For a projection $\Pi$, is $\text{tr}(\Pi X)\leq \text{tr}(X)$?

Question

All matrices are finite dimensional symmetric positive semidefinite matrices in this question.

Let $\Pi$ be projection i.e. in its eigenbasis, it is the the identity matrix with some diagonal elements replaced by $0$. Let $X$ be an arbitrary symmetric positive definite matrix. Is it true that

$$\text{tr}(\Pi X)\leq \text{tr}(X)$$

Using the answer here, I see that it is indeed true that $\text{tr}(\Pi X)\leq \text{rank}(\Pi)\text{tr}(X)$ but I was hoping the rank term could also be dropped.

Answer 1

Since $X$ is symmetric positive definite, it has a symmetric positive square root. Then, on the Löwner order, $\Pi\leq I$, and $$ X^{1/2}\Pi X^{1/2}\leq X^{1/2} X^{1/2} =X. $$ Then $$ \operatorname{Tr}(\Pi X)=\operatorname{Tr} (X^{1/2}\Pi X^{1/2})\leq\operatorname{Tr}(X). $$

Answer 2

Yes.

You can always consider $\Pi$ diagonal, because $\Pi = QDQ^{-1}$ with $D$ diagonal and $Q$ orthogonal, so $$ Tr(\Pi X) = Tr(QDQ^{-1}X) = Tr(D Q^{-1}XQ) \\ Tr(Q^{-1}XQ) = Tr(X) $$ and $Q^{-1}XQ$ is still positive semidefinite.

Now, $D$ is just a diagonal of 0 and 1, so for any $Y$ positive semidefinite you have that $Tr(DY)$ is a sum of some diagonal elements of $Y$. But the diagonal elements of $Y$ are all nonnegative since $Y_{i,i} = e_i^TYe_i\ge 0$, so $Tr(DY)\le Tr(Y)$.