Differential Equations Solutions 15

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

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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 Frst 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) DeFne v = z α e 1 where α = e ± z ± . (4) Let u = v / ± v ± , and deFne the Householder transformation by b Q = I 2 uu . (5) Apply the transformation to
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This note was uploaded on 01/21/2012 for the course MAP 3302 taught by Professor Dr.robin during the Fall '11 term at University of Florida.

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