If the product of 2 matrices is symmetric is one of them diagonalizable? Let S be a diagonal matrix with entries > 0. Let A be a matrix with diagonalelements <= 0 and off-diagonalelements >=0. A's largest eigenvalue is 0. That is all that is known about A.
We assume that SA is symmetric and thus can be diagonalized by an orthonormal set of eigenvectors. Given this fact, can we say anything about the diagonalizability of A? E.g. could we state that diagonlizability of A follows from diagonalizability of SA? Could it be that both statements are equivalent?
Any help is deeply appreciated ! 
George
 A: Let $\langle \cdot,\cdot \rangle$ be the Euclidean inner product and define $(x,y)=\langle x,Sy \rangle$. From the given properties, $S$ is positive definite, therefore this function is an inner product, which I will call the $S$-inner product.
Now given a matrix $A$, $SA$ is symmetric if and only if $A$ is self-adjoint with respect to the $S$-inner product. Therefore we can apply the spectral theorem to $A$ and the $S$-inner product. In the process we $S$-orthogonally diagonalize $A$. This diagonalization is in general not $I$-orthogonal (i.e. Euclidean-orthogonal), but it is still a diagonalization. 
By the way, this property has useful applications. For example, consider an irreducible Markov chain on a finite state space, with generator $L$. (In discrete time, we have a transition matrix $P$ and let $L=P-I$). Such a chain has a unique stationary distribution. Let $S$ be the diagonal matrix whose diagonal entries are the probabilities in the stationary distribution. We say the chain is reversible if $L$ is self-adjoint with respect to the $S$-inner product. This property has physical meaning: it means that if the system is at equilibrium, then each possible transition is equilibrated by its reverse transition.
If we have a reversible Markov chain then, by the above discussion, $SL$ is symmetric. Therefore, if we are interested in the eigenproblem for $L$, we can instead consider the eigenproblem for $S^{1/2} L S^{-1/2}$. This matrix is symmetric (since $SL$ is symmetric), which makes its eigenproblem easier, and it is easy to match up eigenpairs for $S^{1/2} L S^{-1/2}$ with eigenpairs for $L$.
