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

My answer:

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

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

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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.

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