Q:

$T: V \rightarrow V$ a linear transformation such that $T^2 = I$ and $H_1= \{v \in V | T(v) = v\}\ $ and $H_2= \{v \in V|T(v) = -v\}\ $

Let V a vector space and $T: V \rightarrow V$ a linear transformation such that $T^2 = I$ and $H_1= \{v \in V | T(v) = v\}\ $ and $H_2= \{v \in V|T(v) = -v\}\ $ then $V = H_1 \bigoplus H_2$
I stuck in this problem please some help.

A:

First of all it is clear that if $v \in H_1 \cap H_2$, then $v = T(v) = -v$ and so we must have $v = 0$. It follows that $H_1 \cap H_2 = 0$. 
Now let $v \in V$, by definition of $T$, $T^2(v)= T(T(v))=v$ which implies that $T(v) \in H_1$, furthermore by linearity of $T$ holds $T(v-T(v)) = T(v)-T^2(v) = -(v-T(v))$ which shows that $v -T(v)\in H_2$. So $v = T(v)+(v-T(v)) \in H_1 + H_2$. Since it is obvious that $H_1 + H_2 \subset V$, the proof is done.

