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 Post subject: Division RingPosted: Sat Jun 25, 2016 11:45 am
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Joined: Mon Nov 09, 2015 1:52 pm
Posts: 426
Prove that the set $\displaystyle{\mathbb{H}=\left\{\begin{pmatrix} z & w\\ -\bar{w} & \bar{z} \end{pmatrix} : z,w\in\mathbb{C}\right\}\subseteq \mathbb{M}_{2}\,\left(\mathbb{C}\right)}\,$ equipped with the usual operations of addition and multiplication of matrices, is a division ring, known as Tetranion division ring of $\displaystyle{\rm{Hamilton}}$ .

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 Post subject: Re: Division RingPosted: Sat Jun 25, 2016 11:46 am
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Joined: Mon Nov 09, 2015 1:52 pm
Posts: 426
1st part : $\displaystyle{\mathbb{H}=\left\{\begin{pmatrix} \,\,\,\,z && w\\ -\overline{w} && \overline{z} \end{pmatrix} : z\,,w\in\mathbb{C}\right\}\subseteq \mathbb{M_2}\,(\mathbb{C})}$

Obviously, $\displaystyle{\mathbb{H}\neq \varnothing}$ cause

$\displaystyle{\mathbb{O}= \begin{pmatrix} 0 && 0\\ 0 && 0 \end{pmatrix}\in \mathbb{H}\,\,,I_2=\begin{pmatrix} 1 && 0\\ 0 && 1 \end{pmatrix}=1_{\mathbb{M_2}\,(\mathbb{C})}\in\mathbb{H}}$

Consider now

$\displaystyle{A=\begin{pmatrix} \,\,\,\,z_1 && w_1\\ -\overline{w_1} && \overline{z_1} \end{pmatrix}\in\mathbb{H}\,\,,B=\begin{pmatrix} \,\,\,\,z_2 && w_2\\ -\overline{w_2} && \overline{z_2} \end{pmatrix}\in \mathbb{H}}$

where $\displaystyle{z_i\,,w_i\in\mathbb{C}\,,i\in\left\{1,2\right\}}$ .

$\displaystyle{A+B=\begin{pmatrix} z_1+z_2 && w_1+w_2\\ -\overline{w_1}-\overline{w_2} && \overline{z_1}+\overline{z_2} \end{pmatrix}=\begin{pmatrix} \,\,z_1+z_2 && w_1+w_2\\ -(\overline{w_1+w_2}) && \overline{z_1+z_2} \end{pmatrix}\in\mathbb{H}}$

and $\displaystyle{-A=\begin{pmatrix} -z_1 && -w_1\\ -(-\overline{w_1}) && -\overline{z_1} \end{pmatrix}=\begin{pmatrix} -z_1 && -w_1\\ -(\overline{-w_1}) && \overline{-z_1} \end{pmatrix}\in\mathbb{H}}$

Also,

$\displaystyle{A\cdot B=\begin{pmatrix} \,\,z_1\,z_2-w_1\,\overline{w_2} && z_1\,w_2+w_1\,\overline{z_2}\\ -z_2\,\overline{w_1}-\overline{z_1\,w_2} && -w_2\,\overline{w_1}+\overline{z_1\,z_2}\end{pmatrix}}$

where :

$\displaystyle{ -z_2\,\overline{w_1}-\overline{z_1\,w_2}=-(z_2\,\overline{w_1}+\overline{z_1\,w_2})=-\overline{z_1\,w_2+w_1\,z_2}}$

and

$\displaystyle{-w_2\,\overline{w_1}+\overline{z_1\,z_2}=\overline{z_1\,z_2-w_1\,\overline{w_2}}}$

so, $\displaystyle{A\cdot B\in\mathbb{H}}$ .

Therefore, the triplet $\displaystyle{\left(\mathbb{H},+,\cdot\right)}$ is a ring with $\displaystyle{\mathbb{O}= \begin{pmatrix} 0 && 0\\ 0 && 0 \end{pmatrix}}$

as the zero-element and

$\displaystyle{I_2=\begin{pmatrix} 1 && 0\\ 0 && 1 \end{pmatrix}}$

as its unity.

