Distinct eigenvalues and matrices problem Let $V$ be a real vector space and $T: V \rightarrow V$ be a linear transformation. 
It is given that  if $v_1, . . . , v_n$ are eigenvectors
for distinct eigenvalues $λ_1, . . . λ_n$ then $\{v_1, . . . , v_n\}$ is linearly independent.
Now let $\{v_1, ..., v_n\}$ be a basis of eigenvectors with distinct eigenvalues
$λ_1, . . . , λ_n$. Prove that for any linear map $S : V → V$ with $ST = TS$, $v_i$ is also an eigenvector of $S$ for each $i = 1, . . . , n$
I have been going round in circles with this for ages.
$T(v)=\lambda v$
$\iff STv=\lambda Sv$
$\iff T(Sv)=\lambda Sv$
There is more pointless algebra but I can't get further than than the previous line where you can deduce $Sv$ is also an eigenvector for $T$
I havent used the fact the question gave at the start which I suspect is what's missing: "if $v_1, . . . , v_n$ are eigenvectors
for distinct eigenvalues $λ_1, . . . λ_n$ then $\{v_1, . . . , v_n\}$ is linearly independent."
I am struggling to find a way to use this however.
 A: $Sv_i$ is an eigenvector for $T$ with eigenvalue $\lambda_i$. The dimension of each eigenspace of $T$ is 1 since the eigenvalues are distinct, so $Sv_i = c_iv_i$ for some $c_i$.
A: You have stated that $\{ v_1,v_2,\cdots,v_n\}$ is a basis of $V$ consisting of eigenvectors of $T$ with distinct eigenvalues
$\{\lambda_1,\lambda_2,\cdots,\lambda_n\}$. And you have assumed that $S$ is another linear operator on $V$ that commutes with $T$.
Because $TS=ST$, then
$$
           TSv_j = STv_j = \lambda_j Sv_j.
$$
Because $\{ v_1,v_2,\cdots,v_n \}$ is a basis, then
$$
     Sv_j = \alpha_1 v_1 + \alpha_2 v_2 + \cdots + \alpha_n v_n, \\
     Sv_j - \alpha_j v_j = \sum_{k=1,k\ne j}^{n}\alpha_k v_k
$$
The left side is an eigenvector of $T$ with eigenvalue $\lambda_j$. Therefore,
$$
            0 = (T-\lambda_j I)(Sv_j-\alpha_j v_j) = \sum_{k=1,k\ne j}^{n}\alpha_k(\lambda_k-\lambda_j)v_k.
$$
Because the vectors on the right are linearly independent and $\lambda_j \ne \lambda_k$ for $j \ne k$, then $\alpha_k = 0$ for all $k$ such that $k \ne j$. Therefore
$$
                Sv_j - \alpha_j v_j = 0 \implies Sv_j = \alpha_j v_j.
$$
This proves that $v_j$ is an eigenvector of $S$ for all $1 \le j \le n$.
