Should a reflection matrix of a vector have the same form as a rotation matrix?

Question

According to the book:

I know that it is not possible to write a reflection as a rotation, but from the text it seems that the matrix of the form

$$ A=\left[ \begin{array}{ c c } a & -b \\ b & a \end{array} \right] $$

is the most general form of $O(2)$ since the way it results in $A$ doesn't show any specification of rotation; on the other hand, I can't find any pair of $a,b$ such that

$$ A=\left[ \begin{array}{ c c } 1 & 0 \\ 0 & -1 \end{array} \right]. $$

Why is that?

Thank you.

Answer 1

HINT: $a^2+b^2=1$ and $T(e_1) =(a,b)^t$, give you two solutions not one: $T_1(e_2) =(-b,a)^t$ and $T_2(e_2) =(b,-a)^t$.

Answer 2

All reflections in the plane have matrices of the form $$ \left( \begin{array}{cc} \cos \alpha & \sin \alpha \\ \sin \alpha & - \cos \alpha \end{array} \right) $$ or, for any $a^2 + b^2 = 1,$ $$ \left( \begin{array}{cc} a & b \\ b & - a \end{array} \right) $$

Answer 3

on the other hand, I can't find any pair of a,b such that...Why is that?

You cannot find such a pair because your assumption

but from the text it seems that the matrix of the form ..is the most general form of O(2)

is not true: As you noted every rotation

$$ R=\left[ \begin{array}{cc} a & -b \\ b & a \end{array} \right] $$ has determinant $\det(R)=a^2+b^2=+1$ but the typical reflection $$ A=\left[ \begin{array}{ c c } 1 & 0 \\ 0 & -1 \end{array} \right]. $$ has determinant $-1$.

Note that the most general form of an Element $A$ of $O(2)$ maps $e_1=(1,0)^t$ to $(a,b)^t$ where $a^2+b^2=1$ (length is preserved) and because $Ae_2$ must be orthogonal you have $$ A=\left[ \begin{array}{cc} a & -b \\ b & a \end{array} \right] \quad \text{or }\quad A=\left[ \begin{array}{cc} a & b \\ b & -a \end{array} \right] $$