Let $\displaystyle{A=\begin{pmatrix} \,\,\,\,z && w\\ -\overline{w} && \overline{z} \end{pmatrix}\in\mathbb{H}-\left\{\mathbb{O}\right\}}$. We have that :

$\displaystyle{z\neq 0}$ or $\displaystyle{w\neq 0}$ and

$\displaystyle{\det\,(A)=z\,\overline{z}+w\,\overline{w}=\left|z\right|^2+\left|w\right|^2>0}$, so the matrix $\displaystyle{A}$

is invertible. There is $\displaystyle{A^{-1}\in\mathbb{M}_{2}\,(\mathbb{C})}$ such that $\displaystyle{A\cdot A^{-1}=I_{2}=A^{-1}\cdot A}$ .

We''ll prove that $\displaystyle{A^{-1}\in\mathbb{H}}$ and then we get the desired. It's known that

$\displaystyle{A\cdot \mathrm{adj}(A)=\det\,(A)\,I_{2}=\mathrm{adj}(A)\cdot A}$ , so

$\displaystyle{A^{-1}=\dfrac{1}{\det\,(A)}\,\mathrm{adj}(A)}$

where:

$\displaystyle{\mathrm{adj}(A)=\begin{pmatrix} \overline{z} && -w\\ \overline{w} && z \end{pmatrix}}$

so, $\displaystyle{A^{-1}=\begin{pmatrix} \overline{z/a} && -w/a\\ -(\overline{-w/a}) && \overline{\overline{z/a}} \end{pmatrix}\in\mathbb{H}\,\,,a=\det\,(A)}$

Finally, $\displaystyle{\left(\mathbb{H},+,\cdot\right)}$ is a division ring but not a field because

$\displaystyle{\begin{pmatrix} i && 0\\ 0 && -i \end{pmatrix}\in\mathbb{H}\,\,,\begin{pmatrix} 0 && 1\\ -1 && 0 \end{pmatrix}\in\mathbb{H}}$ and

$\displaystyle{\begin{pmatrix} i && 0\\ 0 && -i \end{pmatrix}\cdot \begin{pmatrix} 0 && 1\\ -1 && 0 \end{pmatrix}=\begin{pmatrix} 0 && i\\ i && 0 \end{pmatrix} }$

but

$\displaystyle{ \begin{pmatrix} 0 && 1\\ -1 && 0 \end{pmatrix}\cdot \begin{pmatrix} i && 0\\ 0 && -i \end{pmatrix}= \begin{pmatrix} 0 && -i\\ -i && 0 \end{pmatrix}\neq \begin{pmatrix} 0 && i\\ i && 0 \end{pmatrix}}$

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 Post subject: Re: Division RingPosted: Sat Jun 25, 2016 11:48 am
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Joined: Tue Nov 10, 2015 8:25 pm
Posts: 313
After requesting permission from Vaggelis, i am adding a similar question to the above (for those interested), instead of opening a new topic. Here it is:

Show that the set
$\displaystyle S = \left\{ m + ni\sqrt{3} \in \mathbb{C} | \text{ either both m,n } \in \mathbb{Z} \text{ or both } m,n \in \mathbb{Q} \text{ such that } 2m,2m \text{ are odd integers } \right\}$
is a (commutative) ring ( where by ring we mean an associative ring with multiplicative identity ). Find, if possible, $\displaystyle U(R)$, that is the group of invertible elements in $\displaystyle R$.

