From http://www.ueltschi.org/AZschool/notes/EricCarlen.pdf, I have known that $FA^{-1}F^T$ is jointly convex for $(F,A) \in \mathbf{M}_n \times \mathbf{H}_n^+$. Here $\mathbf{M}_n$ is the space of $n \times n$ matrices and $\mathbf{H}_n^+$ is the space of symmetric positive matrices. Here we only consider the case all the entries are in $\mathbf{R}$.
Then is $\mbox{Tr}(FA^{-1}F^T)$ convex for $(F,A) \in \mathbf{M}_n \times \mathbf{H}_n^+$? is it strictly convex? If not, how about for $F$ invertible?
And is $\mbox{Tr}(FA^{-1}F^T)$ rank-one convex for $F$ being rank one matrix?
At any point $(F_0,A_0)$ where $F_0$ is rectangular, $A_0$ is positive definite and the product $FA$ makes sense, the second-order directional derivative of the trace along any direction $(F,A)$, where $A$ is Hermitian, is \begin{align*} &\operatorname{tr}(FA_0^{-1}F^T - FA_0^{-1}AA_0^{-1}F_0^T - F_0A_0^{-1}AA_0^{-1}F^T + F_0A_0^{-1}AA_0^{-1}AA_0^{-1}F_0^T)\\ &=\|FA_0^{-1/2} - F_0A_0^{-1}AA_0^{-1/2}\|_F^2 \ge0. \end{align*} ($\|\cdot\|_F$ means Frobenius norm.) Therefore the trace is convex in $(F,A)$.
The function $\operatorname{tr}\left((1+t)F\,\left((1+t)A\right)^{-1}\,(1+t)F^T\right)$ is affine in $t$. Therefore $(F,A)\mapsto\operatorname{tr}\left(FA^{-1}F^T\right)$ is not strictly convex.