la07-ee

# la07-ee - LINEAR ALGEBRA W W L CHEN c W W L Chen 1982 2005...

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LINEAR ALGEBRA W W L CHEN c W W L Chen, 1982, 2005. This chapter originates from material used by the author at Imperial College, University of London, between 1981 and 1990. It is available free to all individuals, on the understanding that it is not to be used for financial gain, and may be downloaded and/or photocopied, with or without permission from the author. However, this document may not be kept on any information storage and retrieval system without permission from the author, unless such system is not accessible to any individuals other than its owners. Chapter 7 EIGENVALUES AND EIGENVECTORS 7.1. Introduction Example 7.1.1. Consider a function f : R 2 R 2 , defined for every ( x, y ) R 2 by f ( x, y ) = ( s, t ), where s t = 3 3 1 5 x y . Note that 3 3 1 5 3 1 = 6 2 = 2 3 1 and 3 3 1 5 1 1 = 6 6 = 6 1 1 . On the other hand, note that v 1 = 3 1 and v 2 = 1 1 form a basis for R 2 . It follows that every u R 2 can be written uniquely in the form u = c 1 v 1 + c 2 v 2 , where c 1 , c 2 R , so that A u = A ( c 1 v 1 + c 2 v 2 ) = c 1 A v 1 + c 2 A v 2 = 2 c 1 v 1 + 6 c 2 v 2 . Note that in this case, the function f : R 2 R 2 can be described easily in terms of the two special vectors v 1 and v 2 and the two special numbers 2 and 6. Let us now examine how these special vectors and numbers arise. We hope to find numbers λ R and non-zero vectors v R 2 such that 3 3 1 5 v = λ v . Chapter 7 : Eigenvalues and Eigenvectors page 1 of 12

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Linear Algebra c W W L Chen, 1982, 2005 Since λ v = λ 1 0 0 1 v = λ 0 0 λ v , we must have 3 3 1 5 λ 0 0 λ v = 0 . In other words, we must have (1) 3 λ 3 1 5 λ v = 0 . In order to have non-zero v R 2 , we must therefore ensure that det 3 λ 3 1 5 λ = 0 . Hence (3 λ )(5 λ ) 3 = 0, with roots λ 1 = 2 and λ 2 = 6. Substituting λ = 2 into (1), we obtain 1 3 1 3 v = 0 , with root v 1 = 3 1 . Substituting λ = 6 into (1), we obtain 3 3 1 1 v = 0 , with root v 2 = 1 1 . Definition. Suppose that (2) A = a 11 . . . a 1 n . . . . . . a n 1 . . . a nn is an n × n matrix with entries in R . Suppose further that there exist a number λ R and a non-zero vector v R n such that A v = λ v . Then we say that λ is an eigenvalue of the matrix A , and that v is an eigenvector corresponding to the eigenvalue λ . Suppose that λ is an eigenvalue of the n × n matrix A , and that v is an eigenvector corresponding to the eigenvalue λ . Then A v = λ v = λI v , where I is the n × n identity matrix, so that ( A λI ) v = 0 . Since v R n is non-zero, it follows that we must have (3) det( A λI ) = 0 . In other words, we must have det a 11 λ a 12 . . . a 1 n a 21 a 22 λ a 2 n . . . . . . . . . a n 1 a n 2 . . . a nn λ = 0 . Note that (3) is a polynomial equation. Solving this equation (3) gives the eigenvalues of the matrix A . On the other hand, for any eigenvalue λ of the matrix A , the set (4) { v R n : ( A λI ) v = 0 } is the nullspace of the matrix A λI , a subspace of R n .
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