Eigenvalues of the product of a positive semi-definite matrix with an invertible matrix with positive eigenvalues.

Question

Given a Laplacian matrix of a connected graph, $\mathcal{L} \succeq 0 \in \mathbb{S}^n$, and an invertible matrix $V \in \mathbb{R}^{n \times n}$ such that $\lambda_i(V) >0$ for $i = 1, ... ,n$. Is it possible to prove that $\lambda_i(\mathcal{L}V) \geq 0$ for $i = 1, ... ,n$? And can we determine the number of zero eigenvalues of the product?

Answer 1

The answer to the first question is no. A randomly generated counterexample:

$$ \mathcal L = \left[\begin{matrix}2 & -1 & -1\\-1 & 2 & -1\\-1 & -1 & 2\end{matrix}\right], \quad V = \left[\begin{matrix}-58.337 & -432.225 & -100.272\\-29.695 & -206.429 & -48.576\\161.894 & 1158.166 & 270.765\end{matrix}\right]. $$ The entries of $V$ here are exact. The eigenvalues of $V$ are approximately $0.788, 2.129, 3.082$. The eigenvalues of $\mathcal L V$ are approximately $-705.188, 7.894, 0$.

Because $V$ is invertible, the number of zero eigenvalues in the product up to geometric (but not necessarily algebraic) multiplicity is equal to the number of zero eigenvalues in $\mathcal L$, which is equal to the number of connected components in the underlying graph.

For any given $\mathcal L$, it is necessarily possible to find an invertible $V$ for which the eigenvalues of the product $\mathcal L V$ are all equal to zero. I am not sure, however, if such a $V$ with only positive eigenvalues can be found.