Briefly evaluate the quality of this bridge design. For the rest of this problem we will focus on central conics for which the matrix M is invertible. relating an unknown column vector X = (b) Does your matrix M obey any special properties? Find its eigenvalues. y a 2 × 2 matrix M . use the same trick to rewrite your answer to part (a) in the form Y T M Y = g. (c) Your equation in part (a) above should be be quadratic in X. m
Being very careful that you are now dealing with matrices.(d) Find an orthogonal matrix P such that M P = P D where D is a diagonal matrix. Our aim is to analyze the solutions to this equation using matrices. 12. M . Recall that if m = 0. its transpose X T . (a) Rewrite the above quadratic equation as one of the form XT M X + XT C + CT X + f = 0 x . a constant column vector C and the constant f .
If all has gone well. What equation do z and w obey? (Hint. In fact.. Find an expression for Y T M Y in terms of Z and the eigenvalues of M . hyperbolae or a pair of straight lines. and let M be a matrix for L (with respect to some basis for V and some basis for W ). define our final unknown vector Z by Y = P Z. iii. i. Show that 0 is not an eigenvalue of M . L has an inverse if and only if M has an inverse: (a) Suppose that L is bijective (i. iii. Show that M is an invertible matrix. (d) Why is the angle between vectors V and W is not changed when you replace them by P V and P W for P any orthogonal matrix? (e) Explain how to choose an orthogonal matrix P such that M P = P D where D is a diagonal matrix. µ. suppose that M is an invertible matrix. (f) For the choice of P above. (g) Call Z = z . Show that 0 is not an eigenvalue of M . We know that L has an inverse if and only if it is bijective. ii. Show that dim V = rank L = dim W .
238
. ellipses. and we know a lot of ways to tell whether M has an inverse. Our next aim is to rotate the coordinate axes to produce a readily recognizable equation.)
(h) Central conics are circles. Show that L is surjective. Show that L is injective. one-to-one and onto). you have found a way to shift coordinates for the original conic equation to a new coordinate system with its origin at the center of symmetry. g) which produce each of these cases. i. Give examples of values of (λ. ii. 13. Let L : V → W be a linear transformation between finite-dimensional vector spaces V and W . µ and g.e. (b) Now. write w your answer using λ.
in year two she has two pairs of doves etc. F0 = 0.. Thus F4 = 3. (I. (e) Diagonalize M . (c) Let us introduce a column vector Xn = Fn .) (f) Find a simple expression for M n in terms of P . Assume no doves die and that the same breeding pattern continues well into the future. the Queen has no doves. F1 = 1 and F2 = 1. D and P −1 . (h) The number √ 1+ 5 ϕ= 2 is called the golden ratio.. Compute X1 Fn−1 and X2 . this pair of doves breed and produce a pair of dove eggs. (a) Compute F5 and F6 . In year one she has one pair of doves. For example. Captain Conundrum is very happy because now he will never need to buy the Queen a present ever again! Let us say that in year zero. write M as a product M = P DP −1 where D is diagonal.14. Verify that these vectors obey the relationship X2 = M X1 where M = (d) Show that Xn+1 = M Xn . (g) Show that Xn+1 = M n X1 . Then F3 = 2 because the eggs laid by the first pair of doves in year two hatch. 239 1 1 1 0 . Call Fn the number of pairs of doves in years n. male and female for her birthday. two pairs of eggs are laid (by the first and second pair of doves). One year later these eggs hatch yielding a new pair of doves while the original pair of doves breed again and an additional pair of eggs are laid.e. Captain Conundrum gives Queen Quandary a pair of newborn doves. (b) Explain why (for any n ≥ 2) the following recursion relation holds Fn = Fn−1 + Fn−2 . Notice also that in year three.
.. Write the eigenvalues of M in terms of ϕ. After one year.
(c) Find the reduced row echelon form for this augmented matrix. y)-plane as well as their least squares fit. state the condition required for the inverse to exist). (d) Calculate M T M and (M T M )−1 (for the latter computation. (f) What value does Captain Conundrum predict for y when x = 2? 18. (b) Arrange the unknowns (m. 241
. (c) For a generic data set. Draw a rough sketch of the three data points in the (x. (f) The least squares method determines a vector X that minimizes the length of the vector V − M X. b and c. b) in a column vector X and write your answer to (a) as a matrix equation MX = V . would you expect your system of equations to have a solution? Briefly explain your answer. Be sure to give explicit expressions for the matrix M and column vector V . Indicate how the components of V − M X could be obtained from your picture. (b) Write the augmented matrix for this system of equations. (a) Write down a linear system of equations you could use to find the slope m and constant term b.(a) Write down a system of four linear equations for the unknown coefficients a. Suppose you have collected the following data for an experiment x x1 x2 x3 y y1 y2 y3
and believe that the result is well modeled by a straight line y = mx + b . (d) Are there any solutions to this system? (e) Find the least squares solution to the system. (e) Compute the least squares solution for m and b.
one on the right and one going around the outside of the circuit).
(h) Yes. because otherwise it would not be invertible. V = 120 Volts. (2)
The above equations are easily solved (either using an augmented matrix and row reducing. The result is I = −5 Amps. M must have 0 as an eigenvalue. or by substitution). (a) m. 2. This relies on kerM = 0 because if M T M had a non-trivial kernel. Both junctions give the same equation for the currents I + J + 13 = 0 .Solutions
1. 3. (b) n. There are three voltage loops (one on the left. all symmetric matrices have a basis of eigenvectors. 242
. J = 8 Amps. then there would be a non-zero solution X to M T M X = 0. This in turn implies M X = 0 which contradicts the triviality of the kernel of M . You can find the definitions for all these terms by consulting the index of these notes. (e) m × m. But then by multiplying on the left by X T we see that ||M X|| = 0. (f) Yes. (c) Yes. Respectively. they give the equations 60 − I − 80 − 3I = 0 80 + 2J − V + 3J = 0 60 − I + 2J − V + 3J − 3I = 0 . (g) Yes because M T M
T
= M T (M T )T = M T M . (d) n × n. (j) Since the kernel of L is non-trivial. (i) No.
for λ = 2 3 1 1 1 2 −3 −2 2 1 0 2 1 −1 1 0 ∼ 0 − 5 5 2 2 4 4 1 5 3 1 0 4 −5 −2 0 2 4
s = 0 is arbitrary. In this case. NASA would be very pleased in this case. −1 0 . the satellite returns to the origin O. The same will happen for all subsequent orbits. the approximations used by the engineers in their calculations probably fail and a new computation will be needed). with the satellite moving a factor 3/2 further away from O each orbit (in reality. the elite engineers will pat themselves on the back. again. It will continue this wobbling motion indefinitely. For all subsequent orbits it will again return to the origin. s 1 −2. 0 0 0 0 0
s So we find the eigenvector s where s = 0 is arbitrary. after several orbits. Since this is a stable situation.. then Y = −X. then If the mistake X is in the direction of the eigenvector 1 Y = 0. After next orbit will move back to X. 1 1 . I. if the mistake X is in the direction 1 3 move to a point Y = 2 X which is further away from the origin.
1 0 −1 0 0 0 ∼ 0 1 −1 0 .e. the satellite will be lost in outer space and the engineers will likely lose their jobs!
247
. −s So we find the eigenvector 0 where s 3 Finally. Hence the If the mistake X is in the direction 1 satellite will move to the point opposite to X. the satellite will Finally.
Hence Y = X + M −1 C and g = C T M C − f . the result is a pair of hyperbolæ. so neither does the angle between V and W . w)-plane. These shapes all come from cutting a cone with a plane. but λ and µ have opposite signs. we have the equation for an ellipse. µ and g are postive. 250
. Hence none of the dot products in the above formula changes. then Y T M Y = Z T P T M P Z = Z T P T P DZ = Z T DZ λ 0 where D = . and are therefore called conic sections.e. since M times the eigenvectors yields just the eigenvectors back again multiplied by their eigenvalues. (f) If Y = P Z.) Then. V VW W V TV WTW
So replacing V → P V and W → P W will always give a factor P T P inside all the products. so that (X T + C T M −1 )M (X + M −1 C) = C T M C − f . (d) The cosine of the angle between vectors V and W is given by √ V TW V W =√ . (h) When λ = µ and g/λ = R2 . 0 µ (g) Using part (f) and (c) we have λz 2 + µw2 = g . and put them in a matrix P (as columns) then P will be an orthogonal matrix. but P T P = I for orthogonal matrices. normalize them (i. When λ. then we also need to make sure the eigenvectors spanning the two dimensional eigenspace corresponding to λ are orthogonal. Vanishing g along with λ and µ of opposite signs gives a pair of straight lines. it follows that M P = P D where D is the diagonal matrix made from eigenvalues. divide them by their lengths). (e) If we take the eigenvectors of M . we get the equation for a circle radius R in the (z. When g is non-vanishing. (If it happens that λ = µ.(c) The trick is to write X T M X+C T X+X T C = (X T +C T M −1 )M (X+M −1 C)−C T M −1 C .
ii. so dim W = dim V = rank L = dim L(V ). i. But LX is the same as M X. dim V = L + rank L = rank L. so the only solution to M X = 0 is X = 0V .13. (b) Now we suppose that M is an invertible matrix. Hence L = dim ker L = 0. We show that L is bijective if and only if M is invertible. the only solution to LX = 0 is X = 0V . the kernel of L is the zero vector alone. L(V ) = W. Since L is surjective. By the Dimension Formula. Since M is invertible. Since M X = 0 has no non-zero solutions. So L is injective. 251
. Since L is injective. iii. Since M is invertible. i. the matrix M is square so we can talk about its eigenvalues. So by the Dimension Formula. So L does not have zero as an eigenvalue. we have dim V = L + rank L and since ker L = {0V } we have L = dim ker L = 0. Thus rank L = dim L(V ) = dim W. the system M X = 0 has no non-zero solutions. So M does not have zero as an eigenvalue. Thereby dim V = rank L = dim W. we must have that dim V = dim W . That is. Since LX = 0 has no non-zero solutions. But LX is the same as M X. so the only solution to LX = 0 is X = 0V . its kernel is the zero vector alone. Since L is injective. its kernel consists of the zero vector alone. iii. (a) We suppose that L is bijective. the matrix M is invertible. ii. Since dim V = dim W .
(a) We suppose that L is bijective. the matrix M is invertible. i. (b) Now we suppose that M is an invertible matrix. 18. i. we must have that dim V = dim W . We show that L is bijective if and only if M is invertible. Since LX = 0 has no non-zero solutions. dim V = L + rank L = rank L. That is. So rank L = dim L(V ) = dim W. . Since M X = 0 has no non-zero solutions.. Since L is injective. So dim V = rank L = dim W. its kernel is the zero vector alone. the kernel of L is the zero vector alone. L(V ) = W. the system M X = 0 has no non-zero solutions. 17.16. the only solution to LX = 0 is X = 0V . .. Since dim V = dim W . its kernel consists of the zero vector alone. But LX is the same as M X. so the only solution to M X = 0 is X = 0V . so the only solution to LX = 0 is X = 0V . So M does not have zero as an eigenvalue. Since L is surjective. Since L is injective. iii. ii. iii. So by the Dimension Formula. ii. So L is injective. So L does not have zero as an eigenvalue. Since M is invertible. By the Dimension Formula. the matrix M is square so we can talk about its eigenvalues.. we have dim V = L + rank L 254
. So L = dim ker L = 0. Since M is invertible.. But LX is the same as M X.
since the size of B is equal to dim W .and since ker L = {0V } we have L = dim ker L = 0. So L(V ) = span B = W. pick a basis B of L(V ). Since L(V ) is a subspace of W with the same dimension as W . it must be equal to W .. Each element of B is a vector in W . Therefore B is a basis of W .
255
. 19.. So L is surjective. so the elements of B form a linearly independent set in W . To see why. so dim W = dim V = rank L = dim L(V ). .
where In is the n × n identity matrix. in the direction v = (v 1 . and while we can do this in a purely formal (mathematical) sense. . then it does not carry meaning in terms of Rn but just exists in a formal sense. . 0 1 Note that we keep the last coordinate fixed. v 2 . pn . . which is non-linear since it moves the origin. v n . . Vectors
This is an expanded explanation of this remark. . we "lift" Rn up to Rn+1 (sometimes written as Rn ) by stating that all tuples p = (p1 . We can "subtract" two points which gives us the vector between as done which describing choosing the origin. and it is stronger. and this concept is highly used in rendering computer graphics. but that is for a different course) or scale a point. We can also act on Rn in a somewhat non-linear fashion by taking matrices of the form ∗ ∗ 0 1 and this still fixes the last coordinate. a translation is what is known an 256
.D
Points Vs. However we can project it down to a point by 1 scaling by w . There is a notion of a point in Rn representing a vector. Note that if the last coordinate w is not 0 or 1. To make all of this mathematically (and computationally) rigorous. so we move points to points and vectors to vectors. p2 . We also do a similar procedure for all matrices acting on Rn by the following. . . People might interchangeably use the term point and vector in Rn . 0) ∈ Rn+1 correspond to a vector v ∈ Rn . . . v n ) by the following matrix Tv = In v 0 1 . Note that this is an invertible matrix with determinant 1. Naturally (as we should be able to) we can add vectors to points and get a point back. Let A be a k × n matrix. . however these are not quite the same concept. we get the following (k + 1) × (n + 1) matrix A 0 A= . we can represent it as a vector (based at the origin O) by taking v = P − O. For example we can also represent a translation. thus if we take any point P . we really cannot add two points together (there is the related notion of this using convex combinations. . 1) ∈ Rn+1 correspond to a point p ∈ Rn and v = (v 1 . then when we lift. .
and translation. so it is neither a point nor a vector). and scale points and vectors as described and generates nonsense when we can't (i. reflection about a single line.e. you can get all of them by rotation about the origin. A good exercises to try are to check that lifting R2 to R3 allows us to add. Another good exercise is to describe all isometries of R2 . adding two points gives us a 2 in the last coordinate. subtract.isometry on Rn (note it is not an isometry on Rn+1 ). an operator where Tv x = x for all vectors x ∈ Rn . As hint.
257
.
Definition The dual space V ∗ of a vector space V is the vector space of all bounded linear functionals on V . ˆ There exists an inverse g −1 ∈ G for all g ∈ G.2
Groups
Definition A group is a set G with a single operation · which satisfies the axioms:
ˆ Associativity (a · b) · c = a · (b · c) for all a. c ∈ G. Let V be a vector space over F. In all cases below. which is 1 if i = j and 0 otherwise.
Groups can be finite or infinite.
