Maximizing trace subject to two equality constraints

Question

I am looking at the following optimization problem $$\begin{align} \underset{{\bf X}}{\text{maximize}} \qquad&\mathrm{tr}({\bf AX})\\ \text{subject to} \qquad& \mathrm{tr}({\bf X}) = 1,\\ &\mathrm{tr}({\bf BX})=0,\\ &{\bf X}\succeq 0, \end{align}$$ where ${\bf A}$ is a positive semidefinite matrix, and ${\bf B}$ is a symmetric matrix. I've noticed (through experiments) that when the solution is finite, it always has a rank 1. Any idea why that is the case? Can we prove that a rank-1 solution always exists? Is there a closed-form solution to this problem that I am missing?

Note: It is clear that when the second constraint $(\mathrm{tr}({\bf BX})=0)$ is not there, the problem has a rank-1 solution, and a maximizer is ${\bf X}^* = {\bf vv}^T$, where ${\bf v}$ is the eigenvector of ${\bf A}$ corresponding to the largest eigenvalue. In the question, I am wondering about the case when there are two equality trace constraints.

Answer 1

If ${\rm rank} \, X>1$, there exists a two-dimensional space $\mathcal{L}$ such that the restriction of $X$ to $\mathcal{L}$ is positive definite. The space of Hermitian operators, which vanish on $\mathcal{L}^\perp$, is 4-dimensional (3—dimensional in the real case), thus, it contains a non-zero operator $Y$ for which ${\rm tr} \, Y={\rm tr} \, BY=0$. The operator $X+sY$ is positive semidefinite for real $s$ close to 0. We may suppose that ${\rm tr} \,AY\geqslant 0$, otherwise change $Y$ to $-Y$. Consider the maximal $s$ for which $X+sY$ is positive semidefinite (it exists, since $Y$ itself is not positive semidefinite because of ${\rm tr} \,Y=0$). Replace $X$ to $X+sY$. The value of ${\rm tr} \,AX$ increased (maybe not strictly), and the rank of $X$ strictly decreased. After finitely many such changes we get a matrix with rank at most 1.

The set over which you maximize is compact, thus the maximum is achieved, unless this set is empty. The latter happens if and only if $\pm B$ is positive definite.