Let $D\in\mathbb{R}^{n\times n}$ be a real diagonal matrix where $\sum_i D_{ii}<0$. Let also $R\in\mathbb{R}^{n\times n}$ and $L\in\mathbb{R}^{n\times n}$ be real (possibly) non-symmetric matrices (irreducible) with right and left eigenvector $\mathbf{1}$ (associated with eigenvalue 1). It is clear that $$ \frac{\mathbf{1}^T}{\sqrt{n}}\bigg(\underbrace{LR-D}_{\triangleq Q}\bigg)\frac{\mathbf{1}}{\sqrt{n}}>1. $$
I know that Q is NOT symmetric, thus one cannot relate its numerical range to its spectral radius, but given that $\frac{\mathbf{1}}{\sqrt{n}}$ is eigenvector associate with real eigenvalue $1$ of $R$ and $L$ and a combination of eigenvectors of $D$ (which are $e_i$'s), can one claim anything about its spectral radius?
I want to show that the spectral radius is larger than one.
Example: taking $\epsilon>0$ and matrices $$ L=\begin{bmatrix}0 & 0\\1 & 1\end{bmatrix},\quad R=\begin{bmatrix}1 & 0\\1 & 0\end{bmatrix},\quad D=\begin{bmatrix}-\epsilon & 0\\0 & 0\end{bmatrix}, $$ we get $$ Q=LR-D=\begin{bmatrix}\epsilon & 0\\2 & 0\end{bmatrix} $$ with the spectral radius $\epsilon$ being arbitrarily close to zero.