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Unformatted text preview: 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 : R2 → R2 , defined for every (x, y) ∈ R2 by f (x, y) = (s, t), where s t Note that 3 1 3 5 3 −1 = 6 −2 =2 3 −1 and 3 1 3 5 1 1 = 6 6 =6 1 1 . = 3 1 3 5 x y . On the other hand, note that v1 = 3 −1 and v2 = 1 1 form a basis for R2 . It follows that every u ∈ R2 can be written uniquely in the form u = c1 v1 + c2 v2 , where c1 , c2 ∈ R, so that Au = A(c1 v1 + c2 v2 ) = c1 Av1 + c2 Av2 = 2c1 v1 + 6c2 v2 . Note that in this case, the function f : R2 → R2 can be described easily in terms of the two special vectors v1 and v2 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 ∈ R2 such that 3 1 Chapter 7 : Eigenvalues and Eigenvectors 3 5 v = λv. page 1 of 12 Linear Algebra c W W L Chen, 1982, 2005 Since λv = λ we must have 3 1 In other words, we must have (1) 3−λ 1 3 5−λ v = 0. 3 5 − λ 0 0 λ v = 0. 1 0 0 1 v= λ 0 0 λ v, In order to have non-zero v ∈ R2 , we must therefore ensure that det 3−λ 1 3 5−λ = 0. Hence (3 − λ)(5 − λ) − 3 = 0, with roots λ1 = 2 and λ2 = 6. Substituting λ = 2 into (1), we obtain 1 1 3 3 v = 0, with root v1 = 3 −1 . Substituting λ = 6 into (1), we obtain −3 1 Definition. Suppose that (2) a11 . . A= . an1 ... a1n . . . 3 −1 v = 0, with root v2 = 1 1 . . . . ann is an n × n matrix with entries in R. Suppose further that there exist a number λ ∈ R and a non-zero vector v ∈ Rn such that Av = λv. 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 Av = λv = λIv, where I is the n × n identity matrix, so that (A − λI)v = 0. Since v ∈ Rn is non-zero, it follows that we must have (3) In other words, we must have a11 − λ a21 det . . . an1 a12 a22 − λ an2 ... .. a1n a2n . . . = 0. det(A − λI) = 0. . . . . ann − λ 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 ∈ Rn : (A − λI)v = 0} is the nullspace of the matrix A − λI, a subspace of Rn . Chapter 7 : Eigenvalues and Eigenvectors page 2 of 12 Linear Algebra c W W L Chen, 1982, 2005 Definition. The polynomial (3) i...
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This note was uploaded on 06/13/2009 for the course TAM 455 taught by Professor Petrina during the Fall '08 term at Cornell University (Engineering School).

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