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 $K$?

  • $\begingroup$ 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


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.

  • $\begingroup$ 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
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    $\begingroup$ 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
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    $\begingroup$ @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

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