Suppose $\mathcal{A}$ and $\mathcal{B}$ are finite collections of symmetric idempotent matrices, and that $x$ is a fixed vector. If
$$ A^* \in \underset{A \in \mathcal{A}}{\operatorname{argmin}} x^T A x, \qquad B^* \in \underset{B \in \mathcal{B}}{\operatorname{argmin}} x^T B x $$
is the following true?
$$(A^* , B^*) \in \underset{A \in \mathcal{A}, B \in \mathcal{B}}{\operatorname{argmin}} x^T AB x $$
Any insights would be much appreciated!
Here's a simple counterexample: take $x = (1,1,1,1,1)^T$, $\mathcal A = \{A,M\}$, $\mathcal B = \{B,M\}$, where
$$ A = \pmatrix{1\\&1\\&&0\\&&&0\\&&&&0}, M = \pmatrix{0\\&0\\&&1\\&&&0\\&&&&0}, B = \pmatrix{0\\&0\\&&0\\&&&1\\&&&&1}. $$ We have minimizers $A^* = B^* = M$, but $x^TA^*B^*x = 1$, which is greater than $x^TABx = 0$.
If you want a counterexample with disjoint sets, you can replace the element $M$ of $B$ with $\epsilon I + (1 - \epsilon)M$ for some sufficiently small $\epsilon > 0$.