description/proof of that ring of \(n \times n\) quaternion matrices is 'rings - homomorphisms' isomorphic to subring of \(2 n \times 2 n\) complex matrices
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 name% matrices space.
- The reader knows a definition of %category name% isomorphism.
- The reader knows a definition of ring of \(n \times n\) ring matrices.
- 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.
- The reader admits the proposition that for any ring, the multiplications of any matrices over the ring are associative.
- The reader admits the proposition that for any ring, the multiplications of any matrices over the ring are distributive with respect to addition.
- The reader admits the proposition that the multiplication of any matrix made of any same size blocks by any matrix made of blocks of any multiplicable (with blocks of the former matrix) same size is blocks-wise.
- The reader admits the proposition that the range of any ring homomorphism is a subring of the codomain.
- The reader admits the proposition that any bijective ring homomorphism is a 'rings - homomorphisms' isomorphism.
Target Context
- The reader will have a description and a proof of the proposition that the ring of the \(n \times n\) quaternion matrices is 'rings - homomorphisms' isomorphic to a subring of the \(2 n \times 2 n\) complex matrices, via the between-the-quaternions-division-associative-algebra-and-a-subring-of-\(2 \times 2\)-complex-matrices 'rings - homomorphisms' isomorphism.
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
\(f\): \(: \mathbb{H} \to R\), \(= \text{ the 'rings - homomorphisms' isomorphism }\) cited in the proposition that the quaternions division associative algebra is 'rings - homomorphisms' isomorphic to a subring of the \(2 \times 2\) complex matrices ring
\(n\): \(\in \mathbb{N} \setminus \{0\}\)
\(M_n (\mathbb{H})\): \(= \text{ the ring of the } n \times n \text{ quaternions matrices }\) with the canonical addition and multiplication
\(M_{2 n} (\mathbb{C})\): \(= \text{ the ring of the } 2 n \times 2 n \text{ complex matrices }\) with the canonical addition and multiplication
\(f'\): \(: M_n (\mathbb{H}) \to M_{2 n} (\mathbb{C})\) which replaces each component of each element of \(M_n (\mathbb{H})\) with the image of \(f\)
\(f''\): \(: M_n (\mathbb{H}) \to f' (M_n (\mathbb{H}))\) as the codomain restriction of \(f'\)
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Statements:
\(f'' \in \{\text{ the 'rings - homomorphisms' isomorphisms }\}\)
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2: Proof
Whole Strategy: Step 1: see that \(M_n (\mathbb{H})\) and \(M_{2 n} (\mathbb{C})\) are some rings; Step 2: see that \(f'\) is a ring homomorphism; Step 3: see that \(f'\) is an injection; Step 4: conclude the proposition.
Step 1:
\(M_n (\mathbb{H})\) and \(M_{2 n} (\mathbb{C})\) are some rings, by Note for the definition of ring of \(n \times n\) ring matrices.
Step 2:
\(f' (0) = 0\), because \(f (0) = 0\).
\(f' (1) = 1\), because \(f (1) = 1\) and \(f (0) = 0\).
Let \(M_1, M_2 \in M_n (\mathbb{H})\) be any.
\(f' (M_1 + M_2) = f' (M_1) + f' (M_2)\), because for each \((M_1 + M_2)^j_l\), \(f ((M_1 + M_2)^j_l) = f ({M_1}^j_l + {M_2}^j_l) = f ({M_1}^j_l) + f ({M_2}^j_l)\), because \(f\) is a ring homomorphism, which is the \((j, l)\)-th \(2 \times 2\) block of \(f' (M_1 + M_2)\), but the corresponding \(2 \times 2\) block of \(f' (M_1) + f' (M_2)\) is \(f ({M_1}^j_l) + f ({M_2}^j_l)\).
As for \(f' (M_1 M_2)\), for each \((M_1 M_2)^j_l\), \(f ((M_1 M_2)^j_l) = f ({M_1}^j_m {M_2}^m_l) = \sum_{m \in \{1, ..., n\}} f ({M_1}^j_m {M_2}^m_l) = \sum_{m \in \{1, ..., n\}} f ({M_1}^j_m) f ({M_2}^m_l)\), because \(f\) is a ring homomorphism, which is the \((j, l)\)-th \(2 \times 2\) block of \(f' (M_1 M_2)\), but the corresponding \(2 \times 2\) block of \(f' (M_1) f' (M_2)\) is \(\sum_{m \in \{1, ..., n\}} f ({M_1}^j_m) f ({M_2}^m_l)\), by the the proposition that the multiplication of any matrix made of any same size blocks by any matrix made of blocks of any multiplicable (with blocks of the former matrix) same size is blocks-wise, so, \(f' (M_1 M_2) = f' (M_1) f' (M_2)\).
So, \(f'\) is a ring homomorphism.
Step 3:
\(f'\) is an injection, because for each \(M_1 \neq M_2\), \({M_1}^j_l \neq {M_2}^j_l\) for a \((j, l)\), and \(f ({M_1}^j_l) \neq f ({M_2}^j_l)\), because \(f\) is a 'rings - homomorphisms' isomorphism, so, \(f' (M_1) \neq f' (M_2)\), because the corresponding \(2 \times 2\) blocks are different.
Step 4:
\(f' (M_n (\mathbb{H})) \subseteq M_{2 n} (\mathbb{C})\) is a subring, by the proposition that the range of any ring homomorphism is a subring of the codomain.
So, \(f''\) is a map from a ring into a ring.
\(f''\) is a ring homomorphism, because it is just a codomain restriction of ring homomorphic \(f'\).
\(f''\) is a bijection, obviously.
So, \(f''\) is a 'rings - homomorphisms' isomorphism, by the proposition that any bijective ring homomorphism is a 'rings - homomorphisms' isomorphism.
