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Hi I would like to know whether the trace of the inverse of a symmetric positive definite matrix $\mathrm{trace}(S^{-1})$ is convex.

Actually I know that the trace of a symmetric positive definite matrix $S\in M_{m,m}$ is convex since we can find $B\in M_{n,m}$ such that $S=B^T\times B$ then we can write the trace as the sum of scalar quadratic forms, i.e. $\mathrm{trace}(S)=\mathrm{trace}(B^T\times B)=\sum_{j=1}^mb_j^T\times b_j$ where $b_j$ is the $j^{th}$ column of $B$.

for instance if we have $trace([\begin{array}{cc} 1 & 2 \\ 3 & 4 \\ \end{array}] \times [\begin{array}{cc} 1 & 3 \\ 2 & 4 \\ \end{array}])= [\begin{array}{cc} 1 & 2 \\ \end{array}]\times [\begin{array}{c} 1 \\ 2 \\ \end{array}]+ [\begin{array}{cc} 3 & 4 \\ \end{array}]\times [\begin{array}{c} 3 \\ 4 \\ \end{array}]=30$

And so I wonder if $\mathrm{trace}(S^{-1})$ is convex too..

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  • $\begingroup$ So you are asking if the function $S\longmapsto \mbox{trace}\;S^{-1}$ is convex on symmetric positive definite matrices? $\endgroup$ – Julien Feb 8 '13 at 0:48
  • $\begingroup$ yes I would like to know if that function is convex $\endgroup$ – user2987 Feb 8 '13 at 0:51
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    $\begingroup$ The fact that the trace of the matrix itself is convex is obvious, because the trace is linear. That stuff about $B^T B$ is irrelevant (and wrong, since you want to look at a function of $S$, not of $B$). $\endgroup$ – Robert Israel Feb 8 '13 at 1:12
  • $\begingroup$ yes I want to look at the function itself but remember that a function can be convex just in a specific interval. And in this case I guess that this interval is the symmetric positive definite matrices.. $\endgroup$ – user2987 Feb 9 '13 at 22:24
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Yes, it is. Consider $S(t) = A + t B$ where $A$ is symmetric positive definite and $B$ is symmetric. It is enough to show that $$\left.\dfrac{d^2}{d t^2} \text{Tr}(S(t)^{-1})\right|_{t=0} \ge 0$$ Now $$ S(t)^{-1} = (A (I + t A^{-1} B))^{-1} = A^{-1} - t A^{-1} B A^{-1} + t^2 A^{-1} B A^{-1} B A^{-1} + \ldots$$ so $$ \left. \dfrac{d^2}{\partial t^2} \text{Tr}(S(t)^{-1}) \right|_{t=0} = 2 \text{Tr}(A^{-1} B A^{-1} B A^{-1})$$ But $A^{-1} B A^{-1} B A^{-1} = C A^{-1} C^T$ where $C = A^{-1} B$ and $A^{-1}$ is positive definite, so $C A^{-1} C^T$ is positive semidefinite, and therefore $\text{Tr}(CA^{-1} C^T) \ge 0$.

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    $\begingroup$ Very nice Robert, +1. $\endgroup$ – Julien Feb 8 '13 at 1:50
  • $\begingroup$ So can we found the scalar quadratic form for $\mathrm{trace}(S^{-1})$ like I did for $\mathrm{trace}(S)$? $\endgroup$ – user2987 Feb 11 '13 at 21:17
  • $\begingroup$ What scalar quadratic form? $\endgroup$ – Robert Israel Feb 11 '13 at 21:21
  • $\begingroup$ like the one that I have written for $\mathrm{trace}(S)=\mathrm{trace}(B^T\times B)=\sum_{j=1}^mb_j^T\times b_j$ where $b_j$ is the $j^{th}$ column of $B$. $b_j^T\times b_j$ is scalar quadratic form for $Q=I_{m\times m}$. $\endgroup$ – user2987 Feb 12 '13 at 0:55
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    $\begingroup$ A twice differentiable function $f$ on an open subset $U$ of a vector space $V$ is convex iff $$ \left.\dfrac{d^2}{dt^2} f(u + tv)\right|_{t=0} \ge 0$$ for all $u \in U$, $v \in V$. The point is that other values of $t$ (for which $u+tv \in U$) correspond to other choices for $u$ at $t=0$. $\endgroup$ – Robert Israel Nov 19 '13 at 4:31

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