Quaternions form Algebra

From ProofWiki
Jump to navigation Jump to search

Theorem

The set of quaternions $\Bbb H$ forms an algebra over the field of real numbers.

This algebra is:

$(1): \quad$ An associative algebra, but not a commutative algebra.
$(2): \quad$ A normed division algebra.
$(3): \quad$ A nicely normed $*$-algebra.


Proof

The quaternions $\Bbb H$ are formed by the Cayley-Dickson Construction from the complex numbers $\C$.

From Complex Numbers form Algebra, we have that $\C$ forms:

$(1): \quad$ An associative algebra
$(2): \quad$ A commutative algebra
$(3): \quad$ A normed division algebra
$(4): \quad$ A nicely normed $*$-algebra.

From Cayley-Dickson Construction forms Star-Algebra, $\Bbb H$ is a $*$-algebra.

From Cayley-Dickson Construction from Nicely Normed Algebra is Nicely Normed, $\Bbb H$ is a nicely normed $*$-algebra.

From Cayley-Dickson Construction from Commutative Associative Algebra is Associative, $\Bbb H$ is an associative algebra.


Now suppose $\Bbb H$ formed a commutative algebra.

Then from Cayley-Dickson Construction from Real Algebra is Commutative, that would mean $\C$ is a real algebra.

But from Complex Numbers form Algebra it is explicitly demonstrated that $\C$ is not a real algebra.

So $\Bbb H$ can not be a commutative algebra.


Proof of Normed Division Algebra

Consider the element $\left({1, 0}\right)$ of $\C^2$.

We have:

\(\displaystyle \) \(\) \(\displaystyle \left({x_1, x_2}\right) \times \left({1, 0}\right)\)
\(\displaystyle \) \(=\) \(\displaystyle \left({x_1 \times 1 - 0 \times x_2, x_1 \times 0 + x_2 \times 1}\right)\)
\(\displaystyle \) \(=\) \(\displaystyle \left({x_1, x_2}\right)\)

As $\times$ is commutative on $\C$, it follows that $\left({1, 0}\right) \times \left({x_1, x_2}\right) = \left({x_1, x_2}\right)$.

So $\left({1, 0}\right) \in \C^2$ functions as a unit.

That is, $\left({\C^2, \times}\right)$ is a unitary algebra.


We define a norm on $\left({\C^2, \times}\right)$ by:

$\forall \mathbf a = \left({a_1, a_2}\right) \in \C^2: \left \Vert {\mathbf a} \right \Vert = \sqrt {a_1^2 + a_2^2}$

This is a norm because:

$(1): \quad \left \Vert \mathbf x \right \Vert = 0 \iff \mathbf x = \mathbf 0$
$(2): \quad \left \Vert \lambda \mathbf x \right \Vert = \left \vert \lambda \right \vert \left \Vert x \right \Vert$
$(3): \quad \left \Vert x - y \right \Vert \le \left \Vert x - z \right \Vert + \left \Vert z - y \right \Vert$

It also follows that:

$\left \Vert x \times y \right \Vert = \left \vert x \times y \right \vert = \left \vert x \right \vert \times \left \vert y \right \vert = \left \Vert x \right \Vert \times \left \Vert y \right \Vert$

and so $\left({\C^2, \times}\right)$ is a normed division algebra.

$\blacksquare$


Sources