To prove: A linear operator T on a finite-dimensional vector space V is diagonalizable if and only if there exists an ordered basis $\beta$ for V consisting of eigenvectors of T. If $\beta=\{v_1,\ldots,v_n\}$, the corresponding eigenvalues are $\lambda_1,\ldots,\lambda_n$. then $[T]_\beta$=$\begin{pmatrix} \lambda_1 & \dots & 0 \\ 0 & \ddots & \\ 0 & \cdots & \lambda_n \end{pmatrix}$
Definition: A linear mapping T on a finite-dimensional vector space V is diagonalizable if there is an ordered basis $\beta$ for V such that $[T]_\beta$ is a diagonal matrix.
proof: $\Rightarrow$ suppose $\beta=\{v_1,\ldots,v_n\}$ is a basis for V for which $[T]_\beta$ is diagonal, then by definition , we have $[T]_\beta= \begin{pmatrix} \lambda_1 & \dots & 0 \\ 0 & \ddots & \\ 0 & \cdots & \lambda_n \end{pmatrix}$, then $[T(v_i)]_\beta= [\lambda_i v_i]_\beta$ So $T(v_i)=\lambda_i v_i$ and $\beta$ consists of eigenvectors of T for which $v_i$ $\neq$ 0 by definition.
$\Leftarrow$: suppose $B=\{v_1,\ldots,v_n\}$ is a basis consisting of eigenvectors,for V and $T(v_i)$=$\lambda_i v_i$ for all i $(\lambda_i$ $\in F)$, then $T(v_i)=\lambda_i v_i$ for all i $(\lambda_i \in F)$ ,then $T(v_i)=0 v_1+...+\lambda_i v_i+ \dots 0 v_n$ for all i, so $[T]_\beta=\begin{pmatrix} \lambda_1 & \dots & 0 \\ 0 & \ddots & \\ 0 & \cdots & \lambda_n \end{pmatrix}$.
Question: Wondering if this makes sense, especially the expression of notations.