For any primitive matrix $A$ in $\mathcal{M}_n(\mathbb{R})$, we denote $r$ the spectral radius of $A$, $L = xy^\top$ where $x > 0$ is a direction vector of $\operatorname{Ker}(A - rI_n)$ and $y > 0$ is a direction vector of $\operatorname{Ker}(A^\top - rI_n)$ with $y^\top x = 1$. We set $B = A - rL$. Let $\lambda$ be a nonzero eigenvalue of $B$ and let $z$ be an associated eigenvector. By contradiction, we assume $\rho(B) = r$. We can therefore choose $\lambda$ such that $|\lambda| = r$. Show that then $\lambda = r$ then $Lz = z$ and reach a contradiction. Conclude.
For any primitive matrix $A$ in $\mathcal{M}_n(\mathbb{R})$, we denote $r$ the spectral radius of $A$, $L = xy^\top$ where $x > 0$ is a direction vector of $\operatorname{Ker}(A - rI_n)$ and $y > 0$ is a direction vector of $\operatorname{Ker}(A^\top - rI_n)$ with $y^\top x = 1$. We set $B = A - rL$.
Let $\lambda$ be a nonzero eigenvalue of $B$ and let $z$ be an associated eigenvector. By contradiction, we assume $\rho(B) = r$. We can therefore choose $\lambda$ such that $|\lambda| = r$. Show that then $\lambda = r$ then $Lz = z$ and reach a contradiction. Conclude.