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Solution07_121

# Solution07_121 - Solutions to PS 7(Math 121 J Scholtz...

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Solutions to PS 7 (Math 121) J. Scholtz November 18, 2006 Question 1: 5.2/12 1. Suppose that v E λ , then T ( v ) = λv and hence by taking the inverse of both sides we get: v = λT - 1 ( v ) and so v E λ - 1 . Hence the two eigenspaces are equal. 2. Suppose that A = [ T ] is diagonalizable then there exist D = Q - 1 AQ such that D is diagonal. Then by taking an inverse of each side we get D - 1 = Q - 1 A - 1 Q . Note, that since D is diagonal it is invertible and our last steps is hence justified. Question 2: 5.2/14a Find a solution to the system of differential equations: x = x + y and y = 3 x - y . We need to diagonalize the matrix A = 1 1 3 - 1 , which gives us D = 2 0 0 - 2 and Q = - 2 0 0 2 And so our new system looks like p = - 2 p and q = 2 q , therefore p = Ce - 2 t and q = De 2 t . Now we only need to convert to the old basis where p = - x +3 y and q = x + y and so y = 1 4 { Ce - 2 t + De 2 t } , whereas x = - 1 4 { Ce - 2 t - 3 De 2 t } . Question 3: 5.2/17 Two operators T and U are simultaneously diagonalizable iff there exists a basis β such that both [ T ] β and [ U ] β are diagoanal. Similarly two matrices A, B are simultaneously diagonalizable if there exists an invertible Q such that both Q - 1 AQ and Q - 1 BQ are diagonal matrices. 1. Show that if T and U are simultaneously diagonalizable then the matrices [ T ] β and [ U ] β are also simultaneously diagonalizable. If both T and U are both simultaneously diagonalizable then there exists a basis γ such that both [ T ] γ and [ U ] γ are diagonal. But this means that

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