Markov chains by hand If I have a starting point: $A_T=[0,1]$ at $T=1$ and a one step transition matrix of:
$B=\left[ \begin{align} &\frac34 & \frac14& \\& \frac1{20}& \frac {19}{20} &\end{align} \right]$
I can compute $A_nB=A_{n+1}$
And keep reiterating this, but it is slow by hand, I assume there is a faster way? Would diagonalizing the matrix help somehow?

$B$ can be diagonalized as $B = PDP^{-1}$, where
$P = \begin{bmatrix}-5 & 1\\1 & 1\end{bmatrix}$, $P = \begin{bmatrix}7/10 & 0\\0 & 1\end{bmatrix}$, $P^{-1} = \begin{bmatrix}-1/6 & 1/6\\1/6 & 5/6\end{bmatrix}$.
Then $B^n = PD^nP^{-1} = \begin{bmatrix}-5 & 1\\1 & 1\end{bmatrix}\begin{bmatrix}(7/10)^n & 0\\0 & 1\end{bmatrix}\begin{bmatrix}-1/6 & 1/6\\1/6 & 5/6\end{bmatrix}$.
$$\boxed{\therefore B^n = \dfrac{1}{6}\begin{bmatrix}
5\left(\dfrac{7}{10}\right)^n + 1 & 5 - 5\left(\dfrac{7}{10}\right)^n\\
 1 - \left(\dfrac{7}{10}\right)^n &     \left(\dfrac{7}{10}\right)^n + 5
\end{bmatrix}}$$
As $B$ is a right stochastic matrix (because the row sums are $1$, not the column sums), we need to right multiply by $B$:
$A_n = A_0 B^n = \begin{matrix}[0 \quad 1]\\\mbox{} \end{matrix} \times
\dfrac{1}{6}
\begin{bmatrix}
5\left(\dfrac{7}{10}\right)^n + 1 & 5 - 5\left(\dfrac{7}{10}\right)^n\\
 1 - \left(\dfrac{7}{10}\right)^n &     \left(\dfrac{7}{10}\right)^n + 5
\end{bmatrix}\\
\mbox{}\\
\boxed{A_n = \dfrac{1}{6}\left[1 - \left(\dfrac{7}{10}\right)^n \qquad     \left(\dfrac{7}{10}\right)^n + 5\right]}$
