MTH 212 Linear Algebra Course Guide PDF
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Uploaded by AdvancedTonalism
National Open University of Nigeria
2024
Prof. Saheed O. Ajibola
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This document is a course guide for MTH 212 Linear Algebra, offered by the National Open University of Nigeria (NOUN). It outlines the course's structure, including study units, presentation schedule, assessment methods, and provides information on how to get the most from the course. The course covers key concepts in linear algebra such as vector spaces, matrices, and linear transformations.
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COURSE GUIDE MTH 212 LINEAR ALGEBRA Course Team Prof. Saheed O. Ajibola (Course Developer/Editor) - NOUN Dr. Christie Y. Ishola (Reviewer) – NOUN NATIONAL OPEN UNIVERSITY OF NIGERIA MTH 212 COURSE GUIDE © 2024 by...
COURSE GUIDE MTH 212 LINEAR ALGEBRA Course Team Prof. Saheed O. Ajibola (Course Developer/Editor) - NOUN Dr. Christie Y. Ishola (Reviewer) – NOUN NATIONAL OPEN UNIVERSITY OF NIGERIA MTH 212 COURSE GUIDE © 2024 by NOUN Press National Open University of Nigeria Headquarters University Village Plot 91, Cadastral Zone Nnamdi Azikiwe Expressway Jabi, Abuja Lagos Office 14/16 Ahmadu Bello Way Victoria Island, Lagos e-mail : [email protected] URL: www.nou.edu.ng All rights reserved. No part of this book may be reproduced, in any form or by any means, without permission in writing from the publisher. First Printed 2012 Reviewed 2022, 2024 ISBN: ii MTH 212 COURSE GUIDE CONTENTS PAGE Introduction………………………………………………. iv Course Competencies…………………………………….. iv Working through this Course……………………………… v Study Unit…………………………………………………. v Presentation Schedule……………………………………… vi Assessment…………………………………………………. vii Assignments………………………………………………… vii Examination………………………………………………… vii How to get the Most from the Course……………………… viii Facilitation…………………………………………………. viii Learner Support……………………………………………. ix Course Information………………………………………… ix iii MTH 212 COURSE GUIDE INTRODUCTION Linear algebra is a branch of mathematics that deals with linear equations and their representations in the vector space using matrices. In other words, the study of linear vectors and functions is what linear algebra is all about. It is one of the most important mathematical issues. The majority of contemporary geometrical ideas are based on linear algebra. Engineering and physics both rely heavily on linear algebra since it makes it easier to simulate a wide range of natural events. The three most critical elements of this topic are vector spaces, matrices, and linear equations. The many concepts related to linear algebra will be covered in more detail in this article. Students are introduced to the fundamentals of linear algebra in elementary linear algebra. Simple matrix operations, different computations that can be performed on a system of linear equations, and certain characteristics of vectors are all included in this. The following list of key terms related to basic linear algebra includes Scalars – A scalar is a quantity that only has magnitude and not direction. It is an element that is used to define a vector space. In linear algebra, scalars are usually real numbers. Vectors – A vector is an element in a vector space. It is a quantity that can describe both the direction and magnitude of an element. Vector Space – The vector space consists of vectors that may be added together and multiplied by scalars. Matrix – A matrix is a rectangular array wherein the information is organized in the form of rows and columns. Most linear algebra properties can be expressed in terms of a matrix. Matrix Operations – These are simple arithmetic operations such as addition, subtraction, and multiplication that can be conducted on matrices. COURSE COMPETENCIES The Course As a 3-credit unit course, 11 study units grouped into 4 modules of 4 units in module 1, 3 units in module 2, 2 units in module 3 and 2 units in module 4. A quick synopsis of the entire course materials is provided in this course guide. Vector spaces and linear transformations, which are primarily iv MTH 212 COURSE GUIDE concerned with finite dimensional vector spaces over the set of real numbers, ℝ, or set of complex numbers, ℂ, are the two basic building blocks of linear algebra. Adding vectors and multiplying them by numbers or scalars gives required linear combinations. The study of certain mappings between two vector spaces, called linear transformations are also initiated in this course. We shall also demonstrate how a linear transformation can be used to obtain a matrix associated with it, and vice versa. Additionally, by focusing on the corresponding matrix, certain aspects of a linear transformation can be analyzed more clearly. You will find, for instance, that it is frequently simpler to extract a matrix’s characteristic roots than a linear transformation. WORKING THROUGH THIS COURSE You are required to read the study units, set books and other materials provided by the National Open University to complete the course. You will also need to work through practical and self- assessed exercises and submit assignments for assessment purposes. The course will take you about 60 hours to complete at the end of which you will write a final examination. This course consists of the following eleven study units: Having gone the course content, what then is a matrix? STUDY UNITS Module 1 Vector Spaces Unit 1 Vector Spaces Unit 2 Linear Combinations Unit 3 Linear Transformation I Unit 4 Linear Transformation II Module 2 Matrices Unit 1 Matrices I Unit 2 Matrices II Unit 3 Matrices III v MTH 212 COURSE GUIDE Module 3 Determinants Unit 1 Determinants I Unit 2 Determinants II Module 4 Eigenvalues and Eigenvectors Unit 1 Eigenvalues and Eigenvectors Unit 2 Characteristic and Minimal Polynomials The first four units deal with vector spaces and linear transformation which are the two fundamental elements that form the basis of linear algebra which is an area of mathematics that deals with the common features of algebraic systems made up of sets, as well as a logical concept of a “linear combination” of element in the set. The remaining seven units involve matrix theory which occupies an important position in pure as well as applied mathematics. Each study unit involves specific objectives, how to study the reading materials, references to set books and other related sources and summaries of vital points and ideas. The units direct you to work on exercises related to the require reading and to carry out solutions to some exercises where appropriate. A number of self-tests are associated with each unit. These tests give you an indication of your progress. The exercises as well as the tutor-marked assignments will help you in achieving the stated learning outcomes of each unit and of the course. PRESENTATION SCHEDULE The weekly activities are presented in Table 1 while the required hours of study and the activities are presented in Table 2. This will guide your study time. You may spend more time in completing each module or unit. Table I: Weekly Activities Week Activity 1 Orientation and Course Guide 2 Module 1 Unit 1 3 Module 1 Unit 2 4 Module 1 Unit 3 5 Module 1 Unit 4 6 Module 2 Unit 1 7 Module 2 Unit 2 8 Module 2 Unit 3 9 Module 3 Unit 1 10 Module 3 Unit 2 vi MTH 212 COURSE GUIDE 11 Module 4 Unit 1 12 Module 4 Unit 2 13 Response to Exercises 14 Revisions 15 Examination The activities in Table I include facilitation hours (synchronous and asynchronous), class works and assignments. How do you know the hours to spend on each? A guide is presented in Table 2. Table 2: Required Minimum Hours of Study Hour Hour per S/N Activity per Semester Week Synchronous Facilitation (Video 1 2 26 Conferencing) Asynchronous Facilitation (Read and 2 respond to posts including Facilitator’s 4 52 comments, self-study) Assignments, mini-project, laboratory 3 1 13 practical and portfolios Total 7 91 ASSESSMENT Table 3 presents the mode you will be assessed. Table 3: Assessment S/N Method of Assessment Score (%) 3 Computer-Based Tests 30 4 Final Examination 70 Total 100 ASSIGNMENTS Take the assignment and click on the submission button to submit. The assignment will be scored, and you will receive feedback. EXAMINATION Finally, the examination will help to test the cognitive domain. The test items will be mostly application, and evaluation test items that will lead to creation of new knowledge/idea. vii MTH 212 COURSE GUIDE HOW TO GET THE MOST FROM THE COURSE To get the most in this course, you: Need a personal laptop. The use of mobile phone only may not give you the desirable environment to work. Need regular and stable internet. Need to install the recommended software. Must work through the course step by step starting with the programme orientation. Must not plagiarize or impersonate. These are serious offences that could terminate your studentship. Plagiarism check will be used to run all your submissions. Must do all the assessments following given instructions. Must create time daily to attend to your study. FACILITATION There will be two forms of facilitation–synchronous and asynchronous. The synchronous will be held through video conferencing according to weekly schedule. During the synchronous