Suppose $T$ is an operation on a vector space $V$ and $T^2 = T$. Prove that $V = N(T) \oplus R(T)$

Question

Suppose $T$ is an operation on a vector space $V$ and $T^2 = T$. Prove that $V$ is the direct some of the null space and range of $T$, namely $V = N(T) \oplus R(T)$

I'm not sure how to start on this proof, how do I use the assumption of $T^2 = T$ in this proof?

Answer 1

Suppose $T:V\rightarrow V$. Let $v\in V$, then $T(T(v)-v)=0$ i.e $v-T(v)\in \ker T$ .Because by assumption $T(T-I)=0$ .Now write $v=(v-Tv)+Tv$. Clearly $Tv\in range V$ And as we saw $v-Tv\in \ker T$. So every vector can be written as sum of two vectors in kernel and image of $T$

Answer 2

Hint: $v= (v-Tv)+Tv$

Show that if $v\in \ker (T) \cap T(V)$, then $v=0$