Finding the symbolic expression for a particular matrix raised to the $n^{\text{th}}$ power Consider the following matrix, with $a$ being a real number between $0$ and $1$.
\begin{equation}
\text{M}=\left[
\begin{array}{cc}
 1-\frac{a^2}{3} & \frac{1}{3} \left(2 a-a^2\right) \\
 \frac{2 a^2}{3} & 1-\frac{2}{3}  \left(2 a-a^2\right) \\
\end{array}
\right].
\end{equation}
For a general integer $n$, I am trying to compute the closed form expression for the entries of:
\begin{equation}
\text{FinalVector}= \text{M}^{n} \left(
\begin{array}{c}
 1\\
 1 \\
\end{array}
\right).
\end{equation}

It is very easy to compute what $\text{FinalVector}$ is like for a few specific values of $n$, but is there a way to find a formula for a general $n$?
Here are the values for $\text{FinalVector}$ for the first few $n$ (from Mathematica.)
$n = 1$
$$\left(
\begin{array}{c}
 -\frac{2 a^2}{3}+\frac{2 a}{3}+1 \\
 \frac{4 a^2}{3}-\frac{4 a}{3}+1 \\
\end{array}
\right)$$
$n = 2$
$$\left(
\begin{array}{c}
 -\frac{2 a^4}{9}+\frac{10 a^3}{9}-\frac{20 a^2}{9}+\frac{4 a}{3}+1 \\
 \frac{4 a^4}{9}-\frac{20 a^3}{9}+\frac{40 a^2}{9}-\frac{8 a}{3}+1 \\
\end{array}
\right)$$
$n = 3$
$$\left(
\begin{array}{c}
 -\frac{2 a^6}{27}+\frac{2 a^5}{3}-\frac{22 a^4}{9}+\frac{122 a^3}{27}-\frac{14 a^2}{3}+2 a+1 \\
 \frac{4 a^6}{27}-\frac{4 a^5}{3}+\frac{44 a^4}{9}-\frac{244 a^3}{27}+\frac{28 a^2}{3}-4 a+1 \\
\end{array}
\right)$$
$n = 4$
$$\left(
\begin{array}{c}
 -\frac{2 a^8}{81}+\frac{26 a^7}{81}-\frac{16 a^6}{9}+\frac{440 a^5}{81}-\frac{812 a^4}{81}+\frac{308 a^3}{27}-8 a^2+\frac{8 a}{3}+1 \\
 \frac{4 a^8}{81}-\frac{52 a^7}{81}+\frac{32 a^6}{9}-\frac{880 a^5}{81}+\frac{1624 a^4}{81}-\frac{616 a^3}{27}+16 a^2-\frac{16 a}{3}+1 \\
\end{array}
\right)$$


There is no need to do any eigendecomposition. As observed by the others, $M$ is a rank-one update of the identity matrix:
$$
M=\pmatrix{
1-\frac{a^2}{3}&\frac{1}{3}(2a-a^2)\\
\frac{2a^2}{3}&1-\frac{2}{3}(2a-a^2)}
=I+\pmatrix{1\\ -2}
\pmatrix{-\frac{a^2}{3}&\frac{1}{3}(2a-a^2)}
=I+uv^T.
$$
Therefore, when $n\ge1$,
\begin{aligned}
M^n
&=(I+uv^T)^n\\
&=\sum_{k=0}^n\binom{n}{k}(uv^T)^k\\
&=I+\sum_{k=1}^n\binom{n}{k}(uv^T)^k\\
&=I+\sum_{k=1}^n\binom{n}{k}(v^Tu)^{k-1}uv^T\\
&=\begin{cases}
I+\frac{(1+v^Tu)^n-1}{v^Tu}uv^T&\text{if }v^Tu\ne0,\\
I+nuv^T&\text{if }v^Tu=0.\\
\end{cases}
\end{aligned}
It follows that
$$
M^n\mathbf1
=\begin{cases}
\mathbf1+\left[\frac{(1+v^Tu)^n-1}{v^Tu}(v^T\mathbf1)\right]u&\text{if }v^Tu\ne0,\\
\mathbf1+n(v^T\mathbf1)u&\text{if }v^Tu=0.\\
\end{cases}.
$$