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 Post subject: Re: Division RingPosted: Sat Jun 25, 2016 11:49 am
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Joined: Mon Nov 09, 2015 1:52 pm
Posts: 426
Some thoughts :

We define $\displaystyle{f:\mathbb{H}\longrightarrow \mathbb{R}^{4}\,\,,f\,\left(\begin{pmatrix} \,\,\,\,z && w\\ -\overline{w} && \overline{z} \end{pmatrix}\right)=\left(Re(z),Im(z),Re(w),Im(w)\right)}$

The function $\displaystyle{f}$ is well defined . Let

$\displaystyle{A=\begin{pmatrix} \,\,\,\,z_1 && w_1\\ -\overline{w_1} && \overline{z_1} \end{pmatrix}\,\,,B=\begin{pmatrix} \,\,\,\,z_2 && w_2\\ -\overline{w_2} && \overline{z_2} \end{pmatrix}\in\mathbb{H}}$ .

Then,

\displaystyle{\begin{aligned}f\,(A+B)&=\left(Re(z_1+z_2),Im(z_1+z_2),Re(w_1+w_2),Im(w_1+w_2)\right)\\&=\left(Re(z_1)+Re(z_2),Im(z_1)+Im(z_2),Re(w_1)+Re(w_2),Im(w_1)+Im(w_2)\right)\\&=\left(Re(z_1),Im(z_1),Re(z_2),Im(z_2)\right)+\left(Re(z_2),Im(z_2),Re(w_2),Im(w_2)\right)\\&=f\,(A)+f\,(B)\end{aligned}}

Now, let $\displaystyle{A=\begin{pmatrix} \,\,\,\,z_1 && w_1\\ -\overline{w_1} && \overline{z_1} \end{pmatrix}\,\,,B=\begin{pmatrix} \,\,\,\,z_2 && w_2\\ -\overline{w_2} && \overline{z_2} \end{pmatrix}\in\mathbb{H}}$

with $\displaystyle{f\,(A)=f\,(B)}$. Then,

$\displaystyle{\left(Re(z_1),Im(z_1),Re(w_1),Im(w_1)\right)=\left(Re(z_2),Im(z_2),Re(w_2),Im(w_2)\right)}$,

so :

$\displaystyle{Re(z_1)=Re(z_2)\,\land Im(z_1)=Im(z_2)\,\land Re(w_1)=Re(w_2)\,\land Im(w_1)=Im(w_2)}$

and thus :

$\displaystyle{z_1=z_2\,\land w_1=w_2\implies A=B}$ , which means that $\displaystyle{f}$ is $\displaystyle{1-1}$ at $\displaystyle{\mathbb{H}}$

Consider $\displaystyle{\left(a,b,x,y\right)\in\mathbb{R}^4}$.

Setting $\displaystyle{z=a+i\,b\,,w=x+i\,y\in\mathbb{C}}$, we have that

$\displaystyle{\begin{pmatrix} \,\,\,\,z && w\\ -\overline{w} && \overline{z} \end{pmatrix}\in\mathbb{H}}$ and

$\displaystyle{f\,\left(\begin{pmatrix} \,\,\,\,z && w\\ -\overline{w} && \overline{z} \end{pmatrix}\right)=\left(Re(z),Im(z),Re(w),Im(w)\right)=\left(a,b,x,y\right)}$

which means that $\displaystyle{f}$ is onto.

Therefore, $\displaystyle{\left(\mathbb{H},+\right)\simeq \left(\mathbb{R}^4,+\right)}$ .

Also,

$\displaystyle{f\,(I_2)=f\,(1_{\mathbb{M_{2}}\,(\mathbb{C}}))=\left(Re(1),Im(1),Re(0),Im(0)\right)=\left(1,0,0,0\right)}$

If $\displaystyle{A=\begin{pmatrix} \,\,\,\,z_1 && w_1\\ -\overline{w_1} && \overline{z_1} \end{pmatrix}\,\,,B=\begin{pmatrix} \,\,\,\,z_2 && w_2\\ -\overline{w_2} && \overline{z_2} \end{pmatrix}\in\mathbb{H}}$ ,

then :

$\displaystyle{A\,B=A\cdot B=\begin{pmatrix} \,\,z_1\,z_2-w_1\,\overline{w_2} && z_1\,w_2+w_1\,\overline{z_2}\\ -z_2\,\overline{w_1}-\overline{z_1\,w_2} && -w_2\,\overline{w_1}+\overline{z_1\,z_2}\end{pmatrix}}$

and if $\displaystyle{z_1=a_1+i\,b_1\,,w_1=c_1+i\,d_1\,,z_2=a_2+i\,b_2\,,w_2=c_2+i\,d_2}$, then :

