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Let $\mathbf{u}$ and $\mathbf{v}$ be two 3D unit vectors. The transform that performs reflection in the plane normal to $\mathbf{u}$ is given by $$ T_{\mathbf{u}}(\mathbf{x}) = \mathbf{x} - 2(\mathbf{x} \cdot \mathbf{u})\mathbf{u} $$ and similarly, reflection in the plane normal to $\mathbf{v}$ is performed by $$ T_{\mathbf{v}}(\mathbf{x}) = \mathbf{x} - 2(\mathbf{x} \cdot \mathbf{v})\mathbf{v} $$ Let $\theta$ be the angle between $\mathbf{u}$ and $\mathbf{v}$, and let $\mathbf{n}$ be the unit vector in the direction of $\mathbf{u} \times \mathbf{v}$. So, then we know that $\cos\theta = \mathbf{u} \cdot \mathbf{v}$, and $\mathbf{u} \times \mathbf{v} = (\sin\theta)\mathbf{n}$.

The composition of these two reflections is a rotation around $\mathbf{n}$ by an angle of $2\theta$ (I believe), and that rotation is given by Rodrigues' formula: $$ R(\mathbf{x}) = (\cos2\theta)\mathbf{x} + (1 - \cos2\theta)(\mathbf{x} \cdot \mathbf{n})\mathbf{n} + (\sin 2\theta)(\mathbf{x} \times \mathbf{n}) $$ It seems to me that we ought to be able to prove from first principles that $$ T_{\mathbf{u}}\big( T_{\mathbf{v}}(\mathbf{x}) \big) = R(\mathbf{x}) $$ I've slogged through pages of vector algebra for a few hours, but to no avail. It's depressing -- I used to be good at this stuff, but apparently not any more. I'd like a proof that uses nothing but elementary vector arithmetic, and I'd like it to be coordinate-free, please.

Edit

As a couple of people have mentioned, it seems sensible to work in the $\mathbf{u}\text{-}\mathbf{v}\text{-}\mathbf{n}$ coordinate system. This doesn't violate my "coordinate free" requirement as long as we don't start writing out explicit coordinates for $\mathbf{u}$, $\mathbf{v}$ and $\mathbf{n}$. The vector $R(\mathbf{x}) - \mathbf{x}$ should be entirely in the $\mathbf{u}\text{-}\mathbf{v}$ plane, so all of its $\mathbf{n}$ terms must vanish, and we should be left with an expression that involves only $\mathbf{u}$ and $\mathbf{v}$, which (I hope) will give us the link to the reflections. The algebraic grunt-work involved is what's giving me trouble.

bubba
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    You could decompose $\mathbf{x}$ as, say, $a \mathbf{u} + b \mathbf{v} + c \mathbf{n}$. After a bit of tedium expanding $T_{\mathbf{u}}(T_{\mathbf{v}}(\mathbf{x}))$ and $R(\mathbf{x})$, you'll get equality. (I'll admit that this approach isn't purely "coordinate-free", since $(a,b,c)$ gives $\mathbf{x}$ in the $\mathbf{u}\mathbf{v}\mathbf{n}$ coordinate system ... but I think it's acceptable if you think of "$a$", "$b$", and "$c$" as abbreviating appropriate dot product expressions. :) – Blue Mar 08 '14 at 18:38
  • Hi @Blue. That's a good idea. And, as long as we don't write out the components of $\mathbf{u}$, $\mathbf{v}$, $\mathbf{n}$, I would regard it as "coordinate free". In fact, nothing much happens in the $\mathbf{n}$ direction, so all the hard work is probably in the $\mathbf{u}$ and $\mathbf{v}$ components, as you can see in the proof of Rodrigues' formula. I know that this ought to work. It's the "tedium" part that's giving me trouble. If you have gone through the tedium, that's what I'd like to see. – bubba Mar 08 '14 at 23:50
  • @bubba The determinant of a reflection in a plane is $-1$ (since its eigenvalues are all $1$ except for one $-1$.) Then the product of two of these reflections has determinant $1$. A (nonreflective) rotation is an isometry with determinant $1$. – rschwieb Aug 20 '14 at 14:01
  • @rschwieb -- Thanks. That's basically the same argument given in David's answer below. But, as I said in my comment, I'm looking for a more concrete first-principles argument using nothing more than 3D vector arithmetic. – bubba Aug 21 '14 at 01:46
  • @bubba you mentioned you were getting bogged down in computations, and this is the solution to that. I'm not sure what you're hoping to see using vector computations anyway... – rschwieb Aug 21 '14 at 10:01
  • @rschwieb -- what I wanted is exactly what's in Blue's answer. I got stuck with the vector computations, but I wanted to get through them, not avoid them. – bubba Aug 21 '14 at 10:41

4 Answers4

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Write $c := \cos\theta$, $s := \sin\theta$, $\mathbf{w} := \mathbf{u}\times\mathbf{v} = s\mathbf{n}$, and $T := T_\mathbf{u}\left(T_\mathbf{v}\right)$, so that we have ...

