Why is it natural to define reflections in terms of its hyperplane?

Question

There are two ways to reflect a vector $v$ by another vector $u$. The first is to 'directly' reflect $v$ through $u$:

where $v'=2P_uv-v$ with $P_u=\frac{uu^T}{|u|^2}$. The second way is to reflect $v$ through the normal plane spanned by $u$ like this:

This time the reflection is $-1$ times the other convention i.e. $v'=v-2P_uv$. To me the first way seems more intuitive but if I look at this article https://en.wikipedia.org/wiki/Geometric_algebra#Reflection they use the second convention. Is it common in math to use the second convention? Are there any benefits compared to the first convention?

Answer 1

Reflection typically means “I have an eigenspace for 1 with dimension $n-1$ and an eigenspace for $-1$ of dimension 1”. The former eigenspace is the hyperplane, the latter is the subspace spanned by the normal.

It seems to me the reason you think the two are similar is compounded by the fact your example is in 2-d.

The first example you are giving is “reflecting through a line.” This works in 2 dimensions, but in $n$ dimensions, if you reflect all vector across a fixed vector in the plane they generate, you can get something else, namely a rotation. I hesitate to call this a reflection but as it happen some authors (see last big paragraph of the lede ) are willing to admit any nonidentity involutive isometry with an affine eigenspace for 1, of dimension less than $n$ as a reflection.

The choice of thinking in terms of the hyper plane or its normal are both still there in $n$ dimensions, but the different “nature” of each type is a little more obvious. Each one has the data needed to determine the other one. This is a duality. But they are only both lines in 2-d.

Answer 2

The geometric algebra reflection expression is based on the following illustration

Observe that the projection of $ \mathbf{v} $ along the direction $ \hat{\mathbf{u}} $ (assumed to be a unit vector for simplicity) is $$(\hat{\mathbf{u}} \cdot \mathbf{v}) \hat{\mathbf{u}},$$ and the component of $ \mathbf{v} $ perpendicular to $ \hat{\mathbf{u}} $ (the rejection) is: $$\mathbf{v} - (\hat{\mathbf{u}} \cdot \mathbf{v}) \hat{\mathbf{u}}.$$

From the diagram, it is clear that the reflection is just $$\begin{aligned}\mathbf{v}' &= \mathbf{v} - 2 \left( { \mathbf{v} - (\hat{\mathbf{u}} \cdot \mathbf{v}) \hat{\mathbf{u}} } \right) \\ &= 2 (\hat{\mathbf{u}} \cdot \mathbf{v}) \hat{\mathbf{u}} - \mathbf{v}.\end{aligned}$$

There is no use of geometric algebra in this. A geometric algebra is a vector space where the elements of the space are products of vectors, subject to an additional ``contraction identity'' $\mathbf{x}^2 = \mathbf{x} \cdot \mathbf{x}$. One can form various special product combinations that have useful applications. In particular, it can be shown that a symmetric sum is related to the dot product $$\mathbf{x} \cdot \mathbf{y} = \frac{1}{{2}} \left( { \mathbf{x} \mathbf{y} + \mathbf{y} \mathbf{x} } \right).$$ We can also form an antisymmetric sum $$\mathbf{x} \wedge \mathbf{y} = \frac{1}{{2}} \left( { \mathbf{x} \mathbf{y} - \mathbf{y} \mathbf{x} } \right),$$ where we call $ \wedge $ the wedge operator, an operator that happens to be related to the cross product in 3D. The concepts above can be used to decompose a vector into projective and rejective components as follows $$\begin{aligned}\mathbf{v}&= \mathbf{v} \hat{\mathbf{u}} \hat{\mathbf{u}} \\ &= \left( { \mathbf{v} \hat{\mathbf{u}} } \right) \hat{\mathbf{u}} \\ &= \left( { \mathbf{v} \cdot \hat{\mathbf{u}} + \mathbf{v} \wedge \hat{\mathbf{u}} } \right) \hat{\mathbf{u}} \\ &= \left( { \mathbf{v} \cdot \hat{\mathbf{u}} } \right) \hat{\mathbf{u}} + \left( { \mathbf{v} \wedge \hat{\mathbf{u}} } \right) \hat{\mathbf{u}}.\end{aligned}$$ The first term we recognize as the projection, allowing us to identify the remainder as the rejection. It's also straightforward to show that $$\left( { \left( { \mathbf{v} \cdot \hat{\mathbf{u}} } \right) \hat{\mathbf{u}} } \right) \cdot \left( { \left( { \mathbf{v} \wedge \hat{\mathbf{u}} } \right) \hat{\mathbf{u}} } \right) = 0,$$ satisfying our expections for the projection and rejection.

We can utilize this to compute the reflection, which is $$\begin{aligned}\mathbf{v}'&= \mathbf{v} - 2 \left( { \mathbf{v} \wedge \hat{\mathbf{u}} } \right) \hat{\mathbf{u}} \\ &= \mathbf{v} \hat{\mathbf{u}} \hat{\mathbf{u}} - 2 \left( { \mathbf{v} \wedge \hat{\mathbf{u}} } \right) \hat{\mathbf{u}} \\ &= \left( { \mathbf{v} \hat{\mathbf{u}} - 2 \left( { \mathbf{v} \wedge \hat{\mathbf{u}} } \right) } \right) \hat{\mathbf{u}} \\ &= \left( { \mathbf{v} \hat{\mathbf{u}} - \left( { \mathbf{v} \hat{\mathbf{u}} - \hat{\mathbf{u}} \mathbf{v} } \right) } \right) \hat{\mathbf{u}} \\ &= \hat{\mathbf{u}} \mathbf{v} \hat{\mathbf{u}}.\end{aligned}$$ This sandwich of $ \mathbf{v} $ between $ \hat{\mathbf{u}} $ is the usual way that reflection is expressed in geometric algebra. As for benefits, this approach leads to the sandwich reflection identity above, which is probably about as simple as a reflection can be expressed.