Reflection and rotation of a vector.

Question

Considerer a vector $(x,y)$ in the cartesian plane. The matrix that represents, first, a reflection of this vector in relation to a line passing through the origin and forming an angle $\alpha$ with the axis $x$ positive and, after, a rotation around the origin by an angle $\beta$ is?

Comments I considered a rotation of the system by an angle $\alpha$. Then we obtain the matrix \begin{bmatrix} \cos\alpha & \sin\alpha \\ -\sin\alpha & \cos\alpha \end{bmatrix} In the sequence, in this system I reflected on the axis $x$ multiplying by the matrix \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix} and so I got the matrix \begin{bmatrix} \cos\alpha & \sin\alpha \\ \sin\alpha & -\cos\alpha \end{bmatrix}.

I don't know if this reasoning is correct and I don't know what it means "a rotation around the origin by an angle $\beta$".

I thank you for your help.

Answer 1

you just use the fact that this is the composition of linear transformations: Denoting by $R(x,y)$ the reflection, and $T(x,y)$ the rotation, the transformation of your exercise is $(T \circ R)(x ,y)$. It is a known fact of linear algebra that the matrix of the composition of linear transformations is the product of the matrices, i.e. $[T \circ R] = [T][R]$ (assuming to use the canonical basis of $\mathbb{R}^2$) so just do the multiplication of the rotation and reflection matrices.

Answer 2

Reading this, it seems like you were trying to get the reflection by rotating the plane and reflecting across the x-axis. That's a valid strategy... as long as you reverse the rotation afterwards!

I'm assuming you're using transformation matrices multiplied on the right of row vectors, since that matches your rotation matrix format.

The angle $\alpha$ is used to specify the angle of the diagonal in this picture:

You'd need to work out what matrix reflects the plane over that line. I suppose there are a lot of ways to do it.

Following your idea, we could rotate by $-\alpha$, reflect across the $x$-axis, then rotate back $\alpha$ to achieve the reflection.

If you do those transformations in that order:

$$ \begin{bmatrix} \cos\alpha & -\sin\alpha \\ \sin\alpha & \cos\alpha \end{bmatrix}\begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}\begin{bmatrix} \cos\alpha & \sin\alpha \\ -\sin\alpha & \cos\alpha \end{bmatrix}=\begin{bmatrix} \cos2\alpha & \sin2\alpha \\ \sin2\alpha & -\cos2\alpha \end{bmatrix} $$

Another one I remember is to first figure out the unit normal and then compute its Householder reflection.

It's not hard to work out that the unit normal is $(-\sin\alpha, \cos\alpha)$, and then its reflection is

$$ I_2-2\begin{bmatrix}-\sin \alpha \\ \cos\alpha\end{bmatrix}\begin{bmatrix}-\sin \alpha & \cos\alpha\end{bmatrix}=\begin{bmatrix}1-2\sin^2\alpha & 2\sin\alpha\cos\alpha \\ 2\sin\alpha\cos\alpha &1-2\sin^2\alpha\end{bmatrix} $$

Considering the half-angle identities, this matches the first matrix!

It's odd you say you don't understand "a rotation around the origin by an angle " because you were apparently comfortable writing the matrix for a rotation around the origin by an angle $\alpha$.

In the notation we're using, that matrix would be $\begin{bmatrix} \cos\beta & \sin\beta \\ -\sin\beta & \cos\beta \end{bmatrix}$. Applying this transformation after the previous one would achieve the matrix that's the goal of the problem.