Prove that $A$ is reflection operator

Question

Prove that if $A^2=I$ and $A$ isn't scalar operator (Operator $A$ is called scalar if $Ax=\lambda x$) then $A$ is reflection operator. I have no clue how to do this problem. The only hint my teacher gave me is the fact that $A_J^2=I$ ($A_J$ is Jordan form of A)

Answer 1

I think that the definition for reflection operator on V is:

For $U,W\subseteq V$ s.t. $U+W=V$ and $U\cap W=\left\{ 0\right\} $

($\iff U\oplus W=V$)

Then $\forall v\in V:v=u+w$ while $u\in U,w\in W$

If $f:V\rightarrow V$, $f\left(v\right)=f\left(u+w\right)=u-w$, then $f$ is a reflection operator.

The proof:

Let $U=\ker\left(f-\text{Id}\right),\,W=\ker\left(f+\text{Id}\right).$

If $v\in U\cap W$ then $f\left(v\right)-v=0=f\left(v\right)+v\Rightarrow v=0\Rightarrow U\cap W=\left\{ 0\right\} .$

Now we should prove $V=U+W:$

For $v\in V$, let $u=\frac{v+f\left(v\right)}{2}$ and $w=\frac{v-f\left(v\right)}{2}.$ Then $u+w=\frac{v+f\left(v\right)}{2}+\frac{v-f\left(v\right)}{2}=v.$

Additionaly, $f\left(u\right)=f\left(\frac{v+f\left(v\right)}{2}\right)=\frac{f\left(v\right)+f\left(f\left(v\right)\right)}{2}=\frac{f\left(v\right)+v}{2}=u\Rightarrow u\in U.$

And $f\left(w\right)=f\left(\frac{v-f\left(v\right)}{2}\right)=\frac{f\left(v\right)-f\left(f\left(v\right)\right)}{2}=\frac{f\left(v\right)-v}{2}=w\Rightarrow w\in W$.

(Using $f^{2}=\text{Id}$)

Hence, $U\oplus W=V$. Now, for all $v\in V$, $v=u+w$.

Finally, $f\left(v\right)=f\left(u+w\right)=f\left(u\right)+f\left(w\right)=u-w$, i.e. $f$ is a reflection operator.

(Translated from my school's solutions)