A problem concerning two symmetric matrices - MathOverflow

A problem concerning two symmetric matrices

2010-03-30T12:32:16Z

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.

Answer by Thomas Kragh for A problem concerning two symmetric matrices

2010-03-30T12:53:25Z

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.

Answer by Portland for A problem concerning two symmetric matrices

2010-04-19T18:47:33Z

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 $AB=0$?