CourseNotes - MATH 2040 COURSE NOTES ROB NOBLE The following course notes are based closely on[Poole2006 4 Eigenvalues and Eigenvectors 4.1 Introduction

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MATH 2040 COURSE NOTES ROB NOBLE The following course notes are based closely on [Poole2006]. 4. Eigenvalues and Eigenvectors 4.1. Introduction to Eigenvalues and Eigenvectors. Definition 1 (Eigenvalues and Eigenvectors) . A scalar λ 2 R is called an eigenvalue of the matrix A 2 R n n provided A x = λ x for at least one nonzero vector x 2 R n . Any such vector x 2 R n is referred to as an eigenvector associated to λ . Example 1. Show that v = [2 , - 1 , 1] T is an eigenvector of the matrix A = 2 6 4 3 0 0 0 1 - 2 1 0 1 3 7 5 and determine the associated eigenvalue. Solution. We need to verify that A v is a scalar multiple of v , and determine the multiple. We compute A v = 2 6 4 3 0 0 0 1 - 2 1 0 1 3 7 5 2 6 4 2 - 1 1 3 7 5 = 2 6 4 6 - 3 3 3 7 5 = 3 2 6 4 2 - 1 1 3 7 5 = 3 v . Therefore A v is indeed a scalar multiple of v , and the multiple is 3. Thus v is an eigenvector of A with associated eigenvalue λ = 3 . For A 2 R n n , and λ 2 R , we have the following chain of equivalences: x is an eigenvector of A associated to λ () x 6 = 0 and A x = λ x () x 6 = 0 and A x - λ x = 0 () x 6 = 0 and ( A - λ I ) x = 0 () x is a nonzero vector in the null space of A - λ I . Date : March 19, 2014. 1
2 ROB NOBLE The eigenvectors of A associated to λ are therefore the nonzero vectors in the null space of A - λ I . This particular null space associated to λ is of great importance. Definition 2 (Eigenspace) . Let λ 2 R be an eigenvalue of A 2 R n n . By the eigenspace associated to λ , denoted E λ , we mean the null space of A - λ I . Therefore, the eigenspace E λ consists of all vectors on which A acts as scalar multiplication by λ . That is E λ = { x 2 R n | A x = λ x } consists of the zero vector together with all the eigenvectors of A associated to λ . In particular, if we know an eigenvalue λ of A 2 R n n , we can find the associated eigenvectors by calculating the null space of the matrix A - λ I . Example 2. Show that λ = - 1 is an eigenvalue of the matrix A = 2 6 4 1 0 2 - 1 1 1 2 0 1 3 7 5 and determine the associated eigenspace. Solution. We need to show that A v = ( - 1) v for at least one nonzero vector v and then determine the eigenspace E - 1 consisting of all vectors v (including 0 ) in R 3 for which A v = ( - 1) v . Since it will be easier, we will first compute the null space of A - ( - 1) I in order to find E - 1 , and then verify that - 1 is indeed an eigenvalue of A by choosing a nonzero vector in E - 1 and demonstrating that A transforms this vector into its negative. We compute A - ( - 1) I = 2 6 4 2 0 2 - 1 2 1 2 0 2 3 7 5 2 6 4 2 0 2 - 1 2 1 0 0 0 3 7 5 2 6 4 2 0 2 0 2 2 0 0 0 3 7 5 2 6 4 1 0 1 0 2 2 0 0 0 3 7 5 2 6 4 1 0 1 0 1 1 0 0 0 3 7 5 . The null space in question is therefore E - 1 = span [ - 1 , - 1 , 1] T . All that remains is to verify that - 1 is in fact an eigenvalue of A . We need to produce a nonzero vector that A sends to its negative. Any nonzero vector in E - 1 will do, and so we choose v = [ - 1 , - 1 , 1] and complete the solution with the computation A v = 2 6 4 1 0 2 - 1 1 1 2 0 1 3 7 5 2 6 4 - 1 - 1 1 3 7 5 = 2 6 4 1 1 - 1 3 7 5 = - v .
MATH 2040 COURSE NOTES 3 The question arises: “How do we determine the eigenvalues of a matrix?” Well, when A is a matrix whose action on vectors is known (e.g. a rotation matrix or reflection matrix), then the eigenvectors