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Unformatted text preview: Math 416  Abstract Linear Algebra Fall 2011, section E1 Homework 8 solutions Section 4.2 2.3. (2 pts) Eigenvalues of A : det( A λI ) = 4 λ 3 1 2 λ = (4 λ )(2 λ ) 3 = 8 6 λ + λ 2 3 = λ 2 6 λ + 5 = ( λ 1)( λ 5) λ 1 = 1 ,λ 2 = 5. Let us find the eigenvectors. λ 1 = 1 : A λ 1 I = A I = 3 3 1 1 ∼ 1 1 0 0 Take v 1 = 1 1 . λ 2 = 5 : A λ 2 I = A 5 I = 1 3 1 3 ∼ 1 3 Take v 2 = 3 1 . Taking S = v 1 v 2 and D = λ 1 λ 2 , we obtain the diagonalization A = SDS 1 = 1 3 1 1 1 0 0 5 1 3 1 1 1 from which we obtain A 2004 = SD 2004 S 1 = 1 3 1 1 1 2004 5 2004 1 3 1 1 1 = 1 (3)5 2004 1 (1)5 2004 1 3 1 1 1 = 1 (3)5 2004 1 5 2004 1 4 1 3 1 1 = 1 4 1 + (3)5 2004 3 + (3)5 2004 1 + 5 2004 3 + 5 2004 . 1 2.4. There is a unique matrix A satisfying the prescribed conditions, because the conditions are the effect of A on the basis { v 1 = 1 2 ,v 2 = 1 1 } . Taking S = v 1 v 2 , we have AS = Av 1 Av 2 = v 1 3 v 2 = 1 3 2 3 ⇒ A = 1 3 2 3 S 1 = 1 3 2 3 1 1 2 1 1 = 1 3 2 3 1 1 1 1 2 1 = 1 3 2 3 1 1 2 1 = 5 2 4 1 . 2.9. (1 pt check) a. ϕ n +2 ϕ n +1 = ϕ n +1 + ϕ n ϕ n +1 = 1 1 1 0 ϕ n +1 ϕ n The matrix we are looking for is A = 1 1 1 0 ....
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This note was uploaded on 12/16/2011 for the course MATH 416 taught by Professor Frankland during the Fall '11 term at University of Illinois at Urbana–Champaign.
 Fall '11
 FRANKLAND
 Math, Linear Algebra, Algebra, Eigenvectors, Vectors

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