# Eigenvalue of product of matrices

1. Jul 10, 2013

### mnov

I have two real symmetric matrices $A$ and $B$ with the following additional properties. I would like to know how the eigenvalues of the product $AB$, is related to those of $A$ and $B$? In particular what is $\mathrm{trace}(AB)$?

$A$ contains only 0s on its diagonal. Off diagonal terms are either 0 or 1.
$B$ also contains only 0s on its diagonal. Its off diagonal terms are positive real numbers.

If equalities don't exist, some bounds would also be helpful.

Thanks.

2. Jul 10, 2013

### Mandelbroth

Let $a_{i,j}$ be the element of matrix A from row i and column j, and let $b_{i,j}$ be the element of matrix A from row i and column j. Then,
$$\operatorname{tr}(\textbf{AB})=\sum_{i}\sum_{j}a_{j,i}b_{i,j}.$$
Thus, from the fact that the non-diagonal terms of A are either 0 or 1, we obtain the bounds that
$$0 \leq \operatorname{tr}(\textbf{AB}) \leq \sum_{i}\sum_{j}b_{i,j}.$$

3. Jul 10, 2013

### mnov

Thanks.
I forgot to mention that A is sparse, i.e., most of its off diagonal terms are zero. So the above bound would be pretty bad.

4. Jul 10, 2013

### Mandelbroth

Do you have bounds on the sparsity of the matrix? I could probably do a little better with an idea of how dense the matrix is. Also, is there any idea as to the size of the matrices?

I have nothing to do and I want something to work on. :tongue:

5. Jul 10, 2013

### mnov

A is an n x n matrix. m = constant * n of the off diagonal terms are 1. n is large.

6. Jul 10, 2013

### Office_Shredder

Staff Emeritus
So you have a bound like the trace is smaller than
$$m* max_{i,j}\left{ b_{ij} \right}$$