Is it possible that $\ker(T) = \operatorname{im}(T)$ for some linear transformation $T:V \to V$? Help would be very appreciated.
It it possible that $\ker(T) = \operatorname{im}(T)$ for some linear transformation $T:V \to V$?
 A: 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.
A: Consider the linear transformation $T : \Bbb R^2 \to \Bbb R^2$ so that:
$$
T(x, y) = (y, 0)
$$
We have:
$$
\ker T = \{(x, 0) : x \in \Bbb R\} = \operatorname{im} T
$$
I arrived at this example by noticing that such a linear transformation must have $T^2 = 0$.
A: Since $ \dim \operatorname{ker} T + \dim \operatorname{im} T = n$, we must have $n=2k$.
Now, take any basis of $V$, say $v_i$
Define $T(v_i) = v_{k+i}$ for $i=1$ to $k$, and $T(v_i) = 0 $ for $i=k+1 $ to $2k$. Extend this out as a linear functional to $V$. Then , $\operatorname{im} V = \operatorname{ker} V$.
