How would you prove the limit in the following theorem? Here $x_k$ is a column vector.
Let $P$ be a primitive projection matrix for a population model $x_k = Px_{k-1}$, with spectral radius $r$ and right and left eigenvectors $u$ and $v^T$ normalised so $v^Tu=1$. Then: $$\lim_{k \to \infty}\frac{x_k}{r^k} = (v^T x_0)u$$
I'm assuming it's through the use of the Perron-Frobenius theorem. We also have $x_k = P^kx_0$ if that helps.
I am assuming that a primitive projection matrix is just a primitive matrix. Since $P$ is a primitive matrix, the Perron-Frobinius Theorem applies. That is, $$\lim_{k\to\infty}\,\left(\frac{1}{r^k}\,P^k\right)=u\,v^\top\,.\tag{*}$$ Recall that $x_k=P^kx_0$ for every integer $k\geq 0$. That is, $$\lim_{k\to\infty}\,\left(\frac{1}{r^k}\,x_k\right)=\lim_{k\to\infty}\,\left(\frac{1}{r^k}\,P^kx_0\right)=\left(\lim_{k\to\infty}\,\frac{1}{r^k}\,P^k\right)\,x_0\,.$$ From (*), we get $$\lim_{k\to\infty}\,\left(\frac{1}{r^k}\,x_k\right)=(u\,v^\top)\,x_0=u\,(v^\top\,x_0)\,.$$ Because $v^\top\,x_0$ is a scalar, we obtain $$\lim_{k\to\infty}\,\left(\frac{1}{r^k}\,x_k\right)=(v^\top\,x_0)\,u\,.$$