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Prove that the function $$ f(X) = \operatorname{trace}(X^{−1}) $$ is convex on the domain $S^n_{++}$.

I was given the hint to try using line restriction. So I am trying to prove that $$ g(t) = f(x+ty) $$ is convex for all x $\epsilon $ $ S^{n}_{++} $ and y $\epsilon$ $ R^n $ such that $ \{ t :x+ty \ \epsilon \ S^{n}_{++} \} $. I have tried calculating the hessian matrix but I am getting confused on how to take the gradient of a matrix and the trace function.

Diglett
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kann
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1 Answers1

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For $p,\ q\in S^+$, then a linear combination $ c(t)=tp +(1-t)q$ is symmetric. If $p,\ q$ are close, then $c(t)$ is positive definite.

Note that $c(t)=p+tV$, where $V$ is an any symmetric matrix, is a shortest path. In further, $$ c (t)^{-1}= p^{-1}-tp^{-1}Vp^{-1} -t^2 p^{-1} Vp^{-1} Vp^{-1} +\cdots $$

Hence $$ \frac{d^2}{dt^2}\ f\circ c\ (t) =-2\cdot {\rm trace}\ p^{-1} Vp^{-1} Vp^{-1} \leq 0$$ since $ p^{-1} Vp^{-1} Vp^{-1} $ is non-negative definite : If $\langle\ ,\ \rangle$ is an inner product, and $u$ is a vector, then $$ \langle p^{-1} Vp^{-1} Vp^{-1} u,u\rangle = \langle p^{-1} Vp^{-1} u,Vp^{-1}u\rangle \geq 0 $$

HK Lee
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    There is a flaw in $c(t)^{-1}$, exactly $c(t)^{-1} = p^{-1} - tp^{-1}Vp^{-1} + t^2p^{-1}Vp^{-1}Vp^{-1} + \cdots$, then $$ f^{\prime\prime}(c(t)) = 2 tr(p^{-1} V p^{-1} V p^{-1}) \ge 0 $$ hence $f(c(t))$ is convex – Chia Dec 07 '22 at 08:56