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Thread: Hermitian positive definite matrix and invertibility

  1. #1
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    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?
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  2. #2
    MHF Contributor Bruno J.'s Avatar
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    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?
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  3. #3
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    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 $\displaystyle x_{n+1}$ .
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