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Differential Equations Solutions 15

Differential Equations Solutions 15 - 25 and v 2 =(z eT(z...

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25 and v 2 = ( z α e T 1 )( z α e 1 ) = z z α z 1 αz 1 + α α = z z e z e ζ e z e ζ + z 2 = 2 z z 2 z ζ. Then Qz = ( I 2 uu ) z = ( I 2 v 2 vv ) z = z 2 v z v 2 v = z v = α e 1 . (b) Let the second column of A 1 be [ a, v 1 , . . . , v n 1 ] T . Use the vector v to form the Householder transformation. Then the product QA 1 leaves the first column of A 1 unchanged and puts zeros below the main diagonal in the second column. (c) Assume m > n . (The other case is similar and left to the reader.) Initially, let R = A and Q = I (dimension m ). for j = 1 : n , (1) Let z = [ r jj , . . . r mj ] T . (2) Let the polar coordinate representation of z 1 be e ζ . (3) Define v = z α e 1 where α = e z . (4) Let u = v / v , and define the Householder transformation by Q = I 2 uu . (5) Apply the transformation to R by setting
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