Diagonalizable Linear Operator - Is $Ker(T)=Ker(T^2)$? Been spending a lot of time on this one.
Given a linear operator T, in a vector space V, having a finite dimension and is diagonalizable - Is $Ker(T)=Ker(T^2)$?
One way is trivial, apply $T$ on $T(v)$ to get $0$, but I cannot find the other way around. 
As always, TIA.
Edit:
Proving $Ker(T) \subset Ker(T^2)$:
Since $[T]_B[T(v)]_B=[T^2(v)]_B$ then $[T]^2_B[v]_B=[T^2(v)]_B$
Now given the fact that T is diagonalizable, $T^2$ is the result of squaring its diagonal elements - which we assume are not all equal to zero.
Having that in mind, let $v\in Ker{T^2}$, then $[T^2(v)]_B=0$.
But since
$[T^2(v)]_B=[T^2]_B[v]_B=[T]^2_B[v]_B=[T]_B[T]_B[v]_B$ then $[T]_B[T]_B[v]_B=0$
Since we established that $T$ is not zero we are left to conclude that $[T]_B[v]_B=0$ namely $v\in Ker{T^2}$.

Hagen's answer is already complete, but you can also look at this as follows:
$$x\in\ker T^2\implies T^2x=0$$
Since $\;T\;$ is diagonalizable, then there exists a basis $\;\{x_1,...,x_n\}\;$ of $\;V\;$ such that $\;Tx_i=\lambda_ix\;$ , with $\;\lambda_i\;$ the corresponding eigenvalues, and then
$$x=\sum_{k=1}^n a_kx_k\implies0=T^2=T\left(\sum_{k=1}^na_kTx_k\right)=T\left(\sum_{k=1}^na_k\lambda_kx_k\right)=$$
$$=\left(\sum_{k=1}^na_k\lambda_kTx_k\right)=\left(\sum_{k=1}^na_k\lambda_k^2x_k\right)\iff a_k\lambda^2_k=0\;\;\forall\,k\implies a_k\lambda_k=0$$
$$Tx=\left(\sum_{k=1}^na_k\lambda_kx_k\right)=0$$