facilitation: There will be two hours of online real time contact per week making a total of 26 hours for thirteen weeks of study time. At the end of each video conferencing, the video will be uploaded for view at your pace. You are to read the course material and do other assignments as may be given before video conferencing time. The facilitator will concentrate on main themes. The facilitator will take you through the course guide in the first lecture at the start date of facilitation. For the asynchronous facilitation, your facilitator will: Present the theme for the week. Direct and summarise forum discussions. Coordinate activities in the platform. Score and grade activities when need be. Support you to learn. In this regard personal mails may be sent. Send you videos and audio lectures, and podcasts if need be. Read all the comments and notes of your facilitator especially on your assignments, participate in forum discussions. This will give you opportunity to viiiocialize with others in the course and build your skill for teamwork. You can raise any challenge encountered during your study. To gain the maximum benefit from course facilitation, prepare a viii MTH 212 COURSE GUIDE list of questions before the synchronous session. You will learn a lot from participating actively in the discussions. LEARNER SUPPORT You will receive the following support: Technical Support: There will be contact number(s), email addresses and chat bot on the Learning Management System (LMS) where you can chat or send message to get assistance and guidance any time during the course. 24/7 communication: You can send personal mail to your facilitator and the study centre at any time of the day. You will receive answer to you mails within 24 hours. There is also opportunity for personal or group chats at any time of the day with those that are online. You will receive guidance and feedback on your assessments, academic progress, and receive help to resolve challenges facing your studies. COURSE INFORMATION Course Code: MTH212 Course Title: Linear Algebra Credit Unit: 3 units Course Status: Compulsory Course Blub: This is a basic course designed to help students master the contents of a first course in Linear Algebra. Its availability, make it widely used for self-study especially independent student in an online programme. The materials in this course are standard in that the topic covered are vector spaces, linear maps and transformations, matrices, determinants, and eigenvalues and eigenvectors. Another standard is the audience’ friendly attribute of the material as well as the numerous examples following each topic Semester: Second Semester Course Duration: 13 Weeks Required Hours for Study: 91 hours ix MAIN COURSE CONTENTS Module 1 Vector Spaces………………………………….. 1 Unit 1 Vector Spaces…………………………………… 1 Unit 2 Linear Combinations …………………………… 13 Unit 3 Linear Transformation I………………………… 26 Unit 4 Linear Transformation II………………………... 53 Module 2 Matrices ………………………………………… 75 Unit 1 Matrices I ……………………………………….. 75 Unit 2 Matrices II……………………………………….. 100 Unit 3 Matrices III………………………………………. 112 Module 3 Determinants…………………………………… 137 Unit 1 Determinants I…………………………………… 137 Unit 2 Determinants II…………………………………… 153 Module 4 Eigenvalues and Eigenvectors………………….. 162 Unit 1 Eigenvalues and Eigenvectors……………………. 162 Unit 2 Characteristic and Minimal Polynomials…………. 181 MTH 212 LINEAR ALGEBRA MODULE 1 This module shall define the mathematical object which experience has shown to be the most useful abstraction of this type of algebraic system. There are two aspects to linear algebra. Abstractly, it is the study of vector spaces over fields and their linear maps and bi-linear forms. Concretely, it is matrix theory, since matrices occur in all parts of mathematics and its applications and the diagonalization of a real symmetric matrix is a skill that is required of everyone working in the mathematical sciences or associated fields. Therefore, it is important to discuss both the abstract and concrete aspects in a course of this kind, even when applications are not covered in great length. Unit 1 Vector Spaces Unit 2 Linear Combination Unit 3 Linear Transformation I Unit 4 Linear Transformation II UNIT 1 VECTOR SPACES Unit Structure 1.1 Introduction 1.2 Learning Outcomes 1.3 Vector Spaces 1.3.1 Definitions and Examples of Vector Space 1.3.2 Spaces Associated with Vector Spaces 1.3.3 Definitions and Examples of Vector Subspace 1.4 Summary 1.5 References/Further Readings 1.1 Introduction Vector spaces and linear transformations are two fundamental elements that form the basis of linear algebra. The strength of mathematics typically comes from the ability to abstractly formulate a wide range of situations, and that is exactly what shall be done throughout this module. The area of mathematics known as linear algebra deals with the common features of algebraic systems made up of sets, as well as a logical concept of a "linear combination" of element in the set. Vector spaces, linear maps, and bi-linear forms are dealt with on the theoretical side. Vector spaces over a field F are particularly alluring 1 MTH 212 LINEAR ALGEBRA algebraic objects, since each vector space is completely determined by a single number, its dimension (unlike groups, for example, whose structure is much more complicated). On the practical side,the subject only concerns matrix. A matrix must be used to express any calculation involving a linear map or a bilinear form. As a result, matrices can represent a variety of objects. 1.2 Learning Outcomes By the end of this unit, you will be able to: Define vector space over a field Define vector subspace Cite examples of vector spaces and subspaces State and prove theorems involving vector spaces and subspaces 1.3 Vector Spaces 1.3.1 Definitions and Examples of Vector Space This unit will mainly be concerned with finite dimensional vector spaces over the set of real numbers, ℝ, or set of complex numbers, ℂ. Note that the real and complex numbers have the property that any pair of elements can be added, subtracted or multiplied. Division is also allowed by a non-zero element. Such sets in mathematics are called field, thus the sets of rational numbers, Q, real numbers, R, and complex numbers, C, are examples of field and they have infinite number of elements. But, in mathematics, we do have fields that have only finitely many elements. For example, consider the sets 𝑍4 = {0,1,2,3} and 𝑍5 = {0,1,2,3,4}. In 𝑍4 ,𝑍5 , we define addition and multiplication, respectively, as + 0 1 2 3 × 0 1 2 3 0 0 1 2 3 0 0 0 0 0 1 1 2 3 0 and 1 0 1 2 3 2 2 3 0 1 2 0 2 0 2 3 3 0 1 2 3 0 3 2 1 2 MTH 212 LINEAR ALGEBRA + 0 1 2 3 4 × 0 1 2 3 4 0 0 1 2 3 4 0 0 0 0 0 0 1 1 2 3 4 0 1 0 1 2 3 4 and 2 2 3 4 0 1 2 0 2 4 1 3 3 3 4 0 1 2 3 0 3 1 4 2 4 4 0 1 2 3 4 0 4 3 2 1 Then, we see that the elements of both 𝑍4 and 𝑍5 can be added, subtracted and multiplied. Thus, 𝑍4 and 𝑍5 indeed behave like a field. So, in this module, F will represent a field. Examples of fields are the set of complex numbers, the set of real numbers, the set of rational numbers, and even the finite set of “binary numbers”, {0,1}. Definition 1.3.1: A field is an algebraic system consisting of a non-empty set F equipped with two binary operations + (addition) and · (multiplication) satisfying the conditions: (Recall from MTH103 that the vectors in 𝑅2 and 𝑅3 satisfy the conditions) 1) Vector Addition: To every pair𝑢, 𝑣 ∈ 𝑅3 , there corresponds a unique element 𝑢 + 𝑣 ∈ 𝑅3 (Called the addition of vectors) such that a) 𝑢 + 𝑣 = 𝑣 + 𝑢 (Commutative law) b) (𝑢 + 𝑣 ) + 𝑤 = 𝑢 + (𝑣 + 𝑤) (Associative Law) c) 𝑅3 has a unique element, denoted 0 , called the zero vector that satisfies 𝑢 + 0 = 𝑢, for every 𝑢 ∈ 𝑅3 (called the additive identity). d) For every𝑢 ∈ 𝑅3 , there is an element 𝑤 ∈ 𝑅3 that satisfies𝑢 + 𝑤 = 0. 2) Scalar Multiplication: For each 𝑢 ∈ 𝑅3 and𝛼 ∈ 𝑅, there corresponds a unique element 𝛼 ⋅ 𝑢 ∈ 𝑅3 (Called the scalar multiplication) such that a) 𝛼 ⋅ (𝛽 ⋅ 𝑢) = (𝛼 ⋅ 𝛽) ⋅ 𝑢, for every 𝛼, 𝛽 ∈ 𝑅 and 𝑢 ∈ 𝑅3 b) 1 ⋅ 𝑢 = 𝑢 for every 𝑢 ∈ 𝑅3 ; where 1 ∈ 𝐼𝑅. 3) Distributive Laws: Relating vector addition with scalar multiplication For any 𝛼, 𝛽 ∈ 𝑅 and𝑢, 𝑣 ∈ 𝑅3 ; the following distributive laws hold: a) 𝛼 ⋅ (𝑢 + 𝑣 ) = (𝛼 ⋅ 𝑢 ) + (𝛼 ⋅ 𝑣 ) b) (𝛼 + 𝛽 ) ⋅ 𝑢 = (𝛼 ⋅ 𝑢 ) + (𝛽 ⋅ 𝑣 ) 3 MTH 212 LINEAR ALGEBRA For the above properties to hold for any collection of vectors, we have the following definitions: Definition 1.3.2: A vector space 𝑉 over 𝐹, denoted by 𝑉 (𝐹 ) or in short 𝑉 (if the field 𝐹is clear from the context), is a non-empty set, in which vector addition, scalar multiplication can be defined. In other words, a vector space is composed of three objects; - a set and two operations. Further, with these definitions, the properties of vector addition, scalar multiplication and distributive laws (see items 1, 2 and 3 above) are satisfied. Remarks: A. The elements of F are called scalars. B. The elements of 𝑉 are called vectors. C. The zero element of 𝐹is denoted by 𝟎 whereas the zero element of 𝑉 is also denoted by 0 D. Note that the condition 1d) above implies that for every𝑢 ∈ 𝑉, the vector 𝑤 ∈ 𝑉 such that𝑢 + 𝑤 = 0holds, is unique. i. For if, 𝑤1 , 𝑤2 ∈ 𝑉with𝑢 + 𝑤𝑖 = 0, for 𝑖 = 1,2. then by commutativity of vector addition, we see that 𝑤1 = 𝑤1 + 0 = 𝑤1 + (𝑢 + 𝑤2 ) = (𝑤1 + 𝑢) + 𝑤2 = 0 + 𝑤2 = 𝑤2 ii. Hence, we represent this unique vector by −𝑢 and call it the additive inverse. E. If 𝑉 is a vector space over 𝑅then 𝑉is called a real vector space. F. If 𝑉 is a vector space over 𝐶 then 𝑉is called a complex vector space. G. In general, a vector space over 𝑅or 𝐶is called a linear space. For better understanding of the conditions given above, the following theorem and proof are presented: Theorem 1.3.1: Let V be a vector space over F. Then, i. 