\displaystyle{\begin{aligned}z_1\,z_2-w_1\,\overline{w_2}&=\left(a_1+i\,b_1 \right )\,\left(a_2+i\,b_2 \right )-\left(c_1+i\,d_1 \right )\,\left(c_2-i\,d_2 \right )\\&=\left[(a_1\,a_2-b_1\,b_2)+i\,(a_2\,b_1+a_1\,b_2) \right ]-\left[(c_1\,c_2+d_1\,d_2)+i\,(d_1\,c_2-c_1\,d_2) \right ]\\&=\left(a_1\,a_2-b_1\,b_2-c_1\,c_2-d_1\,d_2 \right )+i\,(a_2\,b_1+a_1\,b_2-d_1\,c_2+c_1\,d_2) \end{aligned}}

and

\displaystyle{\begin{aligned}z_1\,w_2+w_1\,\overline{z_2}&=\left(a_1+i\,b_1 \right )\,\left(c_2+i\,d_2 \right )+\left(c_1+i\,d_1 \right )\,\left(a_2-i\,b_2 \right )\\&=\left[(a_1\,c_2-b_1\,d_2)+i\,(d_2\,a_1+b_1\,c_2) \right ]+\left[(c_1\,a_2+d_1\,b_2)+i\,(d_1\,a_2-c_1\,b_2) \right ]\\&=\left(a_1\,c_2-b_1\,d_2+c_1\,a_2+d_1\,b_2 \right )+i\,(d_2\,a_1+b_1\,c_2+d_1\,a_2-c_1\,b_2) \end{aligned}}

Therefore,
\begin{align*}
A\,B\stackrel{f}{\mapsto} &(a_1\,a_2-b_1\,b_2-c_1\,c_2-d_1\,d_2,a_2\,b_1+a_1\,b_2-d_1\,c_2+c_1\,d_2,a_1\,\\
\end{align*}
If we define multiplication on $\displaystyle{\mathbb{R}^4}$ such that :

$\displaystyle{a=\left(a_1,b_1,c_1,d_1\right)\,,b=\left(a_2,b_2,c_2,d_2\right)\in\mathbb{R}^4}$, then :
\begin{align*}
a\cdot b&=\big(a_1\,a_2-b_1\,b_2-c_1\,c_2-d_1\,d_2,a_2\,b_1+a_1\,b_2-d_1\,c_2+c_1\,d_2,a_1\,\\
\end{align*}
we have that (easy but painful) the triplet $\displaystyle{\left(\mathbb{R}^4,+,\cdot\right)}$ is a ring with $\displaystyle{1=\left(1,0,0,0\right)}$ as its unity,

($\displaystyle{\left(1,0,0,0\right)=f\,(I_2)=f\,(1_{\mathbb{M_{2}}\,(\mathbb{C}))}}$ ) ,

isomorphic to $\displaystyle{\left(\mathbb{H},+,\cdot\right)}$

because the function $\displaystyle{f}$ maintains the operations of addition and multiplication ( painful) and additionally, is one to one and onto.

Therefore, $\displaystyle{\left(\mathbb{R}^4,+,\cdot\right)}$ is a division ring but not a field.

Let $\displaystyle{\left(a,b,c,d\right)\in\mathbb{R}^4-\left\{0\right\}}$ .

Then, $\displaystyle{\left(a,b,c,d\right)=f\,\left(\begin{pmatrix} a+i\,b && c+i\,d\\ -c+i\,d && a-i\,b \end{pmatrix}\right)}$ .

