Finding the limit of a matrix Suppose that $A=\begin{pmatrix}4&1 & 5\\ 2& 7& 1\\ 2& 2& 6\end{pmatrix}.\;$ How can I find $\;\displaystyle \lim_{n\to\infty}A^n$?
What theorem(s) should I use to solve this?
It's harder than what it looks like..... 
I know it's stochastic, right?
 A: Hint: Can you diagonalize the matrix and then take advantage of that result?
Are you familiar with the Jordan Normal Form?
A: Hint:  Matrix $A$ is not stochastic but $A^t$ is, so take the transpose of $\lim_{n\to\infty}(A^t)^n$, and there you go!
A: The eigenvalues of $A$ are $2, 5, 10$. So there is no limit for $A^n$.
For any matrix norm, Gelfand's spectral radius formula yields
$$
10=\rho(A)=\lim_{n\rightarrow +\infty}\|A^n\|^\frac{1}{n}\quad\Rightarrow \quad \lim_{n\rightarrow+\infty}\|A^n\|=+\infty.
$$
But $A^n/10^n$ does have a limit. This will be three copies of the stationary probability vector of the stochastic matrix $A/10$. And if you don't know Perron-Frobenius theory, you can simply diagonalize $A$.
A: $A$ is $10$ times a row stochastic matrix. Therefore the spectral radius of $A^n$ is $10^n$. Consequently, $\lim_{n\to\infty}A^n$ does not exist.
Suppose you are going to find $\lim_{n\to\infty}B^n$ instead, where $B=\frac{A}{10}$. If you are not familiar with Perron-Frobenius theorem, simply diagonalise $B$ as $PDP^{-1}$ and compute the limit of the power as $P\,(\lim_{n\to\infty}D^n)P^{-1}$.
If you are familiar with Perron-Frobenius theorem, here is the most straightforward way to solve the problem, which requires the least amount of calculations. Since $B$ is row stochastic and entrywise positive, it is irreducible. By Perron-Frobenius theorem, $1$ is a simple eigenvalue of $B$ and $\lim_{n\to\infty}B^n=\mathbf{1}\frac{v^T}{\color{red}{\sum_iv_i}}$, where $\mathbf{1}=(1,\ldots,1)^T$ and $v$ is a stationary vector of $B$. So, you need to solve $v^T(B-I)=0$ or $v^T(A-10I)=0$. Since $1$ is a simple eigenvlue of $B$, $10$ is a simple eigenvalue of $A$. Therefore the adjugate matrix of $A-10I$ has rank $1$ and you may take $v=\color{red}{(C_{1j},C_{2j},C_{3j})}$ for whichever $j$ that makes $v$ nonzero, where $C_{ij}$ denotes the $(i,j)$-th cofactor of $A-10I$. Note that the expressions in red above are the only ones that require nontrivial calculations and these calculations are very straightforward.
A: Let $B=\det A.$ $B=100$.
If there exists $\lim\limits_{n\to \infty}A^n$, then exists $\lim\limits_{n\to \infty}B^n$.
Neccessary condition is $|B|\leqslant 1$.
I think there is reason to consider scaled matrix
$$\color{#AA0000}{A=\begin{pmatrix}
.4& .1& .5\\ 
.2& .7& .1\\ 
.2& .2& .6
\end{pmatrix}},\tag{1}$$
where sum of elements in each row equals to 1.
If there exists limit
$$
M = \lim_{n\to \infty} A^n = 
\begin{pmatrix}
a& b& c\\ 
d& e& f\\ 
g& h& k
\end{pmatrix},
$$
then
$$
A\cdot M = M,\tag{2}
$$
$$
M\cdot A = M.\tag{3}
$$
(2) implies
$$
\left\{\begin{array}{c}
0.4a+0.1d+0.5g=a;\\
0.2a+0.7d+0.1g=d;\\
0.2a+0.2d+0.6g=g;\\
\end{array}
\right.
$$
$$
\left\{\begin{array}{c}
-0.6a+0.1d+0.5g=0;\\
0.2a-0.3d+0.1g=0;\\
0.2a+0.2d-0.4g=0;\\
\end{array}
\right.
$$
then $a=d=g$.
Same way we get $b=e=h$, $c=f=k$.
So, 
$$
M =  
\begin{pmatrix}
a& b& c\\ 
a& b& c\\ 
a& b& c
\end{pmatrix}.
$$
Now, (3) implies
$$
\left\{\begin{array}{c}
0.4a+0.2b+0.2c=a;\\
0.1a+0.7b+0.2c=b;\\
0.5a+0.1b+0.6c=c;\\
\end{array}
\right.
$$
$$
\left\{\begin{array}{c}
-0.6a+0.2b+0.2c=0;\\
0.1a-0.3b+0.2c=0;\\
0.5a+0.1b-0.4c=0;\\
\end{array}
\right.
$$
then $b=1.4a$, $c=1.6a$.
If we'll use the rule, that sum of elements in each row equals to 1, then
$a+1.4a+1.6a=1$ $\implies$ $a=0.25$ $\implies$ $(a,b,c)=(0.25,\; 0.35,\; 0.4)$.
Answer (with asumption (1)):
$$
\color{#AA0000}{
\lim_{n\to\infty} A^n = 
\begin{pmatrix}
0.25& 0.35& 0.4\\ 
0.25& 0.35& 0.4\\ 
0.25& 0.35& 0.4
\end{pmatrix}}.
$$
