How can I show that the kernel of $f-id_V$ is equal to the image of $f$?

Question

How can I show that $\ker (f-id_V)=\Im f$ given that $f:V\longrightarrow V$ is a linear transformation such that $f\circ f=f$? Thank you.

Answer 1

$$v\in\ker(f-\operatorname{id}_V)\iff f(v)=v\implies v\in\operatorname{im}f$$ and $$v=f(w)\implies f(v)-v = f(f(w))-f(w)=f(w)-f(w)=0$$

Answer 2

Suppose that $v$ is in the kernel of $f-1$. Then $fv-v=0$. This gives $v=fv$, thus $v$ is in ${\rm im}\; f$; for it equals $fv$. Now suppose $v$ is in the image of $f$, whence $v=fw$ for some $w$. Then $(f-1)v=(f-1)fw=ffw-fw$?

ADD A linear transformation with $ff=f$ is called a projection. It can be proven that if $f:V\to V$ is a projection then $\ker f\oplus{\rm im}\; f=V$, and converesely for every pair of complementary subspaces $U\oplus W=V$ we can construct the projection onto $U$ through $W$, namely the one that has has image $U$ and as kernel $W$. The above tells you that if $f$ is the projection onto $U$ through $W$ then $1-f$ (not $f-1$!) is the projection onto $W$ through $U$.