$\bullet$ First we will prove that there exist $m, M>0$ such that \begin{equation} m\leq \frac{\|x_{k+1}\|}{\|x_k\|}\leq M \label{ineq:Golub0} \end{equation}for all $k$ with $m$ and $M$ independent of the sequence. In order to show this we use the primitivity of the matrices $A_k$ and $A_\infty$ Primitivity of $A_\infty$ implies that there exists $l\in\mathbb{N}$ such that $$ A^l_\infty\gg 0 $$ By continuity, this implies that there exists $k_0$ such that, for all $k\geq k_0$ we have $$ A_{k+l}A_{k+l-1}\cdots A_k\gg 0 $$ Let us then write $x_{k+l+1}$ as a function of $x_k$ $$ x_{k+l+1}=A_{k+l}\cdots A_kx_k $$ We thus have \begin{equation} \|x_{k+l+1}\|\leq C^{l+1} \|x_k\| \label{ineq:Golub1} \end{equation}But since the matrices $A_{k+l}$ ...,$A_k$ are strictly positive for $k\geq k_0$ there exists a $\varepsilon>0$ such that each component of these matrices is superior or equal to $\varepsilon$ From this we deduce that $$ \|x_{k+l+1}\|\geq \varepsilon \|x_k\| $$ for all $k\geq k_0$ Applying relation (), we then have that $$ C^{l} \|x_{k+1}\|\geq \varepsilon \|x_k\| $$ which yields $$ \|x_{k+1}\|\geq \frac{\varepsilon}{C^l}\|x_k\| $$ for all $k\geq 0$ and so we indeed have relation ().

$\bullet$ Let us denote by $e_k$ the (normalised) Perron eigenvector of $A_k$ Thus $$ A_ke_k=\lambda_ke_k\quad \|e_k\|=1 $$ Let us denote by $\pi_k$ the projection on the supplementary space of $\{e_k\}$ invariant by $A_k$ Choosing a proper norm, we can find $\varepsilon>0$ such that $$ \left| A_k\pi_k \right| \leq (\lambda_k-\varepsilon) $$ for all $k$

$\bullet$ We shall now prove that $$ \frac{\left<e_{k+1}^*,x_{k+1}\right>}{\left<e_k^*,x_k\right>}\to \lambda_\infty \textrm{ when } k\to\infty $$ In order to do this, we compute the inner product of the sequence $x_{k+1}=A_kx_k$ with the $e_k$ s: \begin{eqnarray*} \left<e_{k+1}^*,x_{k+1}\right>&=& \left<e_{k+1}^*-e_k^*,A_kx_k\right>+\lambda_k \left<e_k^*,x_k\right> \\ &=& o\left(\left<e_k^*,x_k\right> \right)+ \lambda_k \left<e_k^*,x_k\right> \end{eqnarray*}Therefore we have $$ \frac{\left<e_{k+1}^*,x_{k+1}\right>}{\left<e_k^*,x_k\right>}=o(1)+\lambda_k $$

$\bullet$ Now assume $$ u_k=\frac{\pi_k x_k}{\left<e_k^*,x_k\right>} $$ We will verify that $u_k\to 0$ when $k\to\infty$ We have \begin{eqnarray*} u_{k+1}&=& (\pi_{k+1}-\pi_k)A_k\frac{x_k}{\left<e_{k+1}^*,x_{k+1}\right>} + \frac{\left<e_k^*,x_k\right>}{\left<e_k^*,x_{k+1}\right>} A_k\pi_k\frac{x_k}{\left<e_k^*,x_k\right>} \end{eqnarray*}and so $$ |u_{k+1}|\leq |\pi_{k+1}-\pi_k|C'+\frac{\left<e_k^*,x_k\right>}{\left<e_{k+1}^*,x_{k+1}\right>} (\lambda_k-\varepsilon) |u_k| $$ We deduce that there exists $k_1\geq k_0$ such that, for all $k\geq k_1$ $$ |u_{k+1}| \leq \delta_k+(\lambda_\infty-\frac{\varepsilon}{2}) |u_k| $$ where we have noted $$ \delta_k=(\pi_{k+1}-\pi_k)C' $$ We have $\delta_k\to 0$ when $t\to\infty$ we thus finally deduce that $$ |u_k|\to 0\textrm{ when }k\to\infty $$ Remark that this also implies that $$ z_k=\frac{\pi_kx_k}{\|x_k\|}\to 0\textrm{ when } k\to\infty $$

$\bullet$ We have $z_k\to 0$ when $k\to\infty$ and $x_k/\|x_k\|$ can be written $$ \frac{x_k}{\|x_k\|}=\alpha_ke_k+z_k $$ Therefore, we have $\alpha_ke_k\to 1$ when $k\to\infty$ which implies that $\alpha_k$ tends to 1, since we have chosen $e_k$ to be normalised (i.e.,$\|e_k\|=1$ .

We then can conclude that $$ \frac{x_k}{\|x_k\|}\to e_\infty \textrm{ when }k\to\infty $$ and the proof is done.