linear algebra - A problem concerning two symmetric matrices

Zhaoliang,

Maybe you wanted to ask this question:

Let $A$ and $B$ be two $n times n$ real symmetric matrices such that
$$ det(I_n-xA)det(I_n-yB) = det(I_n - xA-yB)$$
holds for all real values of $x$ and $y$.
Then $A B = 0$.

There are many proof, my favorite is probably a short proof in the paper On a matrix theorem of A. T. Craig and H. Hotelling by Olga Taussky.

You can also assume only that $forall xin mathbb{R}, det(I_n-xA)det(I_n-xB) = det(I_n - xA-xB)$, then you still have $AB=0$, but this is not in Taussky's article.

For those of you interested, here is a variant:
If $mathcal{S}subset mathbb{R}$ such that $|mathcal{S}|=n^2$, and $forall xin mathcal{S}, det(I_n-xA)det(I_n-xB) = det(I_n - xA-xB)$, do we necessarily have $AB=0$?