Is it true that for a fixed positive definite $C\in\mathbb{R}^{n\times n}$ and $M\in\mathbb{R}^{n\times n}$ and a positive definite $T\in\mathbb{R}^{n\times n}$ ($M$ and $T$ are to be chosen), \begin{align*} \text{Tr}(C(MCM^\top T^{-1}+I)^{-1}) \end{align*} is minimized when the eigenvalues of $MT^{-1}M^\top$ are maximized.
My intuition:
- If we were to simplify the matrices to just single variables (i.e., $1\times1$ matrices), the goal would simply become minimizing $\frac{C}{M^2C/T+1}=\frac{1}{M^2/T+1/C}$, which holds when $\frac{M^2}{T}$ is maximized.
- If they are all diagonal matrices, then the above is also true.
- I also thought of using Woodbury matrix identity to simplify the above equation but got stuck: \begin{align*} \text{Tr}(C(MCM^\top T^{-1}+I)^{-1}) &=\text{Tr}(CT(MCM^\top+T)^{-1}) \\ &=\text{Tr}(CT(T^{-1}-T^{-1}M(C^{-1}+M^\top T^{-1}M)^{-1}M^\top T^{-1})) \\ &=\text{Tr}(C(I-M(C^{-1}+M^\top T^{-1}M)^{-1}M^\top T^{-1})) \\ &=\text{Tr}(C)-\text{Tr}(CM(C^{-1}+M^\top T^{-1}M)^{-1}M^\top T^{-1}) \\ &=\text{Tr}(C)-\text{Tr}(CM^\top T^{-1}M(C^{-1}+M^\top T^{-1}M)^{-1}). \end{align*} Then minimizing the above is just maximizing $\text{Tr}(CM^\top T^{-1}M(C^{-1}+M^\top T^{-1}M)^{-1})$ since $C$ is fixed. I got stuck at this point.