What is a reflection of a vector on the plane from the active point of view

Question

Rotations of vectors on the plane can be seen either as active or passive. In the former we rotate the vector by an angle $\theta$ whereas in the latter we rotate the coordinate basis by $-\theta$. We end up with the same components $(a,b)$ for the rotated vector.

The group of rotations on the plane $SO(2)$ excludes reflections about an axis. For example, from a passive point of view, the reflection $S(y):(\hat i,\hat j)\rightarrow (-\hat i,\hat j)$ reflects the $x$ axis about the $y$ axis. The vector $\vec v=(1,1)^T$ would then be written as $\vec v'=(-1,1)^T$. The point is that there is no passive rotation able to do that, so $s$ does not belong to $SO(2)$.

Then it comes my doubts. Let us adopt the active point of view. The equivalent rotated vector would be $\vec v'=(-1,1)^T$ but then this is simply an active rotation by $\pi/2$ which in fact belongs to $SO(2)$.

Perhaps my questions are very basic: Is my assumption on what an active reflection about an axis is correct? If so, then why does every possible active reflection seems to be an element of $SO(2)$ whereas the same does not happen for the passive point of view.

Edit: As a concrete example: Take an arbitrary vector $(a,b)^T$. The reflected (about $y$) vector is $(-a,b)^T$. However this can also be obtained by an active rotation, namely $R(\pi-2\arctan(b/a))$ which has determinant one

Answer 1

It seems that you have a bit of confusion about active and passive transformations.

As a first point, an active and a passive rotation, applied to the same vector gives different results. As ana example consider the rotation of $\frac{\pi}{2}$ in $\mathbb{R}^2$. As an active transformation it is represented by the matrix (that has determinant $=1$) $$R_{\pi/2}= \begin{pmatrix} 0&-1\\1&0 \end{pmatrix} $$ and, as a passive transformation by the inverse matrix $$ R_{\pi/2}^{-1}=\begin{pmatrix} 0&1\\-1&0\end{pmatrix} $$ so, for the same vector $\vec v=(a,b)^T$ we have: $$ R_{\pi/2} \vec v= (-b,a)^T \qquad R_{\pi/2}^{-1} \vec v= (b,-a)^T $$ (For e general example you can see : Active and passive transformations in Linear Algebra .

A reflection in not a rotation and it is represented by a matrix that has determinant $=-1$. The (active) reflection on the $y$axis, e.g., is represented bu the matrix: $$ S_{\pi/2}= \begin{pmatrix} -1&0\\0&1 \end{pmatrix} $$ and the inverse (that represents the passive reflection) is the same matrix ( for a reflection we have $S=S^{-1}$). So for a generic vector $\vec v$ we have $$ S_{\pi/2} \vec v = S_{\pi/2}^{-1} \vec v= (-a,b)^T $$

Note tht this is different from the active or passive rotation.

Answer 2

After thinking a little bit more I might have come to an answer.

The matrices

$$S(y)= \begin{pmatrix} -1&0\\0&1 \end{pmatrix}, \quad\mathrm{and}\quad R(\pi-2\phi)= \begin{pmatrix} \cos(\pi-2\phi)&-\sin (\pi-2\phi)\\ \sin(\pi-2\phi)&\cos (\pi-2\phi) \end{pmatrix} $$ definitely takes the unit vector $\vec v$ from $(\cos\phi,\sin\phi)$ to $(-\cos\phi,\sin\phi)$. The thing is that those two matrices do not represent the same operator. When acting on a different vector $\vec w=(\cos\phi',\sin\phi')$ they will not, in general, give the same result.

So I should say that a matrix $M$ is a rotation if and only if there is single matrix $R\in SO(2)$ such that $M\vec v=R\vec v$ for any $\vec v\in\mathbb R^2$. Of course this is not the case for $S(y)$.