MGSLS - matrix Let he thin QR factorization of A,b be ± A...

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Least Squares with Gram-Schmidt Recall the Gram-Schmidt QR factorization: A = QR, where Q R m × n satisfies Q t Q = I and R R n × n is upper triangular. The cost is 2 mn 2 + O( mn ) flops. If A is overwritten by Q , then only 1 2 n 2 + O( n ) words of memory are required. If ˜ Q and ˜ R are the computed versions of Q and R , then there exists δA R m × n with A + δA = ˜ Q ˜ R , where k δA k = k A k O( μ ), and k ˜ Q t ˜ Q - I k = κ ( A )O( μ ). Now to solve the least squares problem min x k Ax - b k 2 we can use back substitution to solve Rx = Q t b . The condition number of R is κ ( A ). Now let’s look at the right hand side: when we write Q t b we are assuming that Q t Q = I . This is true in exact arithmetic, but the result above says that in finite precision, the orthogonality of Q depends on κ ( A ). Unfortunately, this – combined with the conditioning of R – gives a [ κ ( A )] 2 factor in the backward error for x . Here we will show how to avoid this to get a backward error result for MGS which is equivalent to that of the Householder QR . We do what you did when you first learned G.E.: we augment the coefficient
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Unformatted text preview: matrix. Let he thin QR factorization of [ A,b ] be ± A b ² = ± Q w ² ³ R z β ´ , where w t Q = 0 ( w is just q n +1 ). Note that A = QR and b = Qz + βw . Then the rhs for (LS), Q t b , is Q t b = Q t ( Qz + βw ) = z , and we can solve (LS) using backward substitution on Rx = z. Now do you really think that this removes the κ ( A ) factor which came from the orthogonality errors in Q ? Why should it? I applaud you for your skepticism. The answer lies in the (substantial) differences in behavior between MGS and CGS. Explicitly computing Q t b as Z = Q t * b is the CGS way, but MGS adapts to the errors made in each inner product, giving a z which has, (to the extent that it can), “accounted for” any nonorthogonality in the columns of Q ....
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