Let $V$ a finite dimensional vector space over $\mathbb{F}$ and let $\mathcal{F}$ a family of diagonalizable operators such that, for all $T,G \in \mathcal{F}$ we have $T\circ G = G \circ T$. Show that there is a basis $B$ of $V$ where $[T]_{B}$ is diagonal for all $T \in \mathcal{F}$.
Things that I figure out about this exercise:
If $T \circ G = G \circ T$, then $E_\lambda=\{v \in V: Tv= \lambda v\}$ is $G-$invariant.
Indeed, let $v \in E_\lambda$ then $T(G(v)) = G(T(v)) = G(\lambda v) = \lambda G(v)$. So we have $G(v) \in E_\lambda$.
When $\mathcal{F}$ is the family of multiples of identity operator then the result is obvious.
But in the general, how to solve that?
My professor sad to use induction em $\dim V$.
Your professor's hint is good.
Assume first that $\dim V=1$. Then, the statement is obvious.
Now assume that we've proven the statement for $\dim V=n$. Then, for $\dim V=n+1,$ as you've noted, the statement is trivial if every $T\in \mathcal{F}$ is a multiple of the identity. Accordingly, we can assume $T\in\mathcal{F}$ with at least two distinct Eigenvectors. Hence, $V=\oplus_{\lambda\in \sigma(T)} E_{\lambda}$ and each $E_{\lambda}$ has dimension at most $n$.
Since each $E_{\lambda}$ is $G$-invariant for every $G\in \mathcal{F}$, we can apply the induction hypothesis to $E_{\lambda}$ and get that each $E_{\lambda}$ admits a basis $\{v^{\lambda}_j\}_{1\leq j\leq \dim (E_{\lambda})}$ such that $v^{\lambda}_j$ is an Eigenvector of every $G\in \mathcal{F}$. Then, of course, $\{v^{\lambda}_j\}_{\lambda\in \sigma(T),1\leq j\leq \dim(E_\lambda)}$ is a basis of $V$ and we're done.