We can therefore only nd two linearly independent

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Unformatted text preview: ot v2 = 1 . 0 2 0 Note that the eigenspace corresponding to the eigenvalue 3 is a line through the origin. On the other hand, the matrix 1 −1 0 1 −1 0 0 0 2 Chapter 7 : Eigenvalues and Eigenvectors page 4 of 12 Linear Algebra c W W L Chen, 1982, 2005 has rank 2, and so the eigenspace corresponding to the eigenvalue 1 is of dimension 1 and so is also a line through the origin. We can therefore only find two linearly independent eigenvectors, so that R3 does not have a basis consisting of linearly independent eigenvectors of the matrix A. Example 7.1.6. Consider the matrix 3 A = 1 1 −3 −1 −3 2 2. 4 To find the eigenvalues of A, we need to find the roots of 3−λ det 1 1 −3 −1 − λ −3 2 2 = 0; 4−λ in other words, (λ − 2)3 = 0. The eigenvalue is therefore λ = 2. An eigenvector corresponding to the eigenvalue 2 is a solution of the system 1 (A − 2I)v = 1 1 −3 −3 −3 2 2 v = 0, 2 2 v1 = 0 −1 3 v2 = 1 . 0 with roots and Note now that the matrix 1 1 1 −3 −3 −3 2 2 2 has rank 1, and so the eigenspace corresponding to the eigenvalue 2 is of dimension 2 and so is a plane through the origin. We can therefore only find two linearly independent eigenvectors, so that R3 does not have a basis consisting of linearly independent eigenvectors of the matrix A. Example 7.1.7. Suppose that λ is an eigenvalue of a matrix A, with corresponding eigenvector v. Then A2 v = A(Av) = A(λv) = λ(Av) = (λv) = λ2 v. Hence λ2 is an eigenvalue of the matrix A2 , with corresponding eigenvector v. In fact, it can be proved by induction that for every natural number k ∈ N, λk is an eigenvalue of the matrix Ak , with corresponding eigenvector v. Example 7.1.8. Consider the matrix 4 6. 3 1 0 0 5 2 0 To find the eigenvalues of A, we need to find the roots of 1−λ det 0 0 5 2−λ 0 4 6 = 0; 3−λ in other words, (λ − 1)(λ − 2)(λ − 3) = 0. It follows that the eigenvalues of the matrix A are given by the entries on the diagonal. In fact, this is true for all triangular matrices. Chapter 7 : Eigenvalues and Eigenvectors page 5 of 12 Linear Algebra c W W L Chen, 1982, 2005 7.2. The Diagonalization Problem Example 7.2.1. Let us return to Examples 7.1.1 and 7.1.2, and consider again the matrix A= 3 1 3 5 . We have already shown that the matrix A has eigenvalues λ1 = 2 and λ2 = 6, with corresponding eigenvectors v1 = 3 −1 and v2 = 1 1 respectively. Since the eigenvectors form a basis for R2 , every u ∈ R2 can be written uniquely in the form (5) and (6) Write c= c1 c2 , u= x y , Au = s t . Au = 2c1 v1 + 6c2 v2 . u = c1 v1 + c2 v2 , where c1 , c2 ∈ R, Then (5) and (6) can be rewritten as (7) and (8) respectively. If we write P = 3 −1 1 1 and D= 2 0 0 6 , s t = 3 −1 1 1 2c1 6c2 = 3 −1 1 1 2 0 0 6 c1 c2 x y = 3 −1 1 1 c1 c2 then (7) and (8) become u = P c and Au = P Dc respectively, so that AP c = P Dc. Note that c ∈ R2 is arbitrary. This implies that (AP − P D)c = 0 for every c ∈ R2 . Hence we must have AP = P D. S...
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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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