E. and notice that not alls element in a group must commute (i. This is material is more advanced but will be interesting to anybody wanting a deeper understanding of the underlying mathematical structures behind linear algebra. we assume that the given set is closed under the operation(s) introduced. we can associate V ∗ with row vectors wT as a functional by the matrix multiplication wT v for vectors v ∈ V . ˆ There exists an identity 1 ∈ G. i
E. So the basis for V ∗ is eT or ei . There is a natural basis {Λi } for V ∗ by Λi (ej ) = δij where {ej } is the canonical (standard) basis for V and δij is the Kronecker delta.e. v for vectors v ∈ V . b.1
Dual Spaces
Definition A bounded operator is a linear operator φ : V → W such that φv W ≤ C v V where C > 0 is a fixed constant. and a functional is a function φ : V → F. Here are some examples of groups: 258
. Alternatively we can associated V ∗ with vectors in V as a functional by taking the usual dot product.E
Abstract Concepts
Here we will introduce some abstract concepts which are mentioned or used in this book. the order of multiplication can matter).. Concretely for a finite dimensional vector space V .
all of the above mean that you have notions of +. ˆ All permutations of [1. . There exists a multiplicative identity 1. 2. M4. There exists an additive identity 0. ˆ Any vector space under addition. b.ˆ Non-zero real numbers under multiplication.
E. × and ÷ just as for regular real numbers. Addition is commutative a + b = b + a. ˆ All real numbers under addition. A3. ˆ All n × n real matrices of determinant 1. Roughly. . . c ∈ F the following axioms are satisfied: A1. A4. There exists a multiplicative inverse a−1 if a = 0. R. . There exists an additive inverse −a. A2. We note that every one of the these examples are infinite.3
Fields
Definition A field F is a set with two operations + and · that for all a. −. and C. however this does not necessarily have to be the case. Addition is associative (a + b) + c = a + (b + c). M1. The distributive law holds a · (b + c) = ab + ac. Multiplication is associative (a · b) · c = a · (b · c). M2. M3. Fields are a very beautiful structure. some examples are Q. Multiplication is commutative a · b = b · a. ˆ All invertible n × n real matrices. n] under compositions.
Note that all real numbers under multiplication is not a group since 0 does not have an inverse. D. Let q ≥ 0 and let Zq be the set of remainders 259
.
Now if p is a prime number. A3. An important example of a ring is F[x]. 2 · 2 = 4 ≡ 1. b. For Z5 we have 2−1 = 3 since 2 · 3 = 6 ≡ 1 and 4−1 = 4 since 4 · 4 = 16 ≡ 1. but none of the multiplicative ones must. Recall that you can do everything you want in a field except divide polynomials. Definition A ring R is a set with two operations + and · that for all a. but rings in general are not nearly as nice (for example. then people will write Fq to reinforce that it is a field. Often when q = pn where p is a prime.
E. We call such polynomials irreducible. There exists an additive identity 0. then Zp is a field (often written as Zp ). which is the ring of all polynomials in one variable x with coefficients in a field F. In other words. 260
. then the two equations in axiom D are equivalent. D. 2} with 1 + 1 = 2. These are known as rings. you can get a field. The distributive law holds a·(b+c) = a·b+a·c and (a+b)·c = a·c+b·c. Note that when we have axiom M3. A2. 1. but if you take the modulus with respect to a polynomial which is not a product of two smaller polynomials. c ∈ R the following axioms are satisfied: A1. A4. For rings all of the addition axioms hold. in Z4 two things can be multiplied together to give you 0).of Z (the set of integers) by dividing by q.4
Rings
However Z4 is not a field since 2 · 2 = 4 ≡ 0 and 2 · 3 = 6 ≡ 2. Similarly Z is not a field since 2 does not have a multiplicative inverse. 1} where 1 + 1 = 2 ≡ 0 (these are exactly the bits used in bit matrices) and Z3 = {0. We say Zq is the set of all a modulo q or a mod q for short or a ≡ q where a ∈ Z. So for example we have Z2 = {0. and we define addition and multiplication to be their usual counterparts in Z except we take the result mod q. Clearly all fields are rings. Clearly Z2 is a field. so Z3 is also a field. Addition is commutative a + b = b + a. and from above for Z3 we have 2−1 = 2. Addition is associative (a + b) + c = a + (b + c). There exists an additive inverse −a.
v. S. We note that this is not associative nor commutative under × and that v × v = 0 (so there are in fact no multiplicative inverses). and there is no multiplicative identity. For example. we have D.e. Or in simpler words.e. Another algebra is we can take Mn (R) but take scalars in C and just formally say iM is another element in this algebra. an algebra is a vector space where you can multiply vectors. We have (αv) · (βw) = (αβ)(v · w). Another example is R3 where multiplication is the crossproduct ×.5
Algebras
Definition An algebra A is a vector space over F with the operation · such that for all u.
E.you take a polynomial p and you set p ≡ 0. and check that the field extension Z2 (α) has 4 elements and has characteristic 2 (hence it is not actually Z4 ). Lastly. with real coefficients). The distributive law holds u · (v + w) = u · v + u · w and (u + v) · w = u · w + v · w. these are the central objects in Galois theory and are denoted F(α) where α is a root of p. β ∈ F. Essentially an algebra is a ring that is also a vector space over some field. This is what is known as a field extension. For example. we sum 1 together p times and return to 0).
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. A good exercise is to find an irreducible degree 2 polynomial p in Z2 [x]. so what you get is actually the same field as C since we have x2 + 1 = 0 or perhaps more suggestively x2 = −1. w ∈ A and α. recall that sln defined here is an algebra under [. ]. and in general Zp has characteristic p. all n × n real matrices Mn (R) is a ring but we can let scalars in R act on these matrices in their usual way. For example Z3 has characteristic 3 since 1 + 1 + 1 ≡ 0. One final definition: We say that a field F has characteristic p if p 1 ≡ i=1 0 (i. thus this is just making sure you don't have ab ≡ 0. the polynomial p(x) = x2 + 1 cannot be factored over R (i.
Additionally we can define an inner product on H = L2 (S 1 ) by taking 0
2π
f.
R+ 0 x→∞
In particular. g =
0
f (x)g(x) dx. Note that this is a vector space over R (or C) under addition (in fact it is an algebra under pointwise multiplication) with norm (the length of the vector)
1/p
f
p
=
Ω
|f (x)| dx
p
. take Ω = [0. for example 0 we have ∞ |1|p dx = dx = lim x = ∞. we can take S 1 . and in fact. the unit circle in R2 . sin(nx) =
0 2π
sin(mx) sin(nx) dx cos((m − n)x) − cos((m + n)x) dx 2 262
=
0
. 2π] → R such that f (0) = f (2π) (or more generally for a periodic function f : R → R where f (x) = f (x + 2πn) for all n ∈ Z). Let Lp (Ω) denote the space of all 0 continuous functions f : Ω → R (or C) such that if p < ∞. However 0 note that not every differentiable function is contained in Lp (Ω).
For example. First note that
2π
sin(mx). So the natural question to ask is what is a good 2 basis for H? The answer is sin(nx) and cos(nx) for all n ∈ Z≥0 . and to turn this into a valid integral.
and note that f.F
Sine and Cosine as an Orthonormal Basis
Definition Let Ω ⊆ Rn for some n. then
1/p
|f (x)|p dx
Ω
< ∞. f = f 2 . they are orthogonal.
otherwise |f (x)| < M for some fixed M and all x ∈ Ω. the space L1 (R) is all absolutely integrable functions. 2π) and take functions f : [0.
We also welcome suggestions for other movie themes.1
Introductory Video
Three bears go into a cave. You might even like to try your hand at making your own!
G. two come out. Would you go in?
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.G
Movie Scripts
The authors welcome your feedback on how useful these movies are for helping you learn.
2
What is Linear Algebra: Overview
In this course. . . . . . xm and n given constants a1 . . . . . We end with the problem of finding a least squares fit---find the line that best fits a given data set:
In equation (3) we have n linear functions called f1 . . for some constant λ.G. . . The plot of this is just a straight line through the origin with slope λ
265
. a linear function obeys the linearity property f (a + b) = f (a) + f (b) . xm ) = an . .
(3)
and discuss how to solve them. we start with linear systems f1 (x1 . The solution to this is f (x) = λx . an . . fn (x1 . . . . We need to say what it means for a function to be linear. m unknowns x1 . . . . . . In one variable. fn . . . . xm ) = a1 .
c + d) = f (a. d) . For a linear function of two variables f (x. y) this means f (a + b. We finish with a question. c) + f (b.We should also check that our solution obeys the linearity property. The logic is to start with the left hand side f (a + b) and try to turn it into the right hand side f (a) + f (b) using correct manipulations: f (a + b) = λ(a + b) = λa + λb = f (a) + f (b) . For functions of many variables. The first step here just plugs a + b into f (x). and in the third we recognize that λa = f (a) and λb = f (b). The plot of f (x) = λx+β is a straight line. but does it obey the linearity property?
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. the second is the distributive property. linearity must hold for every slot. This proves our claim.
two dimes. and thus express this problem as the matrix equation 5 10 25 n 65 1 1 1 d = 7 . 65 5 10 25 1 1 1 7 0 0 −1 2 Now to solve it. dimes. and that there are twice as many dimes as quarters. and quarters are in the jar. We can translate this into a system of the following linear equations: 5n + 10d + 25q = 65 n+d+q =7 d = 2q Now we can rewrite the last equation in the form of −d + 2q = 0.
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.G. 0 −1 2 q 0 or as an augmented matrix (see also this script on the notation). Clearly d = 2. we get 45q − 15q = 30q = 65 − 35 = 30 and hence q = 1. and dimes. we have 5n + 20q + 25q = 5n + 45q = 65 n + 2q + q = n + 3q = 7 and by subtracting 5 times the bottom equation from the top. Therefore there are four nickels. and one quarter. and hence n = 7−2−1 = 4. Your friend wants to know how many nickels. using our original set of equations and by substitution. nickels.3
What is Linear Algebra: 3 × 3 Matrix Example
Your friend places a jar on a table and tells you that there is 65 cents in this jar with 7 coins consisting of quarters.
We are asked to find a linear transformation relating this new representation to the one in the lecture.4
What is Linear Algebra: Hint
Looking at the problem statement we find some important information.G. Then write an equation for s using x. y and λ. Hint: Let λ represent the amount of sugar in each apple. 1. This means we must create a system of equations relating the variable x and y to the variables s and f in matrix form. and second that the information about the barrel is recorded as (s. Your answer should be the matrix that transforms one set of variables into the other. use the hint to figure out how much sugar is in x apples. where in the lecture x = the number of apples and y = the number of oranges. and y oranges in terms of λ. f ). where s = units of sugar in the barrel and f = number of pieces of fruit in the barrel. first that oranges always have twice as much sugar as apples. 2. To find the first equation find a way to relate f to the variables x and y.
268
. To find the second equation.
G.
to be equivalent? They are certainly not equal. because they don't match in each component. . because the matrix multiplication can be done in your head. The other augmented matrix represents the system x+0·y = 9 0 · x + y = 18 ? This which clearly has the same solution.6
Gaussian Elimination: Equivalence of Augmented Matrices
Lets think about what it means for the two augmented matrices 1 1 27 2 −1 0 and 1 0 9 0 1 18 .
270
. The first and second system are related in the sense that their solutions are the same. we might want to introduce a new kind of equivalence relation. Notice that it is really nice to have the augmented matrix in the second form. Well we could look at the system of linear equations this represents
x + y = 27 2x − y = 0 ? and notice that the solution is x = 9 and y = 18. but since these augmented matrices represent a system.
We will call two people equivalent if they have the same hair color.
For any x ∈ U .
271
. y and z ∈ U .G.
Firstly remember that an equivalence relation is just a more general version of ''equals''. Lets do a silly example: Lets replace the set of augmented matrices by the set of people who have hair. see Webwork Homework 0. if x ∼ y then y ∼ x. we have x ∼ x.7
Gaussian Elimination: Hints for Review Questions 4 and 5
The hint for Review Question 4 is simple--just read the lecture on Elementary Row Operations. Here we defined row equivalence for augmented matrices whose linear systems have solutions by the property that their solutions are the same. For any x. Problem 4) Show that row equivalence of augmented matrices is an equivalence relation. For any x. if x ∼ y and y ∼ z
(For a more complete discussion of equivalence relations. y ∈ U . So this question is really about the word same. Recall that a relation ∼ on a set of objects U is an equivalence relation if the following three properties are satisfied:
ˆ Reflexive: ˆ Symmetric: ˆ Transitive: then x ∼ z. There are three properties to check:
ˆ Reflexive: This just requires that you have the same hair color as yourself so obviously holds. Question 5 looks harder than it actually is:
Row equivalence of matrices is an example of an equivalence relation.
272
. so this holds too. so the transitive property holds too and we are done. then Bob has the same hair color as Betty. Bob (say) has the same hair color as a second person Betty(say). then it follows that Bob and Brenda have the same hair color.
ˆ Transitive: If Bob has the same hair color as Betty (say) and Betty has the same color as Brenda (say).ˆ Symmetric: If the first person.
so I will switch them. Scalar Multiplication 3. So we get x + 2y = 4 3x + y = 7 The augmented matrix looks like 1 2 4 3 1 7 . Now we can write it in augmented matrix for this way 3 1 7 1 2 4 .
Notice that this won't change the solution of the system. The system 3x + y = 7 x + 2y = 4 is something we learned to solve in high school algebra. In this example there are only two equations. Row Swap 2. Row swap allows us to switch the order of rows.9
Elementary Row Operations: Example
We have three basic rules 1. This will work with a larger system as well. 274
. but you have to decide which equations. but you have to decide which equations to switch.G. This is where we can say that the original augmented matrix is equivalent to the one with the rows swapped.
We can see what these operations allow us to do: 1. This will work with a larger system as well. Row Sum Lets look at an example. or rows to switch. Make sure that you don't forget to switch the entries in the right-most column. but the augmented matrix will look different.
3. In our example we could start with −3x + −6y = −12 3x + y = 7 and replace the first equation with the sum of both equations. which after some simplification is translates to 0 −5 −5 3 1 7 275 . effectively destroying the information contained in the equation. We are not allowed to multiply by zero because it would be like replacing one of the equations with 0 = 0. So we get −3x + 3x + −6y + y = −12 + 7 3x + y = 7. Scalar multiplication allows us to multiply both sides of an equation by a non-zero constant. Row summing allows us to add one equation to another. So if we are starting with x + 2y = 4 3x + y = 7 Then we can multiply the first equation by −3 which is a nonzero scalar. but not the solution of the system.2. and this changes the augmented matrix.
. This operation will give us −3x + −6y = −12 3x + y = 7 which has a corresponding augmented matrix −3 −6 −12 3 1 7 .
Notice that we have multiplied the entire first row by −3.
which makes it much easier to solve for y. Here we replaced the first equation with a sum. notice that the x-terms in the first equation disappeared.When using this row operation make sure that you end up with as many equations as you started with. In the example. Think about what the next steps for solving this system would be using the language of elementary row operations.
276
. but the second equation remained untouched.
you try to demonstrate that the opposite of the statement leads to a contradiction. But that would be weird. in fact this Theorem says that this can never happen! Because this proof comes at the end of the section it is often glossed over.11
Elementary Row Operations: Explanation of Proof for Theorem 3. Then remove all the non-pivot 279
. therefore. we try a proof by contradiction. so maybe that could lead to two different RREF's and in turn two different solution sets for the same linear system. that Alice and Bob do find different RREF augmented matrices called A and B.G. Here's a sketch of what happens in the video:
In words: we start with a linear system and convert it to an augmented matrix.1
The first thing to realize is that there are choices in the Gaussian elimination recipe. That is the method where to show that a statement is true. but it is a very important result. Suppose. because we are studying a uniqueness statement. Here. Then. the opposite statement to the theorem would be to find two different RREFs for the same system.
but it does not ruin row equivalence. a = b. because we left only the pivot columns (plus the first column that differs) we have ˆ A= IN a 0 0 ˆ and B = IN b 0 0 .
