How to find $\lim\limits_{n\to\infty} A^n$? 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 prove that $$\lim\limits_{n\to\infty} A^n=
\begin{bmatrix}
0.25&0.25&0.25&0.25\\
0.25&0.25&0.25&0.25\\
0.25&0.25&0.25&0.25\\
0.25&0.25&0.25&0.25
\end{bmatrix}$$
manually? I tried to find $A^2$, and $A^3$ for a long long time. Any ways to find $\lim\limits_{n\to\infty} A^n$?
 A: The matrix $A$ is almost circulant.  If the third and fourth rows were swapped, then it would be.  So let's consider an alternative matrix: $B=PA$ where $P$ is the permutation matrix
$$P=\begin{bmatrix} 1&0&0&0\\ 0&1&0&0\\ 0&0&0&1\\ 0&0&1&0 \end{bmatrix}.$$
Then $B$ is circulant.  
Since it is circulant, it can be diagonalized with the Discrete Fourier Transform (DFT).  The DFT values of $(0.4, 0.3, 0.2, 0.1)$ are 
$$d = \begin{bmatrix}
1.0000 + 0.0000i \\ 0.2000 - 0.2000i \\ 0.2000 + 0.0000i \\ 0.2000 + 0.2000i
\end{bmatrix}$$
Therefore, $B=F^\ast D F$, where $D=\text{diag}(d)$, where $F$ is the DFT matrix.
Note that $B^2=B\cdot B = F^\ast D F \cdot F^\ast D F$.  But, since $F$ is unitary, $F F^\ast=I$.  Therefore, $B^2=F^\ast D^2 F$.  Hopefully you can now see that $B^n = F^\ast D^n F$.  (This is a common trick.)
Therefore,
$$\lim_{N\rightarrow\infty} B^N = \lim_{N\rightarrow\infty} F^\ast D^N F = F^\ast \text{diag}(1,0,0,0)F.$$
Taking the inverse DFT of $(1,0,0,0)$ yields $(0.25,0.25,0.25,0.25)$.  And making a circulant matrix from this vector yields your answer.
(Another way to think about this is to ask, what circulant matrix would diagonalize with eigenvalues $(1,0,0,0)$; and that matrix is the specified matrix.)
So how does this relate to our original $A$?
I didn't figure it out.  Luckily, there's another person in this thread that appears to have!  https://math.stackexchange.com/a/3114768/24205
A: 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.
