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Unformatted text preview: Sheet1 Page 2 i.e., w1 = v1-2*w1 + w2 = v2 (1/2)*w1 + w2 + w3 = v3 Therefore w1 = v1 w2 = 2*v1 + v2 w3 = -(1/2)*v1 -(2*v1 + v2) + v3 = -(5/2)*v1 - v2 + v3 ( In other words, inv(U) = [ 1 2 -5/2 0 1 -1 0 0 1 ] ) The squared norms of w1, w2 and w3 are given by the diagonal entries of D, i.e., they are equal to 16, 9 and 16 respectively. Dividing each by its norm results in a new set of vectors with unit norm. Therefore w1 = (1/4)*v1 w2 = (2/3)*v1 + (1/3)*v2 w3 = -(5/8)*v1 - (1/4)*v2 + (1/4)*v3 w...
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- Spring '08
- Prime number, Equals sign, Vector Motors, w1 w2