280
. Importantly. Now. Removing columns does change the solution sets. we only performed operations that kept Alice's solution set the same as Bob's. This is a contradiction so the proof is complete. But at all stages.columns from A and B until you hit the first column that differs.
where IN is an identity matrix and a and b are column vectors. by assumption. so A and B have the same solution sets. Record that in the last column and call the results A and B. So if we try to wrote down the solution sets for A and B they would be different.
281
. You should also check that this does not make the whole third line zero.G.R3 −2R1 0 −9 . Therefore we learn that we must avoid a row of zeros preceding a non-vanishing entry after the vertical bar.12
Elementary Row Operations: Hint for Review Question 3
The first part for Review Question 3 is simple--just write out the associated linear system and you will find the equation 0 = 6 which is inconsistent. But if 3 k+6=3−k ⇒k =− . Turning to the system of equations. we first write out the augmented matrix and then perform two row operations 6 1 −3 0 1 0 3 −3 2 k 3−k 1 1 −3 0 6 R2 −R1 . You now have enough information to write a complete solution. 3 3 ∼ 0 k + 6 3 − k −11 Next we would like to subtract some amount of R2 from R3 to achieve a zero in the third entry of the second column. 2 this would produce zeros in the third row before the vertical line.
G. I. Using the dot product in R3 we have N V = d. b. so all the possibilities for C span a plane whose normal vector is N . c).13
Solution Sets for Systems of Linear Equations: Planes
Here we want to describe the mathematics of planes in space. The video is summarised by the following picture:
A plane is often called R2 because it is spanned by two coordinates. then so too does V0 + C whenever C is perpendicular to N . U V = 0 ⇔ U ⊥ V . z) the vector of unknowns and N = (a. c). Remember that when vectors are perpendicular their dot products vanish. and space is called R3 and has three coordinates. This is because N (V0 + C) = N V0 + N C = d + 0 = d . y. This means that if a vector V0 solves our equation N V = d. The equation for a plane is ax + by + cz = d . y. usually called (x. 282
. But C is ANY vector perpendicular to N . Lets simplify this by calling V = (x. z). b.e. Hence we have shown that solutions to the equation ax + by + cz = 0 are a plane with normal vector N = (a.
so the solution set is all of R3 .G. z) is a solution. we must look at two planes. the solution is a plane. z) and label points in R3 . These are often called (x. This is explained in this video or the accompanying script. then any (x. so the solution set will also (usually) be a line:
283
. These usually intersect along a line. y. y.14
Solution Sets for Systems of Linear Equations: Pictures and Explanation
This video considers solutions sets for linear systems with three unknowns. Lets work case by case:
ˆ If you have no equations at all. The picture looks a little silly:
ˆ For a single equation. The picture looks like this:
ˆ For two equations.
there will be no solutions at all. Two different looking equations could determine the same plane. if you had four equations determining four parallel planes the solution set would be empty. If the equations are inconsistent.ˆ For three equations. This looks like this:
284
. most often their intersection will be a single point so the solution will then be unique:
ˆ Of course stuff can go wrong. or worse equations could be inconsistent. For example.
Then since the second equation doesn't depend on µ we can keep the equation x2 = 1. and for a third equation we can write x3 = µ
285
. We will call x1 and x2 the 3 pivot variables. and there is no 3 × 3 identity in the augmented matrix. We'll call x3 the free variable.G. Notice there are more variables than equations and that this means we will have to write the solutions for the system in terms of the variable x3 . let's think about what the solution set looks like 1 0 3 2 0 1 0 1 This looks like the system x1 + 3x3 = 2 x2 = 1 Notice that when the system is written this way the copy of the 1 0 2 × 2 identity matrix makes it easy to write a solution 0 1 in terms of the variables x1 and x2 . The third column does not look like part of 0 an identity matrix.15
Solution Sets for Systems of Linear Equations: Example
Here is an augmented matrix. Let x3 = µ. Then we can rewrite the first equation in our system x1 + 3x3 = 2 x1 + 3µ = 2 x1 = 2 − 3µ.
and many people use summation notation or Einstein's summation convention with the added notation of Mj denoting the j-th row of the matrix. i i
You can see a concrete example after the definition of the linearity property. For example. all you need to show is that M (α · X + β · Y ) = α · (M X) + β · (M Y ) where α.
287
. For the second part. β ∈ R (or whatever field we are using) and y1 y 2 Y = ..G. . . the key is to consider the vector as a n × 1 matrix.16
Solution Sets for Systems of Linear Equations: Hint
For the first part of this problem. for any j we have
k
(M X)j =
i=1
aj x i = aj x i . yk Note that this will be somewhat tedious.
and what I have described is polar coordinates which you should be able to translate back to the usual Cartesian coordinates of (x. things begin to get interesting. but an infinite number which we typically encapsulate as 0 to 2π radians (i. . Luckily when n = 2. We can add vectors by putting them head to toe and we can scale our vectors.G. 0 to 360 degrees). y). z). Now our notion of direction in somewhat more complicated using azimuth and altitude (see Figure G. So why is this system R2 ? The answer comes from trigonometry. and we clearly have a notion of distance.e. just we can now move around in our usual ''3D'' space by basically being able to draw in the air. so make sure you read the note: Points Versus Vectors or Appendix D. It is lucky for us because we can represent this by drawing arrows on paper. x2 . Recall that the length of the vector is also known as its magnitude. y. Just one final closing remark. This is something you should understand well. if n = 1.17
Vectors in Space. You still should be familiar with what things look like here. 288
. this is exactly the same as R2 . xn ) to encapsulate our direction and magnitude and you can just treat vectors in Rn the same way as you would for vectors in R2 . Also we can equivalently write our tuple (x. For example. . y. In short it is the usual vectors we are used to. if we look at this in Cartesian coordinates (x. then it is just the number line where we either move in the positive or negative directions. . Thus for all n ≥ 3. I have been somewhat sloppy through here on points and vectors. but it is secretly still there. and this concept is useful in physics such as Force Vector Diagrams. However what is interesting is that we no longer just have two directions. but ultimately there is nothing really interesting that goes on here. Now for R3 .17 below). So we will just use the tuple to encapsulate the data of it's direction and magnitude. . we just use the tuple (x1 . n-Vectors: Overview
What is the space Rn . z) as x y z so our notation is consistent with matrix multiplication.
5).18
Vectors in Space.Altitude
Azimuth
Figure 2: The azimuth and altitude in spherical coordinates. Thus we write the solution as x = −2λ1 −5λ2 +3 y = λ1 z = λ2 289 y = λ2 . 2.
. In fact this is a system of linear equations whose solutions form a plane with normal vector (1. and d are constants. b. As an augmented matrix the system is simply 1 2 5 3 .
G. c. This is actually RREF! So we can let x be our pivot variable and y. y and z looks like ax + by + cz = d where a. n-Vectors: Review of Parametric Notation
The equation for a plane in three variables x. z be represented by free parameters λ1 and λ2 : x = λ1 . Lets look at the example x + 2y + 5z = 3 .
often called c. the inverse slope is your velocity. In the diagram above c = 1 and corresponds to the lines x = ±t ⇒ x2 − t2 = 0. t) given by ||v||2 = x2 −t2 used in Einstein's theory of relativity. The idea is to plot the story of your life on a plane with coordinates (x. t). z) ∈ R3 ).G. This should get you started in your search for vectors with zero length. The coordinate x encodes where an event happened (for real life situations. The slope of the worldline has to do with your speed. Or to be precise. Einstein realized that the maximum speed possible was that of light.19
Vectors in Space. Therefore you can plot your life history as a worldline as shown:
Each point on the worldline corresponds to a place and time of an event in your life. we must replace x → (x. y. n-Vectors: The Story of Your Life
This video talks about the weird notion of a ''length-squared'' for a vector v = (x.
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. The coordinate t says when events happened.
using our component-wise addition law we have x1 y + 1 x2 y2 := x 1 + x2 y1 + y2 ∈ R2 . Its worth doing this once. y ∈ R obey x + y = y + x. so here we go: Before we start.
This again relies on the underlying real numbers which for any x.
(+ii) Additive commutativity: We want to check that when we add any two vectors we can do so in either order. To do this (unless we invent some clever tricks) we will have to check all parts of the definition. (+i) Additive closure: We need to make sure that when we add and x1 x2
y1 that we do not get something outside the original y2 vector space R2 . (+iii) Additive Associativity: This shows that we needn't specify with parentheses which order we intend to add triples of vectors because their sums will agree for either choice. x1 x2 + y1 y2 =
?
y1 x1 + y2 x2
. i. This fact underlies the middle step of the following computation x1 y + 1 x2 y2 = x1 + y 1 x2 + y 2 = y 1 + x1 y 2 + x2 = y1 y2 + x1 x2 .
292
.
which demonstrates what we wished to show. remember that for R2 we define vector addition and scalar multiplication component-wise.G.20
Vector Spaces: Examples of Each Rule
Lets show that R2 is a vector space. This just relies on the underlying structure of real numbers whose sums are again real numbers so.e. What we have to check is x1 y + 1 x2 y2 + z1 z2 =
?
x1 x2
+
y1 y2
+
z1 z2
.
(Note that it is important to first figure out what 0 is x1 −x1 here!) The answer for the additive inverse of is x2 −x2 because x1 −x1 x1 − x1 0 + = = = 0. x2 −x2 x2 − x2 0
We are half-way done. (+v) Additive Inverse: We need to check that when we have x1 . x2 there is another vector that can be added to it so the sum is 0.e. that we multiply vectors by scalars (i. i.
You can easily check that when this vector is added to any vector. Notice. the answer is 0= 0 0 .
293
. x1 y1 +0= x1 y1 . numbers) but do NOT multiply a vectors by vectors. The computation required is x1 y + 1 x2 y2 = x1 + (y1 + z1 ) x2 + (y2 + z2 ) + z1 z2 = x1 + y1 x2 + y2 + z1 z2 = (x1 + y1 ) + z1 (x2 + y2 ) + z2 y1 y2 + z1 z2 .
=
x1 y + z1 + 1 y1 y2 + z2
=
x1 + x2
(iv) Zero: There needs to exist a vector 0 that works the way we would expect zero to behave.Again this relies on the underlying associativity of real numbers: (x + y) + z = x + (y + z) .e. the result is unchanged.
It is easy to find. now we need to consider the rules for scalar multiplication.
x2 = x1 x2 .x) = (a.b). Just as for addition. I.(·iv) Multiplicative associativity.
Now we are done---we have really proven the R2 is a vector space so lets write a little square to celebrate. Indeed. ''1'' · x1 x2 = x1 x2 .b).
This clearly holds for real numbers a.
There is an obvious choice for this special scalar---just the real number 1 itself.x1 ) a.x1 1. The computation is (a.(b.
295
. (b.b) · x1 x2 =a· b·
?
x1 x2
.e.
which is what we want.x1 (a. this is the requirement that the order of bracketing does not matter.x2 ) = a· b · x1 x2 .(b. We need to establish whether (a.x1 ) (b.b)· x1 x2 = (a.b).x.x2 ) = a. to be pedantic lets calculate 1· x1 x2 = 1.x2 = a. (·v) Unity: We need to find a special scalar acts the way we would expect ''1'' to behave.(b.
).
You can also model the new vector 2J obtained by scalar multiplication by 2 by thinking about Jenny hitting the puck twice (or a world with two Jenny Potters.21
Vector Spaces: Example of a Vector Space
This video talks about the definition of a vector space. plus the vector obtained when they hit the puck together.G. This picture shows vectors N and J corresponding to the ways Nicole Darwitz and Jenny Potter hit a hockey puck. it is actually designed to model a very wide range of real life situations. complicated and abstract. Now ask yourself questions like whether the multiplicative distributive law 2J + 2N = 2(J + N ) make sense in this context.. As an example. Even though the defintion looks long.
296
. You can think about adding vectors by having two players hitting the puck at the same time. The different ways of hitting a hockey puck can all be considered as vectors... consider the vector space V = {all possible ways to hit a hockey puck} .
such as a degree 0 polynomial.22
Vector Spaces: Hint
I will only really worry about the last part of the problem.
297
. and multiplying by i ∈ C. but it is clearly a vector space over Q. As a second hint. and you should take from this that the scalar field matters. The problem can be solved by considering a non-zero simple polynomial. consider Q (the field of rational numbers).G. This will violate one of the vector space rules about scalars. so it is not closed under scalar multiplication. √ √ / This is not a vector space over R since 2 · 1 = 2 ∈ Q. That is R to say we take a vector p ∈ P3 and then considering i · p.
The additive property L(u + v) = L(u) + L(v) is left as a fun exercise. The calculation looks like this: L(cu) = L c x y =L cx cy =c = 2cx − 3cy cx + cy = cL x y = cL(u) . The c is then factored out and we recognize that the vector next to c is just our linear transformation again.23
Linear Transformations: A Linear and A NonLinear Example
This video gives an example of a linear transformation as well as a transformation that is not linear. This verifies the scalar multiplication property L(cu) = cL(u). next comes the rule for multiplying a vector by a number. For a non-linear example lets take the vector space R1 = R with L : R −→ R via x → x + 1.G. via x y → 2 −3 1 1 x y and L(cu) = cL(u) .
Here we focus on the scalar multiplication property L(cu) = cL(u) which needs to hold for any scalar c ∈ R and any vector u.
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. then the rule for the given linear transformation L is used. In what happens below remember the properties that make a transformation linear: L(u + v) = L(u) + L(v) The first example is the map L : R2 −→ R2 .
.
2x − 3y x+y
The first equality uses the fact that u is a vector in R2 .
Now we see the problem.This looks linear because the variable x appears once. but the constant term will be our downfall! Computing L(cx) we get: L(cx) = cx + 1 . but on the other hand cL(x) = c(x + 1) = cx + c . unless we are lucky and c = 1 the two expressions above are not linear.
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. the game is up! x → x + 1 is not a linear transformation. Since we need L(cu) = cL(u) for any c.
Now going to our vector space PR . So to make our image in 3 PR . β ∈ R. take the constant to be non-zero with polynomials which in-fact will make it nonlinear). and we note that this does not necessarily give the same map (ex. Therefore we have to perform a definite x integral instead. Now we first note that for any α. For example I(3x2 ) = x3 + c where c can be any constant. if we take any p(x) = αx3 + 3 βx2 + γx + δ. The other thing we could do is explicitly choose our constant. 0 0 0 0 Similarly now consider the map I where I(f ) = f (x) dx is the indefinite integral on any integrable function f . 4 3 2
and we note that this is outside of PR . and we can write this as the matrix 0 1 0 0 0 0 2 0 D= 0 0 0 3 . then we have I(p) = α 4 β 3 γ 2 x + x + x + δx. Let D denote the usual derivative operator and we note that it is linear. we formally set I(x3 ) = 0.24
Linear Transformations: Derivative and Integral of (Real) Polynomials of Degree at Most 3
For this. so we define I(f ) := 0 f (y) dy.
so I is a linear map on functions. we consider the vector space PR of real coefficient 3 polynomials p such that the degree of deg p is at most 3. we have I(α · p + β · q) = =α α · p(x) + β · q(x) dx p(x) dx + β q(x) dx = αI(p) + βI(q).G. However we note that this is not a well-defined map on vector spaces since the additive constant states the image is not unique. Thus we can now (finally) write this 3
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.