𝑢 + 𝑣 = 𝑢 implies 𝑣 = 0 ii. 𝛼 ⋅ 𝑢 = 0 if and only if either 𝑢 = 0or 𝛼 = 0 iii. (−1) ⋅ 𝑢 = −𝑢, for every 𝑢 ∈ 𝑉. Proof: Part 1: By Condition 1d) and Remarks D above, for each 𝑢 ∈ 𝑉there exists −𝑢 ∈ 𝑉 such that −𝑢 + 𝑢 = 0. Hence 𝑢 + 𝑣 = 𝑢 implies 0 = −𝑢 + 𝑣 = −𝑢 + (𝑢 + 𝑣 ) = (−𝑢 + 𝑢) + 𝑣 =0+𝑣 =𝑣 4 MTH 212 LINEAR ALGEBRA Part 2: As 0 = 0 + 0, using Condition 3, 𝛼 ⋅ 0 = 𝛼 ⋅ (0 + 0) = (𝛼 ⋅ 0) + (𝛼 + 0) By Part 1, 𝛼 ⋅ 0 = 0 for any 𝛼 ∈ 𝐹 Similarly, 0 ⋅ 𝑢 = (0 + 0) ⋅ 𝑢 = (0 ⋅ 𝑢) + (0 ⋅ 𝑢) implies 0 ⋅ 𝑢 = 0, for any 𝑢 ∈ 𝑉. Now, suppose 𝛼 ⋅ 𝑢 = 0. If 𝛼 = 0 then the proof is over. Assume that 𝛼 ≠ 0, 𝛼 ∈ 𝐹, then, (𝛼)−1 ∈ 𝐹 and using 1 ⋅ 𝑢 = 𝑢1 ⋅ 𝑢 = 𝑢 for every vector 𝑢 ∈ 𝑉 (see Condition 2b), we have 0 = (𝛼)−1 ⋅ 0 = (𝛼)−1 ⋅ (𝛼 ⋅ 𝑢) = ((𝛼 )−1 ⋅ 𝛼) ⋅ 𝑢 = 1 ⋅ 𝑢 = 𝑢 Thus, if 𝛼 ≠ 0 and 𝛼 ⋅ 𝑢 = 0 then 𝑢 = 0. Part 3: As 0 = 0 ⋅ 𝑢 = (1 + (−1)) ⋅ 𝑢 = 𝑢 + (−1) ⋅ 𝑢, Then, (−1) ⋅ 𝑢 = −𝑢 The following is one of the two most important definitions in the entire course. Definition 1.3.3: Suppose that V is a set upon which we have defined two operations: (i) vector addition, which combines two elements of V and is denoted by (+), and (ii) scalar multiplication, which combines a complex number C with an element of V and is denoted by (⋅). Then V, along with the two operations, is a vector space over C if the following ten properties hold: i. Additive Closure ii. If 𝑢, 𝑣 ∈ 𝑉, then 𝑢 + 𝑣 ∈ 𝑉 iii. Scalar Closure iv. If 𝛼 ∈ 𝐶and 𝑢 ∈ 𝑉 then 𝛼 ⋅ 𝑢 ∈ 𝑉 v. Commutativity vi. If 𝑢, 𝑣 ∈ 𝑉, then 𝑢 + 𝑣 = 𝑣 + 𝑢 vii. Additive Associativity viii. If 𝑢, 𝑣, 𝑤 ∈ 𝑉, then 𝑢 + (𝑣 + 𝑤) = (𝑢 + 𝑣 ) + 𝑤 ix. Zero Vector x. There is a vector,0, called the zero vector, such that 𝑢 + 0 = 𝑢, for all 𝑢 ∈ 𝑉. xi. Additive Inverses xii. If 𝑢 ∈ 𝑉, then there exists a vector −𝑢 ∈ 𝑉 such that 𝑢 + (−𝑢) = 0 xiii. Scalar Multiplication Associativity: xiv. If 𝛼, 𝛽 ∈ 𝐶and 𝑢 ∈ 𝑉, then 𝛼 ⋅ (𝛽 ⋅ 𝑢) = (𝛼 ⋅ 𝛽) ⋅ 𝑢. xv. Distributivity across Vector Addition: xvi. If 𝛼 ∈ 𝐶and 𝑢, 𝑣 ∈ 𝑉, then 𝛼 ⋅ (𝑢 + 𝑣 ) = 𝛼 ⋅ 𝑢 + 𝛼 ⋅ 𝑣. xvii. Distributivity across Scalar Addition: xviii. If 𝛼, 𝛽 ∈ 𝐶 and 𝑢 ∈ 𝑉, then (𝛼 + 𝛽) ⋅ 𝑢 = 𝛼 ⋅ 𝑢 + 𝛽 ⋅ 𝑣. xix. One: If 𝑢 ∈ 𝑉, then 1 ⋅ 𝑢 = 𝑢. 5 MTH 212 LINEAR ALGEBRA The objects in 𝑉 are called vectors, no matter what else they might really be, simply by virtue of being elements of a vector space. Examples 1. The Euclidean plane 𝑹𝟐 is a real vector space. In other words, two vectors can be added together as well as multiply a vector by a scalar (a real number). There are two approaches to explain these definitions. a) The geometric definition: Think of a vector as an arrow starting at the origin and ending at a point of the plane, then addition of two vectors is done by the parallelogram law (see Figure 1.3.1). The scalar multiple 𝑎𝑣 is the vector whose length is |𝑎| times the length of v, in the same direction if 𝑎 > 0 and in the opposite direction if 𝑎 < 0. Figure 1.3.1: The Parallelogram Law b) The algebraic definition: The points of the plane are represented by Cartesian coordinates (x, y) such that a vector v is just a pair (x, y) of real numbers. Now we define addition and scalar multiplication by (𝑥1 , 𝑦1 ) + (𝑥2 , 𝑦2 ) = (𝑥1 + 𝑥2 , 𝑦1 , 𝑦2 ) 𝑎(𝑥, 𝑦) = (𝑎𝑥, 𝑎𝑦) Let’s check that the rules for a vector space are really satisfied, for instance, we check the law 𝑎(𝑣 + 𝑤) = 𝑎𝑣 + 𝑎𝑤. ( ) ( Let 𝑥1 , 𝑦1 + 𝑥2 , 𝑦2 ) Then, 𝑎(𝑣 + 𝑤) = 𝑎[(𝑥1 , 𝑦1 ) + (𝑥2 , 𝑦2 )] = 𝑎(𝑥1 + 𝑥2 , 𝑦1 + 𝑦2 ) = (𝑎𝑥1 + 𝑎𝑥2 , 𝑎𝑦1 + 𝑎𝑦2 ) = (𝑎𝑥1 , 𝑎𝑦1 ) + (𝑎𝑥2 , 𝑎𝑦2 ) = 𝑎𝑣 + 𝑎𝑤. In the algebraic definition, we say that the operations of addition and scalar multiplication are coordinate-wise, that is, two vectors can be added coordinate by coordinate, and similarly for scalar multiplication. The generalized form of this example using coordinates is given by the next example: 6 MTH 212 LINEAR ALGEBRA 2. The n-tuple, 𝑭𝒏 : Let 𝑛 be any positive integer and 𝐹 be any field, and let 𝑉 = 𝐹 𝑛 , be the set of all n-tuples of elements of 𝐹. Then V is a vector space over 𝐹 where the operations are defined coordinate-wise: Let 𝛼 = (𝑥1 + 𝑥2 , … , 𝑥𝑛 ) of scalars 𝑥𝑖 and 𝛽 = (𝑦1 + 𝑦2 , … , 𝑦𝑛 ) with 𝑦𝑖 ∈ 𝐹. The sum of 𝛼and 𝛽 is defined by 𝛼 + 𝛽 = (𝑥1 + 𝑥2 , … , 𝑥𝑛 ) + (𝑦1 + 𝑦2 , … , 𝑦𝑛 ) = (𝑥1 + 𝑦1 , 𝑥2 + 𝑦2 , … , 𝑥𝑛 + 𝑦𝑛 ) The scalar multiplication condition would be 𝑐𝛼 = (𝑐𝑥1 , 𝑐𝑥2 , … , 𝑐𝑥𝑛 ) The product of a scalar 𝑐 and vector 𝛼 is defined by 𝑐𝛼 = (𝑐𝑥1 , 𝑐𝑥2 , … , 𝑐𝑥𝑛 ) The fact that the vector addition and scalar multiplication satisfy conditions (1) and (2), it is easy to verify, using the similar properties of addition and multiplication of elements of F. 3. The space of (𝒎 × 𝒏) matrices 𝑭𝒎×𝒏 Let 𝐹 be any field and let 𝑚 and 𝑛 be positive integers. Let 𝐹 𝑚×𝑛 be the set of all (𝑚 × 𝑛) matrices over the field F. i. The sum of two vectors A and B in 𝐹 𝑚×𝑛 is defined by (𝐴 + 𝐵 )𝑖𝑗 = 𝐴𝑖𝑗 + 𝐵𝑖𝑗 ii. The product of a scalar c and the matrix (𝐴 + 𝐵 )𝑖𝑗 is defined by 𝑐 (𝐴 + 𝐵)𝑖𝑗 = 𝑐𝐴𝑖𝑗 + 𝑐𝐵𝑖𝑗 Self-Assessment: From what you just read, can you recount the definition of vector space over C and some of its properties? 1.3.2 Spaces Associated with Vector Spaces A) The space of functions from a set to a field: Definition 1.3.4: Let 𝐹 be any field and 𝑆 be any non-empty set. Let 𝑉 be the set of all functions from the set 𝑆 into 𝐹, then, i. The sum of two vectors 𝑓and 𝑔 in 𝑉 is the vector 𝑓 + 𝑔, i.e., the function from 𝑆 into 𝐹 is defined by (𝑓 + 𝑔)(𝑠) = 𝑓 (𝑠 ) + 𝑔 (𝑠 ) ii. The product of the scalar c and the function f is the function 𝑐𝑓defined by (𝑐𝑓)(𝑠) = 𝑐𝑓(𝑠) 7 MTH 212 LINEAR ALGEBRA B) The space of polynomial functions over a field F Definition 1.3.5: Let 𝐹 be a field and let 𝑉 be the set of all functions f from 𝐹 into 𝐹 which have a rule of the form: 𝑓 (𝑥) = 𝑐0 + 𝑐1 𝑥 + 𝑐2 𝑥 2 + ⋯ + 𝑐𝑛 𝑥 𝑛 , where 𝑐0 , 𝑐1 , 𝑐2 , … 𝑐𝑛 are fixed scalars in 𝐹(independent of 𝑥). A function of this type is called a polynomial function on 𝐹. N.B: Let addition and scalar multiplication be defined as in definition 3.2.1. One must observe here that if 𝑓and 𝑔are polynomial functions and 𝑐 ∈ 𝐹, then 𝑓 + 𝑔 and 𝑐𝑓are again polynomial functions. Definition 1.3.6: Let 𝑅 𝑛 = {(𝑎1 , … , 𝑎𝑛 )𝑇 : 𝑎𝑖 ∈ 𝑅, 1 ≤ 𝑖 ≤ 𝑛};𝑢 = (𝑎1 , … , 𝑎𝑛 )𝑇 , 𝑣 = (𝑏1 , … , 𝑏𝑛 )𝑇 ∈ 𝑉 and 𝛼 ∈ 𝑅, we define 𝑢 + 𝑣 = (𝑎1 + 𝑏1 , … , 𝑎𝑛 + 𝑏𝑛 )𝑇 and 𝛼 ⋅ 𝑢 = (𝛼𝑎1 , … , 𝛼𝑎𝑛 )𝑇 (called component- wise operations), then, V is a real vector space. The vector space 𝑅𝑛 is called the real vector space of n-tuples. Definition 1.3.7: Let 𝑚, 𝑛 ∈ 𝑁and 𝑀𝑚+𝑛 (𝐶 ) = {𝐴𝑚×𝑛 = [𝑎𝑖𝑗 ] ∈ 𝐶}, then, with the usual addition and scalar multiplication of matrices, 𝑀𝑚+𝑛 (𝐶 ) is a complex vector space. If 𝑚 = 𝑛, the vector space 𝑀𝑚+𝑛 (𝐶 ) is denoted by 𝑀𝑛 (𝐶 ). Definition 1.3.8: Let 𝑆 be a non-empty set and let 𝑅 𝑆 = {𝑓} such that 𝑓 is a function from 𝑆 to 𝑅. For 𝑓, 𝑔 ∈ 𝑅 𝑆 and 𝛼 ∈ 𝑅, define (𝑓 + 𝛼𝑔)(𝑥) = 𝑓 (𝑥) + 𝛼𝑔(𝑥) for all 𝑥 ∈ 𝑆, then, 𝑅 𝑆 is a real vector space. In particular, for 𝑆 = 𝑁, observe that 𝑅𝑁 consists of all real sequences and forms a real vector space. Let 𝐶 (𝑅, 𝑅) = {𝑓: 𝑅 → 𝑅}, such that 𝑓 is continuous. Then 𝐶 (𝑅, 𝑅) is a real vector space, where (𝑓 + 𝛼𝑔)(𝑥) = 𝑓(𝑥) + 𝛼𝑔(𝑥)∀𝑥 ∈ 𝑅. 1.3.3 Definition and Examples of Vector Subspace Definition 1.3.9: A subspace is a vector space that is contained within another vector space. In other words, every subspace is a vector space in its own right, but it is also defined relative to some other (larger) vector space. The principal definition of subspace is presented below: Definition 1.3.10: Suppose that V and W are two vector spaces that have identical definitions of vector addition and scalar multiplication, and that W is a subset of V, that is, 𝑊 ⊆ 𝑉, then W is a subspace of V. 8 MTH 212 LINEAR ALGEBRA Let us look at an example of a vector space inside another vector space. Vector Subspace Definition 1.3.12: Let V be a vector space over F. Then, a non-empty subset W of V is called a subspace of V if W is also a vector space with vector addition and scalar multiplication in W coming from that in V (compute the vector addition and scalar multiplication in V and then the computed vector should be an element of W). Theorem 1.3.2: Let 𝑉 (𝐹 ) be a vector space and 𝑊 ⊆ 𝑉, 𝑊 ≠ ∅. Then, W is a subspace of V if and only if 𝛼𝑢 + 𝛽𝑣 ∈ 𝑊 whenever 𝛼, 𝛽 ∈ 𝐹 and 𝑢, 𝑣 ∈ 𝑊. Proof: Let W be a subspace of 𝑉 and let 𝑢, 𝑣 ∈ 𝑊. As 𝑢, 𝑣 ∈ 𝑊 is a subspace, the scalar 𝛼, 𝛽 ∈ 𝐹and 𝛼𝑢 + 𝛽𝑣 ∈ 𝑊. Now, we assume that 𝛼𝑢 + 𝛽𝑣 ∈ 𝑊, whenever 𝛼, 𝛽 ∈ 𝐹 and 𝑢, 𝑣 ∈ 𝑊. To show, 𝑊 is a subspace of 𝑉: i. Taking 𝛼 = 0 and 𝛽 = 0 ⇒ 0 ∈ 𝑊. So, W is non-empty. ii. Taking 𝛼 = 1 and 𝛽 = 1, we see that 𝑢 + 𝑣 ∈ 𝑊, for every 𝑢, 𝑣 ∈ 𝑊. iii. Taking 𝛽 = 0, we see that 𝛼𝑢 ∈ 𝑊, for every 𝛼 ∈ 𝐹and 𝑢 ∈ 𝑊. iv. Hence, using Theorem 1.3.1(iii) above, −𝑢 = (−1)𝑢 ∈ 𝑊. v. The commutative and associative laws of vector addition hold as they hold in 𝑉. vi. The conditions related with scalar multiplication and the distributive laws also hold as they hold in 𝑉. Theorem 1.3.3: A non-empty subset 𝑊of 𝑉 is a subspace of 𝑉if and only if for each pair of vectors 𝛼, 𝛽 ∈ 𝑊and each scalar 𝑐 ∈ 𝐹 the vector 𝑐𝛼 + 𝛽 ∈ 𝑊. Proof: Suppose that 𝑊is a non-empty subset of 𝑉 such that 𝑐𝛼 + 𝛽 ∈ 𝑊 for all vector 𝛼, 𝛽 ∈ 𝑊and all scalars 𝑐 ∈ 𝐹. Since 𝑊 is non-empty, there is a vector 𝜙 ∈ 𝑊, which implies that (−1)𝜙 + 𝜙 = 0 is in W. If 𝛼 is any vector in 𝑊, and 𝑐, any scalar, the vector 𝑐𝛼 = 𝑐𝛼 + 0 is in 𝑊. In particular, (−1)𝛼 = −𝛼 is in W. Also, if 𝛼, 𝛽 ∈ 𝑊, then 𝛼 + 𝛽 = 1𝛼 + 𝛽 is in 𝑊. Thus, 𝑊 is a subspace of 𝑉. Conversely, if 𝑊is a subspace of 𝑉, 𝛼, 𝛽 ∈ 𝑊and 𝑐 is a scalar, certainly 𝑐𝛼 + 𝛽 ∈ 𝑊. 9 MTH 212 LINEAR ALGEBRA Examples: 1. If 𝑉 is any vector space and 𝑉 is a subspace of 𝑉; the subset consisting of the zero vector alone is a subspace of 𝑉, called the zero subspace of 𝑉. 2. In 𝐹 𝑛 , the set of n-tuples 𝑥1 , 𝑥2 , … , 𝑥𝑛 with 𝑥𝑖 = 0 is a subspace; however, the set of n-tuples with 𝑥1 = 1 + 𝑥2 is not a subspace (𝑛 ≥ 2). 