If $R$ is an orthogonal reflection, then $\frac{1}{2}(R + I)$ is an orthogonal projection

Question

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...

Answer 1

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 $R=2P-I\,$.

Answer 2

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.