Suppose $A$ and $B$ are two matrices of similar dimensions. When does there exist a matrix $M$ (of appropriate dimensions) s.t for all vectors $v$ (of appropriate dimensions) s.t both $M^\top (v v^\top)A$ and $M^\top (v v^\top)B$ are PSD?
One obvious case is if there exists numbers $\alpha, \beta >0$ s.t for some matrix $X$ we have, $X = \frac{A}{\alpha} = \frac {B}{\beta}$ and then we can choose $M = X$.
Are there more non-trivial examples of this?
A somewhat less trivial example: $A$, $B$ and $M$ are rank-$1$ matrices $a w^T$, $b w^T$, $m w^T$ with the same $w$.
EDIT: Oops: that does make $M^T (v v^T) A$ and $M^T (v v^T) B$ symmetric. But to make them psd, you need $m^T v$, $a^T v$ and $b^T v$ to all have the same sign (when nonzero). But if $a^T v$ and $b^T v$ have the same sign for all $v$, $a$ and $b$ must be collinear, so that takes you back to the first example where $A$ and $B$ are scalar multiples of each other or the trivial example where $M=0$.