3. The space of polynomial functions over the field 𝐹is a subspace of the space of all functions from 𝐹 into 𝐹. 4. An 𝑛 × 𝑛 (square) matrix 𝐴over the field 𝐹is symmetric if 𝐴𝑖𝑗 = 𝐴𝑗𝑖 , for each 𝑖 and 𝑗. The symmetric matrices form a subspace of the space of all 𝑛 × 𝑛 matrices over 𝐹. 5. An 𝑛 × 𝑛 (square) matrix 𝑨 over the field C of complex numbers is Hermitian (or self-adjoint) if 𝐴𝑗𝑘 = 𝐴̄𝑘𝑗 for each 𝑗, 𝑘, the bar denoting complex conjugation. Definition 1.3.13: A 2 × 2 matrix is Hermitian if and only if it has the form 𝑧 𝑥 + 𝑖𝑦 [ ], where 𝑥, 𝑦, 𝑧, 𝑤, are real numbers. 𝑥 − 𝑖𝑦 𝑤 The set of all Hermitian matrices is not a subspace of the space of all 𝑛 × 𝑛matrices over 𝑪. If 𝐴 is Hermitian, its diagonal entries 𝐴11 , 𝐴22 , … , 𝐴𝑛𝑛 are all real numbers, but the diagonal entries of 𝐴𝑖𝑗 are in general not real. On the other hand, it is easily verified that the set of 𝑛 × 𝑛complex Hermitian matrices is a vector space over the field R of real numbers (with the usual operations). Theorem 1.3.4: The solution space of a system of homogeneous linear equations: Let A be an 𝑚 × 𝑛 matrix over F, then the set of all 𝑛 × 1 (column) matrices 𝑋 over 𝐹 such that 𝐴𝑋 = 0 is a subspace of the space of all 𝑛 × 1 matrices over F. To prove this, we must show that 𝐴(𝑐𝑋 + 𝑌) = 0 when 𝐴𝑋 = 0, 𝐴𝑌 = 0, and c is an arbitrary scalar in F. This follows immediately from the following general fact. Lemma: If A is an 𝑚 × 𝑛 matrix over F and B, C are 𝑛 × 𝑝 matrices over F then 𝐴(𝑑𝐵 + 𝐶 ) = 𝑑 (𝐴𝐵) + 𝐴𝐶 for each scalar 𝑑 ∈ 𝐹 Proof: [𝐴(𝑑𝐵 + 𝐶 )]𝑖𝑗 = ∑ 𝐴𝑖𝑘 (𝑑𝐵 + 𝐶 )𝑘𝑗 𝑘 10 MTH 212 LINEAR ALGEBRA = ∑(𝑑𝐴𝑖𝑘 𝐵𝑘𝑗 + 𝐴𝑖𝑘 𝐶𝑘𝑗 ) 𝑘 = 𝑑 ∑ 𝐴𝑖𝑘 𝐵𝑘𝑗 + ∑ 𝐴𝑖𝑘 𝐶𝑘𝑗 𝑘 𝑘 = 𝑑 (𝐴𝐵)𝑖𝑗 + (𝐴𝐶 )𝑖𝑗 = [𝑑 (𝐴𝐵) + 𝐴𝐶 ]𝑖𝑗 Similarly, one can show that (𝑑𝐵 + 𝐶 )𝐴 = 𝑑 (𝐵𝐴) + 𝐶𝐴, if the matrix sums and products are defined. Theorem 1.3.5: Let V be a vector space over the field 𝐹. The intersection of any collection of subspaces of 𝑉 is a subspace of 𝑉. Proof: Let {𝑊𝑎 } be a collection of subspaces of 𝑉, and let 𝑊 = ⋂𝑎 𝑊𝑎 be their intersection. Recall that W is defined as the set of all elements belonging to every 𝑊𝑎. Since each 𝑊𝑎 is a subspace, each contains the zero vector. Thus, the zero vector is in the intersection 𝑊, and 𝑊 is non-empty. Let 𝛼 and 𝛽 be vectors in 𝑊 and let 𝑐 be a scalar. By definition of W, both 𝛼and 𝛽belong to each 𝑊𝑎 , and because each 𝑊𝑎 is a subspace, the vector 𝑐𝛼 + 𝛽 is in every 𝑊𝑎. Thus(𝑐𝛼 + 𝛽) is again in W. Thus, by Theorem 1.3.1, W is a subspace of V. From Theorem 1.3.2, it follows that if S is any collection of vectors in V, then there is a smallest subspace of V which contains S, that is, a subspace which contains S and which is contained in every other subspace containing S. SELF- ASSESSMENT EXERCISE i. By Definition 1.3.3, show that the ten properties hold using the two operations on a vector space over 𝐶. ii. Enumerate any three (3) examples of vector space. 1.4 Summary In this unit we have covered the following points: A vector space is composed of three objects, a set and two operations which satisfy some properties 11 MTH 212 LINEAR ALGEBRA The Euclidean plane𝑅2 , the n-tuple, 𝐹 𝑛 and the space of (𝑚 × 𝑛) matrices 𝐹 𝑚×𝑛 are examples of vector spaces A subspace is a vector space that is contained within another vector space. A vector space 𝑉 over the field 𝐹, denoted by 𝑉 (𝐹 ), is a non-empty set, in which vector addition, scalar multiplication can be defined. The vector space 𝑅𝑛 is called the real vector space of n-tuples. A non-empty subset W of a vector space V over F is called a subspace of V if W is also a vector space with vector addition and scalar multiplication in W coming from that in V. The intersection of any collection of subspaces of V is a subspace of V. 1.5 References/Further Readings Robert A. Beezer (2014). A First Course in Linear Algebra Congruent Press, Gig Harbor, Washington, USA 3(40). Arbind K Lal Sukant Pati (2018). Linear Algebra through Matrices. Peter J. Cameron (2008). Notes on Linear Algebra. 12 MTH 212 LINEAR ALGEBRA UNIT 2 LINEAR COMBINATIONS Unit Structure 2.1 Introduction 2.2 Learning Outcomes 2.3 Linear Combinations 2.3.1 Linear Combination of Column Vectors 2.3.2 Linear Combination and Consistency of a System 2.3.3 Linear Span 2.3.4 Finite and Infinite Dimensional 2.4 Linear Independence 2.4.1 Linearly Independent Vectors 2.4.2 Properties of Linear Independence 2.5 Summary 2.6 References/Further Reading 2.1 Introduction The heart of linear algebra is in two operations, both with vectors. We add vectors to obtain v +w and multiply them by numbers or scalars c and d to get 𝑐𝑣 and 𝑑𝑤.Combining those two operations (adding 𝑐𝑣 to 𝑑𝑤) gives the linear combination 𝑐𝑣 + 𝑑𝑤. Linear combinations are all-important in this subject! Sometimes, one particular combination is required, the specific choice c = 2 and d = 1 that produces 𝑐𝑣 + 𝑑𝑤 = (4,5). Other times, we require all the combinations of v and w (coming from all c and d), the vectors cv lie along a line. When w is not on that line, the combinations 𝑐𝑣 + 𝑑𝑤fill the whole two-dimensional plane ("two- dimensional" because linear algebra allows higher-dimensional planes). For four vectors 𝑢, 𝑣, 𝑤, 𝑧 in four-dimensional space and their combinations: 𝑐𝑢 + 𝑑𝑣 + 𝑒𝑤 + 𝑗𝑧 are likely to fill the space but not always. The vectors and their combinations could even lie on one line. 13 MTH 212 LINEAR ALGEBRA 2.2 Learning Outcomes By the end of this unit, you will be able to: Define Linear Combination of Column Vectors Form Linear Combinations given different scalars Define Finite dimension of a Vector Space Define Consistency of a System Obtain the Solution(s) of a system containing a linear combination of the columns Define Linear Span of a Collection of Vectors 2.3 Linear Combinations 2.3.1 Linear Combinations of Column Vectors Definition 2.3.1: Given n vectors 𝑢1 , 𝑢2 , 𝑢3 , … , 𝑢𝑛 from the column vector 𝐶 𝑚 and n-scalars 𝛼1 , 𝛼2 , 𝛼3 , … , 𝛼𝑛 , their linear combination is the vector 𝛼1 𝑢1 + 𝛼2 𝑢2 + 𝛼3 𝑢3 + ⋯ + 𝛼𝑛 𝑢𝑛. The definition above combines an equal number of scalars and vectors using the two operations (scalar multiplication and vector addition), thus forming a single vector of the same size as the original vectors. Example 1: Linear combinations in 𝐶 5 Let 𝛼1 = −1𝛼2 = −3𝛼3 = 4𝛼4 = 2 and 1 3 2 3 −2 −2 1 6 𝑢1 = 0 𝑢2 = 4 𝑢3 = −2 𝑢4 = −4 1 −3 0 −1 Their linear combination is 𝛼1 𝑢1 + 𝛼2 𝑢2 + 𝛼3 𝑢3 + 𝛼4 𝑢4 1 3 2 3 −2 −2 1 6 = (−1) 0 + (−3) 4 + (4) −2 + (2) −4 1 −3 0 −1 14 MTH 212 LINEAR ALGEBRA −1 −9 8 6 4 2 6 4 12 24 = 0 + −12 + −8 + −8 = −28 −1 9 0 −2 6 [−5] [ −3 ] [ 16 ] [ 2 ] [ 10 ] Other different linear combinations can be formed given different scalars, for instant, given 𝛽1 = 2;𝛽2 = −1;𝛽3 = −3 ; 𝛽4 = 0 We can form a linear combination given by 𝛽1 𝑢1 + 𝛽2 𝑢2 + 𝛽3 𝑢3 + 𝛽4 𝑢4 1 3 2 3 −2 −2 1 6 = (2) 0 + (−1) 4 + (−3) −2 + (0) −4 1 −3 0 −1 4 −3 −6 0 −5 −4 2 −3 0 −5 = 0 + −4 + 6 + 0 = 2 2 3 0 0 5 [ 10 ] [−1] [−12] [−3] 2.3.2 Linear Combination and Consistency of a System Definition 2.3.2: A system 𝐴𝑥 = 𝐵 is consistent if it has a solution and inconsistent if it has no solution. The consistency of the system 𝐴𝑥 = 𝐵 leads to the idea that the vector B is a linear combination of the columns of A. 1 1 2 2 1 1 Example 2: Let 𝐴 = [1 2] and 𝐵 =. Then, = + 1 3 4 4 1 3 2 This implies that is a linear combination of the vectors in 𝑃 = 4 1 1 { , } 1 3 8 Similarly, the vector is a linear combination of the vectors in P as 18 8 1 1 3 = 3 + 5 = 𝐴 [ ] 5 18 1 3 Thus, a formal definition of linear combination is given below: Definition 2.3.3: Let V be a vector space over F and let𝑃 = {𝑢1 , … , 𝑢𝑛 } ⊆ 𝑉. Then, a vector𝑢 ∈ 𝑉 is called a linear combination of elements of P 15 MTH 212 LINEAR ALGEBRA if we can find 𝛼1 , … , 𝛼𝑛 , such that 𝑢 = 𝛼1 𝑢1 + 𝛼2 𝑢2 + ⋯ + 𝛼𝑛 𝑢𝑛 = ∑𝑛𝑖−1 𝛼𝑖 𝑢𝑖. Or equivalently, any vector of the form ∑𝑛𝑖−1 𝛼𝑖 𝑢𝑖 where 𝛼1 , … + 𝛼𝑛 ∈ 𝐹 is said to be a linear combination of the elements of P. Thus, the system 𝐴𝑥 = 𝐵 has a solution, meaning that B is a linear combination of the columns of A. Or equivalently, B is a linear combination means the system 𝐴𝑥 = 𝐵 has a solution. So, recall that when we were solving a system of linear equations, we looked at the point of intersections of lines or plane etc. But here it leads us to the study of whether a given vector is a linear combination of a given set P or not? Or in the language of matrices, is B a linear combination of columns of the matrix A or not? Examples 3: a) Is (4, 5, 5) a linear combination of (1, 0, 0), (2, 1, 0) and (3, 3, 1)? Solution: 1 2 3 4 Let 𝐴 = [0 1 3] and 𝐵 = 0 0 1 5 1 2 3 4 9 + (-10) + 5 = 0 0 1 5 Hence (4, 5, 5) a linear combination of (1, 0, 0), (2, 1, 0) and (3, 3, 1) and 1 2 3 9 4 = [0 1 3] [−10] = 0 0 1 5 5 9 4 𝐴 [−10] = of the form 𝐴𝑥 = 𝐵 5 5 Thus, 𝑥 = 9 −10 5]𝑇 is a solution. [ b) Find condition(s) on 𝑥, 𝑦, 𝑧 ∈ 𝑅 such that i. (𝑥, 𝑦, 𝑧) is a linear combination of (1, 2, 3), (−1, 1, 4) and (3, 3, 2). ii. (𝑥, 𝑦, 𝑧) is a linear combination of (1, 2, 1), (1, 0, −1) and (1, 1, 0). iii. (𝑥, 𝑦, 𝑧) is a linear combination of (1, 1, 1), (1, 1, 0) and (1, 1, 0). Solution: 1 −1 3 3 i. + [ 1 ] + = 3 4 2 9 16 MTH 212 LINEAR ALGEBRA 1 1 1 3 ii. + [ 0 ] + = 1 −1 0 0 1 1 1 3 iii. + + = 1 0 0 1 Self-Assessment: Now, in simple sentences, give a simple description of linear combination of column vectors 2.3.3 Linear Span Definition 2.3.4: Let V be a vector space over F and S a subset of V. We now look at ’linear span' of a collection of vectors. So, here we ask;“What is the largest collection of vectors that can be obtained as linear combination of vectors from S” Or equivalently, what is the smallest subspace of V that contains S? Example 4 Let𝑆 = {(1,0,0), (1,2,0)} ⊆ 𝑅3. Let us find the largest possible subspace of 𝑅3 which contains vectors of the form 𝛼(1,0,0), 𝛽 (1,2,0) and (1,0,0), (1,2,0) for all possible choices of𝛼, 𝛽 ∈ 𝑅. Note that a. 