If $R$ is an orthogonal reflection, then $\frac{1}{2}(R + I)$ is an orthogonal projection I consider a Hilbert space $H$ over $\mathbb{C}$, and a reflection $R$ in the set of bounded linear operators on $H$ denoted $B(H)$. I want to prove that if $R$ is an orthogonal reflection, then $\frac{1}{2}(R + I)$ is an
orthogonal projection.
An operator $R \in B(H)$ is called an orthogonal reflection if there exists a closed subspace $M\subset H$ such that
$x + R(x) \in M$ and $x − R(x) \in M^{\perp}$ for every $x \in H$.
First I don't really see what to prove, do I need to first find the space of the projection?
I tried to show that $\frac{1}{2}(R + I)\big[\frac{1}{2}(R + I)(x)\big]= \frac{1}{2}(R + I)(x)$, but cannot see why this is true. Maybe my definition isn't the good one to use...
 A: An orthogonal reflection $R$ is a self-adjoint square root of the identity.
Both $\frac{1}{2}(I\pm R)$ are then self-adjoint, and
$$\left(\frac{1}{2}(I\pm R)\right)^2 \:=\: 
\frac14(I\pm 2R +I) \:=\: \frac{1}{2}(I\pm R)$$
proves them to be orthogonal projectors.
As their sum is the identity and their product equals zero, the images
$$M =\operatorname{Im}\frac12(I+R) \quad\text{and}\quad
M^\perp =\operatorname{Im}\frac12(I-R)$$
are complementary and orthogonal, thus exhaust the whole space $H=M\oplus M^\perp$.
This leads to the geometry associated with it: $M$ takes the role of the reflection "plane" (which is descriptive, but $M$ need not be two-dimensional in general). All vectors in $M$ are fixed points of $R$, in other words, $M$ is the eigenspace of $R$ for the eigenvalue $1\,$:
$$R\circ\left(\frac{1}{2}(I\pm R)\right) \;=\; \pm\,\frac{1}{2}(I\pm R)\,,$$
and $M^\perp$ is the eigenspace with respect to eigenvalue $-1$.
Notice that $-R$ is an orthogonal reflection too: it's the twin of $R$ because the roles of $M$ and $M^\perp$ are switched then.
And finally from geometry to operator: each decomposition $V\oplus V^\perp =H$ uniquely determines an orthogonal reflection (operator) in $H$:
If $P$ denotes the orthogonal projector onto $V$, then $R=2P-I\,$.
A: For any $x$,
$$\tag1
0=\langle x-Rx,x+Rx\rangle=\langle x,x\rangle-\langle Rx,x\rangle+\langle x,Rx\rangle-\langle Rx,Rx\rangle.
$$
This allows us to write
$$
\langle (R-R^*)x,x\rangle=\langle x,x\rangle-\langle Rx,Rx\rangle\in\mathbb R.
$$
As this occurs for all $x$, it implies that $R-R^*$ is selfadjoint. That is, $R-R^*=R^*-R$, which gives $R^*=R$. So $R$ is selfadjoint. Going back to $(1)$,
$$\tag2
0=\langle x-Rx,x+Rx\rangle=\langle x,x\rangle-\langle Rx,x\rangle+\langle x,Rx\rangle-\langle Rx,Rx\rangle=\langle x,x\rangle-\langle Rx,Rx\rangle.
$$
This we can write as $$\tag3
\langle (I-R^2)x,x\rangle=0.
$$
Thus $$\tag4R^2=I.$$
Now $P=(I+R)/2$ is selfadjoint, and
$$
P^2=\tfrac14\,(I+R)^2=\tfrac14\,(I+R^2+2R)=\tfrac14\,(2I+2R)=\tfrac12\,(I+R)=P. 
$$
So $P$ is an orthogonal projection.