. by linearity we know that we can separate out the sum. so lets think about how we could write this in the vector notation we use in the class. The third part of the problem is asking us to think about this as a linear algebra problem. . So it makes some sense to think about polynomials this way. We could write a0 2 a1 a0 + a1 t + a2 t as a2 And think for a second about how you add polynomials. so what does that mean in this case? Well. We could also write the output b0 b0 + b1 t + b2 t2 + b3 t3 as b1 b3 b2 302
. and pull out the constants so we get L(a0 + a1 t + a2 t2 ) = a0 L(1) + a1 L(t) + a2 L(t2 ) Just this should be really helpful for the first two parts of the problem. So this means if L is a linear transformation from P2 → P3 that the inputs of L are degree two polynomials which look like a0 + a1 t + a2 t 2 and the output will have degree three and look like b0 + b1 t + b2 t2 + b3 t3 We also know that L is a linear transformation.25
Linear Transformations: Linear Transformations Hint
The first thing we see in the problem is a definition of this new space Pn .G. + an tn where the ai 's are constants. since vector addition is also component-wise. Elements of Pn are polynomials that look like a0 + a1 t + a2 t2 + . you match up terms of the same degree and add the constants component-wise.
Then lets look at the information given in the problem and think about it in terms of column vectors
ˆ L(1) = 4 but we can think of the input 1 = 1 + 0t + 0t2 and the 4 1 0 output 4 = 4 + 0t + 0t2 0t3 and write this as L(0) = 0 0 0 0 0 0 ˆ L(t) = t3 This can be written as L(1) = 0 0 1 ˆ L(t2 ) = t − 1 It might be a little trickier to figure out how to write t − 1 but if we write the polynomial out with the terms in order and with zeroes next to the terms that do not appear. Then look at the first two parts of this problem to help you figure out what the entries should be. first think about what the dimensions of the matrix would be.
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. we can see that −1 1 t − 1 = −1 + t + 0t2 + 0t3 corresponds to 0 0 −1 0 1 So this can be written as L(0) = 0 1 0 Now to think about how you would write the linear transformation L as a matrix.
Now lets make a 4 × 4 matrix.
Now draw a picture where each person is a dot. and then draw a line between the dots of people who are friends. Carl. and David. ˆ Alice and Carl are friends. Make a column and a row for each of the four people. This is an example of a graph if you think of the people as nodes. In this tiny facebook there are only four people. ˆ David and Bob are friends. Suppose we have the following relationships
ˆ Alice and Bob are friends. For example Alice and Carl are friends so we can label the table below. Bob. When two people are friends put a 1 the the row of one and the column of the other. ˆ Carl and Bob are friends. Alice. and the friendships as edges. It will look a lot like a table.G. which is an adjacency matrix for the graph. A B C D A 1 B C 1 D 304
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Matrices: Adjacency Matrix Example
Lets think about a graph as a mini-facebook.
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. The adjacency matrix might not be symmetric then.We can continue to label the entries for each friendship. the same way a multiplication table would be symmetric. Here lets assume that people are friends with themselves. This is because on facebook friendship is symmetric in the sense that you can't be friends with someone if they aren't friends with you too. This is an example of a symmetric matrix. so the diagonal will be all ones. You could think about what you would have to do differently to draw a graph for something like twitter where you don't have to follow everyone who follows you. A B C D A 1 1 1 0 B 1 1 1 1 C 1 1 1 0 D 0 1 0 1
Then take the entries of this 1 1 1 1 1 1 0 1
table as a matrix 1 0 1 1 1 0 0 1
Notice that this table is symmetric across the diagonal.
computing AB and BA we get the same result AB = BA = 1 a+b 0 1 . holds for matrices AB = BA . C= 1 0 a 1 . Lets take a simple 2 × 2 example. matrices usually do not commute. B= 1 b 0 1 . 1 + a2 a a 1 . This says the order of bracketing does not matter and is called associativity. Generally. and the problem of finding those that do is a very interesting one.
?
In fact.27
Matrices: Do Matrices Commute?
This video shows you a funny property of matrices. Now we ask ourselves whether the basic property of numbers ab = ba .
so this pair of matrices do commute. For example numbers obey a(bc) = (ab)c and so do matrices: A(BC) = (AB)C. firstly note that we need to work with square matrices even for both orderings to even make sense. For this. Some matrix properties look just like those for numbers.G. and CA = 1 a a 1 + a2
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. Lets try A and C: AC = so AC = CA and this pair of matrices does not commute. let A= 1 a 0 1 .
e.G. Let V be some vector space and M be some collection of matrices. and we say that M is a left-action on V if (M · N ) ◦ v = M ◦ (N ◦ v) for all M. It should be clear that fR (α · M ) = (αM )R = α(M R) = αfR (M ) for any scalar α. Now all that needs to be proved is that fR (M + N ) = (M + N )R = M R + N R = fR (M ) + fR (N ).29
Matrices: Hint for Review Question 5
The majority of the problem comes down to showing that matrices are right distributive.
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. standard matrix multiplication) and ◦ denotes the matrix is a linear map on a vector (i. M (v)). and you can show this by looking at each entry. There is a corresponding notion of a right action where v ◦ (M · N ) = (v ◦ M ) ◦ N where we treat v ◦ M as M (v) as before. We can actually generalize the concept of this problem.e. and define the map fR : Mk → Mm by fR (M ) = M R where R is some k × m matrix. People will often omit the left or right because they are essentially the same. and note the order in which the matrices are applied. Let Mk is all n × k matrices for any n. and just say that M acts on V . N ∈ N and v ∈ V where · denoted multiplication in M (i.
l i
Finally. ni ml = l i l i
l i i l
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. i l
Equally obvious. (We could have given it any name we liked!). i
Hence starting from the left of the statement we want to prove. we now rename i → l and l → i so mi nl = l i
i l l i
ni ml . we have mi nl . LHS = tr M N = l i
i l
Next we do something obvious. Then MN =
l
mi nl . but the index l is a ''dummy'' index because it is summed over.G. first we can write M = (mi ) j and N = (ni ) j
where the upper index labels rows and the lower one columns. just change the order of the entries mi and nl (they are just numbers) so i l mi nl = l i
i l i l
nl mi . since we have finite sums it is legal to change the order of summations ni ml . l j
where the ''open'' indices i and j label rows and columns. There are some useful things to remember.31
Properties of Matrices: Explanation of the Proof
In this video we will talk through the steps required to prove tr M N = tr N M . Finally the trace is the sum over diagonal entries for which the row and column numbers must coincide tr M =
i
mi .
This expression is the same as the one on the line above where we started except the m and n have been swapped so mi nl = tr N M = RHS . l i
i l
This completes the proof.
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see Lecture 13 Problem 5) and eigenvalues. By a bounded operator. Some other useful properties is for block matrices. Additionally in later chapters we will see that the trace function can be used to calculate the determinant (in a sense it is the derivative of the determinant. we note that sln is a vector space.G. A B C D = tr A + tr D. Using a concept from Chapter 17. however it is an example of an element in the dual-space of all n × n matrices since it is a bounded linear operator to the underlying field F. Additionally we can define the set sln as the set of all n × n matrices with trace equal to 0. and we note that sln is closed under bracket since tr(M N − N M ) = tr(M N ) − tr(N M ) = tr(M N ) − tr(M N ) = 0.e. N ] = M N − N M. and since the trace is linear and a·0 = 0. it should be clear that we have tr and that tr(P AP −1 ) = tr P (AP −1 ) = tr (AP −1 )P = tr(AP −1 P ) = tr(A) so the trace function is conjugate (i. it is basis invariant. then tr(M ) ≤ Cd (I believe C = 1 should work). I mean it will at most scale the length of the matrix 2 (think of it as a vector in Fn ) by some fixed constant C > 0 (this can depend upon n).32
Properties of Matrices: A Closer Look at the Trace Function
This seemingly boring function which extracts a single real number does not seem immediately useful. and for example if the length of a matrix M is d.
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. Additionally we can use the fact tr(M N ) = tr(N M ) to define an operation called bracket [M. similarity) invariant.
Lets look at the example problem. .G. n! 2! 3! 313
.
then we can think about the first few terms of the sequence
∞
e =
n=0
A
An 1 1 = A0 + A + A2 + A3 + . n!
So as matrices we can think about
∞
e =
n=0
A
An .
This means we are going to have an any n. . n! idea of what An looks like for of one of the matrices in the λ 1 .33
Properties of Matrices: Matrix Exponent Hint
This is a hint for computing exponents of matrices. A0 = A1 = 1 0 0 1 1 λ 0 1 1 2λ 0 1 1 3λ 0 1 .
A2 = A · A = A3 = A2 · A =
There is a pattern here which is that An = 1 nλ 0 1 . . Let 1 A= 0
Lets compute An for the first few n. So what is eA if A is a matrix? We remember that the Taylor series for
∞
e =
n=0
x
xn .
.
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.Looking at the entries when we add this we get that the upper leftmost entry looks like this: 1+1+ 1 1 + + . n!
Continue this process with each of the entries using what you know about Taylor series expansions to find the sum of each entry.. = 2 3!
∞
n=0
1 = e1 .
Therefore we assume that M is singular which implies that there exists a non-zero vector X0 such that M X0 = 0. and look at what happens to XV + c · X0 for any c in your field.35
Inverse Matrix: Hints for Problem 3
First I want to state that (b) implies (a) is the easy direction by just thinking about what it means for M to be non-singular and for a linear function to be well-defined. Now assume there exists some vector XV such that M XV = V .G. Lastly don't forget to address what happens if XV does not exist.
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G. 2 2 This must be represented by a matrix. In the lecture. L y = 1 1 0 z z
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. y. 1} 2 and Z2 2 . y. and lets take the example x x 0 1 1 y := AX .36
Inverse Matrix: Left and Right Inverses
This video is a hint for question 4 in the Inverse Matrixlecture 10. 2
These have 8 and 4 vectors. that can be depicted as corners of a cube or square:
Z3 2
or Z2 2
Now lets consider a linear transformation L : Z3 −→ Z2 . only inverses for square matrices were discussed. It helps to look at an example with bits to see why. z)|x. To start with we look at vector spaces Z3 = {(x. but there is a notion of left and right inverses for matrices that are not square. respectively. z = 0.
B must be 2 × 3. 1. 1) → (1. 1. 1) L (0. So B cannot exist. This can 2 be done. 1) (0. It would be 2 × 2.Since we have bits. A left inverse B to the matrix A would obey BA = I and since the identity matrix is square. 0) (1. 0) → (0. 0) → (0. 0. 1) → (1. 1) → (1. However a right inverse C obeying AC = I can. we can work out what L does to every vector.
L L L L
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. 0) → (1. 0. 0) (0. 0. 0. 1. It would have to undo the action of A and return vectors in Z3 to where 2 they started from. 1) L (1. 1) Now lets think about left and right inverses. 0) → (1. 1) L (1. 1) → (0. 0) L (1. this is listed below (0. Its job is to take a vector in Z2 back to 2 one in Z3 in a way that gets undone by the action of A. we see that different vectors in Z3 2 are mapped to the same vector in Z2 by the linear transformation 2 L with matrix A. but not uniquely. But above. 1.
you would find that the first equation has already been solved. This backward substitution makes solving the system much faster. Now the second part of the problem −1 is to solve for x. The augmented matrix you get is 1 0 −5 6 0 −1 1 1 0 0 1 −1 319
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LU Decomposition: Example: How to Use LU Decomposition
Lets go through how to use a LU decomposition to speed up solving a system of equations. and is ready to be plugged into the second equation. Suppose you want to solve for x in the equation M x = b 1 0 −5 6 3 −1 −14 x = 19 1 0 −3 4 where you are given the decomposition of M into the product of L and U which are lower and upper and lower triangular matrices respectively. If you were to write out the three equations using variables.G. The augmented matrix you 0 6 0 19 2 4
This is an easy augmented matrix to solve because it is upper triangular. 1 0 −5 1 0 0 1 0 −5 M = 3 −1 −14 = 3 1 0 0 −1 1 = LU 1 0 −3 1 0 2 0 0 1 First you should solve L(U x) = b would use looks like this 1 0 3 1 1 0 for U x. Try it and in a few steps you should be able to get 1 0 0 6 0 1 0 1 0 0 1 −1 6 This tells us that U x = 1 .
)
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. (We note that as augmented matrices (M |V ) ∼ (M |V ).G. Finally you can easily see that 1 7 2 6 3 = M L2 U2 = 1 −3 −21 4 which solves the problem of L2 U2 X = M X = V . Back to our example. have L1 = I3 and U1 = M . we also need to make the corresponding swap on our vector V to get a V since all of this amounts to changing the order of our two equations. −3 0 1 0 0 10 and note that U2 is upper triangular. and hence 1 0 0 1 7 −3 1 0 0 0 U2 = L2 = 1 0 1 0 −1 So initially we 2 10 . and note that this clearly does not change the solution. Yet this gives us a small problem as L2 U2 = M .38
LU Decomposition: Worked Example
Here we will perform an LU decomposition on the matrix 1 7 2 M = −3 −21 4 1 6 3 following the procedure outlined in Section 11. In our original problem M X = V . in fact it gives us the similar matrix M with the second and third rows swapped. −1
However we now have a problem since 0 · c = 0 for any value of c since we are working over a field. but we can quickly remedy this by swapping the second and third rows of U2 to get U2 and note that we just interchange the corresponding rows all columns left of and including the column we added values to in L2 to get L2 .2. we have 1 0 0 1 7 2 L2 = 1 1 0 U2 = 0 −1 −1 .
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. since the number of swaps we used is two (an even number).
This is an even permutation. σ(3) = 4. The order here is important. σ(2) = 2.G. Suppose we had
This looks like a function σ that has values σ(1) = 3. σ(4) = 1 Then we could write this as 1 2 3 4 1 2 3 4 = σ(1) σ(2) σ(3) σ(4) 3 2 4 1 We could write this permutation in two steps by saying that first we swap 3 and 4. and then we swap 1 and 3.40
Elementary Matrices and Determinants: Permutations
Lets try to get the hang of permutations. A permutation is a function which scrambles things.
The main idea is that the row operations changed the augmented matrices. but we also know how to change a matrix M by multiplying it by some other matrix E. In particular can we find ''elementary matrices'' the perform row operations? Once we find these elementary matrices is is very important to ask how they effect the determinant.
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. so that M → EM . Lets take M= a b c d .3. In this case we have to study the equation L(X) = V because V ∈ Rn .
A good thing to think about is what happens to det M = ad − bc under the operations below. but you can think about that for your own self right now. In particular.
ˆ Row swap:
1 E2 =
0 1 1 0
. In particular we want to think about how the augmented matrix method can give information about finding M −1 .
