TITLE: Is it possible that $\ker(T) = \operatorname{im}(T)$ for some linear transformation $T:V \to V$?
QUESTION [6 upvotes]: Help would be very appreciated.
It it possible that $\ker(T) = \operatorname{im}(T)$ for some linear transformation $T:V \to V$?

REPLY [8 votes]: It will be possible if $\dim(V)$  is even because $\dim(T)+\mathrm{null}(T)=\dim(V)$ so $\dim(V)=2\dim(T)$
And then you will have, for example, $2n$ basis for $V$ as $\{\alpha_1,\ldots,\alpha_n,\alpha_{n+1},\ldots,\alpha_{2n}\}$.
For every $T$ that maps $n$ arbitrary basis elements from $\{\alpha_1,\ldots,\alpha_n,\alpha_{n+1},\ldots,\alpha_{2n}\}$ to zero and maps the others (remains basis ) to the first $n$ basis elements, it will be our answer.