Proving there exists basis $B$ for $R^2$ such that $[T]_B = \begin{bmatrix}0 & 0 \\ 0 & 2 \end{bmatrix}$ I've been stumped by this question for the last 2 days, and couldn't figure how to prove it.

$T:R^2 \to R^2$ is a linear transformation (T$\neq$0) that satisfies $T^2=2T$ and it is known that $T$ is not invertible.
Prove that there exists some basis $B$ of $R^2$ such $[T]_B = \begin{bmatrix}0 & 0 \\ 0 & 2 \end{bmatrix}$

What I don't understand is, if $\begin{bmatrix}0 & 0 \\ 0 & 2 \end{bmatrix}^2=2\begin{bmatrix}0 & 0 \\ 0 & 2 \end{bmatrix}$, doesn't it mean that there will always exist some $B$ for whom their will be a some $T$ such that the requirements are still satisfied?
Thanks in advance :)
Edit:
I forgot to include that $T\neq0$.
Edit 2:
Unfortunately, I suppose we were meant to solve this question without eigenvalues and eigenvectors. I had this idea, but I don't know if its entirely correct.
If $T^2=2T$, then: $T^2-2T=0 \to T(T-2I)=0$. So for any $(x,y) \in R^2$,
$T(T-2I)(x,y)=0$. But because $T\neq0$ is given, then $T-2I=0$ must be true. which leads to: $T=2I$. However, because $T$ is noninvertible, the matrix $[T]_B$ can't be $\begin{bmatrix}2 & 0 \\ 0 & 2 \end{bmatrix}$ (because it invertible). But there are two other similar matrices that satisfy the same requirements:
$[T]_B=\begin{bmatrix}0 & 0 \\ 0 & 2 \end{bmatrix}$ and $[T]_C=\begin{bmatrix}2 & 0 \\ 0 & 0 \end{bmatrix}$. And if for some basis $B=(b_1,b_2)$, the matching matrix is $[T]_B$, then for some other basis $C=(b_2,b_1)$ the matching matrix is $[T]_C$, and in anyway there is some basis whose matrix is $[T]_B$.
Is this line of thinking correct?
 A: You're thinking the wrong way around: you're given $T$ and have to show that there's some basis such that the representation of $T$ with respect to that basis is the given matrix.
But the statement is actually wrong the way it's stated: consider the zero map $T \colon \mathbb{R} \to \mathbb{R}, v \mapsto 0$. This is clearly linear, satisfies $T^2=0=2T$ and isn't invertible. Assume that there was a basis $e_1, e_2$ such that $[T]_B = \pmatrix{0 & 0 \\ 0 & 2}$ then $[T]_B [e_2]_B = \pmatrix{0 \\ 2}$; but $[T]_B [e_2]_B = [T e_2]_B = [0]_B = \pmatrix{0 \\ 0}$ which is a contradiction.
You thus have to assume that $T$ is of rank exactly 1 for this to be true.
How to prove that modified statement: let $v \in \mathbb{R}^2$. Then $T(Tv) = T^2v= 2 (Tv)$ so if $Tv$ isn't zero it's an eigenvector of $T$ with eigenvalue $2$. Since $T$ is of rank 1 there is some $v$ such that $Tv$ is nonzero. Pick that $Tv$ as one basis element and add some other one to complete the basis and you can show that the representation w.r.t. that basis is the one you're after.
A: Let $T\colon \Bbb R^2 \to \Bbb R^2$ be any endomorphism such that $T^2 = 2T$.
Let $P(X) = X^2 - 2X = X(X-2)$ and $\mu_T$ be the minimal polynomial of $T$.
By assumption, $P(T)=0$, and $\mu_T$ must divide $P$.
Three cases can occur:

*

*$\mu_T(X) = X$.
In this case, we have $0 = \mu_T(T) = T$.
Hence $T=0$.
In any basis $B$ of $\Bbb R^2$, we have
$[T]_B = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}$.

*$\mu_T(X) = X-2$.
In this case, we gave $0 = \mu_T(T) = T - 2I_2$.
Hence, $T= 2I_2$, and in any basis $B$ of $\Bbb R^2$, we have
$[T]_B = \begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix}$.

*$\mu_T(X) = P(X)$.
Since the roots of the minimal polynomial $\mu_T$ are precisely the eigenvalues of $T$, it follows that $T$ has two eigenvalues, $0$ and $2$.
Let $e_0$ be an eigenvector of $T$ satifying $Te_0 = 0$, and let $e_2$ be an eigenvector of $T$ satisfying $Te_2 = 2e_2$.
Let $B=(e_0,e_2)$.
Then $B$ is a basis of $\Bbb R^2$, and we have
$[T]_B = \begin{pmatrix} 0 & 0 \\ 0 & 2 \end{pmatrix}$.

Assume that $T\neq 0$ and that $T$ is not invertible.
Then cases $1$ and $2$ cannot occur.
It remains that we are in case $3$, which concludes the proof.
