Optimize a log det function with respect to a matrix, and the saddle point analysis

Question

Suppose I want to to find the local minima of a logdet function $\mathcal{L}$ with respect to a Matrix $\mathbf{A}$, $$ \mathcal{L} = \log\vert \mathbf{I} + \mathbf{A}\mathbf{S} \vert - \mathbf{q}^T(\mathbf{A}^{-1} + \mathbf{S})^{-1} \mathbf{q}, $$ where $\mathbf{S}$ is a Symmetric Positive Semi-Definite Matrix, $\mathbf{q}$ is a column vector, $\mathbf{I}$ is an identity matrix. I could find the local minima of $\mathcal{L}$ by solving, $$ \min_\mathbf{A} \mathcal{L} \qquad \rightarrow \qquad \frac{\partial \mathcal{L}}{\partial \mathbf{A}} = \mathbf{0}. $$ the closed form solution of $\mathbf{A}$ ($\frac{\partial \mathcal{L}}{\partial \mathbf{A}} = \mathbf{0}$) can be derived, $$ \mathbf{A} = \mathbf{S}^{-1}(\mathbf{S} - \mathbf{q}\mathbf{q}^T)\mathbf{S}^{-1} $$

The questions are

1. Given $\mathbf{S}$ a positive definite matrix, $\mathbf{qq}^T$ a rank one positive semi-definite matrix, what's the condition that $\mathbf{A} = \mathbf{S}^{-1}(\mathbf{S} - \mathbf{q}\mathbf{q}^T)\mathbf{S}^{-1}$ being a PD(or PSD) matrix ? Is $\mathrm{Trace}[\mathbf{A}] > 0$ sufficient ?

2. How to analyze the stationary point of the partial derivatives $\frac{\partial \mathcal{L}}{\partial \mathbf{A}} = \mathbf{0}$ ?

Answer 1

As you have used $S^{-1}$ to define $A$, presumably $S$ is positive definite. Now $A$ is p.s.d. if and only if $S-qq^T$ is p.s.d.. Let $Q$ be a real orthogonal matrix with $q/\|q\|$ as its first column. Then $$ B=Q^T(S-qq^T)Q=\pmatrix{\frac{q^TSq}{\|q\|^2}-\|q\|^2 & v^T\\ v&C} $$ for some vector $v$ and some positive definite matrix $C$. Therefore $B$ is p.s.d if and only if $\det(B)\ge0$. Using Schur complement, we get $\det(B)=\det(C)\left(\frac{q^TSq}{\|q\|^2}-\|q\|^2-v^TC^{-1}v\right)$. Therefore $A$ is p.s.d. if and only if $\frac{q^TSq}{\|q\|^2}-\|q\|^2-v^TC^{-1}v\ge0$.

As for the second question, I have no idea what you are asking.

Answer 2

I believe that you made a sign error in deriving your analytical result. It should be $$ \mathbf{A} = -\mathbf{S}^{-1}(\mathbf{S} + \mathbf{q}\mathbf{q}^T)\mathbf{S}^{-1} $$ which is negative definite.