My path for this problem is completely wrong as operators in general are non-diagonizable. Thus, I was wondering if someone can please help me prove it? Thank you!
Let $\beta$ be an ordered basis for a finite-dimensional vector space $V,$ and let $T\colon V\to V$ be linear. Prove that, for any nonnegative integer $k$, $[T^k]_\beta=([T]_\beta)^k.$
$\textbf{Solution:}$ Suppose that if $\beta$ is an ordered basis for a finite dimensional vector space $V$ and $T\colon V\to V$ is linear, then $T$ is diagonalizable such that $$[T]_\beta=\begin{pmatrix} \lambda_1&\alpha_{12} &\dots & 0 \\ 0& \lambda_2 &\dots&0\\ \vdots &\vdots &\cdots &\vdots\\ 0&0&\dots& \lambda_n \end{pmatrix}= \{\lambda_y\colon\lambda_y=0 \text{ for } i\ne j \text{ and } i,j =1,2,\dots, n\}.$$ For non-negative integer $k=1,2,\dots$ $$[T]_\beta=\{\lambda_y^k\colon\lambda_y^k =0\text{ for } i\ne j \} = \{\lambda_y\colon\lambda_y=0 \text{ for } i\ne j\} = [T]_\beta^k.$$
Let $\cal{B} = (b_1,\ldots,b_n)$ be a basis for $V$. Suppose that $T(b_i) = \sum_{j=1}^n\alpha_{i,j}b_j$ for each $1 \leq i \leq n$. Then \begin{align*}T^2(b_i) & = T\left( \sum_{j=1}^n\alpha_{i,j}b_j\right)\\ & = \sum_{j=1}^n\alpha_{i,j}T(b_j)\\ & = \sum_{j=1}^n\alpha_{i,j} \sum_{k=1}^n\alpha_{j,k}b_k\\ & = \sum_{k=1}^n\sum_{j=1}^n\alpha_{i,j} \alpha_{j,k}b_k \end{align*} Note that this means \begin{align*} [T^2(b_i)]_{\cal{B}} = \begin{bmatrix} \sum_{j=1}^n\alpha_{i,j} \alpha_{j,1}\\ \sum_{j=1}^n\alpha_{i,j} \alpha_{j,2}\\ \vdots\\ \sum_{j=1}^n\alpha_{i,j} \alpha_{j,n} \end{bmatrix} \end{align*} Proceed by induction. If $n = 2$, then we have
\begin{align*}[T]_{\cal B}^2 & = \big [ [T(b_1)]_{\cal{B}} \cdots [T(b_n)]_{\cal{B}}\big]^2\\ & = \begin{bmatrix}\alpha_{1,1} & \cdots & \alpha_{n,1}\\ \vdots & \ddots & \vdots\\ \alpha_{1,n} & \cdots & \alpha_{n,n} \end{bmatrix} \begin{bmatrix}\alpha_{1,1} & \cdots & \alpha_{n,1}\\ \vdots & \ddots & \vdots\\ \alpha_{1,n} & \cdots & \alpha_{n,n} \end{bmatrix}\\ & = \begin{bmatrix}\sum_{j=1}^n\alpha_{1,j} \alpha_{j,1} & \cdots & \sum_{j=1}^n\alpha_{n,j} \alpha_{j,1}\\ \vdots & \ddots & \vdots\\ \sum_{j=1}^n\alpha_{1,j} \alpha_{j,n} & \cdots & \sum_{j=1}^n\alpha_{n,j} \alpha_{j,n}\\ \end{bmatrix}\\ & = \big[ [T^2(b_1)]_{\cal{B}} \cdots [T^2(b_n)]_{\cal{B}}\big]\\ & = ([T]_{\cal{B}})^2. \end{align*}
This proves the base case. Let $n\in \mathbb{N}$. Suppose that $[T^n]_{\cal{B}} = ([T]_{\cal{B}})^n$. Then
$([T]_{\cal{B}})^{n+1} = [T]_{\cal{B}}([T]_{\cal{B}})^n = [T]_{\cal{B}}[T^n]_{\cal{B}} = [T \circ T^n]_{\cal{B}} = [T^{n+1}]_{\cal{B}}$. BY P.M.I., this proves the claim.