# Optimizing a function of a matrix

Let \begin{aligned} W= & \underset{X}{\mathrm{maximize}} & & \log \left|X + K_1\right|- \alpha \log \left|X + K_2\right|\\ & \mathrm{subject \; to} & & 0 \preceq X \preceq P, \end{aligned}

in which $X$ and $P$ are symmetric positive semi-definite matrices, and $K_1$, and $K_2$ are symmetric positive definite matrices, $\alpha>1$ is a scalar, and $|\cdot|$ represents the determinant.

I want to know if the following answer is correct or not

\begin{align} X^* =\begin{cases} P, &\mathrm{if} \quad \alpha (P+K_1) \preceq P+K_2 \\ 0, &\mathrm{if} \quad \alpha K_{1} \succeq K_{2}\\ \frac{K_{2}- \alpha K_{1}}{\alpha -1}, &\mathrm{otherwise} \end{cases} \end{align}

Remark: I have solved the above problem by analogy from

\begin{aligned} & \underset{x}{\mathrm{maximize}} & & \log \left(x + k_1\right)- \alpha \log \left(x + k_2\right)\\ & \mathrm{subject \; to} & & 0 \le x \le p, \end{aligned} in which $x\ge 0$, $p\ge0$, $k_1>0$, $k_2>0$, $\alpha>1$ all are scalars.

• I take it your partial ordering is given by "$M \geq N$ if $M − N \geq 0$; i.e., $M − N$ is positive semi-definite". – muaddib Jun 24 '15 at 22:45
• Yes, your answer is correct. – lynn Jun 24 '15 at 23:58
• @muaddib: yes, I mean so. – Bob Jun 25 '15 at 20:14

Bob, it's more complicated. The domain $D$ is defined by $0\leq X\leq P$. If $X$ is in the interior of $D$, then an extremum is reached in $X_0=(K_2-\alpha K_1)/(\alpha-1)$ -if $X_0\in int(D)$- (you must prove it). It remains to consider the edge $K$ of $D$; unfortunately $K$ strictly contains $\{0,P\}$; for instance , if $X\in D$ is not invertible or if $X\in D$ and $P-X$ non-invertible , then $X\in K$.