TITLE: How to find $\lim\limits_{n\to\infty} A^n$?
QUESTION [6 upvotes]: Given $4\times4$ matrix,
$$A=
\begin{bmatrix}
0.4&0.3&0.2&0.1\\
0.1&0.4&0.3&0.2\\
0.3&0.2&0.1&0.4\\
0.2&0.1&0.4&0.3
\end{bmatrix}
.$$
Can we proof $$\lim\limits_{n\to\infty} A^n=
\begin{bmatrix}
0.25&0.25&0.25&0.25\\
0.25&0.25&0.25&0.25\\
0.25&0.25&0.25&0.25\\
0.25&0.25&0.25&0.25
\end{bmatrix}$$
manually? I tried to find $A^2$, and $A^3$ for a long long time. Any ways to find $\lim\limits_{n\to\infty} A^n$?

REPLY [2 votes]: The standard procedure for these sorts of problems is indeed diagonalization. The matrix is almost a circulant; if it were, then it would be diagonalized by the Fourier matrix.
Unfortunately, things don't line up that nicely. We have two of the four diagonal elements equal, not all of them. We would get a circulant matrix if we exchanged the third and fourth rows, but that operation doesn't commute with the 4-cycle that circulants are based on.
That does at least get us one eigenvector. The vector $j$ with all entries $1$ is an eigenvector for all circulants, and it's invariant under exchanging a pair of rows. Here, that eigenvector is associated with the eigenvalue $0.1+0.2+0.3+0.4 = 1$, which is the largest eigenvalue. For the simple question of what the limit of $A^n$ is, that's enough;
$$A^n(c_1j+c_2v_2+c_3v_3+c_4v_4) = c_1j+c_2\lambda_2^nv_2+c_3\lambda_3^nv_3+c_4\lambda_4^nv_4\to c_1j$$
where $\lambda_i$ and $v_i$ are the eigenvalues and eigenvectors. That tells us that $\lim_n A^n=ju^T$ for some row vector $u^T$ with the sum of its elements $1$. This $u^T$ is an eigenvector for multiplication by $A$ on the right; $u^TA=u^T$. Checking, we can easily see that $u^T$ will be equal to a multiple of $j^T$, and $\lim_n A^n =\frac14 jj^T$, the matrix with each entry equal to $\frac14$.
So there's the answer. What happens if we try for a full diagonalization? The characteristic polynomial is $(x-1)(x^3 - 0.2 x^2 - 0.04 x + 0.016)$. That cubic factor is irreducible over $\mathbb{Q}$, with approximate roots $-0.24111389$ and $0.22055694\pm 0.13309139i$. We could find eigenvectors for those, but it'll be ugly. For example, an eigenvector for the real eigenvalue is approximately $(0.1414406,0.37608589,-1.5175265,1)^T$. This is just not worth it.