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 $\\theta$ withthe positive $\\mathbf{x}$ axis. More precisely, we are givena 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 $T$ withrespect to the $x, y$ and $u, v$ coordinates respectively; $[I]_{uv}^{xy}$ is thematrix 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 derivedfrom 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}\\,.

Computing the matrix product (with the help of the double angle identity) yields:

[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 reflectedto $(\\cos 2\\theta,\\sin 2\\theta)$ and $(0,1)$ gets reflected to $(\\sin 2\\theta,-\\cos 2\\theta)$ .Another way is to observe that we can rotate an arbitrary mirror lineonto 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.

Note also that $[T]_{uv}$ and $[T]_{xy}$ are orthogonal matrices,with determinant $-1$ , as expected.

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 reflectionabout 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 reflectionand 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 ofthe 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, sothat 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 justas 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 toa 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.