Abstract description:
Let $\mathbf{A}$ and $\mathbf{B}$ be two $n \times n$ real matrices. Let $\sigma( \mathbf{A B} )$ denote the spectrum of $\mathbf{A B}$.
Assume that
(A1) $\mathbf{A}$ is symmetric and positive definite but not diagonal.
(A2) $\mathbf{B}$ is not symmetric but similar to a symmetric and positive semidefinite (but not positive definite) matrix.
(A3) $\sigma( \mathbf{A B} ) \subset \mathbb{R}$.
Question 1: Do (A1) and (A2) imply (A3)?
Question 2: Can we say something about $\min\sigma( \mathbf{A B} )$?
Detailed description and background:
Let $G$ be a simple graph of order $m$. That is, $G$ is undirected and without loops. $G$ can have vertices with zero degree. Let $H$ denote the graph resulting from G by adding loops to all vertices with zero degree. Let $\mathbf{I}_m$ denote the identity matrix of order $m$ and $\mathbf{O}_m$ the zero matrix of order $m$. Let $\mathbf{S}$ denote the adjacency matrix of $H$. Hence, $\mathbf{S}$ is an $m \times m$ symmetric matrix with nonnegative entries. Let $\mathbf{D}$ be the degree matrix of $H$. Hence, $\mathbf{D}$ is an $m \times m$ diagonal and positive definite matrix. Let $c \in ( -1, 1)$. Note that $\mathbf{I}_m - c \mathbf{D}^{-1} \mathbf{S}$ is nonsingular because $\mathbf{D}^{-1} \mathbf{S}$ is a right stochastic matrix that has spectral radius equal to $1$. Define $\mathbf{M} := ( \mathbf{I}_m - c \mathbf{D}^{-1} \mathbf{S} )^{-1}$.
Conjecture: $\min\sigma( \mathbf{M} + \mathbf{M}' - \mathbf{I}_m ) > 0$ or, equivalently, $\mathbf{M} + \mathbf{M}' - \mathbf{I}_m$ is positive definite.
So far, my thoughts on a proof of this conjecture are as follows:
(P1) $\mathbf{D}^{-1} = \mathbf{P}^{2}$ with $\mathbf{P}$ diagonal and positive definite. That is, $\mathbf{P}$ is the unique square root of $\mathbf{D}^{-1}$.
(P2) $\mathbf{D}^{-1} \mathbf{S}$ is similar to the symmetric matrix $\mathbf{P} \mathbf{S} \mathbf{P}$. Indeed, $\mathbf{P}^{-1} \mathbf{D}^{-1} \mathbf{S} \mathbf{P} = \mathbf{P} \mathbf{S} \mathbf{P}$.
(P3) $\sigma( \mathbf{M} - (1/2) \mathbf{I}_m ) \subset \mathbb{R}$ because of (P2).
(P4) It can be shown that $\sigma( \mathbf{M} - (1/2) \mathbf{I}_m ) \subset ( 0, +\infty )$.
(P5) We have $$\begin{align*} & \sigma( \mathbf{M} + \mathbf{M}' - \mathbf{I}_m )\\ & = \sigma\bigl( \mathbf{M} - (1/2) \mathbf{I}_m + \mathbf{M}' - (1/2) \mathbf{I}_m \bigr)\\ & = \sigma\bigr( \mathbf{P}^{-1} \bigl( \mathbf{M} - (1/2) \mathbf{I}_m + \mathbf{M}' - (1/2) \mathbf{I}_m \bigr) \mathbf{P} \bigr)\\ & = \sigma\bigr( ( \mathbf{I}_m - c \mathbf{P} \mathbf{S} \mathbf{P} )^{-1} - (1/2) \mathbf{I}_m + \mathbf{P}^{-2} ( \mathbf{I}_m - c \mathbf{P} \mathbf{S} \mathbf{P} )^{-1} \mathbf{P}^{2}- (1/2) \mathbf{I}_m \bigr)\\ & = \sigma\Biggl( \biggl( \begin{array}{c}\mathbf{I}_m\\\mathbf{I}_m\end{array} \biggr)' \biggl( \begin{array}{cc}\mathbf{I}_m & \mathbf{O}_{m}\\\mathbf{O}_{m} & \mathbf{P}^2\end{array} \biggr)^{-1} \biggl( \mathbf{I}_2 \otimes \bigl( ( \mathbf{I}_m - c \mathbf{P} \mathbf{S} \mathbf{P} )^{-1} - (1/2) \mathbf{I}_m \bigr) \biggr) \biggl( \begin{array}{cc}\mathbf{I}_m & \mathbf{O}_{m}\\\mathbf{O}_{m} & \mathbf{P}^2\end{array} \biggr) \biggl( \begin{array}{c}\mathbf{I}_m\\\mathbf{I}_m\end{array} \biggr) \Biggr)\\ & \subset \sigma\Biggl( \biggl( \mathbf{I}_2 \otimes \bigl( ( \mathbf{I}_m - c \mathbf{P} \mathbf{S} \mathbf{P} )^{-1} - (1/2) \mathbf{I}_m \bigr) \biggr) \biggl( \begin{array}{cc}\mathbf{I}_m & \mathbf{O}_{m}\\\mathbf{O}_{m} & \mathbf{P}^2\end{array} \biggr) \biggl( \begin{array}{c}\mathbf{I}_m\\\mathbf{I}_m\end{array} \biggr) \biggl( \begin{array}{c}\mathbf{I}_m\\\mathbf{I}_m\end{array} \biggr)' \biggl( \begin{array}{cc}\mathbf{I}_m & \mathbf{O}_{m}\\\mathbf{O}_{m} & \mathbf{P}^2\end{array} \biggr)^{-1} \Biggr)\\ & = \sigma\Biggl( \biggl( \mathbf{I}_2 \otimes \bigl( ( \mathbf{I}_m - c \mathbf{P} \mathbf{S} \mathbf{P} )^{-1} - (1/2) \mathbf{I}_m \bigr) \biggr) \biggl( \begin{array}{cc}\mathbf{I}_m & \mathbf{D}\\\mathbf{D}^{-1} & \mathbf{I}_m\end{array} \biggr) \Biggr)\\ & = \sigma( \mathbf{A} \mathbf{B} ), \end{align*}$$ where $\mathbf{A} := \mathbf{I}_2 \otimes \bigl( ( \mathbf{I}_m - c \mathbf{P} \mathbf{S} \mathbf{P} )^{-1} - (1/2) \mathbf{I}_m \bigr)$ is symmetric and positive definite because of (P4) and $\sigma( \mathbf{M} - (1/2) \mathbf{I}_m ) = \sigma( \mathbf{P}^{-1} ( \mathbf{M} - (1/2) \mathbf{I}_m ) \mathbf{P} ) = \sigma( ( \mathbf{I}_m - c \mathbf{P} \mathbf{S} \mathbf{P} )^{-1} - (1/2) \mathbf{I}_m )$ and $$\mathbf{B} := \biggl( \begin{array}{cc}\mathbf{I}_m & \mathbf{D}\\\mathbf{D}^{-1} & \mathbf{I}_m\end{array} \biggr) = \dfrac{1}{\sqrt{2}} \biggl( \begin{array}{cc}\mathbf{I}_m & -\mathbf{I}_m\\\mathbf{D}^{-1} & \mathbf{D}^{-1}\end{array} \biggr) \biggl( \begin{array}{cc}2\mathbf{I}_m & \mathbf{O}_m\\\mathbf{O}_m & \mathbf{O}_{m}\end{array} \biggr) \biggl( \dfrac{1}{\sqrt{2}} \biggl( \begin{array}{cc}\mathbf{I}_m & -\mathbf{I}_m\\\mathbf{D}^{-1} & \mathbf{D}^{-1}\end{array} \biggr) \biggr)^{-1}$$ is (in general) not symmetric but similar to the symmetric and positive semidefinite matrix $$\biggl( \begin{array}{cc}2\mathbf{I}_m & \mathbf{O}_m\\\mathbf{O}_m & \mathbf{O}_{m}\end{array} \biggr) = \biggl( \begin{array}{cc} 2 & 0\\ 0 & 0 \end{array} \biggr) \otimes \mathbf{I}_m.$$ Note that $\sigma( \mathbf{M} + \mathbf{M}' - \mathbf{I}_m ) \cup \{ 0 \} = \sigma( \mathbf{A} \mathbf{B} )$, so that $\sigma( \mathbf{A} \mathbf{B} ) \subset \mathbb{R}$ because $\sigma( \mathbf{M} + \mathbf{M}' - \mathbf{I}_m ) \subset \mathbb{R}$.
Any ideas as to how to finish the proof? So far, I did not come up with a counter example.
EDIT:
I have reformulated assumption (A2) because it was stated inconsistently. It read as follows: $\mathbf{B}$ is not symmetric but similar to a nondiagonal, symmetric, and positive semidefinite matrix. In addition, I added the condition that $\mathbf{B}$ is not positive definite. I have also made some minor changes in Detailed description and background, which are related to $\mathbf{B}$.
According to the new statement, $A=\begin{pmatrix}20&-4&0\\-4&4&0\\0&0&1\end{pmatrix},B=\begin{pmatrix}1&0&0\\2&2&0\\0&0&0\end{pmatrix}$.
When the next change ?