The quaternionic multiplication table is almost, but not quite, the vector cross product. The only difference is that imaginary quaternions square to a negative number, whereas the cross product of a vector with itself is zero.

This is not a coincidence. Making the obvious identification of vectors $\vv$, $\ww$ with imaginary quaternions $v$, $w$, namely \begin{align} \vv = v_x \ii + v_y \jj + v_z \kk \longleftrightarrow v = v_x i + v_y j + v_x k \end{align} (and similarly for $\ww$), then the imaginary part of the quaternionic product $vw$ is the cross product $\vv\times\ww$, that is \begin{equation} \vv\times\ww \longleftrightarrow \Im(vw) \label{qdot} \end{equation} while the real part is just (minus) the dot product $\vv\cdot\ww$, that is \begin{equation} -\vv\cdot\ww = \Re(vw) \label{qcross} \end{equation} so that the quaternionic product can be thought of as a combination of the dot and cross products! In fact, the use of $\,\ii$, $\jj$, $\kk$ for Cartesian basis vectors originates with the quaternions, which predate the use of vectors [Crowe] .

We define the commutator of two quaternions $p$ and $q$ by \begin{equation} [p,q] = pq - qp \label{commutator} \end{equation} which quantifies the lack of commutativity of the quaternions. For example, we have $[i,j]=2k$. However, the quaternions are associative; it is sufficient to check that \begin{equation} (ij)k = -1 = i(jk) \end{equation} As always, we have distributivity of multiplication over addition.

We define quaternionic conjugation to be the (real) linear map which reverses the sign of each imaginary unit. Thus, the (quaternionic) conjugate $\bar{q}$ of a quaternion $q$ is \begin{equation} \bar{q} = q_1 - q_2 i - q_3 j - q_4 k \end{equation} if $q=q_1+q_2i+q_3j+q_4k$. This leads directly to the norm of a quaternion $|q|$, defined by \begin{equation} |q|^2 = q\bar{q} = q_1^2 + q_2^2 + q_3^2 + q_4^2 \label{qnorm} \end{equation} Again, the only quaternion with norm $0$ is $0$, and every nonzero quaternion has a unique inverse, namely \begin{equation} q^{-1} = {\bar{q} \over  |q|^2} \end{equation}

Quaternionic conjugation satisfies the identity \begin{equation} \bar{pq} = \bar{q}\>\bar{p} \end{equation} from which it follows that the norm satisfies: \begin{equation} |pq| = |p| |q| \end{equation} Squaring both sides and expanding the result in terms of components yields the 4-squares rule, \begin{align} &(p_1q_1-p_2q_2-p_3q_3-p_4q_4)^2 + (p_2q_1+p_1q_2-p_4q_3+p_3q_4)^2 \nonumber\\ &\qquad + (p_3q_1+p_4q_2+p_1q_3-p_2q_4)^2 + (p_4q_1-p_3q_2+p_2q_3+p_1q_4)^2 \nonumber\\ &\qquad\qquad = (p_1^2+p_2^2+p_3^2+p_4^2)(q_1^2+q_2^2+q_3^2+q_4^2) \end{align} which is not quite as obvious as the 2-squares rule. This identity means that the quaternions form a division algebra, that is, not only are there inverses, but there are no zero divisors — if a product is zero, one of the factors must be zero.

Since quaternions are invertible, linear equations such as (8) of §2.3, where now $a,b,c,z\in\HH$, can still be solved for $z$ so long as $a\ne0$. But these are no longer the only linear equations! Consider for instance the equation \begin{equation} d = a x + x b \end{equation} where now $a,b,d\in\CC$ and $x\in\HH$. One solution is clearly \begin{equation} x_0 = {d\over a+b} \label{csol} \end{equation} provided $a+b\ne0$. Note that $x_0$ is complex. Are there any other solutions? Consider the special case $d=0$, $a=i=b$. Then any linear combination of $j$ and $k$ solves the equation!

This turns out to be the generic situation: If $a+\bar{b}\ne0$, then the only solution is the complex solution (\ref{csol}), but if $a+\bar{b}=0$ there are additional quaternionic “homogeneous” solutions, which can be added to the particular solution $x_0$, which is therefore not unique. The situation rapidly becomes more complicated if some or all of $a$, $b$, $d$ are themselves allowed to be quaternionic, rather than complex.