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


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.

  • $\begingroup$ Thanks Emilio. But it still is not clear to me. 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. $\endgroup$ – Diracology Apr 5 '17 at 13:31
  • $\begingroup$ No. The rotation is not a reflection. Take some experiment with different vectors. $\endgroup$ – Emilio Novati Apr 5 '17 at 13:38
  • $\begingroup$ I know it is not and this is clear to me in the passive point of view. But honestly I am not seeing this for active transformations. Does not $R(\pi-2\arctan(b/a))$ take any vector $(a,b)^T$ to $(-a,b)^T$? $\endgroup$ – Diracology Apr 5 '17 at 13:45

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 $S(y)$.


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