3
$\begingroup$

Let $X$ be a positive semidefinite matrix. Is the mapping $X\to X^{-2}$ convex?

Update: or is $Tr[X^{-2} K]$ convex for PSD $X$ and $K$?

$\endgroup$
1
  • $\begingroup$ I don't think so. Maybe checking the Donoghue book on monotone matrix functions could be useful since convex functions are functions whose derivative is increasing. $\endgroup$
    – Bazin
    May 16, 2018 at 19:42

1 Answer 1

4
$\begingroup$

This function is not operator convex. Counterexamples are trivial to find (just pick some random psd matrices, you'll see).

However, that said, it is true that the map $X \mapsto X^{-p}$ is operator convex for $0 < p < 1$, which essentially follows from operator convexity of $X^{-1}$.


EDIT: The updated question asks a much simpler question, whether $\text{tr}(X^{-2}K)$ is convex for psd $K$. I concluded there a bit hastily that using $K=\sum_i v_iv_i^T$, then using linearity of trace the question boils down to verifying convexity of $\text{tr}(X^{-2}vv^T)=v^TX^{-2}v$. But actually, this function is not convex, and again one can find easy numerical counterexamples.

An exception is the choice $K=I$, for which the claim holds.

$\endgroup$
5
  • $\begingroup$ Thanks! In fact what I need is to show that Tr[X^{-2} K] is convex where both X and K are PSD. If there was no K, I could use the convexity of the trace function (en.wikipedia.org/wiki/Trace_inequalities). Any ideas if it is true for a general PSD K? $\endgroup$ May 16, 2018 at 20:40
  • 1
    $\begingroup$ Try taking $K$ a rank $1$ projection. Then, $Tr(X^{-2}K)=Tr(KX^{-2}K)=Tr(K)v^*X^{-2}v$ where $vv^* = K.$ Doesn't that reduce it to the classical case? $\endgroup$ May 16, 2018 at 21:00
  • 1
    $\begingroup$ @SoheilFeizi that is a different question then. Also noting J. E. Pascoe's statement, write $K=\sum_i v_iv_i^T$ for rank-one matrices, and then apply linearity of trace and the same argument in his comment to conclude. $\endgroup$
    – Suvrit
    May 16, 2018 at 21:58
  • $\begingroup$ I oversaw something in my comment; actually, the claim does not hold; I've updated my answer to reflect that. $\endgroup$
    – Suvrit
    May 17, 2018 at 19:00
  • $\begingroup$ Survit, since $Tr[X^{-2}]$ is convex, doesn't Lemma 1.1 of arxiv.org/pdf/1409.0564.pdf imply convexity of $Tr[X^{-2}K]$? $\endgroup$ May 17, 2018 at 21:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.