MidtermSolutionsFall2008 - Math 6643 Midterm Solutions Fall...

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Math 6643 Midterm Solutions Fall 2008 1. [1, 1]' = A = USV'. Find U, S, V. Then A'A = V S^2 V' = 2. Hence S = sqrt(2) = s1, and V = 1 = v1(a constant). Now find u1: A*v1 = s1*u1 so u1 = [1, 1]'/sqrt(2) = U. 3 pts. for V, 3 pts. for S, 4 pts. for U. 2. a. The orthogonal projector is QQ', where the columns of Q are an orthonormal basis for range(A). Here Q = [2, 1]'/sqrt(5). Thus QQ' = 1 5 [ 4 2 2 1 ] . 4 pts. for Q and 2 pts. for QQ'. b. We have Pb = Ax, where Pb = QQ'b = [6/5, 3/5]'. Then x = 3/5. 4 pts. for x computed this way or by solving the normal eqns. A'Ax = A'b. 3. a. q_1 = [1, 1]'/sqrt(2). (normalize first column). Thus R_{11} = sqrt(2). Now R_{12} = projection coefficient of second column of A on q_1 = q_1'*a_2 = 1/sqrt(2). Set v_2 = a_2 – R_{12}*q_1 = [-1/2 , ½]'. Then R_{22} = 1/sqrt(2) and q_2 = [-1, 1]'/sqrt(2). The final answer is Q = 1 2 [ 1 1 1 1 ] ; R = [ 2 1 / 2 0 1 / 2 ] . 3 pts. for q_1, 2 pts. each for R_11, R_12, R_22, 3 pts for q_2.
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This note was uploaded on 12/25/2011 for the course MATH 6643 taught by Professor Staff during the Fall '08 term at Georgia Institute of Technology.

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MidtermSolutionsFall2008 - Math 6643 Midterm Solutions Fall...

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