prove that if $ T^2=T $ then T is diagonalizable operator ( over finite dimension vector space) I know its a very basic question in linear algebra, and I actually solved it myself. But something's tells me there's a lot more ways to prove it, and I'd like to see some other informative ways that I can learn from.
So, let V be a vector space with finite dimension n. and let $ T^2=T $ be a linear operator $ T: V\to V $ (I think its called projection).
We need to prove that T is diagonalizable operator.
What I've done:
Since $ T^2=T $ we can prove that $ V=Im\left(T\right)\oplus Ker\left(T\right) $.
Assume $ \dim Ker\left(T\right)=n_{1} $ 
and 
$ \dim Im\left(T\right)=n_{2} $ 
So, $n_1+n_2=n $  . and notice : $ \ker\left(T\right)=W_{0} $ 
$ Im\left(T\right)=W_{1} $
meaning $KerT $ is the eigenspace of the eigenvalue $ 0 $ and $ImT $ is the eigensapce of the eigenvalue $ 1 $ 
The fact that $ KerT $ is the eigenvalue of $ 0 $ is trivial i guess.
And $ImT$ is the eigenspace of $ 1 $ because:
for any $ v\in V $ $ v\in Im(T) \iff T(v)=v $ 
proof:
$ \Longleftarrow $
Thats trivial.
$ \Longrightarrow $ 
since $ v\in Im(T) $ There's $ u\in V $ such that  $ T(u)=v $ so
$ T\left(v\right)=T\left(T\left(u\right)\right)=T^{2}\left(u\right)=T\left(u\right)=v $ 
Thus, $ ImT $ is the eigenspace of $ 1 $.
since $n_1+n_2=n$ the algebric multiplicity  and geometric multiplicity  of both eigenvalues equal, therefore T is diagonalizable (now at this point im counting on an argument I havent proved myself, but rather accepted as a fact. If anyone can show me the proof, It will be helpful.
Besides, I'd like to see some more ways, if exists, of proving this statement. 
 A: Note that $T$ satisfies $f(x) = x^{2} -x = x(x-1)$.
The minimal polynomial of $T$, $m_{T}(x) = \{x, x-1, f(x) \}$. 
It is a fact that if the minimal polynomial splits into distinct linear factors, then $T$ is diagonalizable.
One can even be more explicit:
$m_{T}(x) = x \implies$ T is similar to the zero matrix.
$m_{T}(x) = x - 1 \implies $ $T$ is similar to the $n \times n$ identity matrix.
$m_{T}(x) = x(x-1) \implies T$ is similar to the matrix gotten from the set of $k$ elements invariant factors: $\{f(x), \cdots, f(x) \}$, where $2k = n$, where $n$ is the dimension of the vector space or the degree of the characteristic polynomial. 
A: Let $p(x) = x^2 - x$. Then $p(T) = T^2 - T = T-T = 0$, so $T$ is a root of $p(x)$. It follows that the minimal polynomial $m_T(x)$ for $T$ divides $p(x)$. If $T$ is a root of $x$, then $T$ is the zero operator, which is diagonalisable. If $T$ is a root of $x-1$, then $T$ is the identity operator, which is diagonalisable. Otherwise, $m_T(x) = p(x) = x^2 - x$. Hence, $T$ is diagonalisable since the minimal polynomial has no repeated roots. 
