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For $x\in\mathbb{R}^n$ and $A\in\mathbb{S_{++}^n}$ (symmetric positive definite), it is very well known that $f(x)=x^TA^{-1}x$ is convex since $A^{-1}$ is positive definite.

I wonder what if we change the function input, i.e.

1) $f(A)=x^TA^{-1}x$,

2) $f(x,A)=x^TA^{-1}x$.

Lee
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    Your second derivative is not correct for (1). It would only be true if it was $A$, not $A^{-1}$, in the middle. – Ninad Munshi Sep 17 '19 at 01:56
  • @NinadMunshi, thanks. I found that $\frac{\partial f(A)}{\partial A}=-A^{-T}xx^TA^{-T}$, but I think finding second derivative is very difficult. – Lee Sep 17 '19 at 02:08
  • As it would be, since it's not a matrix anymore but a rank $4$ tensor – Ninad Munshi Sep 17 '19 at 02:14

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