# show that $V = \text{ker}(f) \oplus \text{im}(f)$ when $f^2 = f$

If $f$ is an endomorphism of $V$, and $f^2 = f$, show that $V = \text{ker}(f) \oplus \text{im}(f)$

$$f(v) = w \in \text{im}(f)\\ f(f(v)) = f(w) = w$$

Then we know that every element that is in our image it's not in our kernel, hence: $$\text{ker}(f) \cap \text{im}(f) = \left\{0\right\}$$ Then: $$V = \text{ker}(f) \oplus \text{im}(f)$$

I don't know if wasn't rigorous enough or I'm wrong... Please correct me if I'm wrong! Thanks

So far, you've only shown that $K := \ker(f)$ and $I:= \text{Im}(f)$ were in direct sum, not that their sum makes up for the all space $V$, i.e. $V = K + I$.
Hint: Let $v \in V$. Then write $v = v - f(v) + f(v)$ and conclude!
Note: A mapping $p : V \rightarrow V$ s.t. $p^2 = p$ is called a projection.