Is the trace of inverse matrix convex?

Question

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..

Answer 1

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 $\text{Tr}(CA^{-1} C^T) \ge 0$.