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Unformatted text preview: Homework 7 Solutions Joshua Hernandez November 16, 2009 5.2  Diagonalizability 2. For each of the following matrices A ∈ M n × n ( R ), test A for diagonalizability, and if A is diagonalizable, find an invertible matrix Q and a diagonal matrix D such that Q1 AQ = D . b. A = 1 3 3 1 Solution: Computing eigenvalues: χ A ( λ ) = det 1 λ 3 3 1 λ = (1 λ ) 2 3 2 This polynomial has roots 1 ± 3 = 4 ,2. Two distinct eigenvalues mean that A is diagonalizable. Then E 4 = N 1 4 3 3 1 4 = N3 3 33 = span 1 1 E2 = N 1 (2) 3 3 1 (2) = N 3 3 3 3 = span 11 . Our diagonalization is therefore A = 1 1 1 1 4 0 2 1 1 1 11 =: QDQ1 . d. A = 7 4 0 8 5 0 6 6 3 Solution: Computing eigenvalues: χ A ( λ ) = det 7 λ4 85 λ 66 3 λ = (7 λ )(5 λ )(3 λ ) (4 · 8(3 λ )) = (3 λ )( λ 2 2 λ 3) = (3 λ )( λ 3)( λ + 1) . This polynomial has roots 3 and 1 (a repeated root means that A might not be diagonalizable). E 3 = N 7 34 85 3 66 3 3 = N 4 4 0 8 8 0 6 6 0 = span 1 1 , 1 1 E1 = N 7 (1)4 85 (1) 66 3 (1) = N 8 4 0 8 4 0 6 6 4 = span 2 4 3 . The two eigenspaces have a total dimension of 3, so A is diagonalizable. Our diagonalization is therefore A = 1 0 2 1 0 4 0 1 3 3 0 0 3 0 0 1 1 0 2 1 0 4 0 1 3 1 =: QDQ1 . f. A = 1 1 0 0 1 2 0 0 3 Solution: Since A is an uppertriangular matrix, we can read its eigenvalues off the diagonal: λ = 1 , 3. Then E 1 = N 1 1 1 1 1 2 3 1 = N 0 1 0 0 0 2 0 0 2 = span 1 . We needn’t bother to compute E 3 . The dimension of E 1 is 1, although the root λ = 1 has multiplicity 2. Therefore A is not diagonalizable. 3b. Let V = P 2 ( R ). Define T : V → V by the mapping T ( ax 2 + bx + c ) = cx 2 + bx + a . If diagonalizable, find a basis β for V such that [ T ] β is a diagonal matrix....
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This note was uploaded on 02/01/2010 for the course MATH math115a taught by Professor Shalom during the Spring '10 term at UCLA.
 Spring '10
 SHALOM
 Linear Algebra, Algebra, Matrices

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