Product of Quaternion with Conjugate

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 $$\mathbf x$$.

Then their product is given by:
 * $$\mathbf x \overline {\mathbf x} = \left({a^2 + b^2 + c^2 + d^2}\right) \mathbf 1 = \overline {\mathbf x} \mathbf x $$

Proof
From the definition of quaternion multiplication:

$$ $$ $$ $$ $$ $$ $$ $$

As $$a, b, c, d \in \R$$ their products commute.

So the terms in $$\mathbf i, \mathbf j, \mathbf k$$ vanish.

The proof that $$\overline {\mathbf x} \mathbf x = \left({a^2 + b^2 + c^2 + d^2}\right) \mathbf 1$$ is similar.

Hence the result.