Commutator on Algebra is Alternating Bilinear Mapping
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Theorem
Let $\struct {A_R, \oplus}$ be an algebra over a ring.
Then the commutator on $\struct {A_R, \oplus}$ is an alternating bilinear mapping:
- $\forall a, b \in A_R: \sqbrk {a, b} = -\sqbrk {b, a}$
Proof
| \(\ds \sqbrk {a, b}\) | \(=\) | \(\ds a \oplus b - b \oplus a\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds -b \oplus a + a \oplus b\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds -b \oplus a - \paren {-\paren {a \oplus b} }\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds -\paren {b \oplus a - a \oplus b}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds -\sqbrk {b, a}\) |
$\blacksquare$