Fix integers $1 \leq k \leq n$. Let $P \in \mathbb{R}^{n \times k}$ be such that $P^T P$ has full rank.

Let $\mathcal{X}$ denote the set of unit trace, real $n \times n$ symmetric positive semidefinite matrices.

**Question:** Given any $X \in \mathcal{X}$ is it possible to find $Y \in \mathcal{X}$ such that $Y$ has rank at most $k$ and such that
$$
P^T X P \preceq P^T YP?
$$

This is obviously true if $k = 1$. Indeed, then for a nonzero $p \in \mathbb{R}^n$, we have for any $X \in \mathcal{X}$, that $Y = \frac{1}{p^T p} pp^T \in \mathcal{X}$ and $$ p^T Y p = p^Tp \geq p^T Xp. $$ Does this hold for $k > 1$?