description/proof of that 'rings - homomorphisms' isomorphism from quaternions division associative algebra onto subring of \(2 \times 2\) complex matrices ring maps quaternion conjugate to Hermitian conjugate
Topics
About: ring
The table of contents of this article
Starting Context
- The reader knows a definition of quaternions division associative algebra.
- The reader knows a definition of ring of \(n \times n\) ring matrices.
- The reader knows a definition of Hermitian conjugate of complex matrix.
- The reader admits the proposition that the quaternions division associative algebra is 'rings - homomorphisms' isomorphic to a subring of the \(2 \times 2\) complex matrices ring.
Target Context
- The reader will have a description and a proof of the proposition that the canonical 'rings - homomorphisms' isomorphism from the quaternions division associative algebra onto the subring of the \(2 \times 2\) complex matrices ring maps the quaternion conjugate of any element to the Hermitian conjugate of the image of the element.
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Structured Description
Here is the rules of Structured Description.
Entities:
\(\mathbb{H}\): \(= \text{ the quaternions division associative algebra }\)
\(R\): \(= \{\begin{pmatrix} c_1 & c_2 \\ \overline{- c_2} & \overline{c_1} \end{pmatrix} \vert c_1, c_2 \in \mathbb{C}\}\) with the canonical addition and multiplication
\(f\): \(: \mathbb{H} \to R, a + b i + c j + d k \mapsto \begin{pmatrix} a + b i & c + d i \\ - c + d i & a - b i \end{pmatrix}\), \(= \text{ the 'rings - homomorphisms' isomorphism }\) defined in the quaternions division associative algebra is 'rings - homomorphisms' isomorphic to a subring of the \(2 \times 2\) complex matrices ring
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Statements:
\(\forall h \in \mathbb{H} (f (\overline{h}) = f (h)^*)\)
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2: Proof
Whole Strategy: Step 1: let \(h = a + b i + c j + d k\) and see that \(f (\overline{h}) = f (h)^*\).
Step 1:
Let \(h = a + b i + c j + d k\).
\(f (\overline{h}) = f (a - b i - c j - d k) = \begin{pmatrix} a - b i & - c - d i \\ c - d i & a + b i \end{pmatrix}\).
\(f (h)^* = \begin{pmatrix} a + b i & c + d i \\ - c + d i & a - b i \end{pmatrix}^* = \begin{pmatrix} a - b i & - c - d i \\ c - d i & a + b i \end{pmatrix}\).
So, \(f (\overline{h}) = f (h)^*\).
