Is the trace of inverse matrix convex?

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

Solution 2:

We can prove that $Tr\{X^{-1}\}$ is convex function by considering an arbitrary line, $X = Z + tV$, where $V \in S^n$ and $Z \in S^n_{++}$. Let $g(t) = f(Z+tV)$ for all $t \in \mathbb R$ such that $Z+tV \in S^n_{++}$. If (for all $X \in S^n_{++}$ and all $V \in S_n$) $g$ is convex then $f$ is convex. Consider \begin{align} g(t) &= Tr\{X^{-1}\} = Tr\{(Z+tV)^{-1}\}\\ & = Tr\{Z^{-1/2}(I+tZ^{-1/2}VZ^{-1/2})^{-1})Z^{-1/2}\} \\ & = Tr\{Z^{-1}(I+tZ^{-1/2}VZ^{-1/2})^{-1})\} \qquad \qquad \because \; Tr\{BA\} = Tr\{AB\} \\ \end{align} Since, $(Z^{-1/2}VZ^{-1/2})^T = Z^{-1/2}VZ^{-1/2}$ as $Z,V \in S^n$. We can write, $ Z^{-1/2}VZ^{-1/2} = Q\Sigma Q^T$, where $Q$ is orthogonal matrix and $\Sigma$ is a diagonal matrix with eigen values $\lambda_i$'s $ Z^{-1/2}VZ^{-1/2}$ of as diagonal entries. Hence,

\begin{align} g(t) & = Tr\{Z^{-1}(I+tQ\Sigma Q^T)\} \quad \because \; Tr\{BA\} = Tr\{AB\} \\ & = Tr\{Z^{-1}Q^T(I+t\Sigma)Q\} \quad \because \; Q^{-1} = Q^T \\ & = Tr\{QZ^{-1}Q^T(I+t\Sigma)\} \\ & = \sum_{i=1}^n (QZ^{-1}Q^T)_{ii} \frac{1}{1+t\lambda_i}\\ \implies g''(t) &= \sum_{i=1}^n (QZ^{-1}Q^T)_{ii} \frac{2\lambda_i^2}{(1+t\lambda_i)^3} \end{align} As, $(1+t\lambda_i)>0$ and $QZ^{-1}Q^T$ is symmetric and positive definite matrix ($\because Z^{-1} \in S^n_{++} $) resulting in the diagonal entries of $QZ^{-1}Q^T$ being positive values. Hence $g''(t)>0$ is always true. This concludes that $f$ is convex.