2

A related question is here.

Let $\Pi$ be a symmetric positive semidefinite projection i.e. $\Pi^2 = \Pi$ and $0\leq \Pi\leq I$ where the inequalities are meant in a positive semidefinite sense. Let $X$ and $A$ be arbitrary symmetric positive definite matrices. Is it true that

$$\text{tr}(X\Pi A\Pi)\leq \text{tr}(XA)?$$

I think it's false but I'm unable to construct a counterexample.

user1936752
  • 1,828
  • 1
  • 15
  • 33

1 Answers1

2

(working over $\mathbb R$)
For reasons of linearity, it suffices to check when $X$ is rank one.

Suppose $A$ is PSD with no zeros on its diagonal, $A\mathbf x =0$ for some $\mathbf x \neq \mathbf 0$ and $X := \mathbf x\mathbf x^T$. Finally, consider $\Pi$ as a diagonal matrix that is all zeros with a single one on its kth diagonal component. Then

$\text{trace}\big(X\Pi A\Pi\big) $
$= \text{trace}\big(\Pi\mathbf x \mathbf x^T\Pi A\big)$
$= x_k^2\text{trace}\big(\mathbf e_k\mathbf e_k^T A \big)$
$= x_k^2\cdot a_{k,k}$
$\gt 0 $
$= \text{trace}\big(A\mathbf x\mathbf x^T\big)$
$ = \text{trace}\big(XA\big)$

user8675309
  • 12,193
  • I don't know what you mean by "for reasons of linearity, it suffices to check when $X$ is rank one." However, since you give a counterexample, it's clear that whichever case you consider is sufficient. – Ben Grossmann Aug 05 '20 at 06:15
  • 1
    Thank you - I will just add an example of such matrices here for explicitness. Let $\Pi = \begin{pmatrix} 1 & 0\ 0& 0\end{pmatrix}, A = \begin{pmatrix} 1 & -1\ -1& 1\end{pmatrix}$ and $X = \begin{pmatrix} 1 & 1\ 1& 1\end{pmatrix}$ – user1936752 Aug 05 '20 at 16:42
  • @BenGrossmann the wording wasn't great, though the idea was: general PSD $X =\sum_{k=1}^n \mathbf x_k\mathbf x_k^T$; by linearity of trace-- $\text{trace}\big(X\Pi A\Pi\big)=\sum_{k=1}^n \text{trace}\big(\mathbf x_k\mathbf x_k^T\Pi A\Pi\big)$ and $\text{trace}\big(X A\big)=\sum_{k=1}^n \text{trace}\big(\mathbf x_k\mathbf x_k^T A\big)$ so a sufficient approach is to check the rank one X case. If you can prove the bound is true in an arbitrary rank 1 case of X, the result comes from summing over the bound On the other hand, for this problem, the rank 1 case generates an easy counterexample. – user8675309 Aug 05 '20 at 16:56
  • @user8675309 Oh sure. In retrospect I should have figured out what you meant – Ben Grossmann Aug 05 '20 at 17:06