Finding the Matrix Power of a matrix and limit Find the matrix power, $A^k$, of 
$$A=\begin{pmatrix}a & 1-a \\ b & 1-b\end{pmatrix}$$
$$D=P^{-1}AP$$
$$A^k=PD^kP^{-1}$$
I think that 
$$P=\begin{pmatrix}1 & \frac{a-1}{b} \\ 1 & 1\end{pmatrix} \ \ \  \text{and}\ \ \ \ P^{-1}=\frac{b}{1+b-a}\begin{pmatrix}1&\frac{1-a}{b}\\-1&1\end{pmatrix}$$
I found $A^k$ to be $\begin{pmatrix}1& (a-1)/b(1-b)^k\\1&(a-b)^k\end{pmatrix}$.
What would the limit of $A^k$ be when $k \rightarrow \infty$?
 A: A = $\begin{bmatrix}1-a \\ b & 1-b\end{bmatrix}$
Finding the eigenvalues and eigenvectors and writing the matrix in Jordan Normal Form yields:
$\displaystyle A = \begin{bmatrix}a & 1-a \\ b & 1-b\end{bmatrix} = P.D.P^{-1} = \begin{bmatrix}1 & \frac{a-1}{b} \\ 1 & 1\end{bmatrix} \cdot \begin{bmatrix}1 & 0 \\ 0 & a-b\end{bmatrix} \cdot \begin{bmatrix}\frac{1}{-\frac{a}{b} +\frac{1}{b} +1} & -\frac{a-1}{{(-\frac{a}{b} +\frac{1}{b} +1})b} \\ -\frac{1}{-\frac{a}{b} +\frac{1}{b} +1} & \frac{1}{-\frac{a}{b} +\frac{1}{b} +1}\end{bmatrix}$
Now, raising $A^k = (P.D.P^{-1})^k$ is just a matter of raising the diagonal matrix to a power (which is straightforward), yielding:
$\displaystyle A^k = \begin{bmatrix}a & 1-a \\ b & 1-b\end{bmatrix}^k = (P.D.P^{-1})^k$
$\displaystyle = \begin{bmatrix}1 & \frac{a-1}{b} \\ 1 & 1\end{bmatrix} \cdot \begin{bmatrix}1 & 0 \\ 0 & a-b\end{bmatrix}^k \cdot \begin{bmatrix}\frac{1}{-\frac{a}{b} +\frac{1}{b} +1} & -\frac{a-1}{{(-\frac{a}{b} +\frac{1}{b} +1})b} \\ -\frac{1}{-\frac{a}{b} +\frac{1}{b} +1} & \frac{1}{-\frac{a}{b} +\frac{1}{b} +1}\end{bmatrix}$
$\displaystyle  = \begin{bmatrix}1 & \frac{a-1}{b} \\ 1 & 1\end{bmatrix} \cdot \begin{bmatrix}1 & 0 \\ 0 & (a-b)^k\end{bmatrix} \cdot \begin{bmatrix}\frac{1}{-\frac{a}{b} +\frac{1}{b} +1} & -\frac{a-1}{{(-\frac{a}{b} +\frac{1}{b} +1})b} \\ -\frac{1}{-\frac{a}{b} +\frac{1}{b} +1} & \frac{1}{-\frac{a}{b} +\frac{1}{b} +1}\end{bmatrix}$
You can handle the rest by multiplying out the three matrices.
