For any primitive matrix $A$ in $\mathcal{M}_n(\mathbb{R})$, the spectral radius $\rho(A)$ is a dominant eigenvalue of $A$ and the associated eigenspace is a line that possesses a strictly positive direction vector $x > 0$. We denote $r$ the spectral radius of $A$. We denote $x$ (respectively $y$) a strictly positive direction vector of the line $D = \operatorname{Ker}(A - rI_n)$ (respectively of the line $\Delta = \operatorname{Ker}(A^\top - rI_n)$). We denote $H = \operatorname{Im}(A - rI_n)$. If necessary by multiplying $y$ by an appropriate strictly positive coefficient, we assume $(y \mid x) = y^\top x = 1$. We denote $L = xy^\top$. Show that $AL = LA = rL$. Deduce: $\forall m \in \mathbb{N}^*, (A - rL)^m = A^m - r^m L$.
For any primitive matrix $A$ in $\mathcal{M}_n(\mathbb{R})$, the spectral radius $\rho(A)$ is a dominant eigenvalue of $A$ and the associated eigenspace is a line that possesses a strictly positive direction vector $x > 0$. We denote $r$ the spectral radius of $A$. We denote $x$ (respectively $y$) a strictly positive direction vector of the line $D = \operatorname{Ker}(A - rI_n)$ (respectively of the line $\Delta = \operatorname{Ker}(A^\top - rI_n)$). We denote $H = \operatorname{Im}(A - rI_n)$. If necessary by multiplying $y$ by an appropriate strictly positive coefficient, we assume $(y \mid x) = y^\top x = 1$. We denote $L = xy^\top$.
Show that $AL = LA = rL$. Deduce: $\forall m \in \mathbb{N}^*, (A - rL)^m = A^m - r^m L$.