TITLE: If $TS=ST$, then $S=\alpha T+\beta$.
QUESTION [5 upvotes]: Let $T=\begin{pmatrix}a&b\\c&d\end{pmatrix}$ be a non-scalar matrix.

If $S=\begin{pmatrix}e&f\\g&h\end{pmatrix}$ be such that $TS=ST$. Why there exists $\alpha,\beta\in \mathbb{C}$ such that
  $$S=\alpha T+\beta I\;?$$

Note that $TS-ST=0$ is equivalent to
$$\begin{bmatrix}bg-fc & af+bh-eb-fd\\
ce+dg-ga-hc & fc-bg\end{bmatrix} = \begin{bmatrix}0 & 0\\0 & 0\end{bmatrix}$$
This implies that
$$\begin{cases}
bg-fc = 0,\\
af+bh-eb-fd = 0,\\
ce+dg-ga-hc = 0,\\
fc-bg = 0.
\end{cases}$$
Since $T$ is non scalar, then $b\neq 0$ or $c\neq 0$ or $a\neq d$. However, I cannot find $\alpha$ and $\beta$.

REPLY [5 votes]: EDIT. We assume that $T\in M_2(K)$, where $K$ is
a field. $C(T)=\{S\in M_2(K);TS=ST\}$ is a vector space containing $\{I,T\}$, that are linearly independent; then it suffices to show that $dim(C(A))\leq 2$, that is, the entries of such a matrix $S=\begin{pmatrix}p&q\\r&s\end{pmatrix}$ depend at most on $2$ parameters.
$\textbf{Proof}$.  There is a vector $u$ s.t. $\{u,Tu\}$ is a basis of $K^2$; otherwise, let $v,w$ be a basis of $K^2$; one has $Tv=av,Tw=bw,T(v+w)=c(v+w)$, that implies that $T=aI$, a contradiction.
In the basis $\{u,Tu\}$, $T$ becomes $\begin{pmatrix}0&a\\1&b\end{pmatrix}$ and
$(TS)_{1,1}=(ST)_{1,1},(TS)_{2,1}=(ST)_{2,1}$ iff 
$q=ar,p=-br+s$; finally, the entries of $S$ depend at most on the $2$ parameters $r,s$ and we are done.