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Here is the question that I want to answer part(c) in it:

Define $E \in GL_{2}(\mathbb{R})$ by $E = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$ and let $\mathcal{R} = \{aI + bE| a,b \in \mathbb{R}\} \subset M_{2}(\mathbb{R}).$

$(a)$ Show that $\mathcal{R} \cong \mathbb{C}$ as rings (so $\mathcal{R}$ is a field). Which matrices correspond to the subgroup $S^{1} \subset \mathbb{C}^{*}$?

$(b)$ Let $\mathbb{H} \subset M_{2}(\mathbb{C})$ be the set of matrices of the form: $$ \begin{pmatrix} z & - \bar{\omega} \\ \omega & \bar{z} \end{pmatrix} \quad \quad z, \omega \in \mathbb{C}$$Show that $\mathbb{H}$ is a division ring. ( $\mathbb{H}$ is called the \textbf{quaternion algebra}).

$(c)$ Find a way to represent $\mathbb{H}$ as a subring of $M_{4}(\mathbb{R}).$\ (Hint: Combine parts $(a)$ and $(b)$)

My question is:

If I am fine with proving $(a)$ and $(b),$ how can I combine parts $(a)$ and $(b)$ to answer $(c)$? I got the following hint **Consider 2 elements $A$ and $B^{-1}$ belong to $H$ if their product $AB^{-1}$ also belong to $\mathbb{H}$ then \mathbb{H} is a subring of $M_{2}(C)$ if field is $\mathbb{R}$ then it would be a subring of $M_{4}(\mathbb{R})$, ** but still I do not know how to show it, could anyone help me please?

  • Consider $z$ and $\omega$ as $2\times2$ real blocks according to (a). – Berci Sep 28 '20 at 21:11
  • See also https://math.stackexchange.com/a/849464/589, which works even if $\mathbb H$ is not commutative. – lhf Sep 28 '20 at 21:11
  • @Berci could you please give me more details ..... I do not understand –  Sep 28 '20 at 21:28
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    Instead of writing elements of $\Bbb C$, use elements of $\mathcal R$ so to obtain a $4\times4$ real matrix. – Berci Sep 28 '20 at 22:47
  • @Berci I will edit my question with the answer and just tell me if it is correct? I understood that we just need to write a $4 \times 4$ matrix to answer this question .... am I correct? –  Sep 29 '20 at 01:45
  • See https://en.wikipedia.org/wiki/Quaternion#Matrix_representations – lhf Sep 29 '20 at 10:37

1 Answers1

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Here is more general approach:

Let $F$ be a field and let $A$ be a finite dimensional algebra over $F$ of dimension $n$.

Then $\mu: A \to End_F(A)$ given by $\mu(a)(x)=ax$ is an injective ring homomorphism.

Choosing a basis for $A$ over $F$ gives a ring isomorphism $\phi:End_F(A) \cong M_n(F)$.

Thus we get an embedding $\phi \circ \mu : A \to M_n(F)$.

Apply this to $F=\mathbb R$, $A=\mathbb H$, and $n=4$. A basis for $\mathbb H$ over $\mathbb R$ is given by $\{1,i,j,k\}$ where $$ 1 = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}, \quad i = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}, \quad j = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}, \quad k = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}, $$

lhf
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  • Adapted from https://math.stackexchange.com/a/849464/589 – lhf Sep 28 '20 at 22:25
  • I am sorry I still do not understand how to solve my question ..... could you please explain that solution? –  Sep 28 '20 at 22:26
  • Is this a theorem or what ? can you give me a name of a book that contains this statement please? how did you know it? –  Sep 28 '20 at 22:35
  • @Smart20, I don't know a reference right now. But it's simple enough to prove. – lhf Sep 29 '20 at 00:04
  • See also https://math.stackexchange.com/questions/2931617/matrix-representation-of-complex-number-is-just-a-trick and https://math.stackexchange.com/questions/2281006/representation-theorem-for-finite-dimensional-algebra-over-fields – lhf Sep 29 '20 at 00:06
  • how the basis is over $\mathbb{R}$ and it contains the complex numbers $i,j,k$ could you clarify this please? –  Sep 29 '20 at 02:45
  • In my problem, what is the $End_{F}(A)$? –  Sep 29 '20 at 02:48