TITLE: If $T^2 = I$, how one can proof that $V = W \oplus U$?
QUESTION [5 upvotes]: Let $K$ be a field with cardinality different of $2$ and let $V$ be a $K-$vector space. Let $T: V \to V$ be a linear operator such that $T^2 = I$.
Let $W = \{ v \in V: \, Tv = v \}$
 and $U = \{ v \in V : Tv = -v \}$.
I'm trying to proof that $V = W \oplus U$.
My attempt:
If $x \in W \cap U$, $Tx = x$ e $Tx = -x$, i.e $x = -x$, hence $x = 0$. Then, $W \cap U = \{0\}$.
If $x \in V$, we have that $x = x - Tx + Tx$ and $x - Tx \in U$, since
$$ T(x - Tx) = Tx - T^2 x = Tx -x = - (x-Tx). $$
However, $Tx$ which is not necessarily in $W$.
Help?

REPLY [12 votes]: I would write $$x=\frac{1}{2}(x+Tx)+\frac{1}{2}(x-Tx)$$