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.

Martin Argerami · Accepted Answer

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

Exodd · Answer

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 $Y_{i,i} = e_i^TYe_ige 0$, so $Tr(DY)le Tr(Y)$.

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

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