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Let $A\in\mathbb{R}^{n\times n}$ be a positive semidefinite matrix and $D\in\mathbb{R}^{n\times n}$ be a diagonal positive definite matrix. Let $\mathrm{diag}(X)\in\mathbb{R}^{n\times n}$ denote the diagonal part of a matrix $X\in\mathbb{R}^{n\times n}$.

My question: Does the following inequality hold true $$ \mathrm{diag}\left[(A+D)^{-1}\right] \ge \left[\mathrm{diag}(A)+D\right]^{-1}\ \ ? $$

For the sake of clarity, I describe below a simple example. Consider $$ A=\begin{bmatrix}2 & 1 \\ 1 & 3\end{bmatrix}, \quad D=\begin{bmatrix}0.5 & 0 \\ 0 & 2\end{bmatrix}. $$ We have $$ \mathrm{diag}\left[(A+D)^{-1}\right] = \begin{bmatrix}0.2927 & 0\\0 & 0.1341 \end{bmatrix}, \quad \left[\mathrm{diag}(A)+D\right]^{-1} = \begin{bmatrix}0.1818 & 0 \\0 & 0.0833 \end{bmatrix}, $$ so that $\mathrm{diag}\left[(A+D)^{-1}\right] \ge \left[\mathrm{diag}(A)+D\right]^{-1}$.

(If true, the above inequality perhaps is just a well-know inequality but I wasn't able to find it in any "classical" matrix theory textbooks.)

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  • $\begingroup$ The inequality is equivalent to $\operatorname{diag}\left(B^{-1}\right) \geq \left(\operatorname{diag} B \right)^{-1}$ for all positive definite $B$, right? $\endgroup$ May 19, 2017 at 15:36
  • $\begingroup$ @darijgrinberg: Yes, if you take $B=A>0$ and $D=0$. Otherwise I don't see the equivalence. $\endgroup$
    – Ludwig
    May 19, 2017 at 15:40
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    $\begingroup$ To get yours from mine, set $B = A + D$. To get mine from yours, set $A = B$ and $D = 0$. Setting $D = 0$ is harmless, since positive semidefinite matrices are limits of positive definite matrices. $\endgroup$ May 19, 2017 at 15:46
  • $\begingroup$ @darijgrinberg: Yes, absolutely right. $\endgroup$
    – Ludwig
    May 19, 2017 at 15:54

1 Answer 1

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This is true. As Darij suggests, denote $B=A+D$, let $e=e_i$ be a basic unit vector. We have to prove $(B^{-1}e,e)\geqslant (Be,e)^{-1}$. Denote further $C=B^{1/2}$, this rewrites $\|Ce\|\cdot \|C^{-1}e\|\geqslant 1$. This follows from Cauchy-Bunyakovsky-Schwarz, as $(Ce,C^{-1}e)=(e,e)=1$.

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