Let $V$ be the inner product space of $n \times n$ symmetric matrices where the inner product is given by $\langle A, B \rangle = trace(AB)$ and let $A, B \in V$ be such that they have eigenvalue $1$ and rank $1$, does there then always exist a matrix $C \in V$ such that $trace(AC)=0, trace(BC)\neq 0$? Or is there a counterexample?
Note: it suffices to only prove the case where $A=diag(1,0,\dots 0)$. If $trace(AB)=0$ everything’s okay of course (take $C=B$).
Motivation: I’d want this to be true as this “strong separation” property of matrices as described above by the space $V$ is a condition I’d want to be true to apply some theorem I’ve proven.
Consider $A$ and $B$ with rank $1$ such that $A = u u^T$ and $B = v v^T$ for som $u,v$. For arbitrary $C = w w^T$ you have $Trace(AC) = trace(u u^T w w^T ) = u^t w \,\, trace(u w^T )$ and $Trace(BC) = v^T w \,\, trace(v w^T )$. Therefore any $w$ that is orthogonal to $u$ but not to $v$ will suffice. Of course if $u$ and $v$ are parallel this does not exist.