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hw7solutions - Homework 7 Solutions Joshua Hernandez...

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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 Q-1 AQ = D . b. A = 1 3 3 1 Solution: Computing eigenvalues: A ( ) = det 1- 3 3 1- = (1- ) 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 = N-3 3 3-3 = span 1 1 E-2 = N 1- (-2) 3 3 1- (-2) = N 3 3 3 3 = span 1-1 . Our diagonalization is therefore A = 1 1 1 -1 4 0 -2 1 1 1 -1-1 =: QDQ-1 . d. A = 7 -4 0 8 -5 0 6 -6 3 Solution: Computing eigenvalues: A ( ) = det 7- -4 8-5- 6-6 3- = (7- )(-5- )(3- )- (-4 8(3- )) = (3- )( 2- 2 - 3) = (3- )( - 3)( + 1) . This polynomial has roots 3 and -1 (a repeated root means that A might not be diagonalizable). E 3 = N 7- 3-4 8-5- 3 6-6 3- 3 = N 4 -4 0 8 -8 0 6 -6 0 = span 1 1 , 1 1 E-1 = N 7- (-1)-4 8-5- (-1) 6-6 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 =: QDQ-1 . f. A = 1 1 0 0 1 2 0 0 3 Solution: Since A is an upper-triangular matrix, we can read its eigenvalues off the diagonal: = 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 neednt bother to compute E 3 . The dimension of E 1 is 1, although the root = 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 for V such that [ T ] is a diagonal matrix....
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hw7solutions - Homework 7 Solutions Joshua Hernandez...

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