$T^3-4T^2+4T=0$: inference about $T$ 
Suppose $T$ is a linear operator on $\mathbb R^2$ such that $T^3-4T^2+4T=\theta$ where $\theta$ is the null transformation. Then, describe $T$, given that $T$ is diagonalizable.

My approach:

$T(T^2-4T+I)=\theta$ implies $T(T-2I)^2=\theta$. Thus $T$ may be the null transformation itself, and it satisfies the condition of diagonalizability.
Consider $f(t)=(t-2)^2$. Then $f$ may be the characteristic polynomial of $T$. Since $T$ is diagonalizable, eigenvalue of $T$ is $2$ and there are two linearly independent eigenvectors of $T$ corresponding to $2$.
Consider $f(t)=t(t-2)$ then $f$ may be the characteristic polynomial of $T$. The eigenvalues are correspondingly $0$ and $2$ and because of diagonalizability, there is one independent eigenvector for $0$ and one eigenvector for $2$.

Is it right?
 A: Your answer looks good, but there are a few missing details (which your solution leads me to believe you know, but left assumed, and which a peer reading the solution might not be able to fill in).  Here is a slightly more complete write up, which is clearer but uses essentially the same logic.
Since $T$ is a linear operator on a two dimensional space, its characteristic polynomial is of degree two, and so it's minimal polynomial is of degree at most two and dividing $f(t)=t^3-4t^2+4t=t(t-2)^2$.  Furthermore, because $T$ is diagonalizable, the minimal polynomial will not have any repeated roots, and is therefore equal to one of $t$, $t-2$, or $t(t-2)$.  Note, if $T$ were not necessarily diagonalizable, then the minimal polynomial could also be $(t-2)^2$.
If the minimal polynomial of $T$ is t, then $T=0$, and if it is $t-2$, then $T=2I$.  If the minimal polynomial is $t(t-2)$, then $0$ and $2$ are both eigenvalues of $T$, and since $T$ is a transform of a two dimensional space, each of these eigenvalues have multiplicity one.
A: If you plug in $T=\operatorname{diag}(\lambda_1,\lambda_2)$ then obviously you have
$$
\lambda_i^3-4\lambda_i^2+4\lambda_i=0
$$
so $\lambda_i \in \{0,2\}$.
If $\operatorname{rk}(T)=0$, then $T=0$ (trivial, or $\lambda_1=\lambda_2=0$) 
If $\operatorname{rk}(T)=1$, then $T$ is projection on a one-dim subspace and multiplication by $2$
If $\operatorname{rk}(T)=2$, then $T$ is $2I$ (because $\lambda_1=\lambda_2=2$)
