soln5 - MATH 245 Linear Algebra 2, Solutions to Assignment...

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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 upper-triangular 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 infinite-dimensional) 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.

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soln5 - MATH 245 Linear Algebra 2, Solutions to Assignment...

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