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:

We have that $a, b, c, d \in \R$.

So from Real Multiplication is Commutative, 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.