Can the trace of a positive matrix increase under a projection?

Question

The question is concerned with symmetric matrices $\mathbb{S}_n$ as a real vector space. Let $X$ be a positive semidefinite symmetric matrix, and let $P : \mathbb{S}_n \to \mathcal{V}$ be a projection onto some subspace $\mathcal{V} \subset \mathbb{S}_n$. Is it always the case that $\mathrm{trace}(P(X)) \leq \mathrm{trace}(X)$?

I can see this being true when $\mathcal{V}$ has an orthonormal basis $\{A_i\}$ consisting of matrices which are all either trace $0$ or positive and trace $\leq 1$. But in general I don't know anything about what kind of basis $\mathcal{V}$ would admit. Is it possible to come up with a counterexample?

Answer 1

Consider the following orthogonal basis of $\mathbb S_2$ with respect to the Frobenius inner product: $$ A=\pmatrix{4&0\\ 0&2},\ B=\pmatrix{1&0\\ 0&-2},\ C=\pmatrix{0&1\\ 1&0}. $$ Let $P$ be the orthogonal projection onto $\mathcal V=\operatorname{span}(A)$ and let $X=A+B=\operatorname{diag}(3,0)$. Then $$ \operatorname{tr}(P(X)) =\operatorname{tr}(P(A+B)) =\operatorname{tr}(A) =6>3 =\operatorname{tr}(X). $$

Answer 2

Recall Von Neumann's Trace Inequality:

For matrices $A,B \in M_{n\times n} (\mathbb{C})$, we have that: $$ |\mathrm{Tr}(AB)| \leq \sum_{ i = 1}^{n} \alpha_i \beta_i $$ where here the the $\alpha_i$ and $\beta_i$ are the singular values of $A,B$ respectively, counted with multiplicity if necessary and are in decreasing order. In particular, if $P$ is a projection, it is diagonalizable with singular values $\alpha_i\in \{0,1\}$. It follows then that if $X$ is positive definite: $$ \mathrm{Tr}(PA) \leq |\mathrm{Tr}(PA)| \leq \sum_{i = 1}^{n}p_ix_i\leq \sum_{i=1}^n x_i = \mathrm{Tr}(A) $$

where here $p_i$ are the eigenvalues of the projection and $x_i$ are the eigenvalues of $X$.