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 = xx has a nontrivial solution. This equation is equivalent to (2 ) . −= x AI 0 Compute 32 20 12 2 38 02 36  =   The columns of A are obviously linearly dependent, so ( 2 ) 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 =− has a nontrivial solution. This equation is equivalent to ( 2 ) . += x 0 Compute 73 2 0 9 3 2 310 2 3 1 + = The columns of A are obviously linearly dependent, so ( 2 ) x 0 has a nontrivial solution, and so 2 is an eigenvalue of A . 3 . Is A x a multiple of x ? Compute 311 1 1 . 384 2 9 4    =≠       λ So 1 4       is not an eigenvector of A . 4 . Is A x a multiple of x ? Compute 21 12 2 14 1 32 −+ = + The second entries of x and A x shows that if A x is a multiple of x , then that multiple must be . + Check + 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 is an eigenvector of A , and the corresponding eigenvalue is . +

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258 CHAPTER 5 • Eigenvalues and Eigenvectors 5 . Is A x a multiple of x ? Compute 37 94 0 45 1 3 0 . 24 410   −− =  So 4 3 1 is an eigenvector of A for the eigenvalue 0. 6 . Is A x a multiple of x ? Compute 367 1 2 1 337 2 4 ( 2 )2 565 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 AI is invertible. 30 1 400 1 0 1 4 23 1 040 2 1 1 34 5 004 3 4 1 = 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 ) x0 : 10 1 0 1 0 1 0 211 0 011 0 0 341 0 044 0 000 0 ∼∼ The equation ( 4 ) has a nontrivial solution, so 4 is an eigenvalue. Any nonzero solution of (4 ) is a corresponding eigenvector. The entries in a solution satisfy 13 0 xx += and 23 0, −− = with 3 x free. The general solution is not requested, so to save time, simply take any nonzero value for 3 x to produce an eigenvector. If 3 1, = x then ( 1 1 1).
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## This note was uploaded on 10/25/2010 for the course EECS 5134531 taught by Professor Harvey,ramesh,liphardt during the Spring '10 term at University of California, Berkeley.

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Chapter_5_Solutions - 5.1 SOLUTIONS Notes: Exercises 16...

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