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Unformatted text preview: MATH 245 Linear Algebra 2, Solutions to Assignment 5 1: (a) Let A = 1 2 1 2 4 2 1 2 1 . Find an orthogonal matrix P and a diagonal matrix D such that P * AP = D . Solution: The characteristic polynomial of A is f A ( ) = det( A I ) = det 1 2 1 2 4 2 1 2 1 = (1 2 + 2 )(4 ) + 8 + 8(  1) + (  4) = 4 9 + 6 2 3 4 + 9 = ( 3 6 2 ) = 2 (  6) so the eigenvalues of A are 1 = 2 = 0 and 3 = 6. For = 0 we have A I = 1 2 1 2 4 2 1 2 1 1 2 1 0 0 0 0 0 0 so a basis for the eigenspace E (which is the same as Null( A )) is given by { v 1 ,v 2 } where v 1 =  1 1 and v 2 =  2 1 . We perform the Gram Schmidt Procedure, letting w 1 = v 1 , and w 2 = v 2 v 2 . w 1  w 1  2 w 1 =  2 1  2 2  1 1 =  1 1 1 and then normalize to get u 1 = w 1  w 1  = 1 2  1 1 and u 2 = w 2  w 2  = 1 3  1 1 1 . For = 6 we have A I =  5 2 1 2 2 2 1 2 5 1 2 5 1 1 1 5 2 1 1 2 5 3 6 0 12 24 1 2 5 0 1 2 0 1 2 1 0 1 0 1 2 0 0 so a unit eigenvector is u 3 = 1 6 1 2 1 . Thus we can take P = ( u 1 ,u 2 ,u 3 ) =  1 2 1 3 1 6 1 3 2 6 1 2 1 3 1 6 and D = 1 2 3 = 0 0 0 0 0 0 0 0 6 . (b) Let A = 2 i 2 + i 1 i 3 . Find a unitary matrix P and an uppertriangular matrix T so that P * AP = T . Solution: The characteristic polynomial is f A ( ) = det( A + I ) = det 2 i  2 + i 1 i 3 = 2 (3 + 2 i ) + 6 i ( 1 + 3 i ) = 2 (3 + 2 i ) + (1 + 3 i ) so the eigenvalues are = (3 + 2 i ) p (3 + 2 i ) 2 4(1 + 3 i ) 2 = (3 + 2 i ) p (5 + 12 i ) (4 + 12 i ) 2 = (3 + 2 i ) 1 2 = 2 + i, 1 + i. When = 1 = 2 + i we have A I = 2 + i 2 + i 1 i 1 i 1 1 0 1 , so u 1 = 1 2 1 1 is a unit eigenvector. We choose u 2 = 1 2 1 1 to extend { u 1 } to the orthonormal basis U = { u 1 ,u 2 } for C 2 . Thus we can take P = ( u 1 ,u 2 ) = 1 2 1 1 1 1 , and T = P * AP = 1 2 1 1 1 1 2 i 2 + i 1 i 3 1 1 1 1 = 1 2 1 1 1 1 2 i 2 + 3 i 2 + i 4 i = 1 2 4 + 2 i 6 4 i 2 + 2 i = 2 + i 3 2 i 1 + i . 2: (a) Let U be a (possibly infinitedimensional) inner product space over R or C , and let L : U U be linear. Show that if L * = L then all of the eigenvalues of L are real, and any two eigenvectors associated to distinct eigenvalues are orthogonal. Solution: Suppose that L has an adjoint and L * = L . We claim that every eigenvalue of L is real. Let be an eigenvalue of L , say L ( u ) = u where 0 6 = u U . Then h u,u i = h u,u i = h L ( u ) ,u i = h u,L * ( u ) i = h u,L ( u ) i = h u,u i = h u,u i ....
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This note was uploaded on 12/08/2011 for the course MATH 245 taught by Professor New during the Fall '09 term at Waterloo.
 Fall '09
 NEW
 Linear Algebra, Algebra

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