0

Let $A\in\mathbb{R}^{n\times n}$ be a positive definite symmetric matrix and consider a decomposition $A=B-C$ such that C is positive definite and symmetric as well. Show that the iteration method $x_{i+1}=(I-B^{-1}A)x_i + B^{-1}b$ converges, where $b\in\mathbb{R}^n$.

I know that the method converges if $\rho(I-B^{-1}A)=\rho(B^{-1}C)<1$ holds, which I'm currently trying to show. In order to do this, most of my approaches have been to try and show that there is a matrix norm on $\mathbb{R}^{n\times n}$such that $\left|\left|B^{-1}C\right|\right|<1$, which would give us the desired result. However, nothing I've tried out so far seems to work.

Thank you very much in advance.

  • I think there's a typo in your question. $I-B^{-1}Ax_i$ doesn't make sense. The first part of the difference is a matrix, the second is a vector. Could you please check and fix that? – jflipp Jun 02 '20 at 19:27
  • @jflipp Thanks for pointing that out, I've fixed it. –  Jun 02 '20 at 19:30

1 Answers1

0

Let me quote three results:

  1. If $D$ and $E$ are matrices, $DE$ and $ED$ have the same nonzero eigenvalues.
  2. Let $D\prec E$ denote that $E-D$ is positive definite. If $D \prec E$, then every eigenvalue of $D$ is smaller than the largest eigenvalue of $E$.
  3. If $D \prec E$, then $XD X \prec X EX$ for every $X$ which positive definite.

For points (2) and (3), a good reference is Chapter 7 of Matrix Theory by Zhang. (Four proofs of (1) are in Section 2.4.)

Notice that $C \prec C + A = B$. Then by (3), $B^{-1/2}CB^{-1/2} \prec B^{-1/2}BB^{-1/2} = I$. By (2), all of the eigenvalues of $B^{-1/2}CB^{-1/2}$ are smaller than $1$. Now apply (1) with $D = B^{-1/2}C$ and $E = B^{-1/2}$ to obtain that $B^{-1/2}B^{-1/2}C = B^{-1}C$ and $B^{-1/2}CB^{-1/2}$ both have the same eigenvalues (all less than $1$ by the previous sentence). Ergo $\rho(B^{-1}C) < 1$.

eepperly16
  • 7,232