Limit of a matrix multiplication How can I calculate this limit:
$\displaystyle\lim_{n\to\infty}\begin{bmatrix}0.9 & 0.2\\0.1 & 0.8\end{bmatrix}^n$
What is the tool that i need to aply? eigenvalues and eigenvectors? diagonalization? canonical form? 
(This came in a contest and was the only problem i cannont have an idea for solve it).
 A: $$A=\begin{bmatrix}\frac{9}{10} & \frac{2}{10}\\\frac{1}{10} & \frac{8}{10}\end{bmatrix}$$
$$A=PDP^{-1}$$
where
$$P=\begin{bmatrix}2 & -1\\1 & 1\end{bmatrix}$$
$$D=\begin{bmatrix}1 & 0\\0 & 7/10\end{bmatrix}$$
So the limit is given by:
$$\lim_{n\to\infty} A^n=\lim_{n\to\infty} PD^nP^{-1}=PD_{\infty}P^{-1}......(1)$$
where
$$D_{\infty}=\begin{bmatrix}1 & 0\\0 & \lim_{n\to\infty}(7/10)^n\end{bmatrix}=\begin{bmatrix}1 & 0\\0 & 0\end{bmatrix}......(2)$$
Substitution of $D_{\infty}$ into (1) leads to:
$$\lim_{n\to\infty} A^n=PD_{\infty}P^{-1}=\begin{bmatrix}\frac23 & \frac23\\\frac13 & \frac13\end{bmatrix}$$
A: Note that your matrix is not an arbitrary matrix --- it is a column stochastic matrix and thus a Markov transition matrix. Hence, from the Perron-Frobenius theorem you will know that each column of the limit matrix will be the normalized eigenvector of your matrix corresponding to the eigenvalue $1$, and as you can check
$$\begin{bmatrix}0.9 & 0.2\\0.1 & 0.8\end{bmatrix}\begin{bmatrix}2/3\\1/3\end{bmatrix}=\begin{bmatrix}2/3\\1/3\end{bmatrix},$$
so that, indeed, $\lim_{n\rightarrow\infty}\begin{bmatrix}0.9 & 0.2\\0.1 & 0.8\end{bmatrix}^n=\begin{bmatrix}2/3 & 2/3\\1/3 & 1/3\end{bmatrix}$. 
This methodology applies generally: If you have a column stochastic matrix $B$ with strictly positive entries, its (power) limit exists and each column of $\lim_n B^n$ is given by the normalized eigenvector of $B$ corresponding to the eigenvalue $1$.
A: Because that is a stochastic matrix you just need to fine the fixed probability vector or steady state vector.
$$(t1,t2)\cdot\left(
\begin{array}{cc}
 0.9 & 0.1 \\
 0.2 & 0.8 \\
\end{array}
\right)=(t1,t2)$$
Notice I have transposed A for convenience and this yields the simultaneous set of equations
$$.9 \cdot t1+.2 \cdot t2=\text{t1}$$
$$.1 \cdot t1+.8 \cdot t2=\text{t2}$$
$$t1 + t2 = 1$$
this is easily solved to get
$t1=\frac{2}{3}$
$t2=\frac{1}{3}$
Because this is a regular Markov chain all the rows of $A^{\infty} $ are equal to $(t1,t2)$, so
$$A^{\infty} = \left(
\begin{array}{cc}
 \frac{2}{3} & \frac{1}{3} \\
 \frac{2}{3} & \frac{1}{3} \\
\end{array}
\right)$$
Transpose A back:
$$A^{\infty} = \left(
\begin{array}{cc}
 \frac{2}{3} & \frac{2}{3} \\
 \frac{1}{3} & \frac{1}{3} \\
\end{array}
\right)$$
A: Diagonalization is precisely the tool you need. 
If you can write $A = PDP^{-1}$, where $D$ is a diagonal matrix, then $A^n = PD^nP^{-1}$, where $D^n$ is also diagonal, and its entries are just the $n$-th power of the entries in $D$. 
Then, $\displaystyle\lim_{n\to\infty}A^n = \lim_{n\to\infty}PD^nP^{-1}$ is easy to compute. 
