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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.

user1936752
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2 Answers2

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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 $$ \operatorname{Tr}(\Pi X)=\operatorname{Tr} (X^{1/2}\Pi X^{1/2})\leq\operatorname{Tr}(X). $$

Martin Argerami
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  • Thank you. Is there a name/reference for the statement "If $A\leq B$, then $X^{1/2}AX^{1/2} \leq X^{1/2}BX^{1/2}$? – user1936752 Jul 22 '20 at 15:36
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    No, you won't find it as anything else than an exercise, since it is so simple. If $x^T(B-A)x\geq0$ for all $x$, then $x^TS^T(B-A)Sx=(Sx)^T(B-A)Sx\geq0$ for any $x$ and any $S$. – Martin Argerami Jul 22 '20 at 17:40
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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 $Tr(DY)\le Tr(Y)$.

Exodd
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