$\displaystyle{\left(\begin{pmatrix} a+i\,b && c+i\,d\\ -c+i\,d && a-i\,b \end{pmatrix}\right)^{-1}=\begin{pmatrix} \dfrac{a-i\,b}{a^2+b^2+c^2+d^2} && -\dfrac{c+i\,d}{a^2+b^2+c^2+d^2}\\ \dfrac{c-i\,d}{a^2+b^2+c^2+d^2} && \dfrac{a-i\,b}{a^2+b^2+c^2+d^2} \end{pmatrix}}$

so, $\displaystyle{\left(a,b,c,d\right)^{-1}=\dfrac{1}{a^2+b^2+c^2+d^2}\,\left(a,-b,-c,-d\right)}$ .

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 Post subject: Re: Division RingPosted: Sat Jun 25, 2016 11:51 am
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Joined: Mon Nov 09, 2015 1:52 pm
Posts: 426

Find, the subset $\displaystyle{Z\,(\mathbb{H})=\left\{r\in\mathbb{H}: r\cdot x=x\cdot r\,,\forall\,x\in\mathbb{H}\right\}}$,

that is the center of $\displaystyle{\left(\mathbb{H},+,\cdot\right)}$ and prove that it is a field.

Also, prove that $\displaystyle{\mathbb{H}}$ is a $\displaystyle{Z\,(\mathbb{H})}$ - vector space and find its dimension.

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 Post subject: Re: Division RingPosted: Sat Jun 25, 2016 11:52 am
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Joined: Mon Nov 09, 2015 1:52 pm
Posts: 426

Let $\displaystyle{x=\left(x_0,x_1,x_2,x_3,x_4\right)\in Z\,(\mathbb{R}^4)}$. Then,

$\displaystyle{x\cdot \left(0,1,0,0\right)=\left(0,1,0,1\right)\cdot x}$ and we have that :

$\displaystyle{\left(-x_1,x_0,x_3,-x_2\right)=\left(-x_1,x_0,-x_3,x_2\right)\implies x_3=x_2=0}$

so : $\displaystyle{x=\left(x_0,x_1,0,0\right)}$ . Also,

$\displaystyle{x\cdot \left(0,0,1,0\right)=\left(0,0,1,0\right)\cdot x}$ ad we get:

$\displaystyle{\left(0,0,x_0,x_1\right)=\left(0,0,x_0,-x_1\right)\implies x_0\in\mathbb{R}\,\land x_1=0}$ .

So, $\displaystyle{x=\left(x_0,0,0,0\right)}$. On the other hand, if $\displaystyle{\left(x,0,0,0\right)\in\mathbb{R}^4}$, then :

for each $\displaystyle{\left(a,b,c,d\right)\in\mathbb{R}^4}$ holds :

$\displaystyle{\left(x,0,0,0\right)\cdot \left(a,b,c,d\right)=\left(a,b,c,d\right)\cdot \left(x,0,0,0\right)=\left(x\,a,0,0,0\right)}$ .

Therefore,

$\displaystyle{Z\,(\mathbb{R}^4)=Z\,(\mathbb{H})=\left\{\left(x,0,0,0\right)\in\mathbb{R}^4: x\in\mathbb{R}\right\}\simeq \left\{\begin{pmatrix} z&0 \\ 0&z \end{pmatrix} : z\in\mathbb{R}\right\}}$

We define $\displaystyle{g:\mathbb{R}\longrightarrow Z\,(\mathbb{R}^4)}$ by $\displaystyle{g(x)=\left(x,0,0,0\right)}$

and we see that $\displaystyle{g}$ is an isomorphism

and thus $\displaystyle{\left(Z\,(\mathbb{R}^4),+,\cdot\right)\simeq \left(\mathbb{R},+,\cdot\right)}$ and it is a field.

