This is a linear algebra question that came up in my research, and I feel like there ought to be either a simple proof or a simple counterexample, but I have been unable to find either.

Assume $V$ and $W$ are real finite-dimensional inner product spaces, so there is an induced inner product on $\text{Hom}(V,W)$ defined via its canonical identification with $V^* \otimes W$. Suppose $\Phi : \text{Hom}(V,W) \to \text{Hom}(V,W)$ is a linear map with the property that

$\langle A , \Phi(A) \rangle > 0$

for every element $A \in \text{Hom}(V,W)$ with **rank 1**. Equivalently, using the identification $\text{Hom}(V,W) = V^* \otimes W$, this condition means

$\langle w , \Phi(v_\flat \otimes w) v \rangle > 0$

for all $v \in V$ and $w \in W$, where $v_\flat := \langle v,\cdot \rangle$.

Does it follow that $\Phi$ is invertible?

My instinct says no: for instance, if we pick bases of $V$ and $W$ and express $\Phi$ in the resulting basis of $\text{Hom}(V,W)$, then the condition says that all diagonal entries of the matrix for $\Phi$ are positive, bus that's certainly not enough to conclude that $\Phi$ is invertible (at least not without some information about the magnitude of the diagonal entries relative to the non-diagonal entries, which I don't have, as far as I know). But the hypothesis seems to say more than this since it is not tied to any specific bases of $V$ and $W$, e.g. the diagonal entries will be positive for *any* basis of $\text{Hom}(V,W)$ consisting of rank 1 elements. My intuition and algebraic knowledge are insufficient to say what that means, but it sounds like it means something.

Or is there a simple counterexample, e.g. where $V$ and $W$ are both 2-dimensional?