I have the following function $f(S)=\mathrm{trace}(S)+m^2\mathrm{trace}(S^{-2})$ where $S\in \mathcal{M}_{m,m}$ symmetric positive definite matrix. I'm trying to prove the convexity of this function and so I'm wondering how to show properly the convexity of $f(S)$.

  • $\begingroup$ what exactly do you mean by $\frac{\partial^2 f}{\partial S^2}$ ? the differential of order two of $f$ is a priori a bilinear form on a space of matrixes, and so should take scalar values $\endgroup$ – Albert Mar 9 '13 at 21:58
  • $\begingroup$ Yes you're right. I have removed that part from my question $\endgroup$ – user2987 Mar 9 '13 at 22:05

Let $S_1$, $S_2$ be two positive definite matrices. Let $\Delta = S_2 - S_1$ and for $t \in [0,1]$, let $$\phi = (S_1 + t\Delta)^{-1} = ((1 - t) S_1 + t S_2)^{-1}$$ We have:

$$\begin{align} & \frac{d}{dt} \phi \;= - \phi \Delta \phi\\ \implies & \frac{d}{dt} \phi^2 \;= - \phi \Delta \phi^2 - \phi^2 \Delta \phi\\ \implies & \frac{d^2}{dt^2} \phi^2 = ( \phi \Delta \phi ) \Delta \phi^2 + \phi \Delta ( \phi \Delta \phi^2 + \phi^2 \Delta \phi ) + ( \phi \Delta \phi^2 + \phi^2 \Delta \phi ) \Delta \phi + \phi^2 \Delta ( \phi \Delta \phi ) \end{align}$$ Taking trace on both side, we get $$\begin{align} \frac{d^2}{dt^2} \operatorname{tr}(\phi^2) &= 2\operatorname{tr}( (\phi\Delta)^2\phi^2 + (\phi\Delta\phi)^2 + \phi^2 (\Delta\phi)^2)\\ &= 2\operatorname{tr}\left( 2 (\sqrt{\phi}\Delta\sqrt{\phi}^3)^T(\sqrt{\phi}\Delta\sqrt{\phi}^3) + (\phi\Delta\phi)^T(\phi\Delta\phi)\right)\ge 0\tag{*} \end{align}$$

Notice for any $t \in [0,1]$, $\phi$ is invertible. This means $\phi\Delta\phi$ is non-zero and hence the R.H.S of $(*)$ is actually positive. As a result,

$$\frac{d^2}{dt^2}\operatorname{tr}\left(((1-t)S_1 + t S_2) + m^2((1 - t) S_1 + t S_2)^{-2}\right) > 0 $$

over $[0,1]$ and hence $\operatorname{tr}(S+m^2 S^{-2})$ is convex over the space of positive definite matrices.


Thinking more about this, it might be cleaner to prove $\operatorname{tr}(S^{-n})$ is convex for all $n \ge 1$ at once.

Let $\psi(t) = S_1 + t\Delta$ and for any $\lambda > 0$, let $Z_{\lambda}(t) = \operatorname{tr}(e^{-\lambda \psi(t)})$, we have:

$$\begin{align} \frac{d}{dt}Z_{\lambda}(t) &= \operatorname{tr}\left( \int_0^1 ds\;e^{-\lambda s\psi(t)}( -\lambda\Delta )e^{-\lambda(1-s)\psi(t)}\right)\\ &= -\lambda \operatorname{tr}\left(e^{-\lambda\psi(t)}\Delta\right)\\ \implies \frac{d^2}{dt^2}Z_{\lambda}(t) &= \lambda^2 \operatorname{tr}\left(\int_0^1 ds\;e^{-\lambda s\psi(t)}\Delta e^{-\lambda(1-s)\psi(t)}\Delta\right)\\ &= \lambda^2 \int_0^1 ds \operatorname{tr}\left( ( e^{-\frac{\lambda s}{2}\psi(t)}\Delta e^{-\frac{\lambda(1-s)}{2}\psi(t)} )^T ( e^{-\frac{\lambda s}{2}\psi(t)}\Delta e^{-\frac{\lambda(1-s)}{2}\psi(t)} ) \right)\\ &> 0 \end{align}$$

So for any $n \ge 1$, we have:

$$\frac{d^2}{dt^2} \operatorname{tr}( \psi(t)^{-n} ) =\frac{d^2}{dt^2} \operatorname{tr}\left(\int_0^{\infty}\frac{\lambda^{n-1}}{n!} e^{-\lambda\psi(t)} d\lambda\right) = \frac{1}{n!}\int_0^{\infty} \lambda^{n-1} \frac{d^2Z_{\lambda}(t)}{dt^2} d\lambda > 0 $$

  • $\begingroup$ Many thanks for your general proof. $\endgroup$ – user2987 Mar 10 '13 at 0:22
  • $\begingroup$ @achille hui: but if we take $m=1$ our function becomes $f(s)= s+\frac{1}{s^2}$ which is not convex and discontinuous function at $0$. $\endgroup$ – user2987 Mar 10 '13 at 20:25
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    $\begingroup$ @Bel, we are working with positive definite matrices. In a base where $S$ is approximately diagonal, all entries on the diagonal will be positive. This corresponds to $s > 0$ in the scalar case and $f(s)$ is convex there. $\endgroup$ – achille hui Mar 10 '13 at 20:49
  • $\begingroup$ yes you're right! $\endgroup$ – user2987 Mar 10 '13 at 20:56
  • $\begingroup$ @achille hui: Do you have any idea which solver can solve properly the minimization of $f(S)$. Actually I tried cvx in matlab but it cannot recognize the function as convex because it considers that there is a convex equality constraint between the input of $\mathrm{trace}$ and the input of $\mathrm{trace}_{inverse}$. That is, $Q=S^2$ inside trace inverse operator which is quadratic and hence convex equality in the convex optimization problem thus the problem is not convex anymore. $\endgroup$ – user2987 Mar 10 '13 at 22:33

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