Let $P : V → V$ be a linear map. Show that $im(P) ⊆ ker(P − I)$ iff $P^2(x) = P(x) $ for all $x ∈ V$.
I need to write a proof for this and I'm really struggling. So here's what I got so far:
$im(P)⊆V$ and $ker(P)⊆V$. So since $V$ is vector space, by closure, $ker (P)+im (P)⊆V $.
It's an iff statement, so I know I need to prove it in both directions.
First Direction: $im(P) ⊆ ker(P − I)$ implies $P^2(x) = P(x) $ for all $x ∈ V$.
Suppose $m∈im(P)$, so $m∈ker(P − I)$. So, $m∈V$. And now I'm stuck.
Second Direction: I think I start with $P^2 = P$. So, $(P-I)∘P = P∘P -I∘P = P∘P-P = P∘P - (P∘P) = 0$ So I now have the kernel. I'm stuck here and fairly confused after that.
I'm having a hard time visualizing and putting together a proper proof. I understand the statement given, but I don't get what the proof is suppose to look like. Could someone please show me how the proof works?
Actually, if $P\times P=P$, then $\text {im}(P)=\text{ker}(P-I)$ For a general proof(you can use this to deal with all things like $f(x)=0$, $V$ equals to the factors' images or kernels) Assume you have know about number theory, we know (x, x-1)=1, then there are polynomials $u(x)$ and $v(x)$ such that $u(x)*x + v(x)*(x-1)=1$. Subtitle $X$ with P(just like Cayley–Hamilton theorem did) $u(P)P+v(P)(P+1)=I$ For any x belong to V, x=u(P)P(x)+v(P)(P-I)(x) Since P(P-I)v(P)(x)=0 and same to another. V=ker(P) + ker(P-I), and two kernels with no joint. Any vector x belong to the joint, then according to $x=u(P)P(x)+v(P)(P-I)(x), x=0$ Finally, you can prove Im(p)=ker(P-I) and Im(P-I)=ker(P) easily. PS: I am just new to here and not good at English, sorry for the casual writing and gammar.