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Let $A$ be HPD. Denote by $D$ the diagonal matrix obtained by observing the diagonal elements of A, i.e. $D = \operatorname{diag}(a_{11},a_{22},\ldots,a_{nn})$.

I would like to show that if the Jacobi method for linear systems converges for the system $Ax = b$, then $2D-A$ must also be HPD.

The inverse is also true; Let $A = D - E - E^{*}$ be the standard decomposition of A to diagonal matrix, upper triangular matrix and lower triangular matrix. The Jacobi matrix can be expressed as $J = D^{-1}(E+E^{*})$. If $(\lambda,v)$ is an eigenpair of $J$, then there holds $v^{*}(E+E^{*})v = \lambda v^{*}Dv$, and we can extract $\lambda$ since $D$ must be HPD, and find a condition so that $|\lambda| < 1$. Then, we need to show the condition holds using the fact that $v^{*}Av* > 0$ and $v^{*}(2D-A)v > 0$, which is rather simple to do.

However, I cannot seem to find a concrete method to prove the original statement. I have came to a conclusion the Jacobi method converges for both $Ax=b$ if and only if it converges for $(2D-A)x = b$, but I do not know how to conclude that $v^{*}(2D-A)v > 0$ for non-zero vectors $v$.

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Welcome to Math.SE! –  Belgi Jun 1 '12 at 17:22
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After some careful thought I was able to prove the claim.

Assume the Jacobi method converges for the linear system $Ax = b$. Then, $|\lambda(J)| < 1$. As before, if $(v, \lambda)$ is an eigenpair of $J$, I can write $v^{*}Dv > 2|Re(v^{*}Ev)|$. From there, one can easily show that all eigenvalues of $2D - A$ are positive. $2D - A$ is hermitian so it has only real eigenvalues, thus it suffices to show none of them can be zero or negative, or it will contradict $v^{*}Dv > 2|Re(v^{*}Ev)|$. Since $2D - A$ is hermitian with only positive eigenvalues, it must be HPD.

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