What Vector Operation Performs $(a,b)*(c,d)=(ac-bd,ad+bc)$?

Question

When you multiply two complex numbers, you get \begin{equation} (a+bi)\times(c+di)=(ac-bd)+(ad+bc)i \end{equation} As betterexplained.com points out, this multiplication of two complex numbers can be understood as a rotation. They give an example of a ship sailing 3 units east for every 4 units north, and ask what is the effect of rotating 45 degrees counterclockwise. Using complex numbers, you can find that \begin{equation} (3+4i)\times(1+i)=-1+7i \end{equation} Therefore using complex numbers, we find that the answer we were looking for: That is, after a 45 degree rotation, the ship is sailing 1 unit west for every 7 units north.

Now as a linear algebra lover, and it seems that the use of $i$ to solve this problem is unnecessary. Rather, it seems like we can just represent what's happening here in terms of vectors in the plane, namely the vectors $(3,4)$ and $(1,1)$. Some operation will take these vectors to the vector $(-1,7)$. In general, the operation would look like this: \begin{equation} (a,b)*(c,d)=(ac-bd,ad+bc) \end{equation} My question is, can this "rotation" operation be represented some other way, i.e. in terms of matrices, and does this operation have a name? Thanks!

Answer 1

$$ \left( \begin{array}{rr} a & b \\ -b & a \end{array} \right) \left( \begin{array}{cc} c & d \\ -d & c \end{array} \right) = \left( \begin{array}{cc} ac-bd & ad+bc \\ -(ad+bc) & ac-bd \end{array} \right) $$

Answer 2

Really, if you want to think about it in terms of vectors, it's easiest to think of one of the pairs $(a,b)$ and $(c,d)$ as fixed, and the other variable (although this obscures the commutativity of the operation). So I'll write $(x,y)$ for $(c,d)$. Then multiplication of complex numbers can be represented by a $2 \times 2$ matrix. That is, if $$x' + y'i = (a + bi)(x + yi),$$ then $$\begin{pmatrix} x' \\ y' \end{pmatrix} = \begin{pmatrix} a & -b\\ b & a \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix}.$$

The linear mapping $(x,y) \mapsto (x',y')$ is a direct similitude of the plane, centred at the origin, with angle $\theta = \arg (a + bi)$ and ratio $k = |a + bi|$. In other words, it's a rotation by an angle of $\theta$ about the origin, followed by a dilation with ratio $k$.

Answer 3

You can put these vectors into the clifford algebra. Use the geometric product, which has the following multiplication table:

$$e_1 e_1 = +1, \quad e_2 e_2 = +1, \quad e_1 e_2 = -e_2 e_1, \quad 1 e_1 = e_1, \quad 1 e_2 = e_2$$

The product is also associative.

Take your two complex numbers and convert them to linear combinations of $1$ and $e_1 e_2$ like so:

$$3+4i \mapsto 3 + 4 e_1 e_2, \quad 1 + i\mapsto 1 + e_1 e_2$$

Notice that, like $i$, $(e_1 e_2)^2 = -1$. The product, then, takes the same form as in complex numbers:

$$(3+4 e_1 e_2)(1 + e_1 e_2) = (3-4) + (3+4) e_1 e_2 = -1 + 7 e_1 e_2$$

What's going on here? What does this "clifford algebra" have to do with complex numbers?

Well, any rotation can be constructed from a couple reflections, and reflections in clifford algebra take the following form: for some unit vector $n$ normal to the object of reflection (a line in 2d, a plane in 3d, etc.), the reflection map $N$ takes the form

$$N(a) = -nan$$

(You can verify this yourself for a few simple cases.)

So, if two reflections yield a rotation, then rotations take the form

$$R(a) = mnanm$$

for some unit vectors $m, n$. The product $mn$ in 2d gives us linear combinations of $1$ and $e_1 e_2$ and thus can be identified with a complex number. Indeed, all such objects are of the algebraic form $mn = \exp(-\theta e_1 e_2/2) = \cos \theta/2 - e_1 e_2 \sin \theta/2$, and the rotation looks like

$$R(a) = \exp(-\theta e_1 e_2/2) a \exp(\theta e_1 e_2/2)$$

This looks like quaternions, doesn't it? Yes, in 3d, this same construction generates objects that are linear combinations of $1, e_1 e_2, e_2 e_3, e_3 e_1$, which can be identified with elements in the quaternion algebra, have the same multiplication laws, and so on.

So, clifford algebra products of vectors are very tightly related to rotations, and the product that you're interested in generalizes to many other kinds of objects, in many useful ways.