Idempotent endomorphism: show $M=\text{Range}(f) \oplus \text{Range}(\text{id} - f)$

Question

let $f$ be in idempotent endomorphism. Show that $M=\text{Range}(f) \oplus \text{Range}(\text{id} - f)$.

I want to show the intersection of both range spaces is $\{0\}$.

I currently have let $v \in \text{Range}(f) \cap {\text{Range}(\text{id} - f)}$

$f(v) = \text{id}(v) - f(v)$

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

and I want to show that $v=0$. But I do not know how to progress. any advice is much appreciated!

Answer 1

Suppose $z = f(y) = x - f(x)$. Then we have $f(x + y) = x$. Then we have $f(f(x + y)) = f(x)$. Since $f$ is idempotent, $f(x + y) = f(x)$. Then $f(y) = 0$. Then $z = 0$. $\DeclareMathOperator{id}{id}$ $\DeclareMathOperator{range}{range}$

To get that the union of the ranges is $M$, just note that $x = x - f(x) + f(x)$. Thus, $M = \range(f) \oplus \range(\id - f)$.