𝑙1 = {𝛼(1,0,0): 𝛼 ∈ 𝑅} gives the x-axis. b. 𝑙2 = {𝛽 (1,2,0): 𝛽 ∈ 𝑅} gives the line passing through (0, 0, 0) and (1, 2, 0). So, we want the largest subspace of 𝑅3 that contains vectors which are formed as sum of any two points on the two lines 𝑙1 and 𝑙2 , or the smallest subspace of 𝑅3 that contains S? 2.3.4 Finite and Infinite Dimensional Definition 2.3.5: Let V be a vector space over F and 𝑆 ⊆ 𝑉, then 1. the linear span of S, denoted by Ls(S), is defined as 𝐿𝑠(𝑆) = {𝛼1 𝑢1 + ⋯ + 𝛼𝑛 𝑢𝑛 | 𝛼1 ∈ 𝐹, 𝑢𝑖 ∈ 𝑆 for 1 ≤ 𝑖 ≤ 𝑛}. This implies that Ls(S) is the set of all possible linear combinations of finitely many vectors of S. If S is an empty set, we define 𝐿𝑠(𝑆) = {0}. 2. V is said to be finite dimensional if there exists a finite set S such that V = L(S). 3. If there does not exist any finite subset S of V such that V = L(S) then V is called infinite dimensional. 17 MTH 212 LINEAR ALGEBRA Example 5: For each set S given below, determine Ls(S). a) 𝑆 = {(1,0)𝑇 , (0,1)𝑇 } ⊆ 𝑅2. Solution: 𝐿𝑠(𝑆) = {𝑎(1,0)𝑇 + 𝑏(1,0)𝑇 |𝑎, 𝑏 ∈ 𝑅 } = {(𝑎, 𝑏)𝑇 |𝑎, 𝑏 ∈ 𝑅} = 𝑅2. Thus, 𝑅 2 is finite dimensional. b) 𝑆 = {(1,1,1)𝑇 , (2,1,3)𝑇 }. What does L(S) represent in𝑅3 ? Solution: 𝐿𝑠(𝑆) = {𝑎(1,1,1)𝑇 + 𝑏(2,1,3)𝑇 |𝑎, 𝑏 ∈ 𝑅 } = {(𝑎 + 2𝑏, 𝑎 + 𝑏, 𝑎 + 3𝑏)𝑇 |𝑎, 𝑏 ∈ 𝑅}. Note that L(S) represents a plane passing through the points (0,0,0)𝑇 , (1,1,1)𝑇 𝑎𝑛𝑑(2,1,3)𝑇. To obtain the equation of the plane, we proceed as follows: Find conditions on x, y and z such that {(𝑎 + 2𝑏, 𝑎 + 𝑏, 𝑎 + 3𝑏) = (𝑥, 𝑦, 𝑧)}. Or equivalently, find conditions on x, y and z such that 𝑎 + 2𝑏 = 𝑥; 𝑎 + 𝑏 = 𝑦 and 𝑎 + 3𝑏 = 𝑧 has a solution for all 𝑎, 𝑏 ∈ 𝑅. The Row-Reduced Echelon Form(RREF) of the augmented matrix equals 1 0 2𝑦 − 𝑥 [0 1 𝑥−𝑦 ] 0 0 𝑧 + 𝑦 − 2𝑥 Thus, the required condition on x, y and z is given by 𝑧 + 𝑦 − 2𝑥 = 0. Hence, 𝐿𝑠(𝑆) = {𝑎(1,1,1)𝑇 + 𝑏(2,1,3)𝑇 |𝑎, 𝑏 ∈ 𝑅 } = {(𝑥, 𝑦, 𝑧)𝑇 ∈ 𝑅3 |2𝑥 − 𝑦 − 𝑧 = 0}. Verify that if 𝑇 = 𝑆 ∪ {(1,1,0)𝑇 }, then 𝐿𝑠(𝑇) = 𝑅3. Hence, 𝑅3 is finite dimensional. In general, for every fixed 𝑛 ∈ 𝑁, 𝑅𝑛 is finite dimensional as 𝑅𝑛 = 𝐿𝑠({𝑒1 , … 𝑒𝑛 }). c) 𝑆 = {1 + 2𝑥 + 3𝑥 2 , 1 + 𝑥 + 2𝑥 2 , 1 + 2𝑥 + 𝑥 3 } Solution: To understand Ls(S), we need to find condition(s) on 𝛼, 𝛽, 𝛾, 𝛿such that the linear system: 𝑎(1 + 2𝑥 + 3𝑥 2 ) + 𝑏(1 + 𝑥 + 2𝑥 2 ) + 𝑐 (1 + 2𝑥 + 𝑥 3 ) = 𝛼 + 𝛽𝑥 + 𝛾𝑥 2 + 𝛿𝑥 3 in the unknowns a, b, c is always consistent. An application of Gauss-Jordan Elimination (GJE) method gives 𝛼 + 𝛽 − 𝛾 − 3𝛿 = 0 as the required condition. Thus, 𝐿𝑠(𝑆) = {𝛼 + 𝛽𝑥 + 𝛾𝑥 2 + 𝛿𝑥 3 ∈ 𝑅 [𝑥]: 𝛼 + 𝛽 − 𝛾 − 3𝛿 = 0}. 18 MTH 212 LINEAR ALGEBRA Note that, for every fixed 𝑛 ∈ 𝑁, 𝑅[𝑥; 𝑛]is finite dimensional as 𝑅[𝑥; 𝑛] = 𝐿𝑠({1, 𝑥, … 𝑥 𝑛 }). 1 0 0 0 1 1 0 1 2 d) 𝑆 = {[0 1 0] , [1 1 2] , [1 0 2]} ⊆ 𝑀3 (𝑅). 0 0 1 1 2 0 2 2 4 Solution: To get the equation, we need to find conditions on 𝑎𝑖𝑗 ′𝑠 such that the system 𝛼 𝛽+𝛾 𝛽 + 2𝛾 𝑎11 𝑎12 𝑎13 [𝛽+𝛾 𝛼+𝛽 2𝛽 + 2𝛾 ] = [ 21 𝑎22 𝑎23 ] 𝑎 𝛽 + 2𝛾 2𝛽 + 2𝛾 𝛼 + 2𝛾 𝑎31 𝑎32 𝑎33 in the unknowns 𝛼, 𝛽, 𝛾 is always consistent. Now, verify that the required condition equals 𝑎22 + 𝑎33 − 𝑎13 𝐿𝑠(𝑆) = {𝐴 = [𝑎𝑖𝑗 ] ∈ 𝑀3 (𝑅): 𝐴 = 𝐴𝑇 , 𝑎11 = , 𝑎12 2 𝑎22 − 𝑎33 + 3𝑎13 𝑎22 − 𝑎33 + 3𝑎13 = , 𝑎23 = } 4 2 In general, for each fixed 𝑚, 𝑛 ∈ 𝑁, the vector space 𝑀𝑚,𝑛 (𝑅) is finite dimensional 𝑀𝑚,𝑛 (𝑅) = 𝐿𝑠({𝑒𝑖𝑗 : 1 ≤ 𝑖 ≤ 𝑚, 1 ≤ 𝑗 ≤ 𝑛}). The vector space R over Q is infinite dimensional. Definition 2.3.6: Let S be a set of vectors in a vector space V. The subspace spanned by S is defined to be the intersection W of all subspaces of V which contain S. When S is a finite set of vectors, 𝑆 = {𝛼1 , 𝛼2 , … , 𝛼𝑛 }, we shall simply call W the subspace spanned by the vectors 𝛼1 , 𝛼2 , … , 𝛼𝑛. Theorem 2.3.1: The subspace spanned by a non-empty subset S of a vector space V is the set of all linear combinations of vectors in S. Proof: Let W be the subspace spanned by S. Then each linear combination 𝛼 = 𝑥1 𝛼1 + 𝑥2 𝛼2 + ⋯ + 𝑥𝑚 𝛼𝑚 of vectors 𝛼1 , 𝛼2 , … , 𝛼𝑚 in S is clearly in W. Thus, W contains the set L of all linear combinations of vectors in S. The set L, on the other hand, contains S and is non-empty. If 𝛼, 𝛽 belong to L then 𝛼is a linear combination, 𝛼 = 𝑥1 𝛼1 + 𝑥2 𝛼2 + ⋯ + 𝑥𝑚 𝛼𝑚 of vectors 𝛼𝑖 in S, and 𝛽is a linear combination 𝛽 = 𝑦1 𝛽1 + 𝑦2 𝛽2 + ⋯ + 𝑦𝑛 𝛽𝑛 of vectors 𝛽𝑗 in S For each scalar c, 𝑐𝛼 + 𝛽 = ∑𝑚 𝑛 𝑖=1(𝑐𝑥𝑖 )𝛼𝑖 + ∑𝑗=1 𝑦𝑗 𝛽𝑗 Hence 𝑐𝛼 + 𝛽belongs to L. Thus, L is a subspace of V. 19 MTH 212 LINEAR ALGEBRA Now we have shown that L is a subspace of V which contains S, and also that any subspace which contains S contains L. It follows that L is the intersection of all subspaces containing S, that is, that L is the subspace spanned by the set S. Definition 2.3.7: If 𝑆1 , 𝑆2 , … , 𝑆𝑘 are subsets of a vector space V, the set of all sums 𝛼1 + 𝛼2 + ⋯ + 𝛼𝑘 of vectors 𝛼𝑖 in 𝑆𝑖 is called the sum of the subsets 𝑆1 , 𝑆2 , … , 𝑆𝑘 and is denoted by 𝑆1 + 𝑆2 + ⋯ + 𝑆𝑘 or ∑𝑘𝑖=1 𝑆𝑖 If 𝑊1 , 𝑊2 , … , 𝑊𝑘 are subspaces of V, then the sum is easily seen to be a subspace of V which contains each of the subspaces 𝑊𝑖. From this it follows, as in the proof of Theorem 3, that W is the subspace spanned by the union of 𝑊1 , 𝑊2 , … , 𝑊𝑘. Example 6: Let F be a subfield of the field C of complex numbers. Suppose 𝛼1 = (1,2,0,3,0), 𝛼2 = (0,0,1,4,0), 𝛼3 = (0,0,0,0,1). By Theorem 2.3.1, a vector 𝛼 is in the subspace W of 𝐹 5 spanned by 𝛼1 , 𝛼2 , 𝛼3 if and only if there exist scalars 𝑐1 , 𝑐2 , 𝑐3 in F such that 𝛼 = 𝑐1 𝛼1 + 𝑐2 𝛼2 + 𝑐3 𝛼3. Thus, W consists of all vectors of the form 𝛼 = 𝑐1 , 2𝑐1 , 𝑐2 , 3𝑐1 + 4𝑐2 , 𝛼3 , where 𝑐1 , 𝑐2 , 𝑐3 are arbitrary scalars in F. Alternatively, W can be described as the set of all 5-tuples 𝛼 = (𝑥1 , 𝑥2 , 𝑥3 , 𝑥4 , 𝑥5 ) with 𝑥𝑖 in F such that 𝑥2 = 2𝑥1 and 𝑥4 = 3𝑥1 + 4𝑥3 Thus (− 3, − 6, 1, 5, 2) is in W, whereas (2, 4, 6, 7, 8) is not. 2.4 Linear Independence 2.4.1 Linearly Independent Vectors Definition 2.4.1: (a) Let V be a vector space over the field K, and let 𝑆 = {𝑣1 , … , 𝑣𝑛 } be a non-empty subset of containing vectors in V. The vectors 𝑣1 , 𝑣2 … , 𝑣𝑛 are linearly independent if, whenever there exists scalars 𝑐1 , … , 𝑐𝑛 satisfying 𝑐1 𝑣1 + 𝑐2 𝑣2 … + 𝑐𝑛 𝑣𝑛 = 0, then necessarily 𝑐1 = 𝑐2 = ⋯ = 𝑐𝑛 = 0𝑛. (b) The vectors 𝑣1 , 𝑣2 … , 𝑣𝑛 are spanning if, for every vector𝑣 ∈ 𝑉, we can find scalars 𝑐1 , 𝑐2 … , 𝑐𝑛 ∈ 𝐾 such that 𝑣 = 𝑐1 𝑣1 + 𝑐2 𝑣2 … + 𝑐𝑛 𝑣𝑛. In this case, we write 𝑉 = (𝑣1 , 𝑣2 … , 𝑣𝑛 ). (c) The vectors 𝑣1 , 𝑣2 … , 𝑣𝑛 form a basis for V if they are linearly independent and spanning. Remarks: Linear independence is a property of a list of vectors. A list containing the zero vector is never linearly independent. Also, a list in which the same vector occurs more than once is never linearly independent. 20 MTH 212 LINEAR ALGEBRA Definition 2.4.2: Let V be a vector space over the field K, then V is finite dimensional if vectors 𝑣1 , 𝑣2 … , 𝑣𝑛 ∈ 𝑉can be found to form a basis for V. Proposition: The following three conditions are equivalent for the vectors 𝑣1 , 𝑣2 … , 𝑣𝑛 of the vector space V over K: a) 𝑣1 , 𝑣2 … , 𝑣𝑛 is a basis; b) 𝑣1 , 𝑣2 … , 𝑣𝑛 is a maximal linearly independent set (that is, if we add any vector to the list, then the result is no longer linearly independent); c) 𝑣1 , 𝑣2 … , 𝑣𝑛 is a minimal spanning set (that is, if we remove any vector fromthe list, then the result is no longer spanning). 2.4.2 Properties of Linear Independence: Theorem 2.4.1 (The Exchange Lemma) Let V be a vector space over K. Suppose that the vectors 𝑣1 , 𝑣2 … , 𝑣𝑛 are linearly independent, and that the vectors 𝑤1 , 𝑤2 … , 𝑤𝑚 are linearly independent, where 𝑚 > 𝑛. Then we can find a number 𝑖 with 1 ≤ 𝑖 ≤ 𝑚 such that the vectors 𝑣1 , … , 𝑣𝑛 , 𝑤𝑖 are linearly independent. A lemma about systems of equations would be used to prove this theorem. Lemma: Given a system 𝑎11 𝑥1 + 𝑎12 𝑥2 + ⋯ + 𝑎1𝑚 𝑥𝑚 = 0 𝑎21 𝑥1 + 𝑎22 𝑥2 + ⋯ + 𝑎2𝑚 𝑥𝑚 = 0 ⋮............. (∗) ⋮ 𝑎𝑛1 𝑥1 + 𝑎𝑛2 𝑥2 + ⋯ + 𝑎𝑛𝑚 𝑥𝑚 = 0 of homogeneous linear equations, where the number n of equations is strictly less than the number m of variables, there exists a non-zero solution (𝑥1 , … , 𝑥𝑚 ) (that is,𝑥1 , … , 𝑥𝑚 are not all zero). Proof: This is proved by induction on the number of variables. If the coefficients 𝑎11 , 𝑎21 … , 𝑎𝑛1 of 𝑥1 are all zero, then putting 𝑥1 = 1and the other variables zero gives a solution. If one of these coefficients is non-zero, then we can use the corresponding equation to express 𝑥1 in terms of the other variables, obtaining (𝑛 − 1) equations in (𝑚 − 1) variables. By hypothesis, 𝑛 − 1 < 𝑚 − 1. So, by the induction hypothesis, these new equations have a non-zero solution. 