Matrices, prove that if $XY=0$ and other conditions, then $\det(X+Y) =\det(X- Y) $

Question

Prove that if $XY=0$ and $\det(X) =\det (Y) =0$, and both $\det(X+Y)$ and $\det(X-Y)$ are positive, then they are equal. I noticed it is enough to prove that $YX=0$.

Thanks

Edit: i mean prove that $\det(X+Y)=\det(X-Y)$

Answer 1

This looks like a contest problem and there might be some nice tricks behind it, but without knowing the tricks, the problem can still be solved in a straightforward manner.

By a change of basis, we may assume that $X$ is already in rational canonical form. Then $X$ cannot contain a non-trivial nilpotent diagonal sub-block, otherwise, if the first diagonal sub-block of $X$ is $$ \pmatrix{\mathbf 0^T&0\\ I_k&\mathbf0}, $$ then the condition $XY=0$ would force the first $k$ rows of $Y$ to be zero, but then both the first rows of $X$ and $Y$ would be zero, contradicting the assumption that $X\pm Y$ is non-singular.

Therefore, by another change of basis, we may assume that $X=D\oplus0_{r\times r}$ for some non-singular matrix $D$. Now the condition $XY=0$ means that $Y=\pmatrix{0_{(n-r)\times(n-r)}&0\\ \ast &Z}$ for some $r\times r$ matrix $Z$. Hence $\det(X+Y)=\det(D)\det(Z)$, $\det(X-Y)=(-1)^r\det(D)\det(Z)$ and the rest is trivial.