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.

  • $\begingroup$ what if rank$ X< n$? I think in this case these points are not optimal $\endgroup$ – Lee Dec 18 '18 at 9:24
  • 1
    $\begingroup$ @Lee a positive definite matrix has full rank. $\endgroup$ – LinAlg Dec 18 '18 at 15:13

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