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# Sol9 - SOLUTION TO ASSIGNMENT 9 MATH 251 WINTER 2007 1...

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SOLUTION TO ASSIGNMENT 9 - MATH 251, WINTER 2007 1. Calculate the characteristic and minimal polynomial of the following matrices with real entries. In each case determine the algebraic and geometric multiplicity of each eigenvalue. Decide which matrix is diagonalizable and for that one, say A , find an invertible matrix M such that M - 1 AM is diagonal. 4 - 2 2 6 - 3 4 3 - 2 3 3 - 2 2 4 - 4 6 2 - 3 5 Solution. The characteristic polynomial of both matrices is Δ = t 3 + 5 t - 4 t 2 - 2 = ( t - 1) 2 ( t - 2) . The minimal polynomial of A = 4 - 2 2 6 - 3 4 3 - 2 3 is t 2 - 3 t +2 = ( t - 1)( t - 2), while the minimal polynomial of B is t 3 + 5 t - 4 t 2 - 2. (Just check that ( B - Id)( B - 2Id) 6 = 0.) We now find the eigenvectors. The matrix A . We have E 1 = Ker( I - A ) = Ker - 3 2 - 2 - 6 4 - 4 - 3 2 - 2 = Span 2 3 0 0 1 1 . E 2 = Ker(2 I - A ) = Ker - 2 2 - 2 - 6 5 - 4 - 3 2 - 1 = Span 1 2 1 . Therefore we have m g (1) = m a (1) = 2 , m g (2) = m a (2) = 1 . The matrix B . We have E 1 = Ker( I - B ) = Ker - 2 2 - 2 - 4 5 - 6 - 2 3 - 4 = Span 1 2 1 . E 2 = Ker(2 I - B ) = Ker - 1 2 - 2 - 4 6 - 6 - 2 3 - 3 = Span 0 1 1 . Therefore we have m g (1) = 1 , m a (1) = 2 , m g (2) = m a (2) = 1 . Let B = Span 2 3 0 0 1 1 , 1 2 1 . With respect to the matrix B , A is represented by diag(1 , 1 , 2). Let M = 2 0 1 3 1 2 0 1 1 . Then M - 1 AM = diag(1 , 1 , 2). 1

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2 SOLUTION TO ASSIGNMENT 9 - MATH 251, WINTER 2007 2. For each matrix N in Exercise 1 (considered as a linear transformation T ) find the Primary Decompo- sition, i.e., the factorization of the minimal polynomial, the kernels of the factors, and for each kernel a matrix representation of T . Solution. For the matrix A , we have m A = ( t - 1)( t - 2), with E 1 = Ker( T - I ) and E 2 = Ker( T - 2 I ) and the transformation is represented by 1 · I 2 on E 1 and by 2 · I 1 on E 2 . The primary decomposition is just E 1 E 2 . For the matrix B we have m A = ( t - 1) 2 ( t - 2). Clearly E 2 = Ker( T - 2 I ) and T is represented by the scalar 2 = 2 · I 1 on E 2 . We now compute Ker( T - I ) 2 . This transformation is represented by the matrix 0 0 0 0 - 1 2 0 - 1 2 , which has kernel equal to W := Span 1 2 1 , 1 0 0 .
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