<s>[INST] Show that $F(G(x))=G(F(x)).$ 
Let $V$ be a finitely dimensional linear space, and $F:V\rightarrow V$
  and $G:V\rightarrow V$ linear transformations that are diagonalizable,
  that is: there exists a base $\mathbf{e}$ for $V$ such that the matrix
  to $F$ in the basis $\mathbf{e}$ is diagonal and a base $\mathbf{f}$
  for $V$ such that the matrix to $G$ is diagonal.
i) Show that if $F$ and $G$ are simultaneously diagonalizable, that is if $\mathbf{e}=\mathbf{f},$ then it follows that
$$F(G(x))=G(F(x)), \quad \forall x\in V. \tag{1}$$
ii) Show that if $(1)$ holds, then there exists a basis $\mathbf{g}$ (possibly different from both $\mathbf{e}$ and $\mathbf{f}$) such that
  the matrices to both $F$ and $G$ in the basis $\mathbf{g}$ are
  diagonal.

Okay, I usually post good questions where I've done my own work before asking for help. But this one has thrown me away. I seriously don't even know how to begin or what I should proove.
This is no homework or assignment, I'm just doing practice problems for my exam at the end of this month. Any tips/tricks/translation of the problem statement is welcome!
 [/INST] Regarding part 1: If $F$ and $G$ are simultaneously diagonalizable, then with respect to some basis of $V$ we can represent $F$ and $G$ as
\begin{align*}
F = \left( \hspace{-0.4cm}
\begin{matrix} 
&\lambda_1 & &0 & &\cdots & &0\\
&0 & &\lambda_2 & &\cdots & &0\\
&\vdots & &\vdots & &\ddots & &\vdots\\
&0 & &0 & &\cdots & &\lambda_n\\
\end{matrix} \right), \quad G = \left( \hspace{-0.4cm}
\begin{matrix} 
&\kappa_1 & &0 & &\cdots & &0\\
&0 & &\kappa_2 & &\cdots & &0\\
&\vdots & &\vdots & &\ddots & &\vdots\\
&0 & &0 & &\cdots & &\kappa_n\\
\end{matrix} \right),
\end{align*}
the point being that both transformations can be represented as diagonal matrices with respect to the same basis. But then, since $F$ and $G$ are diagonal matrices, they commute:
\begin{align*}
FG &= \left( \hspace{-0.4cm}
\begin{matrix} 
&\lambda_1 & &0 & &\cdots & &0\\
&0 & &\lambda_2 & &\cdots & &0\\
&\vdots & &\vdots & &\ddots & &\vdots\\
&0 & &0 & &\cdots & &\lambda_n\\
\end{matrix} \right) \left( \hspace{-0.4cm}
\begin{matrix} 
&\kappa_1 & &0 & &\cdots & &0\\
&0 & &\kappa_2 & &\cdots & &0\\
&\vdots & &\vdots & &\ddots & &\vdots\\
&0 & &0 & &\cdots & &\kappa_n\\
\end{matrix} \right)\\
&= \left( \hspace{-0.4cm}
\begin{matrix} 
&\lambda_1 \kappa_1 & &0 & &\cdots & &0\\
&0 & &\lambda_2 \kappa_2 & &\cdots & &0\\
&\vdots & &\vdots & &\ddots & &\vdots\\
&0 & &0 & &\cdots & &\lambda_n \kappa_n\\
\end{matrix} \right) \\
&= \left( \hspace{-0.4cm}
\begin{matrix} 
&\kappa_1 & &0 & &\cdots & &0\\
&0 & &\kappa_2 & &\cdots & &0\\
&\vdots & &\vdots & &\ddots & &\vdots\\
&0 & &0 & &\cdots & &\kappa_n\\
\end{matrix} \right) \left( \hspace{-0.4cm}
\begin{matrix} 
&\lambda_1 & &0 & &\cdots & &0\\
&0 & &\lambda_2 & &\cdots & &0\\
&\vdots & &\vdots & &\ddots & &\vdots\\
&0 & &0 & &\cdots & &\lambda_n\\
\end{matrix} \right) = GF.
\end{align*}
</s>