Arathi

The following example shows that you have to specify more about the underlying graphs. If A is the adjacency matrix for a graph or digraph on n vertices that has no edges, then A = 0.  If B is the adjacency matrix for any graph or digraph on n vertices. Then AB=0 no matter what edges are in the second graph. 

Now consider any graph or digraph G on n vertices and adjoin it to m isolated vertices. Then A consists of the direct sum of the adjacency matrix for G with an m x m zero matrix.  Let H be any graph or digraph on m vertices, and adjoin n isolated vertices such that B  is the direct sum of the n x n zero matrix with the adjacency matrix for H.  Clearly, AB = 0 no matter what the nonzero eigenvalues of A and B are.

If you know something about the structure of the two graphs as well as the labeling of the vertices in each of the graphs, you might be able to say something.  Otherwise, there are some results that put bounds on the eigenvalues of the product in terms of the eigenvalues of the factors.

Jeff S.

In the present case A and B are symmetric matrices, so eigenvalues are real. Moreover, they commute so they have a common set of eigenvectors. Suppose lamada(i) and mu(i) are eigenvalues of A and B respectively corresponding to eigenvector v(i), then ABv(i) = Amu(i)v(i) = mu(i) Av(i) = mu(i).lamada(i) v(i). So  v(i) is an eigenvector of AB (BA) and lamada(i).mu(i) are its corresponding eigenvalue.

Please see the attachment.

It is known that if B commutes with A then B must be a polynomial in A, so the eigenvalues for B are easy related to the A's eigenvalues

@Peter, please see the attachment.

The condition AB = BA is not necessary in this case.

My previous attachment: AB = B not similar to BA = 0  but it eigenvalues are all zero, since B is a (strictly) upper triangular matrix. 

Thank you all for your answers.

Hypothesis (?)

If AB=BA then any eigenvalue k of AB is a product k=ab of eigenvalues a of A, b of B, where a,b,k correspond to a common eigenvector of the mentioned matrices.

Since k is a simple eigenvalue (multiplicity 1) then there exist non-zero constants a, b such that Av=av andBv=bv. This means, v is a common eigenvector of A, B, AB, BA. The corresponding eigenvalues are a,b,k,k respectively. 

Substitute to the left side of ABv=kv. We obtain A(Bv)=A(bv)=b(Av)=b(av)=abv.

Hence abv=kv for a non-zero vector v , which implies k=ab. 

• Both A, B possess the same eigenspace Xk . Denote a,b the corresponding eigenvalues of A, B respectively. Again, k=ab. (a detailed proof is given as homework :D )
• There exist two different eigenvalues a1,a2 of A such that the corresponding eigenvectors of A belong to Xk. Necessarily there exist eigenvalues b1,b2 of B joined with the same eigenvectors., because the commuting matrices have the same eigenspaces. We can range them so that a1b1=a2b2=k . Moreover, ai, bi with the same index have the same multiplicity. 

However, we can no more assure that each eigenvector of AB is eigenvector of A and of B.

An example illustrating the last situation:

A, B share the same eigenvectors, of course.