# Hermitian positive definite matrix and invertibility

• Apr 22nd 2010, 04:15 PM
math8
Hermitian positive definite matrix and invertibility
Suppose A is a Hermitian positive definite matrix split into $\displaystyle A = C + C^{*} + D$ where $\displaystyle D$ is also Hermitian positive definite.
We show that $\displaystyle B=C+ \omega ^{-1} D$ is invertible. Consider the iteration $\displaystyle x_{n+1} = x_{n} + B^{-1} (b-Ax_{n})$ , with any initial iterate $\displaystyle x_{0}$ . Prove that $\displaystyle x_{n}$ converges to $\displaystyle x= A^{-1}b$ whenever $\displaystyle 0< \omega < 2$.

I suppose to show invertibility, we need to show $\displaystyle det(B) \neq 0$. But I am not sure how to show that. Also for the convergence, do we show $\displaystyle lim_{n \rightarrow \infty } \left\| x_{n}-x \right\|= 0$ ? If yes, how?
• Apr 22nd 2010, 06:36 PM
Bruno J.
You want to show that $\displaystyle B$ is invertible, but $\displaystyle B^{-1}$ appears in the expression which you plan on using to do that... Are you sure you copied the problem properly?
• Apr 23rd 2010, 10:32 AM
math8
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 $\displaystyle x_{n+1}$ .