Let $V$ be a finite dimensional vector space over a field $F$ and $T$ a linear operator on $V$ such that $T^2 = I_V$. 
Let $V$ be a finite dimensional vector space over a field $\mathbb{F}$
  and $T$ a linear operator on $V$ such that $T^2 = I_V$. If $\mathbb{F}
 = \mathbb{R}$ or $\mathbb{C}$, show that $T$ is diagonalizable!


I have no idea how to do this question, the best i did was to have take any basis $B$, we have $$[T^2]_B = [T]_B[T]_B = [I_V]_B$$ which is pretty meaningless.
My answer sheet used a way in which i do not understand, it claimed that any $V$ in this situation will be a direct sum of the eigenspace of $1$ and $-1$.
Anyone has a better proof or direct me to the right direction!
THanks

It suffices to show that every generalised eigenvector of $T$ is an eigenvector of $T.$ Let $v \neq 0$ be a generalised eigenvector of $T$ with corresponding eigenvalue $c.$ By induction, we only need consider the case where $(T-cI)^2(v)=0.$ (why?)  Clearly $c \neq 0$ since if $c=0$ then we'd have $0 = T^2v =v,$ contradiction. Since $c$ is an eigenvalue and $T^2 =I,$ we have $c^2 =1.$ 
Let $w =(T-cI)v$ so that $Tw =cw$. Then we have:
$0=(T^2 -I)(v)= (T+cI)(T-cI)v =(T+cI)w = 2cw$ which implies $w=0,$ that is $Tv =cv$ and $v$ is an eigenvector of $T,$ from which it follows that $T$ is diagonalisable. 
