Internal Direct Sum Questiom I'm posed with the following problem.  Given a vector space $\,V\,$ over a field (whose characteristic isn't $\,2$), we have a linear transformation from $\,V\,$ to itself.  
We have subspaces 
$$V_+=\{v\;:\; Tv=v\}\,\,,\,\, V_-=\{v\;:\; Tv=-v\}$$
I want to show that $\,V\,$ is the internal direct sum of these two subspaces.  I have shown that they are disjoint, but can't seem to write an arbitrary element in V as the sum of respective elements of the subspaces...
Edit:forgot something important! 
$$\, T^2=I\,$$
Sorry!
 A: For $v \in V$ we have 
\begin{align*}
  T(v + Tv) &= Tv + T^2v\\
            &= v + Tv\\
  T(v - Tv) &= Tv - T^2v\\
            &= -v + Tv\\
            &= -(v - Tv)
\end{align*}
Hence, $v + Tv \in V_+$, $v - Tv \in V_-$ and 
$$ v = \frac 12(v + Tv) + \frac 12 (v - Tv) $$
so $V = V_+ + V_-$. As you allready know that $V_+ \cap V_- = \{0\}$, we have shown that $V = V_+ \oplus V_-$.
A: $V_+\cap V_-=\{0\}$ really isn't "disjoint" but I think we all know what you mean.
To show that it generates the whole space, confirm that $v-T(v)\in V_-$. Since $T(v)$ is obviously in $V_+$, you can rearrange and conclude that $v\in V_-+V_+$.
Explicitly, $$v-T(v)=x\in V_-\implies v=T(v)+x\in V_++V_-$$.
A: If you assume you already had a solution $v=v_++v_-$ with $v_+\in V_+$ and $v_-\in V_-$ then $Tv = Tv_++Tv_- = v_+-v_-$. You can solve this for $v_+ = \frac12(v+Tv)$ and $v_-=\frac12(v-Tv)$.
Thus try to set $v_\pm = \frac12(v\pm Tv)$ and observe that indeed $Tv_\pm=\frac12(Tv\pm v)=\pm v_\pm$, hence $v_\pm\in V_\pm$.
A: Another approach: 
Lemma: In arbitrary characteristic, if $V$ is a $K$-vector space and $P:V\to V$ is an endomorphism with $P^2=\lambda P$, $\lambda\ne 0$, then $V=\ker P\oplus\ker (\lambda I-P)$.
Proof: If $v\in \ker P\cap\ker (\lambda I-P)$, then $\lambda v = (\lambda I-P)v+Pv = 0$, hence $v=0$. Thus $\ker P\cap\ker (\lambda I-P)=0$.
For $v\in V$ let $u=\lambda^{-1}Pv$ and $w=v-u$. Then clearly $v=u+w$ and $(\lambda I-P)u =Pv-\lambda^{-1}P^2v = 0$ and $Pw =Pv-Pu = Pv-\lambda^{-1}P^2v=0$, i.e. $u\in\ker(\lambda I-P)$ and $w\in \ker P$. We conclude that $V=\ker P\oplus\ker(\lambda I-P)$.
Corollary: If $\operatorname{char} K\ne 2$, $V$ is a $K$-vector space and $T:V\to V$ is an endomorphism with $T^2=I$, then $V=T_+\oplus T_-$ (with $T_\pm$ defined as in the OP).
Proof: Let $P=T+I$. Then $P^2 = T^2+2T+I^2=I+2T+I=2P$. Since $\lambda:=2\ne0$, the lemma applies. Here $\ker P=T_-$ and $\ker (2 I-P)=\ker(I-T)=T_+$.