1 E2 M =
0 1 1 0
a b c d
=
c d a b
. how it can be used for handling determinants. As a linear transformation L : Rn −→ Rn via Rn X −→ M X ∈ Rn . Lets focus on the first two methods. here is how all these elementary matrices work for a 2 × 2 example. Lets tabulate our names for the matrices that perform the various row operations: Row operation Ri ↔ Rj Ri → λRi Ri → Ri + λRj Elementary Matrix
i Ej Ri (λ) i Sj (λ)
To finish off the video.
G. Notice how when we are applying τi.42
Elementary Matrices and Determinants: Hints for Problem 4
Here we will examine the inversion number and the effect of the transposition τ1. and the inversion number is now 5 since we now also have 4 > 3.4 ν = [3. 2. 2]. So the inversion number of ν is 4 since 3 > 1 and 4 > 1 and 3 > 2 and 4 > 2. Recall that the inversion number is basically the number of items out of order. Now we have τ1. 1. Finally we have τ1. 4. 1.
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.2 and τ2.2 ν = [4.4 ν = [2. 1.j the parity of the inversion number changes. 2] by interchanging the first and second entries. 4] and the resulting inversion number is 2 since 2 > 1 and 3 > 1. 4] whose inversion number is 3 since 3 > 2 > 1. 3. 1. 3.2 τ2. Next we have τ2.4 on the permutation ν = [3.
First remember that the job of an elementary row matrix is to perform row operations. 1 328
.. 1 1 .
ˆ Scalar multiplication Ri (λ): multiplying a row by λ multiplies the determinant by λ. ... 1 0 . Moreover. we also know what row operations do to determinants:
i ˆ Row swap Ej : flips the sign of the determinant. Ej = . The next thing to remember is that the determinant of the identity is 1. 0 1 i . EM is the matrix M with a row operation performed on it. i .
The corresponding elementary matrices are obtained by performing exactly these operations on the identity: 1 .G.
i ˆ Row addition Sj (λ): adding some amount of one row to another does not change the determinant..43
Elementary Matrices and Determinants II: Elementary Determinants
This video will show you how to calculate determinants of elementary matrices. . . λ R (λ) = . . so that if E is an elementary row matrix and M some given matrix. ..
When we 0 0 compute the determinant. If we have a system of equations M x = b and we have the inverse M −1 then if we multiply on both sides we get x = M −1 M x = M −1 b. where it is not possible for both of these equations to be true. the matrix M is still the same. Even if we had a set of contradictory set of equations such as The matrix for this would be M = x+y =1 2x + 2y = 0. 330
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Elementary Matrices and Determinants II: Determinants and Inverses
Lets figure out the relationship between determinants and invertibility. and still has a determinant zero. we could replace the second row with a row of all zeros. Somehow the determinant is able to detect that there is only one equation here. But we 2 2 know that with an elementary row operation. So what could go wrong when we want solve a system of equations and get a solution that looks like a point? Something would go wrong if we didn't have enough equations for example if we were just given x+y =1 or maybe.G. If instead we were given redundant equations x+y =1 2x + 2y = 2 1 1 and det(M ) = 0. If the inverse exists we can solve for x and get a solution that looks like a point. to make this a square matrix M we could write this as x+y =1 0=0 The matrix for this would be M = 1 1 and det(M ) = 0. this row of all zeros gets multiplied in every term.
and what makes the determinant a useful tool is that with this reasonably simple computation we can find out if the matrix is invertible. The last row of all zeros cannot be converted into the bottom row of a 3 × 3 identity matrix. It can be difficult to see when one of the rows of a matrix is a linear combination of the others. and if the system will have a solution of a single point or column vector.Lets look at a three by three example. this matrix has no inverse. where the third equation is the sum of the first two equations.
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.
x+y+z =1 y+z =1 x + 2y + 2z = 2
and the matrix for this is 1 1 1 M = 0 1 1 1 2 2 If we were trying mentary matrices 1 1 1 0 1 1 1 2 2 to find the inverse to this matrix using ele 1 1 1 1 0 0 1 0 0 0 1 0 = 0 1 1 0 1 0 0 0 1 0 0 0 −1 −1 1
And we would be stuck here. and the row of all zeros ensures that the determinant will be zero.
if M is the matrix with rows rj = rj + λri for j = i and ri = ri .e. det(A) = det(U )). and hence we have det(AA ) = det(U U ) =
i
ui uii i uii
i
=
i
ui i
= det(U ) det(U ) = det(A) det(A ).
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. then det(M ) = det(M ) Essentially we have M as M multiplied by the elementary row sum i matrices Sj (λ). then iteratively (increasing k by 1 each i time) for fixed k using the k-th row to set mk → 0 for all i > k.45
Elementary Matrices and Determinants II: Product of Determinants
Here we will prove more directly that the determinant of a product of matrices is the product of their determinants. we note that AA has a corresponding upper-triangular matrix U U . i Now note that for two upper-triangular matrices U = (uj ) and i U = (uij ). Hence we can create an upper-triangular matrix U such that det(M ) = det(U ) by first using the first row to set m1 → 0 for all i > 1.G. i Let A and A have corresponding upper-triangular matrices U and U respectively (i. by matrix multiplication we have X = U U = (xj ) is i upper-triangular and xi = ui uii . First we reference that for a matrix M with rows ri . Also since every permutation would i i contain a lower diagonal entry (which is 0) have det(U ) = i ui .
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. Consider the example 1 2 3 det 3 1 2 = ((1·1·1)+(2·2·0)+(3·3·0))−((3·1·0)+(1·2·0)+(3·2·1)) = −5 0 0 1 We can draw a picture with similar diagonals to find the terms that will be positive and the terms that will be negative. We have a similar pattern for 3 × 3 matrices.46
Properties of the Determinant: Practice taking Determinants
Lets practice taking determinants of 2 × 2 and 3 × 3 matrices.
Now we can look at three by three matrices and see a few ways to compute the determinant. For 2 × 2 matrices we have a formula det a b = ad − bc c d
This formula can be easier to remember when you think about this picture.G.
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.Another way to compute the determinant of a matrix is to use this recursive formula. Here I take the coefficients of the first row and multiply them by the determinant of the minors and the cofactor. Then we can use the formula for a two by two determinant to compute the determinant of the minors 1 2 3 1 2 3 2 3 1 det 3 1 2 = 1 −2 +3 = 1(1−0)−2(3−0)+3(0−0) = −5 0 1 0 1 0 0 0 0 1 Decide which way you prefer and get good at taking determinants. you'll need to compute them in a lot of problems.
so det M = a det a b d g e f − b det + c det c d f i h i . a b c d d −b −c a . Recall. we can use it to immediately compute the inverse 1 det M Lets now think about a 3 × 3 matrix a b d e M= g h M −1 = adjM . i
The first thing to remember is that we can compute the determinant by expanding in a row and computing determinants of minors. When the determinant det M = 0.47
Properties of the Determinant: The Adjoint Matrix
In this video we show how the adjoint matrix works in detail for the 3x3 case.
We can think of this as the product of det det M = a b c − det det 335
a row and column vector a b c d d g .G. that for a 2 × 2 matrix M= the matrix N= had the marvelous property M N = (det M ) I (you can easily check this for yourself).
c f . We call N := adjM . the adjoint matrix of M . f i e f h i
.
Lets put this problem in the language of matrices. x + x3 } Linear combinations where the polynomials are multiplied by scalars in R is fine. one is by finding a r1 .G. x + x3 }
We want to check whether
If you are wondering what it means to be in the span of these polynomials here is an example 2(x2 ) + 5(2x + x2 ) ∈ span{x2 . Since we can write x2 = 0 + 0x + 1x2 + 0x3 we can write it as a column vector. r2 and r3 that work. another is to notice that there are no constant terms in any of the equations and to simplify the system so that it becomes 0 2 1 r1 1 1 1 0 r2 = 0 0 0 1 r3 −1 340
. 2x + x2 . r2 and r3 such that 0 0 0 0 0 2 1 r1 1 1 1 0 r2 = 0 r3 0 0 1 −1 There are two ways to do this. where the coefficient of each of the terms is an entry. We are not allowed to multiply the polynomials together. since in a vector space there is not necessarily a notion of multiplication for two vectors.50
Subspaces and Spanning Sets: Hint for Problem 2
x − x3 ∈ span{x2 . 0 0 0 0 2 1 x2 = . do there exist r1 . 2x + x2 . 2x + x2 = and x + x3 = 1 1 0 0 0 1 0 1 Since we want to find out if x − x3 = is in the span of these 0 −1 polynomials above we can ask.
r2 and r3 that satisfy this equation. If the matrix has an inverse you can say that there are r1 .
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. without actually finding them.From here you can determine if the now square matrix has an inverse.
Now repeat the same logic for W and you will be nearly done.G. since U is a subspace and all subspaces are vector spaces. try drawing an example in R3 :
Here we have taken the subspace W to be a plane through the origin and U to be a line through the origin. The hint now is to think about what happens when you add a vector u ∈ U to a vector w ∈ W . For the first part. Lets suppose that v ∈ U ∩ W and v ∈ U ∩ W . Does this live in the union U ∪ W ? For the second part.
So. This implies v∈U and v ∈U. we take a more theoretical approach. we know that the linear combination αv + βv ∈ U .51
Subspaces and Spanning Sets: Hint
This is a hint for the problem on intersections and unions of subspaces.
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.
This says that α3 and α4 are not pivot variable so are arbitrary. For example. since the vectors v1 . Thus α1 = − 4 71 µ+ ν . The relationship quoted in the notes is just one of those choices.
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. we set them to µ and ν. v3 and v4 are linearly dependent. The pattern here is to keep the vectors that correspond to columns with pivots. setting µ = −1 (say) and ν = 0 in the above allows us to solve for v3 while µ = 0 and ν = −1 (say) gives v4 . explicitly we get 53 4 3 71 v1 + v2 . α4 = ν . 25 25
In fact this is not just one relation. but infinitely many. ν. 25 25 α2 = − 53 3 µ− ν . for any choice of µ.
Thus we have found a relationship among our four vectors − 71 4 µ+ ν v1 + 25 25 − 53 3 µ− ν v2 + µ v3 + µ4 v4 = 0 . v4 = − v3 + v4 . we can try to eliminate some of them. v3 = 25 25 25 25 This eliminates v3 and v4 and leaves a pair of linearly independent vectors v1 and v2 . respectively. v2 . 25 25 α3 = µ . Finally.
e. Now without loss of generality we order our vectors such that c1 = 0. a + b = b + a).53
Linear Independence: Proof of Theorem 16.G. and we can do so since addition is commutative (i.1
Here we will work through a quick version of the proof. Therefore we have
n
c1 v1 = −
i=2 n
ci vi ci vi c1
v1 = −
i=2
and we note that this argument is completely reversible since every ci = 0 is invertible and 0/ci = 0. so i c vi = 0 where there exists some ck = 0. Let {vi } i denote a set of linearly dependent vectors.
345
.
you can either put a 1 or a 0. you can prove that they do by repeating what you did in part (c). check that every vector can be written using only copies of of these two vectors. This means that you have 23 = 8 possibilities for vectors in B 3 . If you don't think it will work you should show why. Which means when you add 1 + 1 you get 0.
346
. there are three places where you have to make a decision between two things. The only two elements are 1 and 0.54
Linear Independence: Hint for Problem 1
Lets first remember how Z2 works. You will need you have enough so that you can make every vector in B 3 using linear combinations of elements in S but you don't want too many so that some of them are linear combinations of each other. I suggest trying something really simple perhaps something that looks like the columns of the identity matrix For part (c) you have to show that you can write every one of the elements as a linear combination of the elements in S. so we can look at an entire vector space. This is kind of neat because it means that the possibilities are finite. you want to think about how many vectors you need. For part (d) if you have two vectors that you think will span the space. When you want to think about finding a set S that will span B 3 and is linearly independent. your only choices are 1 and 0. this will check to make sure S actually spans B 3 .G. Now lets think about B 3 there is choice you have to make for each coordinate. perhaps using an argument that counts the number of possible vectors in the span of two vectors. It also means when you have a vector v ∈ B n and you want to multiply it by a scalar.
. . . . . and show that this causes a contradiction. . . For this to be a contradiction we need to have ci = di for some i. . vn are linearly independent. What we don't yet know is that these c1 .
347
. which is a contradiction. then every vector w ∈ V can be written uniquely as a linear combination of vectors in the basis S: w = c1 v1 + · · · + cn vn .55
Basis and Dimension: Proof of Theorem
Lets walk through the proof of this theorem. .G. . . we know two things
ˆ V = span S ˆ v1 . In order to show that these are unique. cn are unique. . + an vn = 0 this implies that ai = 0 for all i = 1. . . which means that whenever we have a1 v1 + . n.
Since the vi 's are linearly independent this implies that ci − di = 0 for all i. We should remember that since S is a basis for V . . Then look what happens when we take the difference of these two versions of w: 0V = w−w = (c1 v1 + · · · + cn vn ) − (d1 v1 + · · · + dn vn ) = (c1 − d1 )v1 + · · · + (cn − dn )vn . So suppose there exists a second set of constants di such that w = d1 v1 + · · · + dn vn . vn } a basis for a vector space V . We want to show that for S = {v1 . . we will suppose that they are not. this means that we cannot have ci = di .
This first fact makes it easy to say that there exist constants ci such that w = c1 v1 + · · · + cn vn . .
we have to study linear independence.
1 Thus. e4 = . 3. v2 = . To do that. w ∈ Z5 } .
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. 3 2 0 0 1 0 4 1 0 0 0 1 The last four vectors are clearly a basis (make sure you understand this. we will take the vector space V = {(x. 1. e2 = . For fun. e3 = . We want to keep v1 and v2 but find a way to turf out two of the vectors in the canonical basis leaving us a basis of four vectors.. This is like four dimensional space R4 except that the numbers can only be {0. 2. 3 2 4 1
The way to proceed is to add a known (and preferably simple) basis to the vectors given.. y. z. e1 = . w)|x. its a choice of base field designed to make computations go quicker! Now. Don't get too caught up on this aspect. here's the problem we will solve: 1 0 2 3 Find a basis for V that includes the vectors and .) and are called the canonical basis. z. thus we consider 1 0 1 0 0 0 2 3 0 1 0 0 v1 = . y.. 4 = 4 because 4 = 16 = 1 + 3 × 5 = 1.56
Basis and Dimension: Worked Example
In this video we will work through an example of how to extend a set of linearly independent vectors to a basis. 4}. This is like bits. for example. but now the rule is 0 = 5.G. or in other words a linear system problem defined by 0 = α1 e1 + α2 e2 + α3 v1 + α4 v2 + α5 e3 + α6 e4 .
as a check.
350
. note that e1 = v1 + v2 which explains why we had to throw it away.Finally.
351
. Explicitly we have B 1 = {(0).57
Basis and Dimension: Hint for Problem 2
Since there are two possible values for each entry.G. then setting the i-th entry to 1 and all higher entries to 0. (1)} so there is only 1 basis for B 1 . Similarly we have B2 = 0 1 0 1 . . We note that dim B n = n as well. . Now in general we note that we can build up a basis {ei } by arbitrarily (independently) choosing the first i − 1 entries. 0 0 1 1
and so choosing any two non-zero vectors will form a basis. we have |B n | = 2n .
