# Householder reflection in geometric algebra is not working for me

A Householder reflection of a vector $v$ along a direction $n$ is given by the formula \begin{eqnarray} v' = v - 2 \frac{n \cdot v}{|n|^2} n. \end{eqnarray}

If $n$ is unitary then $v'=v-2 (n \cdot v) n$. Clearly for $v=(2,3)$ and $n=(0,1)$ we find that $v'=(2,-3)$ which in fact is the Householder reflection of $(2,3)$ along the vector $(0,1)$.

According to Geometric Algebra (GA) a Housholder reflection is defined as $-nvn^{-1}$. Since $n$ is unitary $n^{-1}=n$ and we can write $v'=-nvn$. However when I compute this formula with the vectors above, I get different results.

Let me explain: \begin{eqnarray} n v = (0,1) \cdot (2,3) + (0,1) \wedge (2,3) = 3 - 2 e_1 \wedge e_2. \end{eqnarray} and so

\begin{eqnarray} v' = -nvn^{-1} = -(3 -2 e_1 \wedge e_2) n = -3n + 2 e_1 \wedge e_2 n. \end{eqnarray}

My concern is that, since $e_1 \wedge e_2$ is equivalent to a 90 degree counterclockwise rotation then we find that the second (last) term here is $(-2,0)$ and so the result is $v'=-3(0,1)+2(-1,0)=(-2,-3)$ and not $(2,-3)$ as expected.

Probably I have a stupid error such a sign somewhere or a deep problem on understanding the geometric algebraic product.

I would appreciate any help on this matter.

Thanks.

The problem is apparently with your interpretation, because the math is working out. $$-nvn^{-1} = -(3 -2 e_1 \wedge e_2) n = -3n + 2 e_1 \wedge e_2 n=-3e_2+2e_1e_2e_2=2e_1-3e_2$$.

I’m not even sure why you were using the product identity with the wedge. You may as well just compute it directly:

$$-nvn =-e_2(2e_1+3e_2)e_2=-2e_2e_1e_2-3e_2e_2e_2=2e_1-3e_2$$

It seems to me that multiplication by $e_1\wedge e_2$ on the left creates a clockwise rotation by 90, since it maps $e_2$ to $e_1$ and $e_1$ to $-e_2$.

• So, is it wrong to say that $e_1 e_2 u$ (I get your point. I can remove the wedge symbol here) is a $\pi/2$ positive rotation on $u$? I thought $e_1 e_2$ behaved as the imaginary complex number $i=\sqrt{-1}$. Thanks. – Herman Jaramillo Apr 29 '18 at 21:22
• $| e_1 e_2 |^2 = (e_1 e_2)^*(e_1 e_2)= e_2 e_1 e_1 e_2 = 1$ so $e_1 e_2 = \sqrt{-1}$. Right? – Herman Jaramillo Apr 29 '18 at 21:39
• Now the problem reduces to the following: Assuming the associative law is valid: $(e_1 e_2) e_2=-e_1$, $e_1(e_2 e_2)=e_1 1=e_1$. Why is the first interpratation wrong? Thanks. – Herman Jaramillo Apr 29 '18 at 21:45
• @HermanJaramillo If you multiply on the right with it, it will send $e_1\to e_2$ and $e_2\to -e_1$. To model complthe x multiplication, you’d have to write numbers with a real part and a complex part using $e_1\wedge e_2$, not as a linear combination of $e_1$ and $e_2$. So I think you are mixing interpretations. – rschwieb Apr 29 '18 at 22:16
• That is correct. Previously I did not understand your notation $\langle, \rangle$ because I was thinking on inner products but you are talking about spanning a space. Now since 1 is a scalar it commutes with $e_1 e_2$, but if instead of 1 you pick a vector $a$ then $\langle a, e_1 e_2 \rangle$ spans a space where only if $e_1 e_2$ is at the right , it resembles a counterclockwise rotation on $a$. On the left would be a clockwise rotation. I am thinking on $\mathbb{R}^2$ – Herman Jaramillo May 2 '18 at 21:35