21 MTH 212 LINEAR ALGEBRA Computing the value of 𝑥1 gives a solution to the original equations. Now we turn to the proof of the Exchange Lemma. Let us argue for a contradiction by assuming that the result is false; that is, assume that none of the vectors wi can be added to the list (𝑣1 , 𝑣2 … , 𝑣𝑛 ) to produce a larger linearly independent list, this means that, for all j, the list (𝑣1 , … , 𝑣𝑛 , 𝑤𝑖 ) is linearly dependent. So, there are coefficients 𝑐1 , … , 𝑐𝑛 , 𝑑, not all zero, such that 𝑐1 𝑣1 + ⋯ + 𝑐𝑛 𝑣𝑛 + 𝑑𝑤𝑖 = 0. We cannot have 𝑑 = 0; for this would mean that we had a linear combination of 𝑣1 , 𝑣2 … , 𝑣𝑛 equal to zero, contrary to the hypothesis that these vectors are linearly independent. So, we can divide the equation through by d, and take 𝑤𝑖 to the other side, to obtain (changing notation slightly) 𝑛 𝑤𝑖 = 𝑎1𝑖 𝑣1 + 𝑎2𝑖 𝑣2 + ⋯ + 𝑎𝑛𝑖 𝑣𝑛 = ∑ 𝑎𝑗𝑖 𝑣𝑗 (𝑖) 𝑗=1 We do this for each value of 𝑖 = 1,... , 𝑚. Now take a non-zero solution to the set of equations (𝑖) above: that is, 𝑚 ∑ 𝑎𝑗𝑖 𝑥𝑖 = 0; 𝑗 = 1,... , 𝑛 (𝑖𝑖) 𝑖=1 Multiplying the formula for 𝑤𝑖 by 𝑥𝑖 and adding, we obtain 𝑛 𝑚 𝑥1 𝑤1 + ⋯ + 𝑥𝑚 𝑤𝑚 = ∑ (∑ 𝑎𝑗𝑖 𝑥𝑖 ) 𝑣𝑗 = 0 (𝑖𝑖𝑖) 𝑗=1 𝑖=1 But the coefficients are not all zero, so this means that the vectors (𝑤1 , 𝑤2 … , 𝑤𝑚 ) are not linearly dependent, contrary to hypothesis. So, the assumption that no 𝑤𝑖 can be added to (𝑣1 , 𝑣2 … , 𝑣𝑛 ) in order to obtain a linearly independent set must be wrong, and the proof is complete. Definition 2.4.3: Let B be a subset of a set A. Then, B is said to be a maximal subset of A having property P if 1. B has property P 2. No proper superset of B in A has property P. Example 7: Let T = {2, 3, 4, 7, 8, 10, 12, 13, 14, 15}. Then, a maximal subset of T of consecutive integers is S = {2, 3, 4}. Other maximal subsets are {7, 8}, {10} and {12, 13, 14, 15}. Note that {12, 13} is not maximal. Why? 22 MTH 212 LINEAR ALGEBRA Definition 2.4.4: Let V be a vector space over F. Then, S is called a maximal linearly independent subset of V if 1. S is linearly independent and 2. no proper superset of 𝑆 ∈ 𝑉 is linearly independent. Example 8: a. In 𝑅3 , the set 𝑆 = {𝑒1 , 𝑒2 } is linearly independent but not maximal as 𝑆 ∪ {(1,0,0)𝑇 } is a linearly independent set containing S. b. In 𝑅3 , set 𝑆 = {(1,0,0)𝑇 , (1,1,0)𝑇 , (1,1, −1)𝑇 } is a maximal linearly independent set as S is linearly independent and any collection of four or more vectors from 𝑅3 is linearly dependent. c. Is the set {1, 𝑥, 𝑥 2 , … } a maximal linearly independent subset of 𝐶 [𝑥] over 𝐶 ? d. Is the set {1 ≤ 𝑖 ≤ 𝑚, 1 ≤ 𝑗 ≤ 𝑛} a maximal linearly independent subset of 𝑀𝑚,𝑛 (𝐶 ) over 𝐶? Theorem 2.4.2: Let 𝑉 be a vector space over 𝐹 and 𝑆 a linearly independent set in 𝑉. Then, 𝑆 is maximal linearly independent if and only if 𝐿𝑠(𝑆) = 𝑉. Proof: Let 𝑣 ∈ 𝑉. As 𝑆 is linearly independent, using Corollary, the set 𝑆 ∪ {𝑣} is linearly independent if and only if 𝑣 ∈ 𝑉\𝐿𝑠(𝑆). Thus, the required result follows. Let 𝑉 = 𝐿𝑠(𝑆) for some set S with |𝑆| = |𝑇|. Then, using the Theorem 2, we see that if 𝑇 ⊆ 𝑉 is linearly independent then |𝑇| ≤ 𝑘. Hence, a maximal linearly independent subset of V can have at most k vectors. Thus, we arrive at the following important result. Theorem 2.4.3: Let 𝑉 be a vector space over 𝐹 and let 𝑆 and 𝑇 be two finite maximal linearly independent subsets of V. Then, |𝑆| = |𝑇|. Proof: By Theorem 2, S and T are maximal linearly independent if and only if 𝐿𝑠(𝑆) = 𝑉 = 𝐿𝑠(𝑇). Now, use the previous paragraph to get the required result. Let V be a finite dimensional vector space. Then, by Theorem 3.4.6, the number of vectors in any two maximal linearly independent set is the same. We would now use this number to now define the dimension of a vector space. Definition 2.4.5: Let V be a finite dimensional vector space over F. Then, the number of vectors in any maximal linearly independent set is called the dimension of V, denoted𝑑𝑖𝑚(𝑉 ). By convention,𝑑𝑖𝑚({0}) = 0 23 MTH 212 LINEAR ALGEBRA Examples 9: a. As {1} is a maximal linearly independent subset of R, dim(R) = 1. b. As {𝑒1 , … , 𝑒𝑛 } is a maximal linearly independent subset in𝑅𝑛 , 𝑑𝑖𝑚(𝑅𝑛 ) = 𝑛. c. As {𝑒1 , … , 𝑒𝑛 } is a maximal linearly independent subset in 𝐶 𝑛 over C, 𝑑𝑖𝑚(𝐶 𝑛 ) = 𝑛. d. Using 9c, {𝑒1 , … , 𝑒𝑛 , 𝑖𝑒1 , … , 𝑖𝑒𝑛 } is a maximal linearly independent subset in 𝐶 𝑛 over 𝑅. Thus, as a real vector space, 𝑑𝑖𝑚(𝐶 𝑛 ) = 2𝑛. e. As {𝑒1𝑗 |1 ≤ 𝑖 ≤ 𝑚, 1 ≤ 𝑗 ≤ 𝑛} is a maximal linearly independent subset of 𝑀𝑚,𝑛 (𝐶 ) over C, 𝑅𝑛 , 𝑑𝑖𝑚 (𝑀𝑚,𝑛 (𝐶 )) = 𝑚𝑛. SELF-ASSESSMENT EXERCISE(S) 1. Use different sets of scalars to construct different vectors. You might build a few new linear combinations of 𝑢1 , 𝑢2 , 𝑢3 , 𝑢4 2. Let𝑆 = {(1,1,1,1)𝑇 , (1, −1,1,2)𝑇 , (1,1, −1,1)𝑇 } ⊆ 𝑅4. Does 𝑇 (1, −1,1,2) ∈ 𝐿𝑠(𝑆)? Determine conditions on x, y, z and u such that (𝑥, 𝑦, 𝑧, 𝑢)𝑇 ∈ 𝐿𝑠(𝑆) 3. Prove that if two vectors are linearly dependent, one of them is a scalar multiple of the other. 4. Are the vectors: 𝑎1 = (1,1,2,4), 𝑎2 = (2, −1, −5,2), 𝑎3 = (1, −1, −4,0), 𝑎4 = (2,1,1,6) linearly independent in 𝑅4 ? 5. Find a basis for the subspace of 𝑅4 spanned by the four vectors of Exercise 2 above. 6. Show that the vectors 𝑎1 = (1,0, −1), 𝑎2 = (1,2,1), 𝑎3 = (0, −3,2)form a basis for 𝑅3. Express each of the standard basis vectors as linear combinations of 𝑎1 , 𝑎2 and 𝑎3. 7. Find three vectors in 𝑅3 which are linearly dependent, and are such thatany two of them are linearly independent. 2.5 Summary The largest collection of vectors that can be obtained as linear combination of vectors is called ’linear span and that {1} is a maximal linearly independent subset of R, dim(R) = 1. Also, if V be a vector space over F and S a linearly independent set in V, then, S is maximal linearly independent if and only if Ls(S) = V. 24 MTH 212 LINEAR ALGEBRA The subspace spanned by 𝑆 is defined to be the intersection 𝑊 of all subspaces of 𝑉 which contain 𝑆. The vector 𝛼1 𝑢1 + 𝛼2 𝑢2 + 𝛼3 𝑢3 + ⋯ + 𝛼𝑛 𝑢𝑛 is the linear combination of n-vectors 𝑢1 , 𝑢2 , 𝑢3 , … , 𝑢𝑛 from the column vector 𝐶 𝑚 and n-scalars 𝛼1 , 𝛼2 , 𝛼3 , … , 𝛼𝑛 The number of vectors in any maximal linearly independent set of a finite dimensional vector space V over F is called the dimension of V. 2.6 References/Further Readings Robert A. Beezer (2014). A First Course in Linear Algebra. Congruent Press Gig Harbor, Washington, USA 3(40). Arbind K Lal Sukant Pati (2018). Linear Algebra through Matrices. Peter J. Cameron (2008). Notes on Linear Algebra. 25 MTH 212 LINEAR ALGEBRA UNIT 3 LINEAR TRANSFORMATIONS I Unit Structure 3.1 Introduction 3.2 Learning Outcomes 3.3 Linear Transformations 3.3.1 Spaces Associated with a Linear Transformation 3.3.2 The Range Space and the Kernel 3.3.3 Rank and Nullity 3.3.4 Some Types of Linear Transformations 3.4 Theorems of Vector Spaces 3.4.1 Isomorphism Theorems of Vector Spaces 3.4.2 Homomorphism Theorems of Vector Spaces 3.5 Summary 3.6 References/Further Reading 3.1 Introduction You have already learnt about vector space and several concepts related to it. In this unit we initiate the study of certain mappings between two vector spaces, called linear transformations. The importance of these mappings can be realized from the fact that, in the calculus of several variables, every continuously differentiable function can be replaced, to a first approximation, by a linear one. This fact is a reflection of a general principle that every problem on the change of some quantity under the action of several factors can be regarded, to a first approximation, as a linear problem. It often turns out that this gives an adequate result. Also, in physics it is important to know how vectors behave under a change of the coordinate system. This requires a study of linear transformations. In this unit we study linear transformations and their properties, as well as two spaces associated with a linear transformation and their properties, as well as two spaces associated with a linear transformation, and their dimensions. Then, we prove the existence of linear transformations with some specific properties, as discuss the notion of an isomorphism between two vector spaces, which allows us to say that all finite-dimensional vector spaces of the same dimension are the “same”, in a certain sense. Finally, we state and prove the Fundamental Theorem of Homomorphism and some of its corollaries, and apply them to various situations. 26 MTH 212 LINEAR ALGEBRA 3.2 Learning Outcomes By the end of this unit, you should be able to: Verify the linearity of certain mappings between vector spaces; Construct linear transformations with certain specified properties; Define the Range and the Kernel of Linear Transformation Calculate the rank and nullity of a linear operator; Prove and apply the Rank Nullity Theorem; Define an isomorphism between two vector spaces; Show that two vector spaces are isomorphic if and only if they have the same dimension; Prove and use the fundamental theorem of homomorphism. 