Show that $V = \mbox{ker}(f) \oplus \mbox{im}(f)$ for a linear map with $f \circ f = f$

Question: Let $V$ be a $K$-Vectorspace and $f: V \rightarrow V$ be linear. It holds that $f \circ f = f$. Show that $V = \mbox{ker}(f) \oplus \mbox{im}(f)$.

My attempt: So i guess that the $\oplus$ denotes a direct sum which means i have to show that (i) $V = \mbox{ker}(f) + \mbox{im}(f)$ and (ii) $\mbox{ker}(f) \cap \mbox{im}(f) = \{0\}$.

I tried to do (ii) first: Let $v \in \mbox{im}(f) \cap \mbox{ker}(f)$

$\Rightarrow \exists u: f(u)=v \wedge f(v) = 0$

(can i put a "Rightarrow" here?) $(f \circ f)(u)=f(f(u)) = f(v) = 0$

As for (i) i am having difficulty with an approach to showing that $V = \mbox{ker}(f) + \mbox{im}(f)$. Should I even be trying to do this in the first place? if so, any advice as to how?

Let's see (ii) first: (Yes, your use of the implication symbol $\Rightarrow$ is appropriate.) You want to use that $f\circ f=f$, which you do not mention explicitly. The point is: In (ii) you need to show that the $v$ you are considering is $0$. You have: $f\circ f=f$, so $v=f(u)=(f\circ f)(u)$ etc, as you did. This gives $v=0$.

For (i), let $v$ be any element of $V$. What can we say about $v-f(v)$? Let's call it $w$. We have $f(w)=f(v-f(v))=f(v)-f(f(v))=f(v)-f(v)=0$. Here, I used that $f$ is linear and (again) that $f\circ f=f$. This shows that $w\in{\rm ker}(f)$. So, $v=(v-f(v))+f(v)=w+f(v)$ is the sum of an element of the kernel (namely, $w$) and one of the image (namely, $f(v)$), and that is precisely what you needed.

(A meta-question here is how would one think of considering $v-f(v)$. Ok, let's try to work backwards. Suppose we already know that $v=a+b$ with $a$ in the kernel and $b$ in the range. Then $f(v)=f(a)+f(b)=f(b)$, since $a$ is in the kernel. Also, $f(v)=f(f(v))$, so it is reasonable, as a first approach, to let $b=f(v)$, and see if we can directly prove that $a=v-b$ is in the kernel. And this is precisely what we did.)

$v = (v - f(v)) + f(v)$