I have two large real-symmetric matrices $A,B\in \mathbb{R}^{N\times N}$, which obey $A \cdot B = 0 = B \cdot A$, meaning that they both commute (and thus share eigenvectors) and anticommute.
I am looking for a simple criterion for checking whether the null spaces of $A$ and $B$ have a non-trivial overlap. Specifically, I would like to prove that the null spaces of my choice of $A$ and $B$ are orthogonal.
For brevity, I speak of "zero eigenvectors" (meaning: elements of the null space) and of "non-zero eigenvectors" (meaning: eigenvectors orthogonal to the null space).
To that end, I recognize that $A\cdot B = 0$ implies that non-zero eigenvectors of $B$ are among the zero eigenvectors of $A$, and likewise $B \cdot A = 0$ means that non-zero eigenvectors of $A$ are among the zero eigenvectors of $B$. However, this still leaves the option that $A$ and $B$ share a non-trivial zero eigenvectors. (By combining the above, one can also show that the sum of the dimensions of the null spaces of $A$ and of $B$ must be equal or larger than the dimension $N$ of the matrices. Orthogonality of the null spaces should be equivalent to this bound is saturated -- which is a kind of a test, but I am looking for something that avoids the explicit computation of the eigensystem.)
Thus, I am wondering: Is there a simple algebraic test for proving that the null spaces of such matrices are orthogonal?