quaternion inverse

The easiest way to get the inverse of a full quaternion $a {\bf 1}+ {\bf v}$ is to go back to its representation by elementary matrices, in the $2x2$ form

$$\begin{eqnarray*} a {\bf 1}+ b {\bf i}+ c {\bf j}+ d {\bf k}& \rightarrow & \... ...}{cc} a + d & b + c \\ - b + c & a - d \end{array} \right]. \end{eqnarray*}$$

The determinant, $(a+d)(a-d)-(c+b)(c-b) = a^2-d^2+b^2-c^2$, which is $a^2-({\bf v},{\bf v})$, could be considered to be a candidate for the square of the norm of a full quaternion, in contrast to the norm of a mere vector. Under that assumption, the formula for the inverse of a $2x2$ matrix gives the inverse quaternion

$$\begin{eqnarray*} \frac{1}{(a^2-\vert{\bf u}\vert^2)} \left[ \begin{array}{cc} ... ...t{\bf u}\vert^2)} (a {\bf 1}- b {\bf i}- c {\bf j}- d {\bf k}). \end{eqnarray*}$$

which is almost the formula for inverting Hamilton's authentic quaternions, except for the way the norm is calculated.

Nevertheless, the difference is important, because null quaternions exist, just as well as null vectors, and they cannot be invertible. Hamiton's quaternions are invertible unless zero. Not only do they constitute a field, albeit noncommutative; they lie amongst the very few examples of algebraically and topologically complete infinite fields. The objects which we have just defined are not quaternions according to Hamilton's definition; neither are Hamiton's own quaternions taken with complex coefficients (look at $({\bf i}- i {\bf j})^2$, which vanishes).

At least we have a quantity which decides the invertibility of a quaternion such as $a {\bf 1}+ {\bf v}$, multiplicative for being a determinant, and consistent with the definition of the vector norm relative to a sign choice.

$$\begin{eqnarray*} \Vert a{\bf 1}+{\bf v}\Vert^2 & = & a^2 - ({\bf v},{\bf v}) \end{eqnarray*}$$