<s>[INST] If $A =\left[ \begin{smallmatrix} 3 & -4 \\ 1 & -1\end{smallmatrix}\right]$, then find $A^n$ 
If $$ A = \begin{bmatrix}
3 &  -4 \\
1 & -1 
\end{bmatrix}$$
Then find $A^n$

I have tried solving it using diagonalization $PDP^{-1}$ but I am getting only one independent eigenvector i.e $$ \begin{bmatrix}
2\\
1
\end{bmatrix}$$
Please tell me the correct method to solve it.
 [/INST] Indeed, $A$ has one and only one eigenvalue: $1$. And $(2,1)$ is an eigenvector corresponding to that eigenvalue. Now, consider the vector $(1,0)$. Then$$A.(1,0)=(3,1)=(2,1)+(1,0).$$So, if $M=\left[\begin{smallmatrix}2&1\\1&0\end{smallmatrix}\right]$, then$$M^{-1}.A.M=\begin{bmatrix}1&1\\0&1\end{bmatrix},$$or$$A=M.\begin{bmatrix}1&1\\0&1\end{bmatrix}\cdot M^{-1}.$$So,\begin{align}A^n&=M.\begin{bmatrix}1&1\\0&1\end{bmatrix}^n.M^{-1}\\&=\begin{bmatrix}2&1\\1&0\end{bmatrix}.\begin{bmatrix}1&n\\0&1\end{bmatrix}.\begin{bmatrix}0&1\\1&-2\end{bmatrix}\\&=\begin{bmatrix}2n+1&-4n\\n&1-2n\end{bmatrix}.\end{align}
Note that the equality $\left[\begin{smallmatrix}1&1\\0&1\end{smallmatrix}\right]^n=\left[\begin{smallmatrix}1&n\\0&1\end{smallmatrix}\right]$ has to do with the fact that$$(\forall x,y\in\mathbb R):\begin{bmatrix}1&x\\0&1\end{bmatrix}.\begin{bmatrix}1&y\\0&1\end{bmatrix}=\begin{bmatrix}1&x+y\\0&1\end{bmatrix}.$$
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