Field Norm of Quaternion is Positive Definite

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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 $\mathbf x$:

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

is positive definite.


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.

$\blacksquare$


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