Are the eigenvalues of Laplacian times Diagonal times Laplacian matrix non-negative?

Question

Consider two (not identical) symmetric Laplacian matrices $L_1$ and $L_2$ and a diagonal matrix $D > 0$. My question is if the eigenvalues of the product $A =L_1 D L_2$ have non-negative real-part. It is well known that the product of two positive semi-definite matrices have eigenvalues with non-negative real part. (see e.g. the book by Bernstein: Matrix Mathematics, Theory, Facts, and Formulas). Extensive simulations show that $A$ has indeed eigenvalues with non-negative real part and that this not true in general if $L_1$ and $L_2$ were `only' positive semi-definite matrices. However, I have been, trying similarity transformations and Sylvester's law of inertia, unable to find a proof. Of course, a counterexample would be very useful as well.

Answer 1

The answer is no. A random counterexample: the eigenvalues of $$ A=\pmatrix{1&0&-1\\ 0&1&-1\\ -1&-1&2} \pmatrix{1\\ &3\\ &&1} \pmatrix{2&-1&-1\\ -1&1&0\\ -1&0&1} =\pmatrix{3&-1&-2\\ -2&3&-1\\ -1&-2&3} $$ are $0$ and $\frac12(9\pm i\sqrt{3})$, so $A$ has non-real eigenvalues. When the matrix sizes are larger (such as $5\times5$), we can even obtain easily a counterexample in which $A$ has a negative real eigenvalue.

Edit. Another counterexample: the matrix product \begin{align*} A&=\pmatrix{ 1& 0& 0&-1& 0\\ 0& 2&-1& 0&-1\\ 0&-1& 2& 0&-1\\ -1& 0& 0& 1& 0\\ 0&-1&-1& 0& 2} \pmatrix{1\\ &3\\ &&3\\ &&&3\\ &&&&3} \pmatrix{ 2&-1& 0& 0&-1\\ -1& 2& 0& 0&-1\\ 0& 0& 1&-1& 0\\ 0& 0&-1& 1& 0\\ -1&-1& 0& 0& 2}\\ &=\pmatrix{ 2&-1& 3&-3&-1\\ -3&15&-3& 3&-12\\ 6&-3& 6&-6&-3\\ -2& 1&-3& 3& 1\\ -3&-12&-3& 3&15}, \end{align*} has an eigenpair $(\lambda,v)$ where $\lambda\approx-0.2111<0$ and $v\approx(0.46902,\,0.30552,\,-0.61103,\,-0.46902,\,0.30552)^T$.