Let $H$ be a Hilbert space and $T:H\rightarrow H$ be bounded, linear, self-adjoint and unitary. Show that $H=H_+ \oplus H_{-}$, where $H_+ = \{x\in H: T(x)=x\}$ and $H_- = \{x\in H:T(x)=-x\}$.
My attempt:
It is easy to see that $H_+$ is closed, and by general results on projecting onto a complete subspace, $H=H_+\oplus H^\perp_+$. Thus, it suffices to show $H^\perp_+=H_-$.
If $T(x)=-x$ and $y\in H_+$, then $\langle x,y \rangle = \langle Tx,Ty \rangle = \langle -x,y \rangle $, so $\langle x,y \rangle =0$. This shows $H_-\subset H^\perp_+$.
The remaining containment is where I am having difficulty. If we let $x\in H_+^\perp$, so that $\langle x,y \rangle =0$ for all $y\in H_+$, then what remains to be shown is $||T(x)+x||=0$. But this is a challenge.. The closest result I've got was $||T(x)+x||^2 = ||T(x+y)+x||^2$. How should I proceed?
Best Answer
$$ (T-I)(T+I)=T^2-I=T^*T-I=I-I=0. $$ Therefore $$ x = \frac{1}{2}(I-T)x+\frac{1}{2}(I+T)x \\ = x_{-}+x_{+} $$ where $x_{-}=\frac{1}{2}(I-T)x$ and $x_{+}=\frac{1}{2}(I+T)x$ satisfy $$ x=x_{-}+x_{+}, \\ Tx_{-} = -x_{-}, \; Tx_{+}=x_{+}. $$ This is an orthogonal decomposition because $$ \langle (I-T)x,(I+T)y\rangle = \langle (I+T)(I-T)x,y\rangle =0. $$ So $H=H_{+}\oplus H_{-}$ is an orthogonal decomposition where $$ H_{+} = \mathcal{R}(I+T),\;\;\; H_{-}=\mathcal{R}(I-T). $$ $T$ is the identity on $H_+$ and is the negative of the identity on $H_-$.