We can also write what happens when L acts on a general vector v ∈ V . Now suppose that L does the following to the basis vectors in V → → a
L
+ c
=: L(→) . Such a v can be written v=x→ + y↑ .
Now we rewrite the right hand side as a matrix acting from the right on the basis vectors in W : L(→) L(↑) = a b c d .
.
The matrix on the right is the matrix of L with respect to this pair of bases.
Now arrange L acting on the basis vectors in a row vector (this will be a row vector whose entries are vectors).58
Eigenvalues and Eigenvectors: Worked Example
Lets consider a linear transformation L : V −→ W where a basis for V is the pair of vectors {→.G. (Don't be afraid that we are using arrows instead of latin letters to denote vectors!) To test your understanding. L(→) L(↑) = a + c b + d .
↑ → b
L
+ d
=: L(↑) . First we compute L acting on this using linearity of L L(v) = L(x → + y ↑) and then arrange this as a row vector (whose entries are vectors) times a column vector of numbers L(v) = L(→) 352 L(↑) x y . ↑} and a basis for W is given by some other pair of vectors { . }. see if you know what dim V and dim W are.
suppose that you want to make a change of basis in W via = → + ↑ and = −→+↑ .
Finally.
Can you compute what happens to the matrix of L?
353
. as a fun exercise.Now we use our result above for the row vector(L(→) L(↑)) and obtain L(v) = a b c d x y .
we need to solve (M − 2I3 )v = 0 or v1 + v2 = 0 v2 + v3 = 0 0 = 0. Next for λ = 2.5. Hence this is the only other eigenvector for M . and thus we choose v 1 = 1.
357
. This is a specific case of Problem 20.and we immediately see that we must have V = e1 . which implies v 2 = −1 and v 3 = 1.
G.61
Eigenvalues and Eigenvectors II: Eigenvalues
Eigenvalues and eigenvectors are extremely important. where v is a column vector and M is an n×n matrix (both expressed in whatever basis we chose for V ). Now here comes a VERY important fact N u = 0 and u = 0 ⇐⇒ det N = 0. Since V is finite dimensional.e. In this video we review the theory of eigenvalues.
358
. we can represent L by a square matrix M by choosing a basis for V . The scalar λ is called an eigenvalue of M and the job of this video is to show you how to find all the eigenvalues of M . this gives (M − λI)v = 0 . Consider a linear transformation L : V −→ V where dim V = n < ∞. I. So the eigenvalue equation Lv = λv becomes M v = λv. Notice how we used the identity matrix I in order to get a matrix times v equaling zero. a square matrix can have an eigenvector with vanishing eigenvalue if and only if its determinant vanishes! Hence det(M − λI) = 0.. The first step is to put all terms on the left hand side of the equation.
However. There is an amazing fact about polynomials called the fundamental theorem of algebra: they can always be factored over complex numbers. To see why. we have z 2 + 1 = (z − i)(z + i) . The necessity for complex numbers is easily seems from a polynomial like z2 + 1 whose roots would require us to solve z 2 = −1 which is impossible for real number z. try a simple 2 × 2 example det a b λ 0 − c d 0 λ = det a−λ b c d−λ = (a − λ)(d − λ) − bc .λn are the eigenvalues of M (or its underlying linear transformation L). Returning to our characteristic polynomial.. For the n × n case. λ2 . The word can does not mean that explicit formulas for this are known (in fact explicit formulas can only be give for degree four or less).. the order n term comes from the product of diagonal matrix elements also. introducing the imaginary unit i with i2 = −1 .
359
.The quantity on the left (up to a possible minus sign) equals the so-called characteristic polynomial PM (λ) := det(λI − M ) ... This means that degree n polynomials have n complex roots (counted with multiplicity). we call on the fundamental theorem of algebra to write PM (λ) = (λ − λ1 )(λ − λ2 ) · · · (λ − λn ) .
which is clearly a polynomial of order 2 in λ. The roots λ1 . It is a polynomial of degree n in the variable λ.
. and a sequence of vectors starting x(0) with v(0) = and defined recursively so that y(0) v(1) = x(1) y(1) =M x(0) y(0)
. . but think what would the eigenvalue have to be? If you found a v(0) with this property would cv(0) for a scalar c also work? Remember that eigenvectors have to be nonzero.63
Eigenvalues and Eigenvectors II: Hint
We are looking at the matrix M .
361
.G. so what if c = 0? For part (c) if we tried an eigenvector would we have restrictions on what the eigenvalue should be? Think about what it means to be pointed in the same direction. Consider that eigenvectors are a possibility. When we think about the question in part (b) when we want to find a vector v(0) such that v(0) = v(1) = v(2) . and the corresponding eigenvectors. You should do the computations to find these for yourself. We first examine the eigenvectors and eigenvalues of M= 3 2 . det 3−λ 2 2 3−λ
=0
By computing the determinant and solving for λ we can find the eigenvalues λ = 1 and 5. 2 3
We can find the eigenvalues and vectors by solving det(M − λI) = 0 for λ. we are looking for a vector that satisfies v = M v.
. . 362
. . consider the space PC of complex 3 polynomials of degree at most 3. Can you find some? This is an example of how things can change in infinite dimensional spaces. Note that we are ignoring infinite cases for simplicity.
which is in the basis {n−1 xn }n (where for n = 0. . . . but if you want to consider infinite terms such as convergent series or all formal power series where there is no conditions on convergence. . . .G. . . 0 ··· 0 ··· 1 ··· . into an infinite Jordan cell with 0 1 0 . . Therefore we note that 1 (constant polynomials) is the only eigenvector with eigenvalue 0 for polynomials since they have finite degree. . . We note that this transforms eigenvalue 0 or 0 1 0 0 0 0 . . .. and so the derivative is not diagonalizable. . Note that this is a nilpotent matrix since D4 = 0. . For a more finite example. . . .64
Diagonalization: Derivative Is Not Diagonalizable
First recall that the derivative operator is linear and that we can write it as the matrix 0 1 0 0 ··· 0 0 2 0 · · · d = 0 0 0 3 · · · .. . but the only nilpotent matrix that is ''diagonalizable'' is the 0 matrix. we just have 1). . and recall that the derivative D can be written as 0 1 0 0 0 0 2 0 D= 0 0 0 3 . dx . . 0 0 0 0 You can easily check that the only eigenvector is 1 with eigenvalue 0 since D always lowers the degree of a polynomial by 1 each time it is applied. there are many eigenvectors.
oranges): Oranges (x. Since this was a linear systems problem. y) = (apples. this representation would label what's in the barrel by a vector x := xe1 + ye2 = e1 e2 x y . 0) and e2 := (0. 1).
Since this is the method ordinary people would use. we can try to represent what's in the barrel using a vector space. The first representation was the one where (x. we will call this the ''engineer's'' method! 363
. y)
Apples
Calling the basis vectors e1 := (1.65
Diagonalization: Change of Basis Example
This video returns to our first example of a barrel filled with fruit
as a demonstration of changing basis.G.
we can use a minus sign. where if money is owed to somebody else. f )
sugar
WARNING: To make sense of what comes next you need to allow for the possibity of a negative amount of fruit or sugar. The vector x says what is in the barrel and does not depend which mathematical description is employed. The vector e1 corresponds to one apple and one orange. The way nutritionists label x is in terms of a pair of basis vectors f1 and f2 : x = sf 1 + f f 2 = f 1 f 2 s f . They would note the amount of sugar and total number of fruit (s. f ): fruit
(s.But this is not the approach nutritionists would use.
Thus our vector space now has a bunch of interesting vectors:
The vector x labels generally the contents of the barrel. The vector e2 is 364
. This would be just like a bank.
Comparing to the nutritionist's formula for the same object x we learn that 1 f1 = − e1 − e2 and f2 = 2e1 − 2e2 . You might remember that the amount of sugar in an apple is called λ while oranges have twice as much sugar as apples. Thus s = λ (x + 2y) f = x+y. this is already our change of basis formula. The vector f1 means one unit of sugar and zero total fruit (to achieve this you could lend out some apples and keep a few oranges). λ Rearranging these equation we find the change of base matrix P from the engineer's basis to the nutritionist's basis: f1 f2 = e1 e2
1 −λ 1 λ
2 −1
=: e1 e2 P .one orange and no apples.
.
Putting this in the engineer's formula for x gives x = e1 e2
1 −λ 1 λ
2 −1
s f
1 = − λ e1 − e2
2e1 − 2e2
s f
. f 1 1 y We can easily invert this to get x y =
1 −λ 1 λ
2 −1
s f
.
We can also go the other direction. Finally the vector f2 represents a total of one piece of fruit and no sugar. Essentially. but lets play around and put it in our notations. First we can write this as a matrix s λ 2λ x = . changing from the nutritionist's basis to the engineer's basis e1 e2 = f1 f2 λ 2λ 1 1 365 =: f1 f2 Q .
Now the matrix M is the matrix of some linear transformation L in the basis of the engineers. It does not act on columns of numbers! We can easily compute M P and find MP = 1 1 2 −1
1 −λ 1 λ
2 −1
=
0 1 3 −λ 5
.
Note that P −1 M P is the matrix of L in the nutritionists basis.these are the objects we have written with a sign on top of them. where you were given that there were 27 pieces of fruit in total and twice as many oranges as apples. Lets convert it to the basis of the nutritionists: Lx = L f1 f2 s f = L e1 e2 P s f = e1 MP e2 s f . but we don't need this quantity right now.
Note here that the linear transformation on acts on vectors -.
366
. In equations this says just x + y = 27 and 2x − y = 0 . (which is in fact how we constructed P in the first place). 1 1 2 −1 . lets consider the very first linear systems problem. Finally. we must have Q = P −1 .
But we can also write this as a matrix system MX = V where M := Note that x = e1 e2 X . X := x y V := 0 27 .Of course. Also lets call v := e1 e2 V .
However we can do the 2 same thing except starting with b2 and get b2 = eψ and b1 = eψ since 1 2 we have just interchanged two basis vectors which corresponds to a reflection which picks up a minus sign as in the determinant.G. eθ } for some θ. 1 2 Also we have eθ 1
2
= eθ 2
2
= sin2 θ + cos2 θ = 1. Thus b1 = eφ and if b2 = eφ . 2π). Now first we need to show that for a fixed θ that the pair is orthogonal: eθ eθ = − sin θ cos θ + cos θ sin θ = 0.67
Orthonormal Bases: Sine and Cosine Form All Orthonormal Bases for R2
We wish to find all orthonormal bases for the space R2 . b2 } and note that b1 forms an angle φ with the vector e1 (which is e0 ).
and hence {eθ .
-sin θ cos θ
cos θ sin θ
θ
370
. To show that every or1 2 thonormal basis of R2 is {eθ . eθ } is an orthonormal basis. cos θ
for some θ ∈ [0. we are done. and they are {eθ . otherwise 1 2 1 b2 = −eφ and it is the reflected version. consider an orthonormal 1 2 basis {b1 . eθ = 2 sin θ − sin θ . eθ } up to reordering where 1 2 eθ = 1 cos θ .
Because this is a basis any v ∈ V can be uniquely expressed as v = c1 v1 + c2 v2 + · · · + v n cn . v2 .
So different vectors in the basis are orthogonal:
However. You are asked to consider an orthogonal basis {v1 . i=j. v1 v1
This should get you started on this problem. and the number n = dim V . To complete the hint. lets use the dot product to compute a formula for c1 in terms of the basis vectors and v. 371
. . Solving for c1 (remembering that v1 v1 = 0) gives c1 = v1 v . vn }. Lecture 21
This video gives a hint for problem 2 in lecture 21.G. the basis is not orthonormal so we know nothing about the lengths of the basis vectors (save that they cannot vanish). Consider v1 v = c1 v1 v1 + c2 v1 v 2 + · · · + cn v1 vn = c1 v1 v1 .68
Orthonormal Bases: Hint for Question 2. . . Since this is an orthogonal basis vi vj = 0 .
Remember that u · v = u v cos(θ). so this is a linear combination of v and u. (a) Is the vector v ⊥ = v − in the plane P ?
Remember that the dot product gives you a scalar not a vector. u·v so if you think about this formula u·u is a scalar.
372
.69
Orthonormal Bases: Hint
u·v u u·u
This video gives a hint for problem 3 in lecture 21. (c) Given your solution to the above. and in particular if they are perpendicular θ = you will get u · v = 0. Do you think it is in the span? (b) What is the angle between v ⊥ and u? This part will make more sense if you think back to the dot product formulas you probably first saw in multivariable calculus. how can you find a third vector perpendicular to both u and v ⊥ ? Remember what other things you learned in multivariable calculus? This might be a good time to remind your self what the cross product does.
π 2
and cos( π ) = 0 2
Now try to compute the dot product of u and v ⊥ to find u v⊥ cos(θ) u·v u u·u u·v u = u·v−u· u·u u·v u·u = u·v− u·u
u · v⊥ = u · v −
Now you finish simplifying and see if you can figure out what θ has to be.G.
and if you get stuck try drawing a sketch of the vectors you have. All you need to do to turn this into an orthonormal basis is turn those into unit vectors.
Try it out. If you did part (c) you can probably find 3 orthogonal vectors to make a orthogonal basis.(d) Construct an orthonormal basis for R3 from u and v.
373
. (e) Test your abstract formulae starting with u= 1 2 0 and v = 0 1 1 .
a direct sum and an orthogonal complement in R3 . their union U ∪ V is usually not a subspace. For that we can keep the plane U but must replace V by a line:
Taking a direct sum we again get the whole space. Remember that even if U and V are subspaces. Given U it provides a space U ⊥ such that the direct sum returns the whole space: U ⊕ U ⊥ = R3 . U ⊕ V = R3 . There is not really a notion of subtraction for subspaces but the orthogonal complement comes close. A picture of this is a pair of planes in R3 :
Here U + V = R3 . lets start with the subspace sum. Firstly.71
Gram-Schmidt and Orthogonal Complements: Overview
This video depicts the ideas of a subspace sum. 376
. the span of their union certainly is and is called the subspace sum U + V = span(U ∪ V ) . This is just the subspace sum for the case when U ∩ V = {0}.G. However. Now we come to an orthogonal complement. You need to be aware that this is a sum of vector spaces (not vectors). Next lets consider a direct sum.
we need to just tilt the line V above until it hits U at a right angle:
Notice.
377
. i.The orthogonal complement U ⊥ is the subspace made from all vectors perpendicular to any vector in U . we can apply the same operation to U ⊥ and just get U back again. ⊥ U⊥ = U .e. Here.
