Hermitian positive definite matrix and invertibility

**math8** · April 22nd 2010, 04:15 PM

Suppose A is a Hermitian positive definite matrix split into $A = C + C^{*} + D$ where $D$ is also Hermitian positive definite.
We show that $B=C+ \omega ^{-1} D$ is invertible. Consider the iteration $x_{n+1} = x_{n} + B^{-1} (b-Ax_{n})$ , with any initial iterate $x_{0}$ . Prove that $x_{n}$ converges to $x= A^{-1}b$ whenever $0< \omega < 2$ .

I suppose to show invertibility, we need to show $det(B) \neq 0$ . But I am not sure how to show that. Also for the convergence, do we show $lim_{n \rightarrow \infty } \left\| x_{n}-x \right\|= 0$ ? If yes, how?

**Bruno J.** · April 22nd 2010, 06:36 PM

You want to show that $B$ is invertible, but $B^{-1}$ appears in the expression which you plan on using to do that... Are you sure you copied the problem properly?

**math8** · April 23rd 2010, 10:32 AM

yes, the problem is copied properly. I think we need to first show that B is invertible (i.e. B inverse exists) to be able to use the expression for $x_{n+1}$ .

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