# The signature of the tensor product of skew-symmetric non-singular matrices

Let $$X$$ be a non-singular (real) symmetric matrix which is the tensor product of two (real) $$n\times n$$ skew symmetric non-singular matrices, i.e. $$X=A\otimes B$$. Then how to see the number of negative eigenvalues of $$X$$ equals to the number of positive eigenvalues of $$X$$.

What I know is the eigenvalues of $$X$$ are the pairwise products of the eigenvalues of $$A$$ and $$B$$ and the eigenvalues of $$X$$ are purely imaginary. I wonder if it suffices to show the claim? If so, is that possible to show me a proof?

Hint: In addition to the fact that eigenvalues of $$X$$ are the pairwise products of the eigenvalues of $$A$$ and $$B$$ and that the eigenvalues of $$A$$ and $$B$$ are purely imaginary, note that the eigenvalues of $$A$$ (and $$B$$) come in conjugate pairs. That is, both $$A$$ and $$B$$ have eigenvalues that come in pairs $$\pm \lambda i$$ (with $$\lambda \in \Bbb R$$).
Alternatively: note that $$A$$ is similar to $$A^T = -A$$. That is, there exists an invertible matrix $$S$$ such that $$SAS^{-1} = -A$$. It follows that $$(S \otimes I)(A \otimes B)(S \otimes I)^{-1} = (S AS^{-1}) \otimes B = (-A)\otimes B= -(A \otimes B).$$ That is, $$A \otimes B$$ is a non-singular symmetric matrix that is similar to $$-(A \otimes B)$$. This is enough to show the claim.