Let $u_0:=v \in R^d$ and $u_{k+1}:=Au_k, A \in M_d(\mathbb{R})$ being a symmetrical matrix with non-negative coefficients such that the sum of its row elements is $1$ for each row. Moreover, one eigenvalue is $1$, the other eigenvalues are $\lambda_i<1$ . I need to prove the following: $$\lim_{k\to \infty} u_k=\left(\frac{1}{d}\sum_{i=1}^{d} v_i~, \cdots ,~\frac{1}{d}\sum_{i=1}^{d} v_i\right)$$
Can somebody provide some hint or prove the statement ?
By Perron-Frobenius, $1$ is also the spectral radius. If the condition could be sharpened to $|λ_i|<1$, then from the properties of the power iteration we know that $u_k$ converges to the eigenvector of the largest eigenvalue. By the matrix properties, $\newcommand{\one}{1\!\!1}\one$, the one-vector, is that eigenvector. But then also $$ \one^Tu_{n+1}=\one^TAu_n=\one^Tu_n=...=\one^Tv. $$ All these properties combined give the claim.