3.3 Linear Transformations By now you are familiar with vector spaces 𝑅2 and 𝑅3. Now consider the mapping 𝑓: 𝑅2 → 𝑅3 |𝑓 (𝑥, 𝑦) = (𝑥, 𝑦, 0). 𝑓 is a well-defined function. Also notice that i. 𝑓((𝑎, 𝑏) + (𝑐, 𝑑 )) = 𝑓((𝑎 + 𝑐, 𝑏 + 𝑑 )) = (𝑎 + 𝑐, 𝑏 + 𝑑, 0) = (𝑎, 𝑏, 0) + (𝑐, 𝑑, 0)for (𝑎, 𝑏), (𝑐, 𝑑 ) ∈ 𝑅2 and ii. For any 𝛼 ∈ 𝑅 and (𝑎, 𝑏) ∈ 𝑅2 , 𝑓((𝛼𝑎, 𝛼𝑏)) = (𝛼𝑎, 𝛼𝑏, 0) = 𝛼𝑓((𝑎, 𝑏)). So, we have a function 𝑓 between two vector spaces such that (𝑖) and (𝑖𝑖) above hold true. i. says that the sum of two plane vectors is mapped under 𝑓 to the sum to sum of their images under 𝑓. ii. says that a line in the plane 𝑅2 is mapped under 𝑓 to a line in 𝑅2. Properties i) and ii) together say that 𝑓is linear, a term that we now define. Definition 3.3.1: Let U and V be vector spaces over a field F. A linear transformation (or linear operator) from U to V is a function𝑇: 𝑈 → 𝑉, such that 𝑳𝑻𝟏 : 𝑇(𝑢1 + 𝑢2 ) = 𝑇(𝑢1 ) + 𝑇(𝑢2 ), for 𝑢1 , 𝑢2 ∈ 𝑈 and 𝑳𝑻𝟐 :𝑇(𝛼𝑢) = 𝛼𝑇(𝑢) for 𝛼 ∈ 𝐹 and 𝑢 ∈ 𝑈. Conditions (𝑖) and (𝑖𝑖) above can be combined to give the following equivalent condition. 𝑳𝑻𝟑 : 𝑇(𝛼1 𝑢1 + 𝛼2 𝑢2 ) = 𝛼1 𝑇(𝑢1 ) + 𝛼2 𝑇(𝑢2 ), for 𝛼1 , 𝛼2 ∈ 𝐹𝑢1 , 𝑢2 ∈ 𝑈. 27 MTH 212 LINEAR ALGEBRA What we are saying is that [𝐿𝑇1 and 𝐿𝑇2 ] implies 𝐿𝑇3. This can be easily shown as follows: We will show that 𝐿𝑇3 → 𝐿𝑇1 ; 𝐿𝑇3 → 𝐿𝑇2. Now, 𝐿𝑇3 is true for all 𝛼1 , 𝛼2 ∈ 𝐹. Therefore, it is certainly true for𝛼1 = 𝛼2 , that is, 𝐿𝑇1 holds. Now, to show that𝐿𝑇2 is true, Consider 𝑇(𝛼𝑢) for any𝛼 ∈ 𝐹and𝑢 ∈ 𝑈. We have 𝑇(𝛼𝑢) = 𝑇(𝛼𝑢 + 0 ⋅ 𝑢) = 𝛼𝑇(𝑢) + 0 ⋅ 𝑇(𝑢) = 𝛼𝑇(𝑢), thus proving that 𝐿𝑇2 holds. You can try and prove the converse now, that is, what the following exercise is all about! E1) Show that the conditions 𝐿𝑇1 and 𝐿𝑇2 together imply 𝐿𝑇3. Before going further, let us note two properties of any linear transformation, 𝑇: 𝑈 → 𝑉, which follow from 𝐿𝑇1 (or 𝐿𝑇2 or 𝐿𝑇3 ). 𝑳𝑻𝟒 :𝑇(0) = 0. Let’s see why this is true. Since (0) = 𝑇(0 + 0) = 𝑇(0) + 𝑇(0) , by 𝐿𝑇1 , we subtract 𝑇(0) from both sides to get 𝑇(0) = 0. 𝑳𝑻𝟓 :𝑇(−𝑢) = −𝑇(𝑢) for all 𝑢 ∈ 𝑈. Why is this so? Well, since, 0 = 𝑇(0) = 𝑇(𝑢 – 𝑢) = 𝑇(𝑢) + 𝑇(−𝑢); We have 𝑇(−𝑢) = −𝑇(𝑢). E2) Can you show how 𝐿𝑇4 and 𝐿𝑇5 will follow from 𝐿𝑇2 ? Now let us look at some common linear transformations. Example 1: Consider the vector space U over a field F, and the function 𝑇: 𝑈 → 𝑉, defined by 𝑇(𝑢) = 𝑢 for all 𝑢 ∈ 𝑈. Show that T is a linear transformation. (This transformation is called the identity transformation, and is denoted by 𝐼𝑢, or just𝐼, if the underlying vector space is understood). Solution: For any 𝛼, 𝛽 ∈ 𝐹 and 𝑢1 , 𝑢2 ∈ 𝑈, we have 𝑇(𝛼𝑢1 + 𝛽𝑢2 ) = 𝛼𝑢1 + 𝛽𝑢2 = 𝛼𝑇(𝑢1 ) + 𝛽𝑇 (𝑢2 ) Hence, 𝐿𝑇3 holds, and T is a linear transformation. Example 2: Let 𝑇: 𝑈 → 𝑉be defined by 𝑇(𝑢) = 0for all 𝑢 ∈ 𝑈. Check that T is a linear transformation. (It is called the Null or Zero Transformation, and is denoted by 0). Solution: For any 𝛼, 𝛽 ∈ 𝐹 and 𝑢1 , 𝑢2 ∈ 𝑈, we have 𝑇(𝛼𝑢1 + 𝛽𝑢2 ) = 0 = 𝛼 ∙ 0 + 𝛽 ∙ 0 Therefore, T is linear transformation. Example 3: Consider the function 𝑃𝑟1 : 𝑅𝑛 → 𝑅, defined by 𝑃𝑟1 [(𝑥1 , … , 𝑥𝑛 )] = 𝑥𝑖. Show that this is a linear transformation. (This is called the projection on the first coordinate). 28 MTH 212 LINEAR ALGEBRA Similarly, we can define 𝑃𝑟𝑖 : 𝑅𝑛 → 𝑅 by 𝑃𝑟𝑖 [(𝑥1 , … , 𝑥𝑖−1 , 𝑥𝑖 … , 𝑥𝑛 )] = 𝑥𝑖 to be the projection on the 𝑖𝑡ℎ coordinate for 𝑖 = 2, … , 𝑛. For instance, 𝑃𝑟2 : 𝑅3 → 𝑅 by 𝑃𝑟2 [(𝑥, 𝑦, 𝑧)] = 𝑦 Solution: We will use 𝐿𝑇3 to show that projection is a linear operator. For 𝛼, 𝛽 ∈ 𝑅 and (𝑥1 , … … … 𝑥𝑛 ), (𝑦1 , … … … 𝑦𝑛 ) ∈ 𝑅𝑛, we have P𝑃𝑟1 [∝ (𝑥1 , … … , 𝑥𝑛 ) + 𝛽(𝑦1 , … … , 𝑦𝑛 )] = 𝑃𝑟1 (𝛼𝑥1 + 𝛽𝑦1 , 𝛼𝑥2 + 𝛽𝑦2 , … … , 𝛼𝑥𝑛 + 𝛽𝑦𝑛 ) = 𝛼𝑥1 + 𝛽𝑦1 = 𝛼𝑃𝑟1 [𝑥1 , ⋯ ⋯ , 𝑥𝑛 ] + 𝛽𝑃𝑟1 [(𝑦1 , ⋯ ⋯ , 𝑦𝑛 )] Thus 𝑃𝑟1 (and similarly 𝑃𝑟𝑖 ) is a linear transformation. Before going to the next example, we make a remark about projections. Remark: Consider the function𝑃: 𝑅3 → 𝑅2 : 𝑃(𝑥, 𝑦, 𝑧) = (𝑥, 𝑦), this is a projection from 𝑅3 on to the xy-plane. Similarly, the functions f and g, from𝑅3 → 𝑅2 , defined by 𝑓 (𝑥, 𝑦, 𝑧) = (𝑥, 𝑧), and 𝑔(𝑥, 𝑦, 𝑧) = (𝑦, 𝑧)are projections from 𝑅3 onto the 𝑥𝑧-plane and the 𝑦𝑧-plane, respectively. In general, any function 𝜃: 𝑅𝑛 → 𝑅𝑚 (𝑛 > 𝑚), which is defined by dropping any (𝑛 – 𝑚) coordinate, is a projection map. Now let us see another example of a linear transformation that is very geometric in nature. Example 4: Let 𝑇: 𝑅 2 → 𝑅2 be defined by 𝑇(𝑥, 𝑦) = (𝑥, −𝑦) ∀ 𝑥, 𝑦 ∈ 𝑅. Show that T is a linear transformation. (This is the reflection in the x-axis that we show in Fig. 2). y P(2,1) 0 x Q(-2,1) Fig 2: Q is the reflection of P in the X-axis. Solution: For 𝛼, 𝛽 ∈ 𝑅 and (𝑥1 , 𝑦1 ), (𝑥2 , 𝑦2 ) ∈ 𝑅2 , we have 𝑇[𝛼(𝑥1 , 𝑦1 ) + 𝛽(𝑥2 , 𝑦2 )] = 𝑇(𝛼𝑥1 + 𝛽𝑥1 , 𝛼𝑦1 + 𝛽𝑦2 ) = (𝛼𝑥1 + 𝛽𝑥2 , −𝛼𝑦1 − 𝛽𝑦2 ) = 𝛼(𝑥1 , −𝑦1 ) + 𝛽 (𝑥2 , −𝑦2 ) = 𝛼𝑇(𝑥1 , 𝑦1 ) + 𝛽𝑇 (𝑥2 , 𝑦2 ) Therefore, T is a linear transformation. So, far we have given examples of linear transformations. Now, we give an example of a very important function which is not linear. This example’s importance lies in its geometric applications. 29 MTH 212 LINEAR ALGEBRA Example 5: Let 𝑢0 be a fixed non-zero vector in U. Define 𝑇: 𝑈 → 𝑈by 𝑇(𝑢) = 𝑢 + 𝑢0 , ∀𝑢 ∈ 𝑈.Show that T is not a linear transformation. (T is called the translation by𝑢0. See Fig 3 for a geometrical view). Solution: T is not a linear transformation since 𝐿𝑇4 does not hold. This is because 𝑇(0) = 𝑢0 ≠ 0 y 4 𝐴′ 𝐵′ B 3 A 2 𝐶′ 𝐷′ 1 C D 0 1 2 3 4 x Fig. 3: 𝑨′𝑩′𝑪′𝑫′ is the transformation of ABCD by (1,1). Now, try the following Exercises. E3)Let 𝑇: 𝑅2 → 𝑅2 be the reflection in the y-axis. Find an expression for T as in Example 4. Is T a linear operator? E4) For a fixed vector(𝑎1, 𝑎2 , 𝑎3 )in𝑅3 , define the mapping 𝑇: 𝑅3 → 𝑅 by 𝑇(𝑥1, 𝑥2 , 𝑥3 ) = a1 𝑥1 + 𝑎2 𝑥2 + 𝑎3 𝑥3. Show that T is a linear transformation. Note that 𝑇(𝑥1, 𝑥2 , 𝑥3 ) is the dot product of (𝑥1, 𝑥2 , 𝑥3 ) and (𝑎1, 𝑎2 , 𝑎3 ). E5) Show that the map 𝑇: 𝑅3 → 𝑅3 defined by 𝑇(𝑎1, 𝑎2 , 𝑎3 ) = (𝑥1 + 𝑥2 − 𝑥3 , 2𝑥1 − 𝑥2 , 𝑥2 + 2𝑥3 ) is a linear operator. Let us consider the real vector space 𝑃𝑛 of all polynomials of degree less than or equal to n. E6) Let 𝑓 ∈ 𝑃𝑛 be given by 𝐹 (𝑥) = 𝛼0 + 𝛼1 𝑥 + 𝑎2 𝑥 2 … … + 𝛼𝑛 𝑥 𝑛 , 𝛼𝑖 ∈ 𝑅 ∀𝑖. We define (𝐷𝑓)(𝑥) = 𝛼1 + 2𝛼2 𝑥 + ⋯ … + 𝑛𝛼𝑛 𝑥 𝑛−1. Show that 𝐷: 𝑃𝑛 is a linear transformation. (Observe that 𝐷𝑓 is nothing but the derivative of f and D is called the differentiation operator). There is also the concept of a quotient space. We now define a very useful linear transformation, using this concept. Example 6: Let W be a subspace of a vector space U over a field F. W gives rise to the quotient space U/W. Consider the map 𝑇: 𝑈 → 𝑈/𝑊defined by 𝑇(𝑢) = 𝑢 + 𝑊. Show that T is a linear transformation. Solution: For any 𝛼, 𝛽 ∈ 𝐹 and 𝑢1 , 𝑢2 ∈ 𝑈, we have 𝑇(𝛼𝑢1 + 𝛽𝑢2 ) = 𝛼𝑢1 + 𝛽𝑢2 + 𝑊 = (𝛼𝑢1 + 𝑊 ) + (𝛽𝑢2 + 𝑊 ) = 𝛼(𝑢1 + 𝑊 ) + 𝛽 (𝑢2 + 𝑊 ) 30 MTH 212 LINEAR ALGEBRA = 𝛼𝑇(𝑢1 ) + 𝛽𝑇 (𝑢2 ) Thus, T is a linear transformation. Now solve the following exercise. Example 7: Let 𝑢1 = (1, – 1), 𝑢2 = (2, −1), 𝑢3 = (4, – 3), 𝑣1 = (1,0), 𝑣2 = (0, 1) and𝑣3 = (1, 1) be six vectors in 𝑅2. Can you define a linear transformation 𝑇: 𝑅2 → 𝑅2 such that(𝑢1 ) = 𝑣𝑖 , 𝑖 = 1,2,3 ? (Hint: Note that 2𝑢1 + 𝑢2 = 𝑢3 and 𝑣1 + 𝑣2 = 𝑣3 ). You have already seen that a linear transformation 𝑇: 𝑈 → 𝑉 must satisfy 𝑇(𝛼1 𝑢1 + 𝛼2 𝑢2 ) = 𝛼1 𝑇(𝑢1 ) + 𝛼2 𝑇(𝑢2 ), for 𝛼1 , 𝛼2 ∈ 𝐹 and 𝑢1 , 𝑢2 ∈ 𝑈. More generally, we can show that, 𝑳𝑻𝟔: 𝑻(𝜶𝟏 𝒖𝟏 + ⋯ + 𝜶𝒏 𝒖𝒏 ) = 𝜶𝟏 𝑻(𝒖𝟏 ) + ⋯ + 𝜶𝒏 𝑻(𝒖𝒏 );where 𝛼 ∈ 𝐹 and 𝑢𝑖 ∈ 𝑈. This shall be shown by induction, that is, we assume the above relation for 𝑛 = 𝑚, and prove it for 𝑚 + 1. Now, 𝑇(𝛼1 𝑢1 + ⋯ 𝛼𝑚 𝑢𝑚 + 𝛼𝑚+1 𝑢𝑚+1 ) = 𝑇(𝑢 + 𝛼𝑚+1 𝑢𝑚+1 ) where,𝑢 = 𝛼1 𝑢1 + ⋯ + 𝛼𝑚 𝑢𝑚 = 𝑇(𝑢) + 𝛼𝑚+1 𝑇(𝑢𝑚+1 ), Since the result holds for 𝑛 = 2 = 𝑇(𝛼1 𝑢1 + ⋯ 𝛼𝑚 𝑢𝑚 ) + 𝛼𝑚+1 𝑇(𝑢𝑚+1 ) = 𝛼1 𝑇(𝑢1 ) + ⋯ + 𝑇(𝛼𝑚 )𝑇(𝑢𝑚 ) + 𝛼𝑚+1 𝑇(𝑢𝑚+1 ) Since we have assumed the result for 𝑛 = 𝑚. Thus, the result is true for 𝑛 = 𝑚 + 1. Hence, by induction, it holds true for all n. Let us now come to a very important property of any linear transformation 𝑇: 𝑈 → 𝑉. In the earlier unit, we mentioned that every vector space has a basis. Thus, U has a basis. We will now show that T is completely determined by its values on a basis of U. More precisely, we have: Theorem 3.1: Let S and T be two linear transformations from U to V, where, 𝑑𝑖𝑚1 𝑈 = 𝑛. Let (𝑒1 , … … , 𝑒𝑛 ) be a basis of U. Suppose 𝑆(𝑒𝑖 )for 𝑖 = 1, … , 𝑛. Then, 𝑆(𝑢) = 𝑇(𝑢) for all 𝑢 ∈ 𝑈. Proof: Let 𝑢 ∈ 𝑈. Since (𝑒1 , … … , 𝑒𝑛 )is a basis of U, u can be uniquely written as: 𝑢 = 𝛼1 𝑒1 + … … + 𝛼𝑛 𝑒𝑛 , where the 𝛼𝑖 are scalars. Then, 𝑆(𝑢) = 𝑆(𝛼1 𝑒1 + … … + 𝛼𝑛 𝑒𝑛 ) = 𝛼1 𝑆(𝑒1 ) + … … + 𝛼𝑛 𝑆(𝑒𝑛 ) by LT6 = 𝛼1 𝑇(𝑒1 ) + … … + 𝛼𝑛 𝑇(𝑒𝑛 ) = 𝛼1 (𝛼1 𝑒1 + … … + 𝛼𝑛 𝑒𝑛 ) by LT6 = 𝑇(𝑢). 