Math121sol8

# Math121sol8 - Math 121: Linear Algebra and Applications...

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Math 121: Linear Algebra and Applications Prof. Lydia Bieri Solution Set 8 Posted: December 13th Written by: Luca Candelori Exercise 1 (5.2/14(b), 16) . (a) We want to diagonalize the matrix A = ± 8 10 - 5 - 7 ² ﬁrst. Its eigenvalues are 3 , - 2 with corresponding eigenvectors ( - 2 , 1), ( - 1 , 1). If we let Q be the matrix whose ﬁrst and second columns are these eigenvectors, respectively, then Q - 1 = ± - 1 - 1 1 2 ² Set x ( t ) = ( x 1 ( t ) ,x 2 ( t )), and let y = Q - 1 x ( t ). Then the system of equations becomes y 0 = Dy , where D = ± 3 0 0 - 2 ² The ﬁrst row gives y 0 1 ( t ) = 3 y 1 ( t ) y 1 ( t ) = c 1 e 3 t and similarly y 2 ( t ) = c 2 e - 2 t . Plugging into the equality Qy ( t ) = x ( t ) we get x 1 ( t ) = - 2 c 1 e 3 t - c 2 e - 2 t x 2 ( t ) = c 1 e 3 t + c 2 e - 2 t (b) We are going to use c ’s to denote the entries of C and f ’s for the entried of Y . Consider the i,j -th entry of ( CY ) 0 . We have ( CY ) 0 ij = ( n X k =1 c ik f ki ) 0 = n X k =1 c ik f 0 ki = ( CY 0 ) ij Exercise 2 (5.4/1) . (a) FALSE. For any T and V , 0 and V are always invariant. 1

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(b) TRUE. Theorem 5.24 (c) FALSE. Following example 4, take V = P 2 ( R ) and T ( f ) = f 0 . Then the T -cyclic subspace generated by x 2 +2 is span( x 2 +2 , 2 x, 2) which is equal to span( x 2 , 2 x, 2), the T -cyclic subspace of x 2 . But certainly x 2 6 = x 2 + 2. (d) FALSE. In the case above, the T -cyclic subspace generated by x 2 is span( x 2 , 2 x, 2), whereas the T -cyclic subspace generated by T ( x 2 ) = 2 x is span(2 x, 2). (e) TRUE. This is the Cayley-Hamilton Theorem. Take
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## This note was uploaded on 10/23/2010 for the course MATH Math 121 taught by Professor Bieri during the Fall '07 term at Harvard.

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Math121sol8 - Math 121: Linear Algebra and Applications...

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