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!


1 Answer 1


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.

  • $\begingroup$ 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"... $\endgroup$
    – Simone256
    Commented 2 days ago
  • $\begingroup$ I do not understand what exactly do you mean $\endgroup$ Commented 2 days ago

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