If ${\rm Im} (T) = \ker (T)$, then $T$ is nilpotent.

Question

I have to prove, that if ${\rm Im} (T) = \ker (T)$, then the transformation matrix is nilpotent.

How can I do this?

I know the Rank–nullity theorem:

If $T: V \to W$, then $\dim{\rm Im}(T) + \dim \ker (T) =\dim V$

In this case: $2 \dim{\rm Im} (T) = 2 \dim \ker (T) = \dim V$

I don't see how to prove that $T$ is nilpotent.

Answer 1

Notice that $T(T(v))=0$ for all $v\in V$. Hence $T$ is nilpotent.

Answer 2

Let's look at $T(T(v))$ for any vector $v$. Since $T(v)$ is in $Im\;T=Ker\;T$, applying $T$ on it will result in $0$ by definition of $Ker\;T$