# C1 pcf5h101pcdbdcecssc0 hptbv1gwcti s t0 e c r 0 r e

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Unformatted text preview: o Hermitian positive deﬁnite, with M = CC ∗ for some C , then (3) is equivalent to [C −1 AC −∗ ]C ∗ x = C −1 b The matrix in brackets is Hermitian positive deﬁnite, so this equation can be solved by conjugate gradient or related iterations At the same time, since C −1 AC −∗ is similar to C −∗ C −1 A = M −1 A, it is enough to examine the eigenvalues of the non-Hermitian matrix M −1 A to investigate convergence 21 / 25 22 / 25 Straight CG iteration converges slowly, achieving about 5-digit residual reduction after 40 iterations Straight CG is an improvement over a direct method √ Take M = diag(A), the diagonal matrix with entries Mii = 0.5 + i √ Set C = M for a new preconditioned CG iteration and with 30 steps it gives convergence to 15-digit residual reduction w \$ 2£ ¦ e e ¦  ¨§¥    ¦ ¨ ©)§    & ' 35 i# ¡ §  w ¨  ¦  ©¨ ¦ ©¨ ¦§ "&%      e © ! ¥  & ' x F   2¥ 1!¥ £  § &     "    ( ¡   ¥ 2¥      " (  30 E 25  20 ( ¨ 15 40 4 10 ¡¡ ¢¡ a 5  ¦ !¥  ¡ ¤5qa 0 ¡¢¢¡ a ¡ "% &9 ¡ T § ¦" #£ § AR w 1ce0 E 10 −16 10 −14 iAS 0 0iS ¦ R 61cTHE11cT6ApQ 10 −12 10 −10 10 −8 10 −6 i1cE1pc6¦A1BE5pS A S T0 0 i S T R Q T 10   I H −4 10 −2 10 0 10 2 Consider a 1000 × 1000 symmetric matrix A whose entries are all zero √ except for Aij = 0.5 + i on the diagonal...
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## This document was uploaded on 02/10/2014.

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