Find a example of a linear map $f:F^2\to F^2$ for which $\ker(f)=\operatorname{Im}(f)$ Find a example of a linear map $f:F^2\to F^2$ for which $\ker(f)=\operatorname{Im}(f)$
I'm not sure whether this is possible for arbitrary spaces or not
 A: It is not possible for odd-dimensional spaces, but it is possible for any even-dimensional space. The reason is that any even-dimensional space can be decomposed into $V\simeq W\oplus W$. Choose a linear automorphism $\phi:W\rightarrow W$ and let $f(w_1,w_2)=(\phi(w_2),0)$. Then $f$ satisfies your condition.
A: Elaborating on Brian's answer, you can do this in an arbitrary space $V$, as long as $\dim V$ is even, since we know that $$\dim V=\dim\operatorname{Ker}(f)+\dim\operatorname{Im}(f)$$
And when they are equal, you should have $\dim V=2\dim\operatorname{Ker}(f)=2\dim\operatorname{Im}(f)$.
When $\dim V$ is even, choose a basis $(v_1,...,v_{2k})$ of $V$ and define 
$$f(v_i)=\left\{\begin{array}{cc}0 & 1\leq i\leq k\\ v_{i-k} & k+1\leq i\leq 2k\end{array}\right.$$
This gives you a linear map (you can easily extend it to all $v\in V$) with $\operatorname{Ker}(f)=\operatorname{Im}(f)=\operatorname{Span}\{v_1,...,v_k\}$
A: How about $f(\langle x,y\rangle)=\langle y,0\rangle$?
A: First of all, $\operatorname{Im}(f) \subseteq \operatorname{Ker}(f)$ if and only if $f^2 = 0$. Also, for the equality $\operatorname{Im}(f) = \operatorname{Ker}(f)$ to hold, the map $f$ must be nonzero. Using rank-nullity, you can show that in our case (the vector space $F^2$) $f^2 = 0$ and $f \neq 0$ is enough for the equality to hold. Hence one example is the map $x \mapsto Ax$, where $A = \left( \begin{matrix}0 & 0 \\ 1 & 0 \end{matrix} \right)$.
