# If $T^2 = I$, how one can proof that $V = W \oplus U$?

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?

• Do you mean $U=\{ v \in V : Tv = -v \}$? – edm Aug 11 '17 at 14:57
• do you mean the field $K$ has "characteristic" different from $2$? – Alvin Lepik Aug 11 '17 at 15:03

I would write $$x=\frac{1}{2}(x+Tx)+\frac{1}{2}(x-Tx)$$