Prove that $ST = TS$ if $S$ and $T$ have same eigenvectors 
Suppose $V$ is finite-dimensional, $T \in \mathcal{L}(V)$ has dim $V$ distinct eigenvalues, and $S \in \mathcal{L}(V)$ has the same eigenvectors as $T$ (not necessarily with the same eigenvalues). Prove that $ST = TS$. 

My general thinking is take $v \in V$ as a linear combination of the eigenvectors $v_1 + \dots + v_n$. Then 
$$S(Tv) = S(\lambda_Tv) = \lambda_S\lambda_Tv = \lambda_T \lambda_S v = T(\lambda_S v) = TSv$$
Something feels incomplete and I think this is the right direction, but it doesn't feel like I am using everything that is given to me. Not sure where to go from here or how to troubleshoot this myself

Let $(e_i)_{i \in \{1, \cdots n\}}$ be the eigenvectors of $T$. They form a basis on $V$ (why?) and both $T,S$ can be simultaneously diagonalized w.r.t this new basis. Write $S = B D_S B^{-1}$ and $T= B D_T B^{-1}$ where $D_T$ and $D_S$ are diagonal matrices. 
Calculating yields $ST = B D_S B^{-1} B D_TB^{-1}= BD_SD_TB^{-1}$ Now as diagonal matrices commute with each other we have: 
$BD_SD_TB^{-1}= BD_TD_SB^{-1}=B D_T B^{-1} B D_SB^{-1} = TS$.
