Kernel and image of a nilpotent linear map

Question

My question is as follows. Let $T : V \to V$ be a linear map of a finite dimensional vector space V. If $ker T = Im T$, then is $dim V$ even and $T^{2} = 0$? I believe this to be false since I have previously concluded that for a nilpotent function $T$ we find that $kerT \cap Im(T)$={0} which would imply that $ker T \neq Im T$. I am struggling to work out an answer in the direction asked for. Thanks!

Answer 1

You said, you have previously concluded that for a nilpotent linear map $T$ we find that $\ker(T) \cap {\rm im}(T)=0$. This is not true. Consider $$ T=\begin{pmatrix} 0 & 1 \cr 0 & 0\end{pmatrix}. $$ Then we have $\ker(T)=\langle e_1 \rangle={\rm im}(T)$ with respect to a basis $(e_1,e_2)$.

In addition, suppose that $\ker(T)={\rm im}(T)$. Then $T(T(x))=0$, so $T^2=0$.