$$\begin{align} T(\mathbf{x}) &=\mathbf{x}-2(\mathbf{x}\cdot\mathbf{u})\mathbf{u}-2(\mathbf{x}\cdot\mathbf{v})\mathbf{v} + 4c(\mathbf{x}\cdot\mathbf{v})\mathbf{u}\\ R(\mathbf{x}) &= (2c^2-1)\mathbf{x} + 2 s^2 (\mathbf{x}\cdot\mathbf{n}) \mathbf{n} + 2 s c (\mathbf{x}\times\mathbf{n}) \\ &= (2c^2-1)\mathbf{x} + 2 (\mathbf{x}\cdot\mathbf{w})\mathbf{w} + 2 c (\mathbf{x}\times\mathbf{w}) \\ \end{align}$$ Decomposing $\mathbf{x}$ as $p\mathbf{u} + q\mathbf{v} + r \mathbf{w}$, we can get fairly directly ... $$\mathbf{x}\cdot\mathbf{u} = p + q c \qquad \mathbf{x}\cdot\mathbf{v}=pc+q \qquad \mathbf{x}\cdot\mathbf{w}=rs^2 \qquad (\star)$$ $$\mathbf{x}\times\mathbf{w} = \mathbf{x}\times \left(\mathbf{u}\times\mathbf{v}\right) = (\mathbf{x}\cdot\mathbf{v})\mathbf{u}-(\mathbf{x}\cdot\mathbf{u})\mathbf{v} \qquad (\star\star)$$

Then it's straightforward to show that the difference of the transformations vanishes:

$$\begin{align} T(\mathbf{x}) - R(\mathbf{x}) &=\mathbf{x}-2(p+qc)\mathbf{u}-2(pc+q)\mathbf{v} + 4c(pc+q)\mathbf{u}\\ &-\left( (2c^2-1)\mathbf{x} + 2 r s^2 \mathbf{w} + 2 c \left( (pc+q)\mathbf{u} - (p+qc)\mathbf{v} \right) \right) \\[6pt] &= (2-2c^2)\;\mathbf{x} + 2 \left(-p-qc+2pc^2+2qc-pc^2-qc\right)\;\mathbf{u} \\ &+ 2\left(-pc-q+pc+qc^2\right)\mathbf{v} - 2 r s^2 \mathbf{w} \\[6pt] &= 2 s^2 \left( \mathbf{x} - p\mathbf{u} - q \mathbf{v} - r\mathbf{w} \right) \\[6pt] &= 0 \end{align}$$

and we conclude that the transformations are equivalent. $\square$

Edit. Without jumping immediately to the decomposition of $\mathbf{x}$, we can use the expansion in $(\star\star)$ to write

$$\begin{align} \frac{T(\mathbf{x})-R(\mathbf{x})}{2s^2} \;\;&=\;\; \mathbf{x} \;-\; \left( \; \frac{\mathbf{x}.( \mathbf{u} - c \mathbf{v} )}{s^2}\;\mathbf{u} \;+\; \frac{\mathbf{x}.( \mathbf{v} - c \mathbf{u} )}{s^2}\;\mathbf{v} \;+\; \frac{\mathbf{x}.\mathbf{w}}{s^2}\;\mathbf{w} \;\right) \end{align}$$

If you can "see" that the coefficients of $\mathbf{u}$, $\mathbf{v}$, $\mathbf{w}$ are the components of $\mathbf{x}$ ---which would be clear for orthogonal $\mathbf{u}$ and $\mathbf{v}$, for which $c=0$ and $s=1$--- then you're done. If not, note that you can arrive at this insight by solving the dot-product equations $(\star)$ for $p$, $q$, $r$.

Blue
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As long as you are not obliged to use your formulae for $T_{\bf u}({\bf x})$ and so on, there is a very easy approach to this.

Any reflection in ${\Bbb R}^3$ is represented by an orthogonal matrix of determinant $-1$. Any rotation in ${\Bbb R}^3$ is represented by an orthogonal matrix of determinant $+1$. Multiply two of the former and you get one of the latter. . .

Not sure if that satisfies the requirements in your last sentence though.

David
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  • Thanks, @David. Yes, I'm aware of the things that we know from abstract algebra. But, it's too well ... abstract. I have a formula for the product of reflections, and I have a formula for rotation. Those two formulae ought to be the same, and I'd like to prove that by brute-force elementary algebra. – bubba Mar 08 '14 at 12:55
  • Yep, tried that. And one can also write $(\mathbf{x} \cdot \mathbf{w}) \mathbf{w} = \mathbf{x} (\mathbf{w} \otimes \mathbf{w})$, where $\mathbf{w} \otimes \mathbf{w} = (\mathbf{w} \mathbf{w}^T)$. That didn't help, either. – bubba Mar 08 '14 at 13:06
  • Sorry -- same thing you said (roughly). – bubba Mar 08 '14 at 13:08
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Restricted to any plane perpendicular to the line of intersection of the reflection planes, the reflections are a pair of orientation reversing isometries fixing the same point. Their composition is thus an orientation preserving isometry fixing one point: a rotation. To see what the rotation angle must be, track the effect of each rotation on angles around the fixed point.

This argument works for pairs of orthogonal reflections across $(n-1)$ dimensional hyperplanes in $n$ dimensional Euclidean space, since the $(n-2)$ dimensions in the intersection of the planes are frozen out leaving a $2$-d problem.

zyx
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Reflection operators $P$ and $Q$ on $\mathbb{R}^n$ satisfy
$\qquad P = P^T$
$\qquad Q = Q^T$
$\qquad P^2 = Q^2 = I$
$\Rightarrow$
$\qquad (PQ)^T PQ = Q^T P^T PQ = Q^T(P^T P)Q = Q^TQ = I$
i.e. $PQ$ is orthogonal. It also has determinant 1, so is a rotation .

(By the way, "Every rotation is the result of ... an even number of reflections" — reflection .)

denis
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  • Thanks. Sure, we know from the properties of determinants that a product of two refelections is a rotation. But that's not the problem. I want to show that the product of two specific reflections $T_{\mathbf{u}}$ and $T_{\mathbf{v}}$ is the reflection $R$, and I want to do it by 3D vector algebra alone. Blue's answer is exactly what I wanted. – bubba Aug 21 '14 at 01:51