Prove $V$ is isomorphic to the direct sum $W^+$ and $W^-$ I got stuck at this question, please anyone help
Let $T:V\to V$ be a linear operator on a real vector space $V$ such
that $T^2=I$. Define subspaces $W^+$ and $W^-$ of $V$ as follows:
$$W^+=\{v\in V:T(v)=v\}$$
$$W^-=\{v\in V:T(v)=-v\}$$
Prove that V is isomorphic to the direct sum $W^+\oplus W^-$.
 A: *

*$W^++W^-=V$.  Let $v\in V$, and let $v^+=\frac{1}{2}(v+T(v))$ and $v^-=\frac{1}{2}(v-T(v))$.  Since $$T(v^+)=\frac{1}{2}(T(v)+T(T(v)))=\frac{1}{2}(T(v)+v)=v^+$$ we see $v^+\in W^+$.  Similarly, $$T(v^-)=\frac{1}{2}(T(v)-T(T(v)))=\frac{1}{2}(T(v)-v)=-v^-$$ so $v^-\in W^-$.  Thus, $v=v^++v^-$ is a decomposition of as a sum of vectors from $W^+$ and $W^-$.  Since $v$ was arbitrary, $W^++W^-=V$.

*$W^+\cap W^-$ is trivial.  Suppose $v\in W^+\cap W^-$.  Then on one hand, $T(v)=v$, and on the other, $T(v)=-v$.  Thus, $T(v)=0$.  Applying $T$ to both sides, $T(T(v))=T(0)=0$, and $T^2=I$, so $v=0$.  Thus, the only vector in the intersection is $0$.

*Therefore, $V=W^+\oplus W^-$.

If you know about eigenvectors, $T^2=I$ implies $\pm 1$ are the only eigenvalues.  By Jordan normal form, one can see $T$ must be diagonalizable.  Thus, $V$ decomposes as a direct sum of the two eigenspaces, namely $W^+$ for $1$ and $W^-$ for $-1$.
A: In this case one can write explicit operators. Note that $T$ is invertible hence $\ker T = \{0\}$
To see how one might come up with these, suppose $Tu=u, Tv=-v$ then
note that $T(u+v) = u-v$ and so we get $u={1 \over 2} ((u+v)+T(u+v))$
and $v={1 \over 2} ((u+v)-T(u+v))$.
So, define $T_+ x = {1 \over 2} (x+Tx)$, $T_- x = {1 \over 2} (x-Tx)$,
then $I = T_++T_-$,
$T (T_+x) = T_+ x$, $T (T_-x) = -T_- x$, and furthermore
$\ker T_+ \cap \ker T_- = \{0\}$.
A: I think you need the assumption that vector space over field $F$ with characteristic not equal to 2. 
So, you have $W^+ \cap W^- = \{0\}$. Indeed, one has $v \in W^+ \cap W^- $, then $T(v) = v$ and $T(v) = -v$, so $2v = 0$ then $v=0$.
Let $f:V \rightarrow W^+ \cap W^-$ such that $f(v) = (\frac{1}{2}(Tv+v),\frac{1}{2}(Tv-v))$.
Note that $T(\frac{1}{2}(Tv+v)) = \frac{1}{2}(Tv+v)$ and $T(\frac{1}{2}(Tv-v)) = -\frac{1}{2}(Tv-v)$.
One can see that $f$ is bijection.
