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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.

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    $\begingroup$ There is a counter-example: let A=diag(2,1,0,0), B=diag(-1,1,-1,0), then A+B=diag(1,2,-1,0) and AB=diag(-2,1,0,0). Perhaps something missing in the formulation. $\endgroup$
    – Sergei Ivanov
    Commented Mar 30, 2010 at 12:47
  • $\begingroup$ Would assuming $X\cap Y=\emptyset$ repair the problem? $\endgroup$
    – Harald Hanche-Olsen
    Commented Mar 30, 2010 at 15:57
  • $\begingroup$ I'll answer my own question. If $X\cap Y=\emptyset$ and $AB=0$ then one of the two matrices is invertible, hence the other is the zero matrix. Hence, if this was the formulation of the problem then the conclusion holds in a trivial sense, which at least makes the problem not so good. I could change my extra condition into $X\cap Y\subseteq\{0\}$, but I'd better stop my idle guessing game now. $\endgroup$
    – Harald Hanche-Olsen
    Commented Mar 30, 2010 at 17:57

2 Answers 2

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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 $AB=0$?

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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.

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  • $\begingroup$ Multiset union would have total multiplicity 2n making the assumption impossible. $\endgroup$
    – Sergei Ivanov
    Commented Mar 30, 2010 at 13:02
  • $\begingroup$ you are right of course, however, regarding this in a based sense and ignoring multiplicities of 0 it does work. $\endgroup$
    – Thomas Kragh
    Commented Mar 30, 2010 at 20:09

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