How to compute the limit of $M^{(n)}/(n+1)$ for the matrices P?

Question

Let $M^{(n)}=\sum_{r=0}^{n}P^r$. How to calculate $\lim_{n\to \infty}\frac{M^{(n)}}{n+1}$ for $P=\begin{bmatrix} 0.5 & 0.5 & 0 & 0 \\ 0.5 & 0 & 0.5 & 0\\ 0 & 0.5 & 0 & 0.5 \\ 0 & 0 & 0.5 &0.5 \end{bmatrix}$ ?

First, I used the method of diagonalization to compute $P^n$. I used Matlab and I got reslut

A =[0.5,0.5,0,0;0.5,0,0.5,0;0,0.5,0,0.5;0,0,0.5,0.5]
[S,D]=eig(A)
A =

    0.5000    0.5000         0         0
    0.5000         0    0.5000         0
         0    0.5000         0    0.5000
         0         0    0.5000    0.5000


S =

    0.2706    0.5000    0.6533    0.5000
   -0.6533   -0.5000    0.2706    0.5000
    0.6533   -0.5000   -0.2706    0.5000
   -0.2706    0.5000   -0.6533    0.5000


D =

   -0.7071         0         0         0
         0   -0.0000         0         0
         0         0    0.7071         0
         0         0         0    1.0000

But I have no idea what to do next.

Answer 1

As you have already shown, the eigenvalues of $P$ are $\lambda_1 = -1/\sqrt{2}$, $\lambda_2 = 0$, $\lambda_3 = 1/\sqrt{2}$ and $\lambda_4 =1$. Therefore, we can write $P$ as $P = SDS^T$ where $D = \text{diag}(\lambda_1, \lambda_2, \lambda_3, \lambda_4)$. We also know that $P^r = SD^r S^T$. Therefore, we have, \begin{align*} \lim_{n \to \infty} M^{(n)} & = \lim_{n \to \infty} \frac{1}{n + 1} \sum_{r = 0}^n P^r \\ & = \lim_{n \to \infty} \frac{1}{n + 1} \sum_{r = 0}^n SD^r S^T \\ & = \lim_{n \to \infty} \frac{1}{n + 1} \sum_{r = 0}^n S \ \text{diag}(\lambda_1^r, \lambda_2^r, \lambda_3^r, \lambda_4^r) \ S^T \\ & = \lim_{n \to \infty} \frac{1}{n + 1} S \ \text{diag} \left( \sum_{r = 0}^n \lambda_1^r, \sum_{r = 0}^n \lambda_2^r, \sum_{r = 0}^n \lambda_3^r, \sum_{r = 0}^n \lambda_4^r \right) \ S^T \\ & = \lim_{n \to \infty} \frac{1}{n + 1} S \ \text{diag} \left( \dfrac{1 - \lambda_1^n}{1 - \lambda_1}, 0, \dfrac{1 - \lambda_3^n}{1 - \lambda_3}, n+1 \right) \ S^T. \end{align*}

Note that since $|\lambda_1|, |\lambda_3| < 1$, $\displaystyle \lim_{n \to \infty} \frac{1}{n + 1} \dfrac{1 - \lambda_i^n}{1 - \lambda_i} = 0$ for $i = \{1, 3 \}$. Thus, \begin{align*} \lim_{n \to \infty} M^{(n)} & = S \ \text{diag}(0,0,0,1) \ S^T = \frac{1}{4} \mathbf{1}_{4 \times 4}, \end{align*} where $\mathbf{1}_{n \times n}$ is the $n \times n$ matrix consisting of all ones.