Consider the rectangular matrices $A \in \mathbb R^{m \times n}$ and $B \in \mathbb R^{n \times m}$. Is it true that, if there exists a square and positive semi-definite matrix $X \succeq 0$ such that $AXB \succeq 0$, then there exists a square and positive definite matrix $Y \succ 0$ such that $AYB \succ 0$?
Here is my initial attempt at proving this: given $X \succeq 0$ such that $AXB \succeq 0$, this implies that, for every $z \in \mathbb R^m$, $z^TAXBz \geq 0$. Because $X \succeq 0$, then its square root $X^{1/2}$ exists, such that $z^TAXBz = z^TAX^{1/2}X^{1/2}Bz = x^Ty \geq 0$, where $x = X^{1/2}A^Tz$ and $y = X^{1/2}Bz$. However, I'm not sure how to proceed from here, so would appreciate any guidance. My next guess would be to let $Y = X + \varepsilon I \succ 0$ for some $\varepsilon > 0$, but I'm not sure.
Your conjecture is unreasonable because the special case $m=n,A=X=I_n$ would imply $$ B\succeq 0\implies(\exists Y\succ 0\quad YB\succ 0)\implies\det B\ne0.$$