$T^2=Id$ then $V=ker(T-Id)\oplus ker(T+Id)$

Question

Let $T:V\to V$ linear, $V$ vectorial space and $T^2=Id$

a) $Im(T+Id)\subset ker(T-Id)$

b) $Im(T-Id)\subset ker(T+Id)$

c) $V=ker(T-Id)\oplus ker(T+Id)$

I already prove a and b. For example, for a) let $v\in Im(T+Id)$ then $v=Tw+w$ some $w$ in $V$, then $(T-Id)(Tw+w)=0$ then $v\in ker(T-Id)$. For b) is similar.

How prove c)?

pd: V need finite dimension?

Answer 1

Let $v\in V$. By a) and b) you have $v=\frac12((v+Tv)+(v-Tv))\in ker(T-id)+ker(T+id)$. To show that the sum is direct, assume that $v$ is in the intersection of these kernels, that is $v+Tv=v-Tv=0$. It follows that $v=Tv=-v$ thus $v=0$.