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# solved09 - S 9.1(P 2.25 v1 =-1 1 1 v2 =[2-1 3 s =[1 2 3(i V...

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_______________ v1 = [-1 1 1].' v2 = [2 -1 3].' s = [1 2 3].' (i) V = [v1 v2] ||v1||^2 = 1+1+1 = 3 ||v2||^2 = 4+1+9 = 13 v1'*v2 = -2 -1 +3 = 0 (v1 and v2 are orthogonal) V'*V = [3 0 0 14] V'*V is diagonal, so its inverse is given by [1/3 0 0 1/14] Now s_hat = c(1)*v1 + c(2)*v2, where (V'*V)*c = V'*s, or c = (V'*V)\(V'*s) Here v1'*s = -1+2+3 = 4 v2'*s = 2-2+9 = 9 and thus c1 = 4/3 and c2 = 9/14 (Since v1 and v2 are orthogonal, the projection of s onto R(V) is the sum of the projections onto v1 and v2.) s_hat = V*c = (1/42)*[-2 29 137].' = [-0.0476 0.6905 3.2619]' (ii) Projection of s onto v1 is q*v1, where q = (v1'*s)/(v1'*v1) = 4/3 (this was computed above). Projection of s_hat onto v1 is m*v1, where m = (v1'*s_hat)/(v1'*v1) Since v1'*s_hat = (1/42)*[-2 29 137]*[-1 1 1].' = 4 v1'*v1 = 3 the same answer is obtained. In three dimensions, we know that the projection of a point which lies above a plane onto a line lying on the plane can be accomplished in two stages: first project onto the plane, then project that point onto the line. The equation V'*s_hat = V'*s

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solved09 - S 9.1(P 2.25 v1 =-1 1 1 v2 =[2-1 3 s =[1 2 3(i V...

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