Sufficient Condition for Quaternion Multiplication to Commute

Theorem

In general, quaternion multiplication does not commute.

But, for $\mathbf x,\mathbf y \in \H$, $\mathbf x \times \mathbf y = \mathbf y \times \mathbf x$ if any one of the following conditions hold:

\(\text {(1a)}: \quad\)

\(\ds \mathbf x, \mathbf y\)

\(\in\)

\(\ds \set {a \mathbf 1 + b \mathbf i + 0 \mathbf j + 0 \mathbf k: a, b \in \R}\)

\(\text {(1b)}: \quad\)

\(\ds \mathbf x, \mathbf y\)

\(\in\)

\(\ds \set {a \mathbf 1 + 0 \mathbf i + c \mathbf j + 0 \mathbf k: a, c \in \R}\)

\(\text {(1c)}: \quad\)

\(\ds \mathbf x, \mathbf y\)

\(\in\)

\(\ds \set {a \mathbf 1 + 0 \mathbf i + 0 \mathbf j + d \mathbf k: a, d \in \R}\)

\(\text {(2a)}: \quad\)

\(\ds \mathbf x\)

\(\in\)

\(\ds \set {a \mathbf 1 + 0 \mathbf i + 0 \mathbf j + 0 \mathbf k: a \in \R}\)

\(\text {(2b)}: \quad\)

\(\ds \mathbf y\)

\(\in\)

\(\ds \set {a \mathbf 1 + 0 \mathbf i + 0 \mathbf j + 0 \mathbf k: a \in \R}\)

\(\text {(3)}: \quad\)

\(\ds \mathbf x\)

\(=\)

\(\ds \paren {a \mathbf 1 + 0 \mathbf i + 0 \mathbf j + 0 \mathbf k} \times \overline {\mathbf y}: a \in \R\)

Proof

Proof of $\paren 1$

It follows directly from Complex Numbers form Subfield of Quaternions and Complex Multiplication is Commutative.

$\Box$

Proof of $\paren 2$

Let $\mathbf x \in \set {a \mathbf 1 + 0 \mathbf i + 0 \mathbf j + 0 \mathbf k: a\in \R}$.

Let $\mathbf y = e \mathbf 1 + f \mathbf i + g \mathbf j + h \mathbf k: e, f, g, h \in \R$.

Then:

\(\ds \mathbf x \times \mathbf y\)

\(=\)

\(\ds a e \mathbf 1 + a f \mathbf i + a g \mathbf j + a h \mathbf k\)

Definition of Quaternion Multiplication; the other terms are $0$

\(\ds \)

\(=\)

\(\ds \mathbf y \times \mathbf x\)

Definition of Quaternion Multiplication

The above is the proof of $\paren {2a}$, and the proof of $\paren {2b}$ is similar.

$\Box$

Proof of $\paren 3$

\(\ds \mathbf x \times \mathbf y\)

\(=\)

\(\ds \paren {a \mathbf 1 + 0 \mathbf i + 0 \mathbf j + 0 \mathbf k} \times \overline {\mathbf y} \times \mathbf y\)

\(\ds \)

\(=\)

\(\ds \paren {a \mathbf 1 + 0 \mathbf i + 0 \mathbf j + 0 \mathbf k} \times \overline {\mathbf y} \times \overline {\paren {\overline {\mathbf y} } }\)

Quaternion Conjugation is Involution

\(\ds \)

\(=\)

\(\ds \paren {a \mathbf 1 + 0 \mathbf i + 0 \mathbf j + 0 \mathbf k} \times \cmod {\overline{\mathbf y} }^2 \mathbf 1\)

Quaternion Modulus in Terms of Conjugate

\(\ds \)

\(=\)

\(\ds \paren {a \mathbf 1 + 0 \mathbf i + 0 \mathbf j + 0 \mathbf k} \times \cmod {\mathbf y}^2 \mathbf 1\)

Quaternion Modulus of Conjugate

\(\ds \)

\(=\)

\(\ds \paren {a \mathbf 1 + 0 \mathbf i + 0 \mathbf j + 0 \mathbf k} \times \mathbf y \times \overline {\mathbf y}\)

Quaternion Modulus in Terms of Conjugate

\(\ds \)

\(=\)

\(\ds \mathbf y \times \paren {a \mathbf 1 + 0 \mathbf i + 0 \mathbf j + 0 \mathbf k} \times \overline {\mathbf y}\)

from $\paren 2$

\(\ds \)

\(=\)

\(\ds \mathbf y \times \mathbf x\)

$\blacksquare$