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hwsolB19 - iii Since v3 is orthogonal to v1 and v2 the...

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Sheet1 Page 1 HW 19 Solution ______________ i) V = [ v1 v2 v3 ] v1'*v2 = 3(a+bj) - 18j + 6j(a-jb) v1'*v2 = 0 for orthogonality, so Re( v1'*v2 ) = 3a + 6b = 0 -> a = -2b Im( v1'*v2 ) = 3b - 18 + 6a = 0 -> b = -2, a = 4 The same terms are obtained in v2'*v3 and v3'*v1, so these inner products are zero also. The square norm of each column then equals 65, and V'*V = diag([65 65 65]) ii) v1'*s = 65 v2'*s = -130 v3'*s = -65 therefore, s = v1 - 2*v2 - v3
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Unformatted text preview: iii) Since v3 is orthogonal to v1 and v2, the projection of s onto the subspace generated by v1 and v2 is simply s_hat = v1 - 2*v2 Thus || s - s_hat||^2 = || -v3||^2 = 65 (iv) x - y = -(3/2)*v1 - 2*v2 + 4*v3 ||x-y||^2 = (9/4)||v1||^2 + 4||v2||^2 + 16||v3||^2 = (9/4 + 4+16)*65 x'*(y - lambda*x) = 0 lambda = (x'*y)/||x||^2 = (1 - 3/4 - 3)/(1/4 + 1/4 + 9) = -11/38...
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