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# solved10 - x(y lambda*x = 0 i.e lambda =(x*y/||x||^2 Since...

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S 10.1 ______ 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 (iii) The projection of y onto x equals lambda*x, where
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Unformatted text preview: x'*(y - lambda*x) = 0 i.e., lambda = (x'*y)/||x||^2 Since x and y are given in terms of the orthogonal vectors v1, v2 and v3, inner products involving x and y can be computed directly using those coefficients (i.e., in the new coordinate system). All the coefficients involved are real, therefore the same is true for the inner products: x'*y = (1 - 3/4 - 3)*65 ||x||^2 = (1/4 + 1/4 + 9)*65 Thus lambda = (1 - 3/4 - 3)/(1/4 + 1/4 + 9) = -11/38...
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