A problem concerning two symmetric matrices Let A , B denote two symmetric matrices of the same order n. and Spec(A)=X , Spec(B)=Y.
If Spec(A+B)=X $\cup$ Y , proof thar AB=0.
here Spec(A) means the set of the engevalues of A.
This is a problem posed in a math forum bbs in China , I find it interesting, so I 
also  put it here.
 A: 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 $\forall x\in \mathbb{R}, \det(I_n-xA)\det(I_n-xB) = \det(I_n - xA-xB)$, 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 $\forall x\in \mathcal{S}, \det(I_n-xA)\det(I_n-xB) = \det(I_n - xA-xB)$, do we necessarily have $AB=0$?
A: Counter example:
take the diagonal matrices:
$A=D(2,1,1,0,0)$ and $B=D(0,0,1,2,0)$
If you want the multiplicity to match (thinking of $X \cup Y$ is a "multiset" union) then it is easy to create an inductive argument proving the assertion.
