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Unformatted text preview: Assignment 9 Math 417 Winter 2009 Due March 27 5.3#16. Since A and B are symmetric, we have A = A T and B = B T . Therefore, ( A + B ) T = A T + B T = A + B , which means that A + B is symmetic too. 5.3#26. Just like in the case of the product of two matrices, the transpose of the product of three matrices can be evaluated by transposing each matrix and multiplying them in the opposite order. (You can see this by using the product rule twice: ( XY Z ) T = ( Y Z ) T X T = Z T Y T X T .) Therefore, ( B ( A + A T ) B T ) T = ( B T ) T ( A + A T ) T B T = B ( A + A T ) T B T . Now note that ( A + A T ) T = A T + ( A T ) T = A T + A = A + A T , where the last equality follows from the fact that matrix addition is commutative. Substi- tuting, we see that ( B ( A + A T ) B T ) T = B ( A + A T ) B T , which means that B ( A + A T ) B T is symmetric. 5.3#28. Here are two different solutions. The first uses the Polarization Identity dis- cussed in class and the second uses properties of orthogonal matrices. First Solution. By the Polarization Identity (discussed in class) we have L ( ~u ) L ( ~v ) = 1 4 ( k L ( ~u ) + L ( ~v ) k 2- k L ( ~u )- L ( ~v ) k 2 ) . Since L is a linear transformation, we always have L ( ~u )+ L ( ~v ) = L ( ~u + ~v ) and L ( ~u )- L ( ~v ) = L ( ~u- ~v ). Therefore, since L preserves the lengths of ~u ~v , L ( ~u ) L ( ~v ) = 1 4 ( k L ( ~u + ~v ) k 2- k L ( ~u- ( ~v ) k 2 ) = 1 4 ( k ~u + ~v k 2- k ~u- ( ~v k 2 ) = ~u ~v. 1 Second Solution. Let A be the matrix associated to the linear transformation L . Then A is a n n orthogonal matrix, which means that A T A = I n by Theorem 5.3.7. Thus, using Theorem 5.3.6 to evaluate dot-products, we have L ( ~u ) L ( ~v ) = ( A~u ) ( A~v ) = ( A~u ) T ( A~v ) = ~u T A T A~v = ~u T I n ~v = ~u T ~v = ~u ~v....
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