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Set $$D=\begin{bmatrix}0\\d_{1}&0\\d_{2}&d_{1}&0\\\vdots&\vdots&\ddots&\ddots\\d_{n}&d_{n-1}&\cdots&d_{1}&0\end{bmatrix}$$ and $$W(\alpha)=(I-\alpha D)^{-1}-\frac{1}{2}I$$ $$\Sigma(\alpha)=W(\alpha)+W^{*}(\alpha)$$ where $\alpha$ is a real number in $[0,1]$, $*$ represents conjugate transpose.

The question is:

suppose $$\Sigma(1)>0$$, if we can think $$\Sigma(\alpha)>0$$ with $\alpha $ changing from $0$ to $1$ ? Thanks a lot

I know that $W(\alpha)$ is also a lower triangular matrix with diagonal elements $\frac{1}{2}$.

Yufang Cui
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  • It is pretty easy to see that $\Sigma(\alpha)$ is positive definite at $\alpha = 0$ by construction, and you "suppose" that $\Sigma(1)$ also is positive definite. The problem thus seems to ask whether intermediate values $0\lt \alpha \lt 1$ will also produce positive definite matrices. If you like, I can tweak the wording of your Question to make that a bit clearer. – hardmath Oct 27 '18 at 15:40
  • Yes, thanks a lot – Yufang Cui Oct 29 '18 at 06:38

1 Answers1

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Let $C=(I-\alpha D)$. Then \begin{align} \Sigma(\alpha) &= (C^\ast)^{-1} + C^{-1} - I\\ &= (C^\ast)^{-1}(C + C^\ast - C^\ast C)C^{-1}\\ &= (C^\ast)^{-1}\left[I-(C^\ast-I)(C-I)\right]C^{-1}\\ &= (C^\ast)^{-1}(I-\alpha^2D^\ast D)C^{-1}\\ \end{align} Since $\Sigma(1)\succ0$, we must have $I\succ D^\ast D$. Hence $I-\alpha^2D^\ast D$ and in turn $\Sigma(\alpha)$ are positive definite for every $-1\le\alpha\le1$.

user1551
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