MAT444acm104hw4 - Alex Gittens February 9 2007 ACM 104 HW 4...

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Alex Gittens February 9, 2007 ACM 104: HW 4 Exercise 1. Find the Jordan canonical form and the corresponding Jordan canonical basis for the following linear operators. (a) T : R 3 R 3 defined by Tx = Ax for x R 3 where A = 11 - 4 - 5 21 - 8 - 11 3 - 1 0 . (b) T : P 2 ( R ) P 2 ( R ) defined by T ( f ) = 2 f - f . Answer .
2 (b) Let β = { 1 , x, x 2 } , then T (1) = 2 T ( x ) = 2 x - 1 T ( x 2 ) = 2 x 2 - 2 x, so A = [ T ] β = 2 - 1 0 0 2 - 2 0 0 2 , so λ = 2 and dim( K λ ) = 3. To find a basis for K λ , first check the dimension of the eigenspace: Ax = 2 x x = c 1 0 0 dim( E λ ) = 1 . Note a basis of K λ cannot consist of an eigenvector and a two cycle of generalized eigenvectors, because the initial vector of the cycle would be an eigenvector, so these three vectors would not be linearly independent. Instead, the basis of K λ must be a three cycle of generalized eigenvectors { ( A - 2 I ) 2 v, ( A - 2 I ) v, v } such that the initial vector is an eigenvector: ( A - 2 I ) 3 v = 0 v = a 0 0 1 + b 0 1 0 + c 1 0 0 . Take v = 0 0 1 ( A - 2 I ) v = 0 - 2 0 , ( A - 2 I ) 2 v = 2 0 0 . Therefore { 2 , - 2 x, x 2 } is a Jordan canonical basis for T and J = 2 0 0 0 - 2 0 0 0 1 - 1 2 - 1 0 0 2 - 2 0 0 2 2 0 0 0 - 2 0 0 0 1 = 2 1 0 0 2 1 0 0 2 is a Jordan canonical form of T .

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