TITLE: Quaternions: why does ijk = -1 and ij=k and -ji=k
QUESTION [26 upvotes]: Currently i am studying quaternions.
I do understand that i, j and k are imaginairy numbers. so $i^2 = j^2 =k^2 = -1$. 
But I could not understand this:
$$\begin{matrix}ij=k,&ji=-k,\\jk=i,&kj=-i,\\ki=j,&ik=-j\end{matrix}$$ 
Why is this? There seems no explanation why that is true. I would like to understand why that is true instead of just assuming that that is true.
Could somebody provide me some help?

REPLY [5 votes]: You can derive the properties of quaternions through clifford algebra and the geometric product of vectors.
The geometric product works like so:
$$e_a e_b = \begin{cases} 1, & a = b \\ -e_b e_a, & a \neq b \end{cases}$$
where $a, b$ can be $x$, $y$, or $z$ as usual.  This captures both the work of the cross product and the dot product in one product of basis vectors.
You can then identify
$$\begin{align*}
i &= -e_y e_z \\
j &= -e_z e_x \\
k &= -e_x e_y
\end{align*}$$
And then the properties of quaternions naturally follow.
$$i^2 = (-e_y e_z)(-e_y e_z) = (e_y e_z)(e_y e_z) = -e_y (e_z e_z) e_y = -e_y e_y = -1$$
And similarly for $j^2$ and $k^2$, as well as the $ijk$ product:
$$ijk = (-e_y e_z)(-e_z e_x)(-e_x e_y) = - e_y e_z e_z e_x e_x e_y = -e_y (e_z e_z)(e_x e_x) e_y = -e_y e_y = -1$$
This allows you to interpret quaternions in a very geometric way:  the $i,j,k$ do not represent vectors, but rather oriented planes.  It's just that in 3d each plane has a unique normal vector, so we often abuse this duality.