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Chapter_5_Solutions

# Chapter_5_Solutions - 5.1 SOLUTIONS Notes Exercises 16...

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257 5.1 SOLUTIONS Notes : Exercises 1–6 reinforce the definitions of eigenvalues and eigenvectors. The subsection on eigenvectors and difference equations, along with Exercises 33 and 34, refers to the chapter introductory example and anticipates discussions of dynamical systems in Sections 5.2 and 5.6. 1 . The number 2 is an eigenvalue of A if and only if the equation 2 A = x x has a nontrivial solution. This equation is equivalent to ( 2 ) . = x A I 0 Compute 3 2 2 0 1 2 2 3 8 0 2 3 6 A I = = The columns of A are obviously linearly dependent, so ( 2 ) A I = x 0 has a nontrivial solution, and so 2 is an eigenvalue of A . 2 . The number 2 is an eigenvalue of A if and only if the equation 2 A = − x x has a nontrivial solution. This equation is equivalent to ( 2 ) . + = x A I 0 Compute 7 3 2 0 9 3 2 3 1 0 2 3 1 A I + = + = The columns of A are obviously linearly dependent, so ( 2 ) A I + = x 0 has a nontrivial solution, and so 2 is an eigenvalue of A . 3 . Is A x a multiple of x ? Compute 3 1 1 1 1 . 3 8 4 29 4      =           λ So 1 4       is not an eigenvector of A . 4 . Is A x a multiple of x ? Compute 2 1 1 2 2 1 2 1 4 1 3 2 − + − + = + The second entries of x and A x shows that if A x is a multiple of x , then that multiple must be 3 2. + Check 3 2 + times the first entry of x : 2 (3 2)( 1 2) 3 2 2 2 1 2 2 + − + = − + + = − + This matches the first entry of , x A so 1 2 1 − + is an eigenvector of A , and the corresponding eigenvalue is 3 2. +

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258 CHAPTER 5 Eigenvalues and Eigenvectors 5 . Is A x a multiple of x ? Compute 3 7 9 4 0 4 5 1 3 0 . 2 4 4 1 0         =             So 4 3 1 is an eigenvector of A for the eigenvalue 0. 6 . Is A x a multiple of x ? Compute 3 6 7 1 2 1 3 3 7 2 4 ( 2) 2 5 6 5 1 2 1     = = −       So 1 2 1 is an eigenvector of A for the eigenvalue 2. 7 . To determine if 4 is an eigenvalue of A , decide if the matrix 4 A I is invertible. 3 0 1 4 0 0 1 0 1 4 2 3 1 0 4 0 2 1 1 3 4 5 0 0 4 3 4 1 A I = = Invertibility can be checked in several ways, but since an eigenvector is needed in the event that one exists, the best strategy is to row reduce the augmented matrix for ( 4 ) A I = x 0 : 1 0 1 0 1 0 1 0 1 0 1 0 2 1 1 0 0 1 1 0 0 1 1 0 3 4 1 0 0 4 4 0 0 0 0 0
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