# Maximize $\det X$, subject to $X_{ii}\leq P_i$, where $X>0$

Given $$P_1,P_2,\cdots,P_N$$.

$$\begin{array}{ll} \text{maximize} & \det X\\ \text{subject to} & \mathrm X_{ii}\leq P_i \\\forall i=1,2,\cdots,n\end{array}$$

$$X\in\mathbb{R}^{n\times n}$$, $$X>0$$ (i.e. positive definite).

For $$n=2$$, maximum is achieved when $$X_{ii}=P_i$$ and $$X$$ is diagonal matrix.

Then for $$n=3$$, Let $$X=\begin{bmatrix}X_1 & X_2\\X_2^T &x_3\end{bmatrix}$$, where $$X_1\in\mathbb{R}^{2\times2}$$.

$$\det(X)=\det(X_1)\times \det(x_3-X_2^TX_1^{-1}X_2)\leq \det(X_1)\times x_3\leq P_1P_2P_3$$.

Then following same logic my conjecture is $$\max\{ \det X\}=P_1\cdots P_n$$, when $$X_{ii}=P_i$$.

Is it correct? If not, can you please point my mistake or give me counter example? Thanks

Your problem is equivalent to maximizing $$\log \det X$$ such that $$X_{ii} \leq P_i$$ and $$X$$ is positive definite. That is convenient, because now you have a convex optimization problem that satisfies the Slater condition, so the KKT conditions are necessary and sufficient. The Lagrangian is $$L(X,\lambda) = \log\det X - \sum_i \lambda_i (X_{ii} - P_i)$$ The derivative of $$\log\det X$$ is $$(X^{-1})^T$$, so the KKT conditions are: $$(X^{-1})_{ii} - \lambda_i = 0$$ $$(X^{-1})_{ij} = 0 \quad (i \neq j)$$ $$\lambda_i (X_{ii} - P_i)=0$$ $$X_{ii} \leq P_i$$ $$\lambda \geq 0.$$ The point you found satisfies these conditions and is therefore optimal.

• what if rank$X< n$? I think in this case these points are not optimal – Lee Dec 18 '18 at 9:24
• @Lee a positive definite matrix has full rank. – LinAlg Dec 18 '18 at 15:13