# derivation of 2D reflection matrix

## Reflection across a line of given angle

Let $\mathbf{x},\mathbf{y}$ be perpendicular     unit vectors  in the plane. Suppose we want to reflect  vectors (perpendicularly) over a line that makes an angle $\theta$ with the positive $\mathbf{x}$ axis. More precisely, we are given a direction direction vector $\mathbf{u}=\cos\theta\,\mathbf{x}+\sin\theta\,\mathbf{y}$ for the line of reflection. A unit vector perpendicular to $\mathbf{u}$ is $\mathbf{v}=-\sin\theta\,\mathbf{x}+\cos\theta\,\mathbf{y}$ (as is easily checked). Then to reflect an arbitrary vector $\mathbf{w}$, we write $\mathbf{w}$ in of its components   in the $\mathbf{u},\mathbf{v}$ axes: $\mathbf{w}=a\mathbf{u}+b\mathbf{v}$, and the result of the reflection is to be $\mathbf{w}^{\prime}=a\mathbf{u}-b\mathbf{v}$. We compute the matrix for such a reflection in the original $x,y$ coordinates.

Denote the reflection by $T$. By the matrix change-of-coordinates formula, we have

 $\displaystyle[T]_{xy}=[I]_{uv}^{xy}\>[T]_{uv}\>[I]_{xy}^{uv}\,,$

where $[T]_{xy}$ and $[T]_{uv}$ denote the matrix representing $T$ with respect to the $x,y$ and $u,v$ coordinates respectively; $[I]_{uv}^{xy}$ is the matrix that changes from $u,v$ coordinates to $x,y$ coordinates, and $[I]_{xy}^{uv}$ is the matrix that changes coordinates the other way.

The three matrices on the right-hand side are all easily derived from the description we gave for the reflection $T$:

 $\displaystyle[I]_{uv}^{xy}=\begin{bmatrix}\cos\theta&-\sin\theta\\ \sin\theta&\cos\theta\end{bmatrix}\,,\quad[T]_{uv}=\begin{bmatrix}1&0\\ 0&-1\end{bmatrix}\,,\quad[I]_{xy}^{uv}=\Bigl{(}[I]_{uv}^{xy}\Bigr{)}^{-1}=% \begin{bmatrix}\cos\theta&\sin\theta\\ -\sin\theta&\cos\theta\end{bmatrix}\,.$
 $[T]_{xy}=\begin{bmatrix}\cos 2\theta&\sin 2\theta\\ \sin 2\theta&-\cos 2\theta\end{bmatrix}\,.$ (1)

For the of the reader, we note that there are other ways of “deriving” this result. One is by the use of a diagram, which would show that $(1,0)$ gets reflected to $(\cos 2\theta,\sin 2\theta)$ and $(0,1)$ gets reflected to $(\sin 2\theta,-\cos 2\theta)$. Another way is to observe that we can rotate an arbitrary mirror line onto the x-axis, then reflect across the x-axis, and rotate back. (The matrix product $[T]_{xy}$ can be seen as operating this way.) We took neither of these two approaches, because to justify them rigorously takes a bit of work, that is avoided by the pure linear algebra approach.

## Reflection across a line of given direction vector

Suppose instead of being given an angle $\theta$, we are given the unit direction vector $u$ to reflect the vector $w$. We can derive the matrix for the reflection directly, without involving any trigonometric functions   .

In the decomposition $\mathbf{w}=a\mathbf{u}+b\mathbf{v}$, we note that $b=\mathbf{w}\cdot\mathbf{v}$. Therefore

 $\mathbf{w}^{\prime}=(a\mathbf{u}+b\mathbf{v})-2b\mathbf{v}=\mathbf{w}-2(% \mathbf{w}\cdot\mathbf{v})\mathbf{v}\,.$

(In fact, this is the formula used in the to draw the diagram in this entry.) To derive the matrix with respect to $x,y$ coordinates, we resort to a trick:

 $\mathbf{w}^{\prime}=I\mathbf{w}-2\mathbf{v}(\mathbf{w}\cdot\mathbf{v})=I% \mathbf{w}-2\mathbf{v}(\mathbf{v}^{\textrm{tr}}\mathbf{w})=I\mathbf{w}-2(% \mathbf{v}\mathbf{v}^{\textrm{tr}})\mathbf{w}\,.$