Then you need to remember that we are searching for a decomposition M = QR where Q is an orthogonal matrix. Thus the upper triangular matrix R = QT M and QT Q = I. Moreover, orthogonal matrices perform rotations. To see this compare the inner product u v = uT v of vectors u and v with that of Qu and Qv: (Qu) (Qv) = (Qu)T (Qv) = uT QT Qv = uT v = u v . Since the dot product doesn't change, we learn that Q does not change angles or lengths of vectors. Now, here's an interesting procedure: rotate v1 , v2 and v3 such that v1 is along the x-axis, v2 is in the xy-plane. Then if you put these in a matrix you get something of the form a b c 0 d e 0 0 f which is exactly what we want for R! Moreover, the vector a 0 0 380
For part (a), we can consider any complex number z as being a vector in R2 where complex conjugation corresponds to the matrix 1 0 . Can you describe z z in terms of z ? For part (b), think ¯ 0 −1 about what values a ∈ R can take if a = −a? Part (c), just compute it and look back at part (a). For part (d), note that x† x is just a number, so we can divide by it. Parts (e) and (f) follow right from definitions. For part (g), first notice that every row vector is the (unique) transpose of a column vector, and also think about why (AAT )T = AAT for any matrix A. Additionally you should see that xT = x† and mention this. Finally for part (h), show that x† M x = x† x x† M x x† x
T
and reduce each side separately to get λ = λ.
384
G.76
Kernel, Range, Nullity, Rank: Invertibility Conditions
Here I am going to discuss some of the conditions on the invertibility of a matrix stated in Theorem 24.6. Condition 1 states that X = M −1 V uniquely, which is clearly equivalent to 4. Similarly, every square matrix M uniquely corresponds to a linear transformation L : Rn → Rn , so condition 3 is equivalent to condition 1. Condition 6 implies 4 by the adjoint construct the inverse, but the converse is not so obvious. For the converse (4 implying 6), we refer back the proofs in Chapter 18 and 19. Note that if det M = 0, there exists an eigenvalue of M equal to 0, which implies M is not invertible. Thus condition 8 is equivalent to conditions 4, 5, 9, and 10. The map M is injective if it does not have a null space by definition, however eigenvectors with eigenvalue 0 form a basis for the null space. Hence conditions 8 and 14 are equivalent, and 14, 15, and 16 are equivalent by the Dimension Formula (also known as the Rank-Nullity Theorem). Now conditions 11, 12, and 13 are all equivalent by the definition of a basis. Finally if a matrix M is not row-equivalent to the identity matrix, then det M = 0, so conditions 2 and 8 are equivalent.
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G.77
Kernel, Range, Nullity, Rank: Hint for 1
Lets work through this problem. Let L : V → W be a linear transformation. Show that ker L = {0V } if and only if L is one-to-one: 1. First, suppose that ker L = {0V }. Show that L is one-to-one. Remember what one-one means, it means whenever L(x) = L(y) we can be certain that x = y. While this might seem like a weird thing to require this statement really means that each vector in the range gets mapped to a unique vector in the range. We know we have the one-one property, but we also don't want to forget some of the more basic properties of linear transformations namely that they are linear, which means L(ax + by) = aL(x) + bL(y) for scalars a and b. What if we rephrase the one-one property to say whenever L(x)− L(y) = 0 implies that x − y = 0? Can we connect that to the statement that ker L = {0V }? Remember that if L(v) = 0 then v ∈ ker L = {0V }. 2. Now, suppose that L is one-to-one. Show that ker L = {0V }. That is, show that 0V is in ker L, and then show that there are no other vectors in ker L. What would happen if we had a nonzero kernel? If we had some vector v with L(v) = 0 and v = 0, we could try to show that this would contradict the given that L is one-one. If we found x and y with L(x) = L(y), then we know x = y. But if L(v) = 0 then L(x) + L(v) = L(y). Does this cause a problem?
386
Suppose v ∈ L(U ) and you have found a vector ups that obeys L(ups ) = v.G. Now suppose we have a solution x where L(x) = v. If we add the equations together L(x) + L(y) = L(x + y) = v + 0 we get another solution for free. Explain why you need to compute ker L to describe the solution space of the linear system L(u) = v. Let L : U → V be a linear transformation. If we have an vector y ∈ ker L then we know L(y) = 0.78
Least Squares: Hint for Problem 1
Lets work through this problem. Now we have two solutions. is that all?
387
. Remember the property of linearity that comes along with any linear transformation: L(ax + by) = aL(x) + bL(y) for scalars a and b. This allows us to break apart and recombine terms inside the transformation.
G. note that v v = v T v. Can you express this in terms of v ? Also you need the trivial kernel only for the last part and just think about the null space of M .79
Least Squares: Hint for Problem 2
For the first part. what is the transpose of a 1 × 1 matrix? For the other two parts.
388
. It might help to substitute w = M x.
ˆ An eigenvector example for lecture 18 by Ashley Coates. The copyright to this work belongs to them as does responsibility for the correctness of any information therein.
389
.H
Student Contributions
Here is a collection of useful material created by students. lecture 2 by Ashley Coates. ˆ A hint for review problem 1. ˆ A hint for review problem 1.
ˆ 4D TIC TAC TOE by Davis Shih. lecture 12 by Philip Digiglio. ˆ Some cartoons depicting matrix multiplication by Asun Oka. | 677.169 | 1 |
Functions and Relations
Ninth graders take notes while the teacher explains what relations are in Algebra. They examine functions using an input and output as a review. They determine whether the "Vertical Line Test" is a function or not. They work with the teacher in guided practice and then independently. | 677.169 | 1 |
Publication Venue
We start with a mathematical definition of a real interval as a closed, connected set of reals. Interval arithmetic operations (addition, subtraction, multiplication, and division) are likewise defined mathematically and we provide algorithms for computing these operations assuming exact real arithmetic. Next, we define interval arithmetic operations on… (More) | 677.169 | 1 |
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Allan G. Bluman taught mathematics and statistics in high school, college, and graduate school for 39 years. He is the recipient of "An Apple for the Teacher Award" for bringing excellence to the learning environment and the "Most Successful Revision of a Textbook" award from McGraw-Hill. Mr. Bluman's biographical record appears in Who's Who in American Education, 5th Edition. He is the author of three mathematics textbooks and several highly successful books in the DeMYSTiFieD series, including Pre-Algrebra DeMYSTiFieD and Business Math DeMYSTiFieD YourYour Professional, 2011. Book Condition: New. Brand New, Unread Copy in Perfect Condition. A+ Customer Service! Summary: The second edition of one of the most successful math word problems books on the market is updated with all-new quizzes and test questions, clearer explanations of the material, and a completely refreshed interior design. Bookseller Inventory # ABE_book_new_0071763864
Book Description McGraw-Hill Professional Publishing. Paperback. Book Condition: New. Paperback. 334 pages. Dimensions: 8.9in. x 7.3in. x 0.8in.Your solution to MATH word PROBLEMS! Find yourself stuck on the tracks when two trains are traveling at different speeds Help has arrived! Math Word Problems Demystified, Second Edition is your ticket to problem-solving success. Based on mathematician George Polyas Its a no-brainer! Youll This item ships from multiple locations. Your book may arrive from Roseburg,OR, La Vergne,TN. Paperback. Bookseller Inventory # 978007176387984611763868 | 677.169 | 1 |
The focus of this book is on a rationale and a set of strategies for problem generation in mathematics. Six chapters attempt to involve readers in the process of posing problems and in understanding why that process is important. Chapter 1 discusses two problem posing perspectives, while chapter 2 looks at the first phase of problem posing. The "what-if-not" problem posing strategy is presented in chapter 3, and depicted in action in chapter 4. Some natural links between problem posing and problem solving are discussed in chapter 5. Finally, conclusions are briefly presented. Each chapter contains illustrative problems as well as suggested questions or directions. In the appendix, a college course on problem posing is described, also with illustrative examples. Finally, a bibliography is provided. (MNS) | 677.169 | 1 |
This
Point
In
R.L. Moore, distinguished mathematician and teacher at the University of Texas, developed a Socratic method of teaching in which students are challenged to construct their own proofs of theorems. While hugely successful in influencing students to pursue advanced studies in mathematics, the Moore method can be intimidating to someone who has only experienced traditional-style mathematics coursesEL With this book instructors can choose what level of discovery learning to incorporate into their course, depending upon the backgrounds and aptitudes of their students. The chapters are nicely organized with definitions, exercises, and propositions preceding the skeleton proofs that followELThis textbook will introduce the Moore method to the wider audience it deserves. ( Walter Richardson, Jr., The University of Texas at San Antonio)
This is a well-motivated guided tour through the main areas of modern general topology. Because of this, I highly recommend it both as a textbook for advanced undergraduates and as a main resource for anyone setting out to learn the subject on their own. ( Paul Bankston, Marquette University)
About the Author:
Aisling McCluskey graduated at Queen's University Belfast with a doctorate in Pure Mathematics (topology) in 1990 and subsequently was awarded a postdoctoral fellowship in Toronto in 1991. She was appointed to a permanent lectureship in Mathematics in the National University of Ireland (NUI), Galway, Ireland in January 1992. Since then, she has established a meaningful and rewarding academic career, maintaining an active research profile in the field whilst holding the teaching and learning of mathematics central to her academic endeavour. In recent years, her research interests have diversified into undergraduate mathematics education. She was awarded an institutional award in 2008 for excellence in teaching and then a national award in 2009 across all higher education institutions in Ireland.
Brian McMaster studied at Queen's University Belfast, graduating with a PhD in 1972, and has served his alma mater department in various capacities including those of Adviser of Studies, Head of Research and Associate Director of Education. His publication profile covers over sixty refereed journal articles, mostly in the area of general topology but incorporating applications in disciplines as diverse as probabilistic metric spaces and decision support theory. He is presently formally retired but continues to deliver a full undergraduate teaching load on a voluntary basis, following his lifelong commitment to and passion for communicating mathematics to students. His teaching interests focus around analysis (real and complex) and set theory and their development into various fields especially that of analytic topology.
Descrizione libro Oxford University Press. Hardback. Condizione libro: new. BRAND NEW, Undergraduate Topology: A Working Textbook, Aisling McCluskey, Brian McMaster, B9780198702337
Descrizione libro Oxford University Press, United Kingdom, 2014. Hardback Oxford University Press, United Kingdom, 2014. Hardback particular foc.Shipping may be from multiple locations in the US or from the UK, depending on stock availability. 176 pages. 0.416. Codice libro della libreria 978 p.Shipping may be from our Sydney, NSW warehouse or from our UK or US warehouse, depending on stock availability. 176 pages. 0.416. Codice libro della libreria 9780198702337 | 677.169 | 1 |
Master the ASVAB Basics--Mathematics Knowledge
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Peterson's provides an in-depth review of Mathematics Knowledge exercises in the ASVAB. Algebra; signed numbers; formulas; circles and quadrilaterals; triangles; powers, roots, and radicals; exponents and sequences problems are included.Tangenta€"a line intersecting a circle at only one point. Circumference The
circumference of a circle is the distance around a circle. (This distance is called
perimeter when it relates to other geometric shapes.) Use this formula to find the
anbsp;...
Title
:
Master the ASVAB Basics--Mathematics Knowledge | 677.169 | 1 |
Never Highlight a Book Again! Just the FACTS101 study guides give the student the textbook outlines, highlights, practice quizzes and optional access to the full practice tests for their textbook.In mathematics and empirical science, it is the act of counting and
measuringthatmaps human sense observations and experiences into members
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Mathematica Navigator: Mathematics, Statistics and GraphicsRuskeepaa gives a general introduction to the most recent versions of Mathematica, the symbolic computation software from Wolfram. The book emphasizes graphics, methods of applied mathematics and statistics, and programming.
Mathematica Navigator can be used both as a tutorial and as a handbook. While no previous experience with Mathematica is required, most chapters also include advanced material, so that the book will be a valuable resource for both beginners and experienced users.
- Covers both Mathematica 6 and Mathematica 7 – The book, fully revised and updated, is based on Mathematica 6 – Comprehensive coverage from basic, introductory information through to more advanced topics – Studies several real data sets and many classical mathematical models
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Description
The theory of abstract groups comes into play in an astounding number of seemingly unconnected areas like crystallography and quantum mechanics, geometry and topology, analysis and algebra, physics, chemistry, and even biology. Readers need only know high school mathematics, much of which is reviewed here, to grasp this important subject. Hundreds of problems with detailed solutions illustrate the text, making important points easy to understand and remember.show more
Table of contents
Sets, Mappings and Binary Operations, Groupoids.Groups and Subgroups.Isomorphism Theorems.Finite Groups.Abelian Groups.Permutational Representations.Free Groups and Presentations.Appendices: A: Number Theory.B: Guide to the Literature.Symbols and Notations.show more | 677.169 | 1 |
CliffsNotes Math Review for Standardized Tests
4.11 - 1251 ratings - Source
Your guide to a higher math score on standardized tests *SAT ACTAr ASVAB GMATAr GREAr CBESTAr PRAXIS IAr GEDAr And More! Why CliffsNotes? Go with the name you know and trust Get the information you needa€"fast! About the Contents: Introduction How to use this book Overview of the exams Part I: Basic Skills Review Arithmetic and Data Analysis Algebra Part II: Strategies and Practice Mathematical Ability Quantitative Comparison Data Sufficiency Each section includes a diagnostic test, explanations of rules, concepts with examples, practice problems with complete explanations, a review test, and a glossary! Test-Prep Essentials from the Experts at CliffsNotesAr For more test-prep help, visit CliffsNotes.comAr *SAT is a registered trademark of the College Board, which was not involved in the production of, and does not endorse, this product.About the Contents: Introduction How to use this book Overview of the exams Part I: Basic Skills Review Arithmetic and Data Analysis Algebra Part II: Strategies and Practice Mathematical Ability Quantitative Comparison Data Sufficiency Each ...
Title
:
CliffsNotes Math Review for Standardized Tests
Author
:
Ph.D., Jerry Bobrow, Jerry Bobrow
Publisher
:
Cliffs Notes - 2010-03-02
ISBN-13
:
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Beginning and Intermediate Algebra
4.11 - 1251 ratings - Source
For Beginning Algebra and Intermediate Algebra courses including lecture-based, self-paced, discussion oriented, and modular classes. This clear, accessible treatment of mathematics features a building-block approach toward problem solving and realistic, diverse applications. Students practice problem solving and decision making with interesting applications throughout the text. The Putting Your Skills to Work and new chapter-end feature, Math in the Media, present students with opportunities to utilize critical thinking skills, analyze and interpret data, and problem solve using applied situations encountered in daily life. The problem solving strategy, highlighted by A Mathematics Blueprint for Problem Solving, helps students determine where to begin the problem-solving process, as well as how to plan subsequent problem-solving steps. Chapter organizers help students focus their study on the concepts and examples. Developing Your Study Skills boxes throughout the text give students tips to help them improve their study skills. These features, together with the applications and emphasis on problem solving, help students to become effective and confident problem solvers.If a new Honda Civic is purchased for Sala#39;s Delivery Service for $16, 750, used for
five years, and then sold for $5, 750, what will be the ... $3500, $3000, $2500,
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Title
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Beginning and Intermediate Algebra
Author
:
John Tobey, Jeffrey Slater
Publisher
:
- 2001-12-01
ISBN-13
:
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An Introduction to the Theory and Practice of Plain and
A physical or electronic signature of the subscriber; 2. To convert from degrees to radians, we multiply by Pi and divide by 180. Trigonometry is used so much in physics that it is as prevalent as simple addition and subtraction for the average person. The student is expected to: (A) determine the domain and range of a linear function in mathematical problems; determine reasonable domain and range values for real-world situations, both continuous and discrete; and represent domain and range using inequalities; (B) write linear equations in two variables in various forms, including y = mx + b, Ax + By = C, and y - y1 = m(x - x1), given one point and the slope and given two points; (C) write linear equations in two variables given a table of values, a graph, and a verbal description; (D) write and solve equations involving direct variation; (E) write the equation of a line that contains a given point and is parallel to a given line; (F) write the equation of a line that contains a given point and is perpendicular to a given line; (G) write an equation of a line that is parallel or perpendicular to the X or Y axis and determine whether the slope of the line is zero or undefined; (H) write linear inequalities in two variables given a table of values, a graph, and a verbal description; and (I) write systems of two linear equations given a table of values, a graph, and a verbal description. (3) Linear functions, equations, and inequalities.
