I'm trying trying to prove diagonalizable criterion for operators and matrices. It's a little lengthy:
Let $V$ have basis $\{e_1,...,e_n\}$ and fix $T\in\scr{L}$$(V)$. I know for any matrix $A(T)$ of $T$, the $\lambda$ is an eigenvalue of $A(T)\iff\lambda$ is an eigenvalue of $T$. Also, $a_1e_1+...+a_ne_n=:v\in V$ is an eigenvector of $T\iff (a_1,...,a_n)\in\mathbb{F}^n$ is an eigenvector of $A(T)$. In particular, we have the following: Since dim$V=$dim$\mathbb{F}^n$, $V\cong\mathbb{F}^n$. Let $\phi$ be the isomorphism $a_1e_1+...a_ne_n\mapsto (a_1,...,a_n)$ between them. Isomorphisms carry linearly independent sets to linearly independent sets. Hence, $\{v_{\lambda_1},...,v_{\lambda_n}\}$ is a basis of eigenvectors of $T\iff \{\phi(v_{\lambda_1}),...,\phi(v_{\lambda_n})\}$ is a basis of eigenvectors of $\mathbb{F}^n$.
\begin{align}\textbf{Operators}\end{align}
1.) It's clear that $T$ has a diagonal matrix $\iff$ $V$ has a basis of eigenvectors of $T$.
2.) If $T$ has $n$ distinct eigenvalues, then $\{v_{\lambda_1},...,v_{\lambda_n}\}$ is a basis (of eigenvectors) of $V$. Hence, $T$ has $n$ distinct eigenvalues $\implies T$ has a diagonal matrix
\begin{align}\textbf{Matrices}\end{align}
3.) $A(T)$ is diagonalizable $\iff$ $A(T)\sim D(T)$ (diagonal matrix) $\iff T$ has basis of eigenvectors of $V\iff$ $A(T)$ has a basis of eigenvectors of $\mathbb{F}^n$, those eigenvectors being precisely as described in the first paragraph.
4.) $A(T)$ has $n$ distinct eigenvalues $\implies$ $A(T)$ is diagonliazable. Proof: $A(T)$ has $n$ distinct eigenvalues $\iff T$ has $n$ distinct eigenvalues (by first paragraph) $\implies$ $T$ has basis of eigenvectors of $V$ (by 2) $\iff A(T)$ has basis of eigenvectors of $\mathbb{F}^n$ (by first paragraph) $\iff A(T)$ is diagonlizable (by 3).
Does this look ok?