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 $v\in Ker{T^2}$.

Express $T$ as diagonal matrix. Then the matrix of $T^2$ has the diagonal entries of $T$ squared. In particular, the number of zeroes on the diagonal is the same.
