Product of Quaternion with Conjugate

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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 $\mathbf x$.


Then their product is given by:

$\mathbf x \overline {\mathbf x} = \paren {a^2 + b^2 + c^2 + d^2} \mathbf 1 = \overline {\mathbf x} \mathbf x$


Proof

From the definition of quaternion multiplication:

\(\ds \mathbf x \overline {\mathbf x}\) \(=\) \(\ds \paren {a^2 - b \paren {-b} - c \paren {-c} - d \paren {-d} } \mathbf 1\)
\(\ds \) \(\) \(\, \ds + \, \) \(\ds \paren {a \paren {-b} + b a + c \paren {-d} - d \paren {-c} } \mathbf i\)
\(\ds \) \(\) \(\, \ds + \, \) \(\ds \paren {a \paren {-c} - b \paren {-d} + c a + d \paren {-b} } \mathbf j\)
\(\ds \) \(\) \(\, \ds + \, \) \(\ds \paren {a \paren {-d} + b \paren {-c} - c \paren {-b} + d a} \mathbf k\)
\(\ds \) \(=\) \(\ds \paren {a^2 + b^2 + c^2 + d^2} \mathbf 1\)
\(\ds \) \(\) \(\, \ds + \, \) \(\ds \paren {-a b + b a - c d + d c} \mathbf i\)
\(\ds \) \(\) \(\, \ds + \, \) \(\ds \paren {-a c + b d + c a - d b} \mathbf j\)
\(\ds \) \(\) \(\, \ds + \, \) \(\ds \paren {-a d - b c + c b + d a} \mathbf k\)


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 $\overline {\mathbf x} \mathbf x = \paren {a^2 + b^2 + c^2 + d^2} \mathbf 1$ is similar.


Hence the result.

$\blacksquare$


Also see


Sources