# Inequality with matrix exponentials. Is it true that $x^\top e^{-xx^\top - AA^\top} x \leq x^\top e^{-xx^\top} x$?

As in the object, I'm looking at the case where $$x \in \mathbb R^d$$ is a generic vector, and $$AA^\top \in \mathbb R^{d \times d}$$ is a p.s.d. matrix.

I'm investigating the following inequality

$$x^\top e^{-xx^\top - AA^\top} x \leq x^\top e^{-xx^\top} x,$$

where I'm considering matrix exponentials. In general, it is not true that if $$BB^\top \geq CC^\top$$ in p.s.d. sense, then we have $$e^{BB^\top} \geq e^{CC^\top}$$. However, here, I'm looking for something weaker, but I'm struggling to prove it / found a counter example!

Alas, this is not correct. You may try $$d=2$$, $$x=(a, a)^\top$$ for certain $$a>0$$, and choose $$A$$ of rank 1 such that $$AA^\top+xx^\top=\operatorname{diag}(p, q)$$. For positive $$p,q$$ such matrix $$A$$ exists iff $$0=\det(-xx^\top+\operatorname{diag}(p, q))=(p-a^2)(q-a^2)=a^4,$$ or $$1/p+1/q=1/a^2$$. Your inequality reads as $$(e^{-p}+e^{-q})a^2\leqslant 2a^2e^{-2a^2}$$. If we denote $$1/p=x,1/q=y$$, then $$x+y=1/a^2$$ is fixed and we need $$f(x)+f(y)$$, where $$f(x)=e^{-1/x}$$, be maximal when $$x=y$$. But $$f$$ is not concave on $$(0,\infty)$$, so this does not hold in general.
• Thank you! Very nice approach! I have a follow up question though. Do you think a similar inequality could be proved, with a correction coefficient that depends on $\| A^\top x \|_2$? A case that breaks the previous inequality, suggested by your proof, is given by using column spaces of $AA^\top$ and $xx^\top$ to be "roughly orthogonal"... Commented 2 days ago