I am trying to solve the following problem
$\min_{T} \operatorname{trace} \left( A(T^T M T + N)^{-1}A^T\right)$,
where $T$ is the matrix I am solving for and $A$ is given, $M\succ0$ and $N\succ0$. $A$ and $N$ are $n\times n$, while $T$ is $m \times n$ and $M$ is $m \times m$.
Any tips and tricks on how to solve this problem would be greatly appreciated!
Let $B=(T^TMT+N)$. Then the function and its differential can be written in terms of the Frobenius (:) product as $$\eqalign{ f &= A:AB^{-1} \cr &= A^TA:B^{-1} \cr\cr df &= A^TA:dB^{-1} \cr &= -A^TA:B^{-1}\,dB\,B^{-1} \cr &= -B^{-T}A^TAB^{-T}:dB \cr &= -B^{-T}A^TAB^{-T}:d(T^TMT) \cr &= -B^{-T}A^TAB^{-T}:(dT^TMT + T^TMdT) \cr &= -(B^{-T}A^TAB^{-T}:dT^TMT) - (B^{-T}A^TAB^{-T}:T^TMdT) \cr &= -(B^{-1}A^TAB^{-1}:T^TM^TdT) - (B^{-T}A^TAB^{-T}:T^TMdT) \cr &= -(MTB^{-1}A^TAB^{-1}:dT) - (M^TTB^{-T}A^TAB^{-T}:dT) \cr &= -(MTB^{-1}A^TAB^{-1} + M^TTB^{-T}A^TAB^{-T}):dT \cr }$$ Since $df=(\frac{\partial f}{\partial T}:dT)\,\,\,$ the gradient must be $$\eqalign{ \frac{\partial f}{\partial T} &= -(MTB^{-1}A^TAB^{-1} + M^TTB^{-T}A^TAB^{-T}) \cr &= -MT(T^TMT+N)^{-1}A^TA(T^TMT+N)^{-1} - M^TT(T^TMT+N)^{-T}A^TA(T^TMT+N)^{-T} \cr\cr }$$ The usual procedure of setting the gradient to zero and solving leads to any $T$ such that $MT=0$, including the trivial solution $T=0$.