Sufficient Condition for Quaternion Multiplication to Commute
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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 $\paren 2$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \mathbf y \times \mathbf x\) |
$\blacksquare$