Pages: 492
Publisher: Nabu Press (February 21, 2014)
ISBN: 1294651781
Mathematics for Self Study
Every clip has real exam questions for students to do - because we all know that to be good at Maths, a student has to do it, not just watch it. "This course is taught so that students will acquire a solid foundation in algebra. Emphasis is placed on understanding the properties of linear, polynomial and rational functions, including inequalities and complex numbers Algebra 2 and Trigonometry read online nssiti.com. A rectangle has dimensions 10 cm by 5 cm. Practical Mathematics; Being the Essentials of Arithmetic, Geometry, Algebra and Trigonometry Volume 2 sadipen.com. You just have to give it the equation with X as variable. There is only integer values for the angles though. Might be the the right program for you if you are uncertain about the precision. ATRIG10: A coventional program, described in the textfile. ATRIG11: This is the same as ATRIG0, exept that it can also handle a precision of half a degree , cited: Logarithmic and trigonometric tables: Prepared under the direction for Earle Raymond Hedrick That's the radius squared times this angle, times 1/2. Well, I think that is exactly the area of this sector. a^2 theta / 2 is the formula for the area of the sector. And this one, this is the vertical elevation. This is the horizontal. a^2 - b^2 is this distance , source: Graphing Calculator Manual for Dwyer / Gruenwalds College Algebra and Trigonometry : A Contemporary Approach" 2ND EDITION Secondly, although it covers algorithms it is not computer oriented. Graph theory has become important precisely because of algorithms. Let me mention two excellent books in order of my preference. Again, thinking of computer science, let me mention another book: A very nice book at the senior-graduate level strictly devoted to generating functions: There are many books on Fibonacci numbers (and the golden ratio) , source: Four-Place Logarithmic and download epub
For example, if the lengths of two sides of a triangle and the measure of the enclosed angle are known, the third side and the two remaining angles can be calculated download. and a direction given by an angle with respect to a reference direction. The component of vector B in the direction of vector A is AB cos θ, where θ is the angle between the positive directions of the vectors. If A were reversed, then the angle would be 180° - θ, and the cosine would change sign. If A is a unit vector (that is, A = 1), then the component of B is just B cos θ Insider's Guide to Teaching with Trigonometry: Trigonometry 9th Edition ISBN 0321530500 download for free. Same way in algebra chapter we know about rules of operations in many topics. And basically algebra is the main chapter for so many branches in mathematics subjects. In this article, we are going to see about brief explain of all the three chapters. Trigonometry is basically all about triangles and its relations. Angleis can be in radians or degrees only 100 Worksheets - Finding Larger Number of 9 Digits: Math Practice Workbook (100 Days Math Greater Numbers Series) (Volume 8) Therefore when using the columns of mean differences for cosines these differences must be subtracted Combo: Trigonometry with read here
HM CLASSPREP: Algebra & Trigonometry, 7E
Student's Solutions Manual for Trigonometry: A Unit Circle Approach
Common questions addressed in the memo: What does a pathway look like to reach calculus? Which assessment will students take if they are accelerated? Standards for suggested compacted courses are included as appendices within the memo to ensure there are no gaps in content or the development of critical skills and understandings unique to the middle grades for students who take Algebra I in middle school Exact Values in Trigonometry: download for free Exact Values in Trigonometry: Five New. The instructor is not there to judge you at the board but he is there to push you to do your best , e.g. Five-place Logarithmic And download pdf It is free to home schooled students upon proof of purchase of the book. Samples are found at BY ACCESSING OR USING THIS SITE, YOU AGREE TO BE BOUND BY THE TERMS AND CONDITIONS SET FORTH IN THE DISCLAIMER. Search this site or the whole web below. Opposite Angle Identities � Sec. 7.1 Sum and Difference Identities � Sec. 7.3 Hint:� Functions are the same; signs are different���������������� Hint:� Signs are same, functions are different Double Angle Identities � Sec. 7.4 Half Angle Identities � Sec. 7.4 The full set of six Trigonometric Ratios is shown in the next section. A Right Triangle has three sides: Hypotenuse, Opposite, and Adjacent. If you do not know how to do this sides labeling, then go and do our previous lesson on this at the link below. The six Trigonometric ratios that we can make for a Right Triangle have special mathematical names as shown in the following Table epub. Ultimate Interactive Trigonometry Calculator contains 13 Modules: Geometry, The Right Triangular Shapes, 30/60 Degrees Triangular Shapes, 45 Degrees Triangular Shapes, Oblique Triangular Shapes, Trigonometry/Geometry Equations, The Ultimate Units Converter, Animated Sin/Cos/Tan Graphs Functions, Interactive Central Angle, Interactive Unit Circle, Circle Functions, Lines, Point & Slopes Plot Functions and the bonus module is: The Arabic Numerals System Algebra & Trigonometry (Your download online projectsforpreschoolers.com.
A treatise of plane trigonometry, and the mensuration of heights and distances To which is prefixed
A new treatise on surveying and navigation, theoretical and practical; with use of instruments, essential elements of trigonometry, and the necessary ... schools, colleges, and practical surveyors
Trigonometrical series
Cookie loves it and Moze initially loved it, but started having trouble with it after her regular teacher was replaced with a student teacher. She was adamant that she did not hate it, though Integrated Algebra and read for free read for free. The idea for this resource woke me up from my sleep at about 4am the other morning! I must have been dreaming about one of Vi Hart's amazing 'maths doodle' videos, in particular the one about The Triangle Party. I'm a big fan of Vi Hart's Blog and the engaging, fast paced videos she produces. I love the way she crams in so much maths in the videos but 'sells' it as just interesting doodling Elements of trigonometry, and read pdf Elements of trigonometry, and! We first study certain important properties of functions, discover how they can be transformed and consider different ways of working with them numerically (in practical terms), algebraically (in abstract theoretical terms) and geometrically (by visual representations) , cited: An Introduction to the Theory and Practice of Plain and Spherical Trigonometry: And the Stereographic Projection of the Sphere: Including the Theory projectsforpreschoolers.com. The operations are carried out in the order that you press the keys. It is therefore vitally important to establish whether your calculator uses everyday or algebraic logic , cited: Rudiments Of Plane Geometry: Including Geometrical Analysis And Plane Trigonometry Who has to take the ALEKS Placement Assessment? Any student who wants to take an entry level calculus course (Math 0120 or Math 0220), Precalculus (Math 0200), or Discrete Math (Math 0400), and who does not have the prerequiste course credit, must take the placement assessment , source: Logarithmic and Trigonometric download for free For the angle A, we have sin A = (axb)x(axc) / Therefore, sin A / sin a = [axb·c]/sin a sin b sin c. The same result is obtained for the other angles, so we have the Law of Sines sin A / sin a = sin B / sin b = sin C / sin c. As the largest side becomes small, we may use the approximations sin a = a and cos a = 1 - a2/2, and similarly for the other sides. Then, the law of cosines becomes c2 = a2 + b2 - 2ab cos C, and the law of sines sin A / a = sin B / b = sin C / c, the analogous formulas for a plane triangle Trigonometry by Gelfand, I.M., read for free A segment is a part of a circle bounded by a chord and the arc which it cuts off. Thus in Fig. 20 the chord PQ divides the circle into two segments. The larger of these PCQ is called a major segment and the smaller, PBQ, is called a minor segment Algebra and Trigonometry Enhanced with Graphing Utilities, Books a la Carte Edition (5th Edition) read online. A focal point of the unit is problem-solving thus application problems are included in the introduction, each lesson, and an entire section devoted to problem solving skills , cited: A treatise on trigonometry. By read epub read epub. There are 360° in a complete revolution, with each degree further divided into 60′ (minutes) and each minute divided into 60″ (seconds) Metrics, Norms and Integrals: An Introduction to Contemporary Analysis aroundthetownsigns.com. For more applications and examples of trigonometry in Interactive Mathematics, check out the many Uses of Trigonometry. In this chapter we start by explaining the basic trigonometric functions using degrees (°), and in the later part of the chapter we will learn about radians and how they are used in trigonometry Senior Middle School High download pdf download pdf. Below you could see some examples of trigonometry with their solutions: You can find topics ranging from simplifying fractions to the cubic formula, from the quadratic equation to Fourier series, from the sine function to systems of differential equations - this is the one stop site for your math needs ref.: Algebra and Trigonometry (Custom) ipp.webtt.ru. | 677.169 | 1 |
Algebra II Is Easy! So Easy
4.11 - 1251 ratings - Source
Rock provides a guide to learning and understanding Algebra II. (Education/Teaching)In other words, a seventh or eighth grade teacher does not have to understand
the Algebra I curriculum. ... the next step in their math career (of course, if the
student has failed seventh and eighth grade math, although promoted, they are ...
The warm-up should be followed by coverage of the standard in association with
the standards-driven worksheet, and then after an ... Students who master
algebra II will gain experience with algebraic solutions of problems in various
content areas, anbsp;...
Title
:
Algebra II Is Easy! So Easy
Author
:
Nathaniel Max Rock
Publisher
:
Team Rock Press - 2006-02-01
ISBN-13
:
Continue
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The best Collection of Math cheat sheetsMany Cheat Sheets in your mobile to have the formulas wherever and whenever you want.With this application you can use your travel time to study, or just have it as a quick reference when needed.NOTE: App can be moved to the SD Card!!Contents:=========Algebra. Elementary techniques for factoring binomials and trinomialsAlgebra. Exponent laws and factoring tipsAlgebra. Solving quadratic equations by completing the squareAlgebra. College Algebra quick referenceAlgebra. Solution of the 3rd degree polynomial equationAlgebra. Solution of the 4th degreee polynomial equationTrig. Basic trig identitiesTrig. Law of sines cosines etc and other triangle formulasTrig. Graphs of the trig functionsTrig. Inverse trig functionsTrig. Power reducing formulas for powers of sines and cosinesTrig. Graph paper for plotting in polar coordinatesTrig. Two unit circles with trig funcion valuesTrig. Single unit circle with trig function valuesCalculus. Basic differentiation formulas and some useful trig identitiesCalculus. Basic differentiation and integration formulasCalculus. Definitions and theorems pertaining to Riemann sums and definite integralsCalculus. A quick reference sheet on Taylor polynomials and seriesCalculus. A summary of convergence testsCalculus. Guidelines for evaluating integrals involving powers of sines and cosinesCalculus. Guidelines for evaluating integrals involving powers of secants and tangentsCalculus. Standard forms for conic sectionsCalculus. Common infinite seriesCalculus. Trigonometric substitutionCalculus. Cylindrical coordinatesCalculus. Spherical coordinatesCalculus. Hyperbolic functionsCalculus. Applications of integralsCalculus. Applications of integralsCalculus. Common ordinary differential equationsCalculus. Common ordinary differential equationsCalculus. Common ordinary differential equationsCalculus. Undetermined coefficients and variation of parametersCalculus. Vector formulasCalculus. Simple summary of cylindrical and spherical coordinatesMisc. Some prime and composite numbersMisc. Sets. Functions lines and sequencesStatistics formula sheet Page 1Statistics Page 2Statistics Page 3----------------------------------------SD Installation supportPERMISSIONS:============Why READ_PHONE_STATE / CHECK_LICENSE permissions?The application verifies it has been been purchased using google licensing library (CHECK_LICENSE). Google recommends to use the telephone identifier (READ_PHONE_STATE) to avoid piracy. Not used for anything else | 677.169 | 1 |
GCSE Mathematics for Edexcel Foundation Homework Book
4.11 - 1251 ratings - Source
4 Section 2: Tables and graphs HOMEWORK 35B 1 Nika tossed a dice 40 times
and got these results. ... Number of patients January 360 February 275 March
190 April 375 May 200 June 210 Draw a vertical line chart to represent the data.
Title
:
GCSE Mathematics for Edexcel Foundation Homework Book
Author
:
Nick Asker, Karen Morrison
Publisher
:
Cambridge University Press - 2015-06-18
ISBN-13
:
Continue
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Best Ways to Pick Up and Learn Algebra in professional fields such as economics, finance, engineering, science and medicine – it also has numerous everyday life applications too.
In short, algebra is important – and if you're having difficulty learning it, there are ways that could be of help.
Understanding the Fundamentals
The first thing you need to do is make sure you fully grasp the fundamentals of algebra. In other words: Do you understand what it is or are you still unsure?
Forget any actual mathematical problems for a minute. On its most basic level, algebra simply consists of using letters or symbols to replace numbers in mathematical equations. Everything else about it stems from that fact, so you need to be sure that you appreciate it.
Assuming you are having problems with this early stage, it would be best to get started by looking up algebra and reading various explanations on it. Different resources tend to explain it in different ways, so by drawing from multiple resources you should eventually find an explanation that makes sense.
Practice Makes Perfect
Once you have a firm grasp of the fundamentals of algebra, the next step is to start looking at basic examples of it being used in equations. More importantly however, you should start looking for basic mathematical problems that you can solve yourself.
Just like most things in mathematics, the one and only way to make sure that you really understand it is to start solving problems. The more problems you solve and the more practiced you are, the more likely it is that you'll be able to tackle more difficult problems.
If you find that you have a set of problems that you can't answer, then you may want to consider getting some help. Check my site algebra homework help for more info on exactly how you can do just that.
Challenge Yourself!
Assuming you find that you are getting quite good at a certain level of algebra problems – start to up the ante. There are numerous principles in algebra that you need to master, so keep learning and applying what you learn.
At any juncture if you find that you've hit a brick wall and need help, you can always just do so. Once you do get help though, you need to be sure you go over the problems again and understand exactly how they were solved.
By constantly learning and slowly but surely improving as you challenge yourself and get better at algebra, you'll gradually begin to master more and more complex problems. Find out more info here on what you need to do to get the help that you need with all your algebra homework problems and issues.... | 677.169 | 1 |
A Guided Tour of Mathematical Methods: For the Physical Sciences
Provides a comprehensive tour of the mathematical methods needed by physical science students.
Booknews
Snieder's (geophysics, Colorado School of Mines) textbook is designed for students in physics or geophysics. All the material is presented in the form of problems, to demonstrate for students the application of mathematical techniques and knowledge to physical problems. Coverage includes vector calculus, linear algebra, Fourier analysis, scale analysis, Green's functions, normal modes, tensor calculus, and perturbation theory. Suitable for undergraduates or lower-level graduate students, as either a standalone text or as a source of problems and examples to supplement other textbooks. The text requires a certain basic knowledge of calculus and linear algebra. Annotation c. Book News, Inc., Portland, OR (booknews.com) | 677.169 | 1 |
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