Optimizing a matrix quadratic form with respect to Loewner order

Question

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 $k > 1$?

Answer 1

This seems easy so maybe I am misreading some condition.

You may assume without loss of generality that $P^TP = I_k$; otherwise orthonormalize its columns / apply Gram-Schmidt to get a new $P$ with this property. (If you want a closed formula, you can set $P \gets P(P^TP)^{-1/2}$.)

Then, for any $X\in\mathbb{R}^{n\times n}$ (even without assuming definiteness/trace conditions), the matrix $$ Y = PP^T X PP^T $$ has rank $k$ and satisfies $P^TYP = P^TXP$.

Answer 2

If you meant that you want that property to hold for all matrices $X$ simultaneously, then the answer is that there exists no trace-1 matrix with that property for $k\geq 2$. Let again $P$ have orthonormal columns without loss of generality, and assume that it has at least two columns $p_1$ and $p_2$. Then testing the condition on $X_1 = p_1 p_1^T$ and $X_2 = p_2 p_2^T$ gives $p_1^T Y p_1 \geq 1$, $p_2^T Y p_2 \geq 1$, and these two inequalities cannot hold together because $1 = \operatorname{Tr} Y \geq p_1^T Y p_1 + p_2^T Y p_2$ (in an orthogonal basis that starts with $p_1, p_2$, the RHS contains the first two elements on the diagonal of $Y$).