<s>[INST] 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.)
$n = 1$
$$\left(
\begin{array}{c}
 -\frac{2 a^2}{3}+\frac{2 a}{3}+1 \\
 \frac{4 a^2}{3}-\frac{4 a}{3}+1 \\
\end{array}
\right)$$
$n = 2$
$$\left(
\begin{array}{c}
 -\frac{2 a^4}{9}+\frac{10 a^3}{9}-\frac{20 a^2}{9}+\frac{4 a}{3}+1 \\
 \frac{4 a^4}{9}-\frac{20 a^3}{9}+\frac{40 a^2}{9}-\frac{8 a}{3}+1 \\
\end{array}
\right)$$
$n = 3$
$$\left(
\begin{array}{c}
 -\frac{2 a^6}{27}+\frac{2 a^5}{3}-\frac{22 a^4}{9}+\frac{122 a^3}{27}-\frac{14 a^2}{3}+2 a+1 \\
 \frac{4 a^6}{27}-\frac{4 a^5}{3}+\frac{44 a^4}{9}-\frac{244 a^3}{27}+\frac{28 a^2}{3}-4 a+1 \\
\end{array}
\right)$$
$n = 4$
$$\left(
\begin{array}{c}
 -\frac{2 a^8}{81}+\frac{26 a^7}{81}-\frac{16 a^6}{9}+\frac{440 a^5}{81}-\frac{812 a^4}{81}+\frac{308 a^3}{27}-8 a^2+\frac{8 a}{3}+1 \\
 \frac{4 a^8}{81}-\frac{52 a^7}{81}+\frac{32 a^6}{9}-\frac{880 a^5}{81}+\frac{1624 a^4}{81}-\frac{616 a^3}{27}+16 a^2-\frac{16 a}{3}+1 \\
\end{array}
\right)$$

 [/INST] $ M = I + B = I + \begin{bmatrix} 1 \\ -2 \end{bmatrix} \begin{bmatrix} \alpha && \beta \end{bmatrix} $
where $\alpha = -\frac{1}{3} a^2 $ and $\beta = \frac{1}{3} (2 a - a^2) $
Simple analysis shows the eigenvalues of $B$ are $0$ with corresponding eigenvector $[- \beta, \alpha]^T$, and $\alpha - 2 \beta $ with corresponding eigenvector $[1, -2]^T$
Now we want to express $[1,1]^T$ as a linear combination of these two eigenvectors.
$[1,1]^T = c_1 [-\beta, \alpha]^T + c_2 [1, -2]^T $
Solving yields,
$c_1 = \dfrac{3}{\alpha - 2 \beta} , \hspace{40pt} c_2 = \dfrac{\alpha+\beta}{\alpha - 2 \beta} $
Hence,
$M^n [1, 1]^T = c_1 (1)^n [-\beta, \alpha]^T + c_2 (\lambda_2)^n [1, -2]^T $
where $\lambda_2 = 1 + \alpha - 2 \beta \lt 1 $
Therefore,
$\displaystyle \lim_{n \to \infty} M^n [1, 1]^T = c_1 [-\beta, \alpha]^T = \dfrac{3}{\alpha-2\beta} \begin{bmatrix} -\beta \\ \alpha \end{bmatrix} $
And this simplifies to
$\displaystyle \lim_{n \to \infty} M^n [1, 1]^T = \dfrac{3}{\frac{a}{3}(a - 4) } \begin{bmatrix} - \frac{a}{3} (2 - a) \\ -\frac{1}{3} a^2 \end{bmatrix} $
And this simplifies further to,
$\displaystyle \lim_{n \to \infty} M^n [1, 1]^T = 3 \begin{bmatrix} \frac{2 -a}{4 - a} \\ \frac{a}{4 - a} \end{bmatrix} $
</s>