Prove, that if $a \in End(V)$ and $a^2 = a$ than $V = Ker(a) \oplus Ker(id_v + a)$

Question

Prove, that if $a \in End(V)$ and $a^2 = a$ than $V = Ker(a) \oplus Ker(id_v + a)$

I was trying to solve, and found that $a(a(v_i)) = v_j$ But this mean that $a(v_i) = v_i$ So first time we going to V and can get to any vector, but we using $a$ again we can get only in dot where we start second time.

Answer 1

I will assume that you mean $Ker(id_v -a)$, first, lets note that $Ker(a)\cap Ker(id_v -a)=\{0\}$ because $v\in Ker(a)$ if and only if $a(v)=0$ and $v \in Ker(id_v-a)$ if and only if $v-a(v)=0$, so $v \in Ker(a)\cap Ker(id_v -a)$ if and only if $a(v)=v-a(v)=0$ this implies $a(v)=0$ and thus $v-a(v)=v-0=v=0$.

Now, we got to proof that $V=Ker(a)+Ker(id_v-a)$ lets take $v \in V$ and lets write it as $v=v-a(v)+a(v)$, now note that $a(v-a(v))=a(v)-a(a(v))=a(v)-a(v)=0$ so $v-a(v)\in Ker(a)$ and $(id_v-a)(a(v))=a(v)-a(a(v))=a(v)-a(v)=0$ so $a(v)\in Ker(id_v-a)$ so we have proved every element in V is sum of an element in $Ker(a)$and one in $Ker(id_v-a)$ and then $V=Ker(a)+Ker(id_v-a)$

And the claim is proved.