Finding a 2x2 Matrix raised to the power of 1000 Let $A= \pmatrix{1&4\\ 3&2}$. Find $A^{1000}$.
Does this problem have to do with eigenvalues or is there another formula that is specific to 2x2 matrices?

Perform an eigenvalue decomposition of $A$, we then get
$$A = 
\begin{bmatrix}
-4/5 & -1/\sqrt2\\
3/5 & -1/\sqrt2
\end{bmatrix}
\begin{bmatrix}
-2 & 0\\
0 & 5
\end{bmatrix}
\begin{bmatrix}
-4/5 & -1/\sqrt2\\
3/5 & -1/\sqrt2
\end{bmatrix}^{-1}
=VDV^{-1}
$$
where $V = \begin{bmatrix}
-4/5 & -1/\sqrt2\\
3/5 & -1/\sqrt2
\end{bmatrix}$ and $D = \begin{bmatrix}
-2 & 0\\
0 & 5
\end{bmatrix}$.
Hence,
$$A^n = \underbrace{\left(VDV^{-1} \right)\left(VDV^{-1} \right)\cdots \left(VDV^{-1} \right)}_{n \text{ times}} = VD^n V^{-1}$$
