MAT444acm104hw4

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 0 . Answer . (a) Since f ( t ) = det( A - tI ) = - t 3 + 3 t 2 - 4 = - ( t + 1)( t - 2) 2 , λ 1 = - 1, dim( K λ 1 ) = 1 and λ 2 = 2, dim( K λ 2 ) = 2. To find a basis for K λ 1 , note Ax = - x x = c 1 3 0 . To find a basis for K λ 2 , first see if it can contain two independent eigenvectors: Ax = 2 x x = c 1 1 1 dim( E λ 1 ) = 1 , so it can’t. Therefore K λ 2 is spanned by a two cycle of generalized eigenvectors { ( A - 2 I ) v,v } such that ( A - 2 I ) v is an eigenvector, so ( A - 2 I ) 2 v = 0 v = a 1 0 2 + b 1 2 0 . Taking v = 1 0 2 yields ( T - 2 I ) v = - 1 - 1 - 1 , so - 1 - 1 - 1 , 1 0 2 is a basis for K λ 2 . Therefore β = 1 3 0 , - 1 - 1 - 1 , 1 0 2 is a Jordan canonical basis for T , and the corresponding Jordan canonical form of T is J = 1 - 1 1 3 - 1 0 0 - 1 2 - 1 11 - 4 - 5 21 - 8 - 11 3 - 1 0 1 - 1 1 3 - 1 0 0 - 1 2 = - 1 0 0 0 2 1 0 0 2 . 1
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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
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MAT444acm104hw4 - Alex Gittens February 9 2007 ACM 104 HW 4...

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