Is the trace of inverse matrix convex? 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..
 A: 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
$\text{Tr}(CA^{-1} C^T) \ge 0$.
A: 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.
