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?
Best Answer
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.