Field Norm of Quaternion is Positive Definite

Theorem

Let $\mathbf x = a \mathbf 1 + b \mathbf i + c \mathbf j + d \mathbf k$ be a quaternion.

Let $\overline {\mathbf x}$ be the conjugate of $\mathbf x$.

The field norm of $\mathbf x$:

$\map n {\mathbf x} := \cmod {\mathbf x \overline {\mathbf x} }$

Proof

 $\displaystyle \map n {\mathbf x}$ $=$ $\displaystyle 0$ $\displaystyle \leadstoandfrom \ \$ $\displaystyle \cmod {\mathbf x \overline {\mathbf x} }$ $=$ $\displaystyle 0$ Definition of Field Norm of Quaternion $\displaystyle \leadstoandfrom \ \$ $\displaystyle a^2 + b^2 + c^2 + d^2$ $=$ $\displaystyle 0$ $\displaystyle \leadstoandfrom \ \$ $\displaystyle a = 0, b = 0, c = 0, d = 0$  $\displaystyle$ $\displaystyle \leadstoandfrom \ \$ $\displaystyle a \mathbf 1 + b \mathbf i + c \mathbf j + d \mathbf k$ $=$ $\displaystyle \mathbf 0$

Hence $n$ is positive definite by definition.

$\blacksquare$