Convergence of matrix with spectral radius 1

Question

The following matrix:

$$M = \begin{pmatrix} 0 & \frac{2}{3} & 0 & \frac{1}{3} \\ \frac{1}{3} & 0 & \frac{1}{3} & \frac{1}{3} \\ \frac{1}{3} & \frac{1}{3} & 0 & \frac{1}{3} \\ \frac{1}{3} & \frac{1}{3} & \frac{1}{3} & 0 \\ \end{pmatrix} $$

has eigenvalues $\lambda_1=1,\lambda_2=-\frac{1}{3},\lambda_3=-\frac{1}{3},\lambda_4=-\frac{1}{3}$. So, $ \lambda_1 $ is the spectral radius. I know that a matrix with a spectral radius $< 1$ converges to $0$ and with a spectral radius $> 1$ to infinity. How can I prove to which value a matrix with spectral radius of $1$ converges?

I found theorem 8 in these slides but don't quite understand how to apply it in the case of the above matrix.

So in other words:

The theorem states that the matrix converges to the lowest order constituent matrix, but not sure how to find this matrix?

Answer 1

Performing a Jordan decomposition for this matrix gives

$$M=VJV^{-1}$$

where

$$V=\left(\begin{array}{cccc} -\frac{1}{16} & -\frac{1}{4} & -\frac{15}{16} & -1\\ -\frac{1}{16} & \frac{1}{12} & \frac{1}{16} & 0\\ -\frac{1}{16} & \frac{1}{12} & \frac{17}{16} & 0\\ -\frac{1}{16} & \frac{1}{12} & \frac{1}{16} & 1 \end{array}\right),\ V^{-1}=\left(\begin{array}{cccc} -4 & -5 & -3 & -4\\ -3 & 9 & -3 & -3\\ 0 & -1 & 1 & 0\\ 0 & -1 & 0 & 1 \end{array}\right)$$

and

$$J=\left(\begin{array}{cccc} 1 & 0 & 0 & 0\\ 0 & -\frac{1}{3} & 1 & 0\\ 0 & 0 & -\frac{1}{3} & 0\\ 0 & 0 & 0 & -\frac{1}{3} \end{array}\right).$$

Clearly, we have that

$$M^n=(VJV^{-1})^n=VJ^nV^{-1}$$

where

$$J^n=\left(\begin{array}{cccc} 1 & 0 & 0 & 0\\ 0 & \left(-\frac{1}{3}\right)^n & n\left(-\frac{1}{3}\right)^{n-1} & 0\\ 0 & 0 & \left(-\frac{1}{3}\right)^n & 0\\ 0 & 0 & 0 & \left(-\frac{1}{3}\right)^n \end{array}\right).$$

Therefore, when $n\to\infty$,

$$J^n\to J^*:=\left(\begin{array}{cccc} 1 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 \end{array}\right).$$

As a result,

$$M^n\to VJ^*V^{-1}=\begin{pmatrix}-\frac{1}{16}\\-\frac{1}{16}\\-\frac{1}{16}\\-\frac{1}{16}\end{pmatrix}\begin{pmatrix} -4 & -5 & -3 & -4\end{pmatrix}=\left(\begin{array}{cccc} \frac{1}{4} & \frac{5}{16} & \frac{3}{16} & \frac{1}{4}\\ \frac{1}{4} & \frac{5}{16} & \frac{3}{16} & \frac{1}{4}\\ \frac{1}{4} & \frac{5}{16} & \frac{3}{16} & \frac{1}{4}\\ \frac{1}{4} & \frac{5}{16} & \frac{3}{16} & \frac{1}{4} \end{array}\right).$$

Answer 2

$M$ can be decomposed into

$$ M = P J P^{-1} $$

where

$ J = \begin{bmatrix} - \frac{1}{3} && 0 && 0 && 0 \\ 0 && - \frac{1}{3} && 1 && 0 \\ 0 && 0 && - \frac{1}{3} && 0 \\ 0 && 0 && 0 && 1 \end{bmatrix} $

and

$ P = \begin{bmatrix} -2 && \frac{1}{4} && \frac{5}{4} && 1 \\ 1 && -\frac{1}{12} && 1 && 1 \\ 1 && - \frac{1}{12} && 0 && 1 \\ 0 && - \frac{1}{12} && 0 && 1 \end{bmatrix} $

One can easily verify that

$\displaystyle \lim_{n \to \infty} J^n = \begin{bmatrix} 0 && 0 && 0 && 0 \\ 0 && 0 && 0 && 0 \\ 0 && 0 && 0 && 0 \\ 0 && 0 && 0 && 1 \end{bmatrix} $

and we also have

$ P^{-1} = \begin{bmatrix} 0 && 0 && 1 && -1 \\ 3 && \frac{15}{4} && \dfrac{9}{4} && -9 \\ 0 && 1 && -1 && 0 \\ \frac{1}{4} && \frac{5}{16} && \frac{3}{16} && \frac{1}{4} \end{bmatrix} $

Hence,

$\displaystyle \lim_{n \to \infty} M^n = \begin{bmatrix} 1 \\1\\1\\1\end{bmatrix} \begin{bmatrix} \frac{1}{4} && \frac{5}{16} && \frac{3}{16} && \frac{1}{4} \end{bmatrix} = \begin{bmatrix} \frac{1}{4} && \frac{5}{16} && \frac{3}{16} && \frac{1}{4} \\ \frac{1}{4} && \frac{5}{16} && \frac{3}{16} && \frac{1}{4} \\\frac{1}{4} && \frac{5}{16} && \frac{3}{16} && \frac{1}{4} \\\frac{1}{4} && \frac{5}{16} && \frac{3}{16} && \frac{1}{4} \end{bmatrix} $