Let A be a matrix with eigenvalue $r=2$ with associated eigenvector $\begin{bmatrix} 1 \\ 0 \\ \end{bmatrix}$ and eigenvalue $r=1$ with associated eigenvector $\begin{bmatrix} 0 \\ 1 \\ \end{bmatrix}$. If $w=\begin{bmatrix} 2 \\ 3 \\ \end{bmatrix}$, calculate $A^{10}w.$
This is incorrect, but how I proceeded:
$$A^{10}w=\left(2\begin{bmatrix} 1 \\ 0 \\ \end{bmatrix},1\begin{bmatrix} 0 \\ 1 \\ \end{bmatrix}\right)^{10}\begin{bmatrix} 2 \\ 3 \\ \end{bmatrix}$$ $$=\left(2^{10}\begin{bmatrix} 1 \\ 0 \\ \end{bmatrix},1^{10}\begin{bmatrix} 0 \\ 1 \\ \end{bmatrix} \right)\begin{bmatrix} 2 \\ 3 \\ \end{bmatrix}$$ $$=\left(\begin{bmatrix} 1024 \\ 0 \\ \end{bmatrix},\begin{bmatrix} 0 \\ 1 \\ \end{bmatrix}\right)\begin{bmatrix} 2 \\ 3 \\ \end{bmatrix}=\begin{bmatrix} 1024&0 \\ 0&1 \\ \end{bmatrix}\begin{bmatrix} 2 \\ 3 \\ \end{bmatrix}=\begin{bmatrix} 2048 \\ 3 \\ \end{bmatrix}$$
Can someone correct this proof and possibly explain the correct "diagonalization" process here? Thanks in advance!
The matrix is $A=PDP^{-1}$ where $D$ has diagonal elements the eigenvalues, and the matrix $P$ has the eigenvectors in the same order. So we have
$$ A= \begin{bmatrix} 2&0\\ 0&1 \end{bmatrix} \implies A^{10} = \begin{bmatrix} 2^{10}&0\\ 0&1^{20} \end{bmatrix} $$