if $g(v) = Av + b$ for A an orthogonal 2 by 2 matrix with $\det(A) = -1$, then $g$ is either a glide reflection or a reflection in $\Bbb R^2$.

Question

I am trying to clarify a proposition in my geometry notes:

Proposition: if $g(v) = Av + b$ for A an orthogonal $2$ by $2$ matrix with $det(A) = -1$, then $g$ is either a glide reflection or a reflection in $\Bbb R^2$.

A given hint to prove this states that the matrix of $A$ represents a reflection in some line about the origin.

Further, We can write $b= proj_n(b) + (b-proj_n(b))$ and where if $proj_n(b)=0$, then we have a reflection, but otherwise we have a glide reflection.

I am confused by the role of $b$. Since $proj_n(b)$ is perpendicular to our line, then if $proj_n(b)=0$, then $b$ must be perpendicular to the normal, and hence on the line?

Any clarifications much appreciated.

Answer 1

The only possible forms for a $2\times2$ orthogonal matrix are $\pmatrix{a&b\\-b&a}$ and $\pmatrix{a&b\\b&-a}$, with $a^2+b^2=1$. The determinant of the former is $1$ and of the latter $-1$, so our matrix $A$ will be of the second form. We have in addition $A^2=I$, hence its eigenvalues are $\pm1$. If we set $a=\cos2\theta$ and $b=\sin2\theta$, a little bit of work produces eigenvectors $(\cos\theta,\sin\theta)^T$ and $(-\sin\theta,\cos\theta)^T$ for the eigenvalues $+1$ and $-1$, respectively. Therefore, $A$ reverses the component of a vector perpendicular to the line through the origin with direction vector $(\cos\theta,\sin\theta)^T$ and leaves the component parallel to this line fixed, i.e., $A$ represents a reflection in this line.

Decompose $\mathbf b$ into its components $\mathbf b_\perp=\operatorname{proj_n}(\mathbf b)$ and $\mathbf b_\parallel=\mathbf b-\operatorname{proj_n}(\mathbf b)$ perpendicular to and parallel to $A$’s line of reflection, respectively. Since $A\mathbf b_\perp=-\mathbf b_\perp$, we have $$A\mathbf v+\mathbf b_\perp=A(\mathbf v-\frac12\mathbf b_\perp)+\frac12\mathbf b_\perp.$$ This means that a reflection in a line through the origin followed by a translation perpendicular to that line is equivalent to a reflection in a parallel line. To sum up, if $\mathbf b_\parallel=0$, we have a pure reflection, whereas if it’s nonzero, we also translate parallel to the reflection line, so we have a glide reflection. This is the opposite of what you say that you have in your notes, so perhaps $\operatorname{proj_n}$ is supposed to represent projection onto $A$’s line of reflection instead of its normal.