Proof check in Linear Algebra with Diagonalizable operator Given $T : V \to V$ linear operator on $V$ space vector such that $T^2 =T$ , prove that $T$ is Diagonalizable,
What i did is $f(x) = x^2-x = x (x-1)$  and so we know that $f(T) = 0$
then it must be that $m_T (x) | f(x)$ where $m_T(x)$ is the minimal polynomial such that $m_T(T) =0$.
so $x(x-1) | m_T(x)$ and $m_T (x) | x(x-1) = f(x)$ and so $m_T(x) = x(x-1)$ and because is the multiplication of linear factors then $T$ is Diagonalizable.
is this proof correct ?!
 A: This proof is not completely correct as $x(x-1) | m_T(x)$ is false.
For example $0$ or $\operatorname{Id}_V$ are such that $0^2=0$ and $\operatorname{Id}_V^2=\operatorname{Id}_V$ but their minimal polynomials are $x$ and $x-1$.
But the proof is almost correct as from $ m_T(x)| x(x-1) $ there is only three different cases:


*

*$m_T(x)=x(x-1)$

*$m_T(x)=x$

*$m_T(x)=(x-1)$

A: One way to show that $T$ is diagonalizable is to show that $V$ has a basis of eigenvectors of $T$, or equivalently, that $V$ can be spanned by eigenvectors of $T$.

Here's a proof using that approach . . .

Let $A=\{a\in V\mid Ta=0\}$.

Let $B=\{b\in V\mid Tb=b\}$.

Then $A,B$ are subspaces of $V$, whose nonzero elements are eigenvectors of $T$, corresponding to the eigenvalues $0$ and $1$, respectively.

To show that $T$ is diagonalizable, it suffices to show that $V=A+B$.

Let $v\in V$.$\;\,$Claim $v\in A+B$.

Let $b'=Tv$,$\;$and let $a'=v-b'$.

Then $v=a'+b'$,$\;$and


*

*$Ta'=T(v-b')=T(v-Tv)=Tv-T^2v=Tv-Tv=0$, so $a'\in A$.$\\[4pt]$

*$Tb'=T(Tv)=T^2v=Tv=b'$, so $b'\in B$.


so $v\in A+B$, as claimed.

It follows that $V=A+B$, hence $T$ is diagonalizable
