Problem 71 2ac for each matrix a find a basis for

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Problem ( § 7.1 # 2.a,c) . For each matrix A , find a basis for each generalized eigenspace of L A consisting of a union of disjoint cycles of generalized eigenvectors. Then find a Jordan canonical form J of A . Proof. Part (a): A = 1 1 - 1 3 . First we calculate the characteristic polynomial. f ( t ) = det( A - tI ) = det 1 - t 1 - 1 3 - t = (1 - t )(3 - t ) + 1 = t 2 - 4 t + 4 = ( t - 2) 2 . Now we compute the dimension of the eigenspace. dim( E 2 ) = 2 - rank ( A - 2 I ) = 2 - rank - 1 1 - 1 1 = 2 - 1 = 1 . This means that a basis for the generalized eigenspace cannot consist of more than one cycle, and hence must be one cycle. Notice that 1 0 is not an eigenvector, and hence will work as the end vector of our cycle (any other non-eigenvector would work here). Therefore β = ( T - 2 I ) 1 0 , 1 0 = - 1 - 1 , 1 0 is a basis for the generalized eigenspace of L A . The Jordan canonical form of L A then looks like J = [ L A ] β = 2 1 0 2 . Part (c): A = 11 - 4 - 5 21 - 8 - 11 3 - 1 0 . First we calculate the characteristic polynomial. f ( t ) = det( A - tI ) = det 11 - t - 4 - 5 21 - 8 - t - 11 3 - 1 - t = (11 - t )( t 2 + 8 t - 11) + 4( - 21 t + 33) - 5(3 t + 3) = - ( t 3 - 3 t 2 + 4) = - ( t - 2) 2 ( t + 1) .
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HOMEWORK 13 SOLUTIONS 3 So the eigenvalues are λ = - 1 , 2 with multiplicities 1 , 2 respectively. For the eigenvalue - 1, we see that is suffices to find a basis for E - 1 . To do so we row-reduce A + I : A + I = 12 - 4 - 5 21 - 7 - 11 3 - 1 1 3 - 1 1 12 - 4 - 5 21 - 7 - 11 3 - 1 1 0 0 - 9 0 0 - 18 3 - 1 0 0 0 1 0 0 0 . Therefore, a basis for K - 1 = E - 1 will be β 1 = 1 3 0 . On the other hand, we row-reduce A - 2 I : A - 2 I = 9 - 4 - 5 21 - 10 - 11 3 - 1 - 2 3 - 1 - 2 9 - 4 - 5 21 - 10 - 11 3 - 1 - 2 0 - 1 1 0 - 3 3 1 0 - 1 0 1 - 1 0 0 0 . Therefore, a basis for E 2 will be b = 1 1 1 . However, we are looking for a basis for K 2 consisting of cycles. Clearly, it can’t contain two cycles (since E 2 is 1-dimensional), so it only has one cycle. To find the end vector of the cycle we have to solve the system ( A - 2 I ) x = b . Going through the row-reductions above, but augmented by b , we obtain ( A - 2 I : b ) = 9 - 4 - 5 : 1 21 - 10 - 11 : 1 3 - 1 - 2 : 1 3 - 1 - 2 : 1 9 - 4 - 5 : 1 21 - 10 - 11 : 1 3 - 1 - 2 : 1 0 - 1 1 : - 2 0 - 3 3 : - 6 1 0 - 1 : 1 0 1 - 1 : 2 0 0 0 : 0 . So (setting x 3 = 0) we find that x = 1 2 0 is a solution to the system. (If you want to double check the computations, just make sure ( A - 2 I ) x = b .) Therefore β 2 = 1 1 1 , 1 2 0 is a basis, consisting of a cycle, for K 2 . The Jordan canonical form will then look like J = - 1 0 0 0 2 1 0 0 2 . Problem ( § 7.1 # 3.a,b,c) . For each linear operator T , find a basis for each generalized eigenspace of T consisting of a union of disjoint cycles of generalized eigenvectors. Then find a Jordan canonical form J of T .
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4 HOMEWORK 13 SOLUTIONS Proof. Part (a): T is the linear operator on P 2 ( R ) defined by T ( f ( x )) = 2 f ( x ) - f ( x ).
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