How to tackle this question related to linear operator? The question is :

Let $T$ be a linear operator on a two dimensional vector space $V$ such that $T  \ne cI$ for any scalar $c$.Let $U$ be a linear operator on $V$ such that $UT = TU$.Then show that $U = g(T)$ for some polynomial $g(t)$.

How can I proceed?Please help me.Thank you in advance.
 A: Here is a rather pedestrian approach:
As an irrelevant aside, since $T$ must satisfy its characteristic equation, we can express $T^k$, $k \ge 2$, as a linear combination of $I,T$, and so the polynomial $g$, if it exists, can be taken to be of degree no larger than one.
(i) Note that $UT=TU$ iff for any invertible $V$, we have
$(V^{-1}U V ) ( V^{-1}T V) = ( V^{-1}T V) ( V^{-1}U V )$.
(ii) Note that for some polynomial $g$, we have
$U = g(T)$ iff for any invertible $V$, we have 
$V^{-1}U V = g(V^{-1}T V) = V^{-1} g(T) V$.
From (i) & (ii), we see that we can presume that $T$ is in a Jordan normal form.
In particular, this means that $T$ has one of the following two forms:
(A) $T = \begin{bmatrix} \lambda & 0 \\ 0 &  -\lambda \end{bmatrix}$, with $\lambda \neq 0$ or
(B) $T = \begin{bmatrix} 0 & \beta \\ 0 &  0 \end{bmatrix}$, with $\beta \neq 0$
(iii) Note that $UT=TU$ iff for any $a,b$ we have $(U+aI)(T+bI) = (T+bI)(U+aI)$, so, by choosing $a,b$ appropriately, we may assume that
$\operatorname{tr} U =  \operatorname{tr} T = 0$.
Suppose $T$ has Form (A):
Let $U = \begin{bmatrix} a & b \\ c & -a \end{bmatrix}$ and 
$T = \begin{bmatrix} \lambda & 0 \\ 0 &  -\lambda \end{bmatrix}$.
Multiplying $UT $ and $TU$ and equating entries shows that
$c \lambda = 0$, $b \lambda = 0$ from which we conclude that $b=c = 0$ (
since $\lambda \neq 0$), and we have $U = {a \over \lambda } T$.
Suppose $T$ has Form (B):
Let $U = \begin{bmatrix} a & b \\ c & -a \end{bmatrix}$ and 
$T = \begin{bmatrix} 0 & \beta \\ 0 &  0 \end{bmatrix}$.
Multiplying $UT $ and $TU$ and equating entries shows that
$\beta c = 0, a\beta = 0$ from which we conclude that $a=b= 0$
and so $U = {b \over \beta} T$.
