On maximizing the matrix determinant

Question

Given a positive $M\times M$ matrix $A$, let $f(A,X) = \det(X^H A X)$ where $X$ is a $M\times L$ matrix with $L \le M$ and each column of $X$ is a unit vector. In order to maximize $f(A,X)$, it is clear that $X$ should be composed of $L$ eigenvectors of $A$ with its largest eigenvalues. But what if there are some limitations on $X$, say some elements of $X$ have to be zeros? Given such kind of patterns of $X$, e.g., $X = \text{Diag}(x_1, x_2, ..., x_L)$ and $x_i$ are vectors, is there still possible to solve this problem? Many thanks in advance.

Answer 1

If $X$ is a (rectangular) diagonal matrix and each column of $X$ is a unit vector, those diagonal elements must have unit moduli. In turn, $\det(X^HAX)$ is always equal to the determinant of the leading principal $L\times L$ submatrix of $A$, regardless of $X$.