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5.3soln

# 5.3soln - Solutions to problems from section 5.3 2 Let P =...

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Solutions to problems from section 5.3 2. Let P = bracketleftbigg 2 - 3 - 3 5 bracketrightbigg , D = bracketleftbigg 1 0 0 1 / 2 bracketrightbigg , and A = PDP - 1 . Find A 4 . Solution: First, since det P = 1 we know that P - 1 = bracketleftbigg 5 3 3 2 bracketrightbigg so that A 4 = ( PDP - 1 ) 4 = PD 4 P - 1 = bracketleftbigg 2 - 3 - 3 5 bracketrightbigg bracketleftbigg 1 4 0 0 (1 / 2) 4 bracketrightbigg bracketleftbigg 5 3 3 2 bracketrightbigg = bracketleftbigg 2 - 3 - 3 5 bracketrightbigg bracketleftbigg 1 0 0 1 / 16 bracketrightbigg bracketleftbigg 5 3 3 2 bracketrightbigg = bracketleftbigg 2 - 3 - 3 5 bracketrightbigg bracketleftbigg 5 3 3 / 16 1 / 8 bracketrightbigg = bracketleftbigg 151 / 16 45 / 8 - 225 / 16 - 67 / 8 bracketrightbigg 6. Find the eigenvalues and a basis for each eigenspace of the matrix 4 0 - 2 2 5 4 0 0 5 = - 2 0 - 1 0 1 2 1 0 0 5 0 0 0 5 0 0 0 4 0 0 1 2 1 4 - 1 0 - 2 Solution: The eigenvalues are 5 and 4. - 2 0 1 , 0 1 0 is a basis for the 5-eigenspace and - 2 4 5 is a basis for the 4-eigenspace. 8. Diagonalize the matrix A = bracketleftbigg 5 1 0 5 bracketrightbigg if possible. Solution: Since A is upper triangular, its eigenvalues are the entries on the main diagonal. Thus the only eigenvalue of A is 5 (with multiplicity 2). Since A - 5 I 2 = bracketleftbigg 0 1 0 0 bracketrightbigg we see that x Nul( A - 5 I 2 ) if and only if x = bracketleftbigg x 1 x 2 bracketrightbigg = bracketleftbigg x 1 0 bracketrightbigg = x 1 bracketleftbigg 1 0 bracketrightbigg .

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