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

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.

Notice that $$T(T(v))=0$$ for all $$v\in V$$. Hence $$T$$ is nilpotent.
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 $$Ker\;T$$
• @Nori The converse is false. Take the matrix $$T =\pmatrix{0 & 1 & 0 \\ 0 & 0& 1 \\ 0& 0 &0}$$ Then the kernel has dimesion $1$ and the image has dimension $2$. May 15, 2020 at 7:22