Consequences of Ker(f)=Im(f)

Question

Suppose we have a Endomorphism on a Vectorspace $V$ whose Kernel is equal to its image:

Since $\textrm{Ker}(f)=f^{-1}(0_{V})$ and $\textrm{Im}(f)=f(V)$ wouldn't that imply, that V can only be equal to ${0_{V}}$ ? Or am I missing something?

Answer 1

One consequence of $ \operatorname{im}(f) =\ker(f)$ is that $\dim V$ is even because $\dim V = \dim \ker(f) + \dim \operatorname{im}(f)$.

More generally, we have

$ \operatorname{im}(f) \subseteq \ker(f)$ iff $f^2=0$

There are $f\ne0$ such that $f^2=0$ in every even dimension. Just take$V=\mathbb R^{2n}$ and the matrix with $n$ diagonal blocks $\pmatrix{ 0 & 1 \\ 0 & 0}$.