Therefore the matrix of the transformation is

 $I-2\mathbf{v}\mathbf{v}^{\textrm{tr}}=\begin{bmatrix}u_{x}^{2}-u_{y}^{2}&2u_{x% }u_{y}\\ 2u_{x}u_{y}&u_{y}^{2}-u_{x}^{2}\end{bmatrix}\,,\quad\mathbf{u}=(u_{x},u_{y})^{% \textrm{tr}}\,,\quad\mathbf{v}=(-u_{y},u_{x})^{\textrm{tr}}\,.$

If $u$ was not a unit vector to begin with, it of course suffices to divide by its magnitude before proceeding. Taking this into account, we obtain the following matrix for a reflection about a line with direction $\mathbf{u}$:

 $\frac{1}{u_{x}^{2}+u_{y}^{2}}\begin{bmatrix}u_{x}^{2}-u_{y}^{2}&2u_{x}u_{y}\\ 2u_{x}u_{y}&u_{y}^{2}-u_{x}^{2}\end{bmatrix}\,.$ (2)

Notice that if we put $u_{x}=\cos\theta$ and $u_{y}=\sin\theta$ in matrix (2), we get matrix (1), as it should be.

## Reflection across a line of given slope

There is another form for the matrix (1). We set $m=\tan\theta$ to be the slope of the line of reflection and use the identities:

 $\displaystyle\cos^{2}\theta$ $\displaystyle=\frac{1}{\tan^{2}\theta+1}=\frac{1}{m^{2}+1}$ $\displaystyle\cos 2\theta$ $\displaystyle=2\cos^{2}\theta-1$ $\displaystyle\sin 2\theta$ $\displaystyle=2\sin\theta\cos\theta=2\tan\theta\cos^{2}\theta=2m\cos^{2}\theta\,.$

When these equations are substituted in matrix (1), we obtain an alternate expression for it in of $m$ only:

 $\frac{1}{m^{2}+1}\begin{bmatrix}1-m^{2}&2m\\ 2m&m^{2}-1\end{bmatrix}\,.$ (3)

Thus we have derived the matrix for a reflection about a line of slope $m$.

Alternatively, we could have also substituted $u_{x}=1$ and $u_{y}=m$ in matrix (2) to arrive at the same result.

## Topology of reflection matrices

Of course, formula (3) does not work literally when $m=\pm\infty$ (the line is vertical). However, that case may be derived by taking the limit $\lvert m\rvert\to\infty$ — this limit operation can be justified by considerations of the topology   of the space of two-dimensional reflection matrices.

What is this topology? It is the one-dimensional projective plane  $\mathbb{RP}^{1}$, or simply, the “real projective line”. It is formed by taking the circle, and identifying opposite points, so that each pair of opposite points specify a unique mirror line of reflection in $\mathbb{R}^{2}$. Formula (1) is a parameterization of $\mathbb{RP}^{1}$. Note that (1) involves the quantity $2\theta$, not $\theta$, because for a point $(\cos\theta,\sin\theta)$ on the circle, its opposite point $(\cos(\theta+\pi),\sin(\theta+\pi))$ specify the same reflection, so formula (1) has to be invariant when $\theta$ is replaced by $\theta+\pi$.

But (1) might as well be written

 $[T]_{xy}=\begin{bmatrix}\cos\phi&\sin\phi\\ \sin\phi&-\cos\phi\end{bmatrix}\,.$ (4)

where $\phi=2\theta$. For this parameterization of $RP^{1}$ to be one-to-one, $\phi$ can range over interval   $(0,2\pi)$, and the endpoints  at $\phi=0,2\pi$ overlap just as for a circle, without identifying pairs of opposite points. What does this mean? It is the fact that $\mathbb{RP}^{1}$ is homeomorphic to the circle $S^{1}$.

The real projective line $\mathbb{RP}^{1}$ is also the one-point compactification of $\mathbb{R}$ (i.e. $\mathbb{RP}^{1}=\mathbb{R}\cup\{\infty\}$), as shown by formula (3); the number $m=\infty$ corresponds to a reflection across the vertical axis. Note that this “$\infty$” is not the same as the usual $\pm\infty$, because here $-\infty$ and $\infty$ are actually the same number, both representing the slope of a vertical line.

Title derivation of 2D reflection matrix DerivationOf2DReflectionMatrix 2013-03-22 15:25:05 2013-03-22 15:25:05 stevecheng (10074) stevecheng (10074) 14 stevecheng (10074) Derivation msc 15-00 RotationMatrix DecompositionOfOrthogonalOperatorsAsRotationsAndReflections DerivationOfRotationMatrixUsingPolarCoordinates