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Let $B$ be an $n\times n$ matrix, and define $f$ to be the function that maps positive semidefinite (PSD) $n\times n$ matrices $A$ to real numbers by

$$ f(A) = \mathrm{trace}( (B^*A^2B)^{1/3}). $$

In other words, $f$ maps $A$ to the sum of $1/3$-powers of the eigenvalues of the PSD matrix $B^*A^2B$.

Is the function $f$ concave over the PSD cone? I.e. is it true that for any two PSD matrices $X$ and $Y$, $f((X + Y)/2) \ge f(X)/2 + f(Y)/2$?

A more general question is whether the function $f(A) = \mathrm{trace}( (B^*A^2B)^{p})$ is concave over the PSD cone for $0< p < 1/2$.

Carlen and Lieb have some closely related results, but I could not find the particular combination of matrix powers in their paper or in other related work on trace inequalities.

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  • $\begingroup$ Do you mean $B^*A^2B$ like in your reference? I think that $BA^2B$ need not be PSD. $\endgroup$ – Brendan McKay Jun 11 '17 at 12:41
  • $\begingroup$ @BrendanMcKay Indeed, this was a typo. Thank you! $\endgroup$ – Sasho Nikolov Jun 11 '17 at 12:43
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Unfortunately, the conjectured function is not concave. Here is a simple simpler counterexample.

\begin{equation*} B = \begin{bmatrix} 1 & 2 \\ 3 & 4\end{bmatrix},\quad A = \begin{bmatrix} 2 & 0 \\ 0 & 3\end{bmatrix},\quad C = \begin{bmatrix} 5 & 0 \\ 0 & 2\end{bmatrix}. \end{equation*}

With this choice, and for $f(X)=\text{tr} (B^*X^2B)^{1/3}$ we see that $f((A+C)/2)- (f(A)+f(C))/2 \approx -0.0616$.


You may find the following necessary condition interesting:

(Prop. 5.1, Hiai, 2013). Assume that $p,s\neq 0$. If $A \in \mathbb{P}_2 \mapsto \text{tr}(B^*A^pB)^s$ is concave for any invertible matrix $B \in \mathbb{C}^{2\times 2}$, then either $0<p\le 1$ and $0<s\le 1/p$, or $-1\le p \le 0$ and $1/p\le s < 0$.

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    $\begingroup$ In the question $f$ is defined on PSD matrices. I.e. I am asking if $f$ is concave on the PSD cone. $C$ does not appear to be PSD as its determinant is -2. $\endgroup$ – Sasho Nikolov Jun 11 '17 at 14:22
  • $\begingroup$ BTW, if you allow non-PSD matrices, then it's easy to see that even the one-dimensional case fails. $\endgroup$ – Sasho Nikolov Jun 11 '17 at 14:42
  • $\begingroup$ I made a typo; just replace the 7 in the diagonal with an 8 or 9; the counterexample still holds; The $2\times 2$ entry of $C$ can be made even bigger if you want an even sharper violation of concavity. $\endgroup$ – Suvrit Jun 11 '17 at 15:48
  • $\begingroup$ Thanks! Can you give a link or a reference to the Hiai characterization? $\endgroup$ – Sasho Nikolov Jun 11 '17 at 18:38
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    $\begingroup$ You're welcome! Added the link now (I had gotten lazy to type it out previously :-) ) $\endgroup$ – Suvrit Jun 11 '17 at 19:11

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