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Let $A \in \mathbb{R}^{n \times n}$ be a positive definite matrix. I am looking for the solution to the problem:

\begin{align} \arg \min_{A} \text{Tr}(A^{-1}) \end{align}

Actually, I know the trace of the inverse of a symmetric positive definite matrix $\mathrm{trace}(A^{-1})$ is convex and I found the proof here: Is the trace of inverse matrix convex?. I was wondering whether there is a exact solution?

Thanks in advance!

SJ93
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    $A^{-1}$ is not defined if $A$ has a zero eigenvalue, so it is not clear what you mean. By choosing $A = tI$ you get $\operatorname{tr} A = {n \over t}$ so the $\inf$ (if you ignore the first detail) is zero. – copper.hat Dec 07 '20 at 18:18
  • let's assume $A$ is positive definite with all eigen values greater than zero. @copper.hat – SJ93 Dec 07 '20 at 18:20
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    See my previous comment. – copper.hat Dec 07 '20 at 18:21
  • so you mean the minimum occurs when $A=tI$ and the minimum value is $n/t$ ? I don't understand what you mean by "the inf is zero"[email protected] – SJ93 Dec 07 '20 at 18:27
  • No. If you consider the function $x \mapsto {1 \over x}$ on $x>0$ then it has no $\min$ but it has an $\inf$. – copper.hat Dec 07 '20 at 18:29
  • you mean there is no min for $\text{Tr}(A^{-1})$ ? but it is proved to be convex. @copper.hat – SJ93 Dec 07 '20 at 18:34
  • It is not defined for semi definite matrices. If you restrict to positive definite matrices no $\min$ exists (the $\inf$ does). Nothing in what I said contradicts convexity. – copper.hat Dec 07 '20 at 18:40
  • considering the $1\times 1$ case, then for $x\in \mathbb R_{\gt 0}$ what is the 'minimum' of $\frac{1}{x}$? For that matter, what is the 'minimum' of $\exp(x)$ for $x\in \mathbb R_{\gt 0}$? Both convex functions, but open sets need not have minimums. – user8675309 Dec 07 '20 at 19:32

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