TITLE: Limit of a matrix multiplication
QUESTION [9 upvotes]: 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).

REPLY [4 votes]: 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:
$$A^{\infty} = \left(
\begin{array}{cc}
 \frac{2}{3} & \frac{2}{3} \\
 \frac{1}{3} & \frac{1}{3} \\
\end{array}
\right)$$