31 MTH 212 LINEAR ALGEBRA What we have just proved is that once we know the values of T on a basis of U, then we can find 𝑇(𝑢) for any 𝑢 ∈ 𝑈. Note: Theorem 3.1 is true even when U is not finite -dimensional. The proof, in this case, is on the same lines as above. Let us see how the idea of Theorem 3.1 helps us to prove the following useful result. Theorem 3.2: Let V be a real vector space and 𝑇: 𝑅 → 𝑉 be a linear transformation. Then there exists 𝑣 ∈ 𝑉 such that 𝑇(𝛼 ) = 𝛼𝑣 , ∀𝛼 ∈ 𝑅. Proof: A basis for R is (1). Let 𝑇(1) = 𝑉 ∈ 𝑉, then, for any 𝛼 ∈ 𝑅, 𝑇(𝛼) = 𝛼𝑇(1) = 𝛼𝑣; 𝑇(𝛼) is a vector space of dimension one, whose basis is [T(1)]. Now try the following exercise, for which you will need Theorem 3.1. E8) We define a linear operator 𝑇: 𝑅2 → 𝑅2 : 𝑇(1,0) = (0,1) and 𝑇(0,5) = (1,0). What is i) 𝑇(3,5)and ii)𝑇(5,3)? Now we shall prove a very useful theorem about linear transformations, which is linked to Theorem 3.1 Theorem 3.3: Let (𝑒1 …. , 𝑒𝑛 ) be a basis of U and let 𝑣1 …. , 𝑣𝑛 be any n vectors in V. Then there exists one and only one linear transformation 𝑇: 𝑈 → 𝑉 such that 𝑇(𝑒1 ) = 𝑣1 ; 𝑖 = 1, … , 𝑛. Proof: Let 𝑢 ∈ 𝑈. Then u can be uniquely written as 𝑢 = 𝛼1 𝑒1 + … … + 𝑇𝛼𝑛 𝑒𝑛. Define 𝑇(𝑢) = 𝛼1 𝑣1 + …. + 𝛼𝑛 𝑣𝑛 T defines a mapping from U to V such that 𝑇(𝑒1 ) = 𝑣1 for all 𝑖 = 1, … , 𝑛 Let us now show that T is linear, Let a, b be scalars and 𝑢, 𝑢′ ∈ 𝑈. Then there exist scalars 𝛼1 , … , 𝛼𝑛 , 𝛽1 , … , 𝛽𝑛 such that 𝑢 = 𝛼1 𝑒1 + ⋯ + 𝛼𝑛 𝑒𝑛 and 𝑢′ = 𝛽1 𝑒1 + ⋯ + 𝛽𝑛 𝑒𝑛 Then, 𝑎𝑢 + 𝑏𝑢′ = (𝑎𝛼1 + 𝑏𝛽1 )𝑒1 + ⋯ + (𝑎𝛼𝑛 + 𝑏𝛽𝑛 )𝑒𝑛 Hence, 𝑇(𝑎𝑢 + 𝑏𝑢′) = (𝑎𝛼1 + 𝑏𝛽1 )𝑣1 + ⋯ + (𝑎𝛼𝑛 + 𝑏𝛽𝑛 )𝑣𝑛 = 𝑎(𝛼1 𝑣1 + ⋯ + 𝛼𝑛 𝑣𝑛 ) + 𝑏(𝛽1 𝑣1 + ⋯ + 𝛽𝑛 𝑣𝑛 ) = 𝑎𝑇(𝑢) + 𝑏𝑇(𝑢′) Therefore, T is a linear transformation with the property that 𝑇(𝑒𝑖 ) = 𝑣𝑖 for all 𝑖 Theorem 3.1 now implies that T is the only linear transformation with the above properties. Let us now see how Theorem 3.3 can be used. 32 MTH 212 LINEAR ALGEBRA Example 8: 𝑒1 = (1,0,0), 𝑒2 = (0,1,0) and 𝑒1 = (0,0,1) form the standard basis of 𝑅3. Let (1,2), (2,3) and (3,4) be three vectors in 𝑅2. Obtain the linear transformation 𝑇: 𝑅3 → 𝑅2 such that 𝑇(𝑒1 ) = (1,2), T(𝑒2 ) = (2,3) and T(𝑒3 ) = (3,4). Solution: By Theorem 3.3, we know that 𝑇: 𝑅3 → 𝑅2 such that T(𝑒1 ) = (1,2), T(𝑒2 ) = (2,3), and T(𝑒3 ) = (3,4). We want to know what 𝑇(𝑥) is, for any 𝑥 = (𝑥1 , 𝑥2 , 𝑥3 ) ∈ 𝑅3. Now, 𝑋 = 𝑥1 𝑒1 + 𝑥2 𝑒2 + 𝑥3 𝑒3 Hence, 𝑇(𝑋 ) = 𝑥1 𝑇(𝑒1 ) + 𝑥2 𝑇(𝑒2 ) + 𝑥3 𝑇(𝑒3 ) = 𝑥1 (1,2) + 𝑥2 (2,3) + 𝑥3 (3,4) = (𝑥1 +2𝑥2 + 3𝑥3 , 2𝑥1 + 3𝑥2 + 4𝑥3 ) Therefore, 𝑇(𝑥1 , 𝑥2 , 𝑥3 ) = (𝑥1 +2𝑥2 + 3𝑥3 , 2𝑥1 + 3𝑥2 + 4𝑥3 ) is the definition of the linear transformation T. E9) Consider the complex field ℂ. It is a vector space over ℝ, a) What is its dimension over ℝ? Give a basis of ℂover ℝ. b) Let 𝛼, 𝛽 ∈ 𝑅. Give the linear transformation which maps the basis elements of ℂ obtained in (a), onto α and 𝛽, respectively. Let us now look at some vector spaces that are related to a linear operator. 3.3.1 Spaces Associated with a Linear Transformation In Unit 1, you found that given any function, there is a set associated with it, namely, it’s range. We will now consider two sets which are associated with any linear transformation, T. These are the range and the kernel of T. 3.3.1.1 The Range Space and the Kernel Let U and V be vector spaces over a field F. Let 𝑇: 𝑈 → 𝑉be a linear transformation. We shall define the range of T as well as the Kernel of T. At first, you will see them as sets. We will prove that these sets are also vector spaces over F. Definition 3.3.2: The range of T, denoted by 𝑹(𝑻), is the set {𝑇(𝑥): 𝑥 ∈ 𝑈} such that the kernel (or null space) of T denoted by Ker T, is the set {𝑥 ∈ 𝑈: 𝑇(𝑥) = 0}. Note that 𝑅(𝑇) ⊆ 𝑉and 𝐾𝑒𝑟 𝑇 ⊆ 𝑈. To clarify these concepts, consider the following examples: Example 9: Let 𝐼: 𝑈 → 𝑉be the identity transformation (see Example1). Find 𝑅(𝐼) and 𝐾𝑒𝑟 𝐼. 33 MTH 212 LINEAR ALGEBRA Solution: 𝑅(𝐼) = {𝐼 (𝑣 ): 𝑣 ∈ 𝑉 } = {𝑣: 𝑣 ∈ 𝑉 } = 𝑉. Also, 𝐾𝑒𝑟 𝐼 = {𝑣 ∈ 𝑉: 𝐼 (𝑣 ) = 0} = {𝑣 ∈ 𝑉: 𝑣 = 0} = {0} Example 10: Let 𝑇: 𝑅3 → 𝑅 be defined by (𝑥1 , 𝑥2 , 𝑥3 ) = 3𝑥1 + 𝑥2 + 2𝑥3. Find 𝑅(𝑇) and 𝐾𝑒𝑟 𝑇. Solution: 𝑅(𝑇) = {𝑥 ∈ 𝑅: 𝑥1 , 𝑥2 , 𝑥3 ∈ 𝑅with 3𝑥1 + 𝑥2 + 2𝑥3 = 𝑥} For example, 0 ∈ 𝑅 (𝑇), Since 0 = 3.0 + 0 + 2.0 = 𝑇(0,0,0) Also, 𝐼 ∈ 𝑅(𝑇), since 𝐼 = 3 ∙ 13 + 0 + 2 ∙ 0 = 𝑇(13, 0,0), or 𝐼 = 3 ∙ 0 + 1 + 2 ∙ 0 = 𝑇(0,1,0) or 𝐼 = 3.0 + 1 + 2.0 = 𝑇(0, 1, 0); or 𝐼 = 1 1 𝑇(0, 0, ½) 𝑜𝑟 𝐼 = 𝑇 ( , , 0). 6 2 Now can you see that R(T) is the whole real line R? This is because, for any 𝛼 ∈ 𝑅, 𝛼 = 𝛼 ∙ 1 = 𝛼𝑇(𝛼3, 0,0) = 𝑇(𝛼3, 0,0) ∈ 𝑅(𝑇) 𝐾𝑒𝑟 𝑇 = {(𝑥1 , 𝑥2 , 𝑥3 ) ∈ 𝑅3 : 3𝑥1 + 𝑥2 + 2𝑥3 = 0} For example, (0,0,0) ∈ ker 𝑇, but(1, 0, 0) ∉ 𝐾𝑒𝑟 𝑇. Therefore, 𝐾𝑒𝑟 𝑇 ≠ 𝑅3. In fact, Ker T is the plane 3𝑥1 + 𝑥2 + 2𝑥3 = 0 in 𝑅3. Example 11: Let𝑇: 𝑅3 → 𝑅3 be defined by 𝑇(𝑥1 , 𝑥2 , 𝑥3 ) = 𝑥1 − 𝑥2 + 2𝑥3 , 2𝑥1 + 𝑥2 , −𝑥1 − 2𝑥2 + 2𝑥3 Find 𝑅(𝑇) and 𝐾𝑒𝑟 𝑇. Solution: To find 𝑅(𝑇), we must find conditions on 𝑦1 , 𝑦2 , 𝑦3 ∈ 𝑅so that (𝑦1 , 𝑦2 , 𝑦3 ) ∈ 𝑅(𝑇), i.e., we must find some (𝑥1 , 𝑥2 , 𝑥3 ) ∈ 𝑅3 so that (𝑦1 , 𝑦2 , 𝑦3 ) = 𝑇(𝑥1 , 𝑥2 , 𝑥3 ) = (𝑥1 − 𝑥 + 2𝑥3 , 2𝑥1 + 𝑥2 , 𝑥1 − 2𝑥2 + 2𝑥3 ) This means 𝑥1 − 𝑥2 + 2𝑥3 = 𝑦1 ……………. (1) 2𝑥1 + 𝑥2 = 𝑦2 ……………. (2) −𝑥1 − 2𝑥2 + 2𝑥3 = 𝑦3 ……………. (3) Subtracting 2 times (1) from (2) and adding (1) and (3) to obtain 3𝑥2 − 4𝑥3 = 𝑦2 − 2𝑦1 ………………… (4) −3𝑥2 + 4𝑥3 = 𝑦1 + 𝑦3 ………………… (5) Adding Equations (4) and (5) we get 𝑦2 − 2𝑦1 + 𝑦1 + 𝑦3 = 0, that is, 𝑦2 + 𝑦3 = 𝑦1 Thus, (𝑦1 , 𝑦2 , 𝑦3 ) ∈ 𝑅(𝑇) ⇒ 𝑦2 + 𝑦3 = 𝑦1. On the other hand, if 𝑦2 + 𝑦3 = 𝑦1 , we can choose 𝑦 −2𝑦 𝑦 −2𝑦 𝑥3 = 0 ; 𝑥2 = 2 1 ; 𝑥1 = 𝑦1 + 2 1 3 3 Then, we see that 𝑇(𝑥1 , 𝑥2 , 𝑥3 ) = (𝑦1 , 𝑦2 , 𝑦3 ) Thus, 𝑦2 + 𝑦3 = 𝑦1 ⇒ (𝑦1 , 𝑦2 , 𝑦3 ) ∈ 𝑅(𝑇) Hence, 𝑅(𝑇) = {(𝑦1 , 𝑦2 , 𝑦3 ) ∈ 𝑅3 : 𝑦2 + 𝑦3 = 𝑦1 } Now (𝑥1 , 𝑥2 , 𝑥3 ) ∈ 𝐾𝑒𝑟 𝑇 if and only if the following equations are true: 𝑥1 − 𝑥2 + 2𝑥3 = 𝑦1 34 MTH 212 LINEAR ALGEBRA 2𝑥1 + 𝑥2 = 𝑦2 gives −𝑥1 − 2𝑥2 + 2𝑥3 = 𝑦3 + 2𝑥3 = 0 2𝑥1 + 𝑥2 = 0 −𝑥1 − 2𝑥2 + 2𝑥3 = 0 Of course, 𝑥1 = 0, 𝑥2 = 0, 𝑥3 = 0 is a solution. Are there other solutions? To answer this, we proceed as in the first part of this example. We see that 3𝑥2 + 4𝑥3 = 0 ⇒⇒ 𝑥3 = 34𝑥2 Also, 2𝑥1 + 𝑥2 = 0 ⇒⇒ 𝑥1 = −12𝑥2 Thus, we can give arbitrary values to 𝑥2 and calculate 𝑥1 and 𝑥3 in terms of 𝑥2. Therefore, 𝐾𝑒𝑟𝑇 = {(−𝛼2, 𝛼, (34)𝛼): 𝛼 ∈ 𝑅}. In this example, we see that finding 𝑅(𝑇) and 𝐾𝑒𝑟 𝑇 amounts to solving a system of equations. In subsequent unit, you will learn a systematic way of solving a system of linear equations by the use of matrices and determinants. The following exercises will help you in getting used to 𝑅(𝑇) and 𝐾𝑒𝑟 𝑇. E10) Let 𝑇 be the zero-transformation given in Example 2. Find 𝐾𝑒𝑟𝑇 and 𝑅(𝑇). Does 𝐼 ∈ 𝑅(𝑇)? E11) Find 𝑅(𝑇) and 𝐾𝑒𝑟𝑇 for each of the following operators: a) 𝑇: 𝑅3 → 𝑅2 : 𝑇(𝑥, 𝑦, 𝑧) = (𝑥, 𝑦) b) 𝑇: 𝑅3 → 𝑅: 𝑇(𝑥, 𝑦, 𝑧) = 𝑧 c) 𝑇: 𝑅3 → 𝑅 3 : 𝑇(𝑥1 , 𝑥2 , 𝑥3 ) = (𝑥1 + 𝑥2 + 𝑥3 , 𝑥1 + 𝑥2 + 𝑥3 , 𝑥1 + 𝑥2 + 𝑥3 )). (Note that the operators in (a) and (b) are projections onto the 𝑥𝑦-plane and the 𝑧-axis, respectively). Now that you are familiar with the sets 𝑅(𝑇) and 𝐾𝑒𝑟𝑇, we will prove that they are vector spaces. Theorem 3.4: Let U and V be vector spaces over a field F. Let 𝑇: 𝑈 → 𝑉 be a linear transformation. Then Ker T is a subspace of U and R(T) is a subspace of V. Proof: Let 𝑥1 , 𝑥2 ∈ 𝐾𝑒𝑟𝑇 ⊆ 𝑈 and 𝛼1 , 𝛼2 ∈ 𝐹. Now, by definition, 𝑇(𝑥1 ) = 𝑇(𝑥2 ) = 0 Therefore, 𝛼1 𝑇(𝑥1 ) + 𝛼2 𝑇(𝑥2 ) = 0 But 𝛼1 𝑇(𝑥1 ) + 𝛼2 𝑇(𝑥2 ) = 𝑇(𝛼1 𝑥1 + 𝛼2 𝑥2 ) Hence, 𝑇(𝛼1 𝑥1 + 𝛼2 𝑥2 ) = 0 This means that 𝛼1 𝑥1 + 𝛼2 𝑥2 ∈ 𝐾𝑒𝑟𝑇. Thus, by Theorem 3.2.3 of Unit 1, KerT is a subspace of U. 35 MTH 212 LINEAR ALGEBRA Let 𝑦1 , 𝑦2 ∈ 𝑅(𝑇) ⊆ 𝑉, and 𝛼1 , 𝛼2 ∈ 𝐹, then, by definition of R(T), there exist 𝑥1 , 𝑥2 ∈ 𝑈 such that 𝑇(𝑥1 ) = 𝑦1 and 𝑇(𝑥2 ) = 𝑦2 So, 𝛼1 𝑦1 + 𝛼2 𝑦2 = 𝛼1 𝑇(𝑥1 ) + 𝛼2 𝑇(𝑥2 ) = 𝑇(𝛼1 𝑥1 + 𝛼2 𝑥2 ) Therefore, 𝛼1 𝑦1 + 𝛼2 𝑦2 ∈ 𝑅(𝑇), which proves that 𝑅(𝑇) is a subspace of 𝑉. Now that we have proved that 𝑅(𝑇) and 𝐾𝑒𝑟 𝑇 are vector spaces, you know, from Unit 1, that they must have a dimension. We shall study these dimensions now. 3.3.1.2 Rank and Nullity Consider any linear transformation, 𝑇: 𝑈 → 𝑉, assuming that dim 𝑈 is finite. Then KerT, being a subspace of 𝑈, has finite dimension and dim(𝐾𝑒𝑟𝑇) ≤ dim 𝑈. Also note that 𝑅(𝑇) = 𝑇(𝑈), the image of U under T, a fact you will need to use in solving the following exercise. E12) Let {𝑒1 , ⋯ , 𝑒𝑛 } be a basis of U. Show that 𝑅(𝑇) is generated by {𝑇(𝑒1 ), … , 𝑇(𝑒1 )}. From E12), it is clear that, if dim 𝑈 = 𝑁, then dim 𝑅(𝑇) ≤ 𝑛. Thus, 𝑑𝑖𝑚 𝑅(𝑇) is finite, and the following definition is meaningful. Definition 3.3.3: The rank of T is defined to be the dimension of R(T), the range space of T. The nullity of T is defined to be the dimension of Ker T, the kernel (or the null space) of T. Thus, 𝒓𝒂𝒏𝒌 (𝑻) = 𝑑𝑖𝑚 𝑅(𝑇) and 𝑛𝑢𝑙𝑙𝑖𝑡𝑦 (𝑇) = 𝑑𝑖𝑚 𝐾𝑒𝑟 𝑇. We have already seen that 𝑟𝑎𝑛𝑘 (𝑇) ≤ 𝑑𝑖𝑚 𝑈and 𝑛𝑢𝑙𝑙𝑖𝑡𝑦 (𝑇) ≤ dim 𝑈. Example 12: Let 𝑇: 𝑈 → 𝑉be the zero-transformation given in example 2. What are the rank and nullity of T? Solution: In Exercise 11, you saw that R(T) = (0) and Ker T = U, Therefore, rank (T) = 0 and nullity (T) = dim 𝑈. Note that rank (T) + nullity (T) = dim 𝑈, in this case. E13) If T is the identity operator on V, find rank (T) and nullity (T). E14) Let D be the differentiation operator in E6). Give a basis for the range space of D and for Ker D. What are rank (D) and nullity (D)? In the above example and exercises you will find that for 𝑇: 𝑈 → 𝑉, then rank (T) + nullity (T) = dim 𝑈. In fact, this is the most important result about rank and nullity of a linear operator. We shall now state and prove this result. 36 MTH 212 LINEAR ALGEBRA Theorem 3.5: Let U and V be vector spaces over a field F and dim 𝑈 = 𝑛 andlet𝑇: 𝑈 → 𝑉be a linear operator. Then rank (T) + nullity (T) = n. Proof: Let 𝑛𝑢𝑙𝑙𝑖𝑡𝑦 (𝑇) = 𝑚, that is, 𝑑𝑖𝑚 𝐾𝑒𝑟 𝑇 = 𝑚, Let {𝑒1 , ⋯ , 𝑒𝑚 }be a basis of Ker T. We know that Ker T is a subspace of U, thus, by a theorem in Unit 1, we can extend this basis to obtain a basis(𝑒1 , ⋯ , 𝑒𝑚 , 𝑒𝑚+1 , ⋯ 𝑒𝑛 )of U. We shall show that {T(𝑒𝑚+1 ), … … , T(𝑒𝑛 ) }is a basis of R(T). Then, our result will follow because dim R(T) will be n – m = n – nullity (T). Let us first prove that {T(𝑒𝑚+1 ), ……, T(𝑒𝑛 )} spans, or generates, R(T). Let 𝑦 ∈ 𝑅(𝑇), then, by definition of R(T), there exists 𝑥 ∈ 𝑈 such that 𝑇(𝑥) = 𝑦. Let = 𝑐1 𝑒1 + ⋯ + 𝑐𝑚 𝑒𝑚 + ?
