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# s3color - ECE 580 Math 587 SPRING 2011 Correspondence 8...

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ECE 580 / Math 587 SPRING 2011 Correspondence # 8 March 3, 2011 Solution Set 3 21. Let us check the four axioms: (i) ( B, A ) = Tr [ B T Q ¯ A ] = Tr [ ¯ B T ¯ QA ] = Tr [ A T ¯ Q T ¯ B ] = ( A, B ), where the third equality holds because Tr [ C ] = Tr [ C T ], and the last one holds because Q is Hermitian. (ii) ( A + B, C ) = Tr [ A T Q C ] + Tr [ B T Q C ] = ( A, C ) + ( B, C ) – holds for all A , B and C . (iii) ( λA, B ) = λ Tr [ A T Q ¯ B ] = λ ( A, B ) – holds for all A and B . (iv) ( A, A ) = Tr [ A T Q ¯ A ] this is positive for all A ̸ = 0 because the matrix Q is Hermitian and positive definite, i.e., all its eigenvalues are positive. Hence, ( A, B ) = Tr [ A T QB ] is indeed an inner product. 22. (a) It is not an inner product on X , because ( x, x ) = 4 1 s 2 x ( s ) ds 2 can be made zero without x being the zero function. (b) Yes , it is an inner product on X , as all four axioms of an inner product are satisfied: (i) ( x, y ) = ( y, x ) for all x, y X . (ii) ( x + y, z ) = ( x, z ) + ( x, z ) for all x, y, z X . (iii) ( λx, y ) = λ ( x, y ) for all real numbers λ and all x, y X . (iv) ( x, x ) = 4 1 t 3 x 2 ( t ) dt 0 and is equal to zero iff x ( t ) = 0 for all t [1 , 4], which is the zero element in X . 23. (a) Here the Hilbert space H = L 2 [ 1 , 2] and M = { m H : m ( t ) = a + bt, a, b IR } . Note that M is a 2-dimensional subspace of H , which is also closed (it is in fact isomorphic to IR 2 ). Hence the problem can be viewed as the optimization problem of minimizing x m over m M , where x ( t ) = t 3 is an element of H . The Projection Theorem directly applies here, leading to the conclusion that there exists a unique m o M that solves this minimization problem, and that x m o M . (b) We have to solve for a and b from the two relationships: x m o 1 ( t 3 m o , 1) = 0 and x m o t ( t 3 m o , t ) = 0 Using m o ( t ) = a + bt , we have 5 4 = a + 1 2 b , 21 5 = 1 2 a + b a = 0 . 2 , b = 2 . 1 (c) The minimum value of F is x m o 2 = ( x m o , x m o ) = ( x m o , x ) = x 2 ( m o , x ) = x 2 − ∥ m o 2 where x 2 = 129 7 and m o 2 = 14 . 61 = min m M F ( m ) = F ( m o ) = 3 . 8187

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24. Here the space L 2 [ 1 , 2] is replaced by the same with only the inner product different: ( x, y ) 2 = 2 1 t 2 x ( t ) y ( t ) dt This is also a Hilbert space (say, H ), and M
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s3color - ECE 580 Math 587 SPRING 2011 Correspondence 8...

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