Besides, since $\displaystyle{\left(\mathbb{R}^4,+,\cdot\right)}$ is a division ring, we have that $\displaystyle{Z\,(\mathbb{R}^4)}$

is a field. Check here: Vector space over the center of a division ring

The commutative additive group $\displaystyle{\left(\mathbb{R}^4,+\right)}$

equuiped with the scalar multiplication $\displaystyle{\star:Z\,(\mathbb{R}^4)\times \mathbb{R}^4\longrightarrow \mathbb{R}^4}$

$\displaystyle{\left(\left(x,0,0,0\right),\left(a,b,c,d\right)\right)\mapsto \left(x,0,0,0\right)\star \left(a,b,c,d\right)\mapsto \left(x\,a,x\,b,x,c,x,d\right)}$

is a left(right) $\displaystyle{S=Z\,(\mathbb{R}^4)}$ -module, or else a $\displaystyle{S=Z\,(\mathbb{R}^4)}$ - vector space, because :

if

$\displaystyle{\left(x,0,0,0\right)\,,\left(y,0,0,0\right)\in S\,\,,\left(a,b,c,d\right)\,,\left(e,f,g,h\right)\in\mathbb{R}^4}$, then :

\displaystyle{\begin{aligned} \left[(x,0,0,0)+(y,0,0,0)\right]\star (a,b,c,d)&=(x+y,0,0,0)\star(a,b,c,d)\\&=((x+y)\,a,(x+y)\,b,(x+y)\,c,(x+y)\,d)\\&=(x\,a,x\,b,x\,c,x\,d)+(y\,a,y\,b,y\,c,y\,d)\\&=(x,0,0,0)\star (a,b,c,d)+(y,0,0,0)\star (a,b,c,d)\end{aligned}}

\displaystyle{\begin{aligned} (x,0,0,0)\star \left[(a,b,c,d)+(e,f,g,h)\right]&=(x,0,0,0)\star (a+e,b+f,c+g,d+h)\\&=(x(a+e),x(b+f),x(c+g),x(d+h))\\&=(x\,a,x\,b,x\,c,x\,d)+(x\,e,x\,f,x\,g,x\,h)\\&=(x,0,0,0)\star (a,b,c,d)+(x,0,0,0)\star (e,f,g,h)\end{aligned}}

\displaystyle{\begin{aligned} \left[(x,0,0,0)\cdot (y,0,0,0)\right]\star (a,b,c,d)&=(xy,0,0,0)\star (a,b,c,d)\\&=((x\,y)\,a,(x\,y)\,b,(x\,y)\,c,(x\,y)\,d)\\&=(x,0,0,0)\star (y\,a,y\,b,y\,c,y\,d)\\&=(x,0,0,0)\star\left[(y,0,0,0)\star (a,b,c,d)\right]\end{aligned}}

$\displaystyle{1_{S}\star (a,b,c,d)=(1,0,0,0)\star (a,b,c,d)=(a,b,c,d)}$ .

Let now $\displaystyle{x=\left(x_0,x_1,x_2,x_3\right)\in\mathbb{R}^4}$. Then,

$\displaystyle{x=(x_0,0,0,0)+(0,x_1,0,0)+(0,0,x_2,0)+(0,0,0,x_3)=\sum_{i=1}^{4}(x_i,0,0,0)\star e_{i}}$, where :

$\displaystyle{e_1=(1,0,0,0)\,,e_2=(0,1,0,0)\,,e_3=(0,0,1,0)\,,e_4=(0,0,0,1)}$ .

Futhermore, if $\displaystyle{\left(a_{i},0,0,0\right)\in S\,,1\leq i\leq 4}$ such that $\displaystyle{\sum_{i=1}^{4}a_{i}\star e_{i}=\left(0,0,0,0\right)}$, then :

$\displaystyle{(a_1,0,0,0)\star (1,0,0,0)+(a_2,0,0,0)\star (0,1,0,0)+(a_3,0,0,0)\star (0,0,1,0)+(a_4,0,0,0)\star (0,0,0,1)=\left(0,0,0,0\right)}$

$\displaystyle{\implies (a_1,0,0,0)+(0,a_2,0,0)+(0,0,a_3,0)+(0,0,0,a_4)=\left(0,0,0,0\right)}$

$\displaystyle{\implies \left(a_1,a_2,a_3,a_4\right)=\left(0,0,0,0\right)}$

$\displaystyle{\implies \forall\,i\in\left\{1,2,3,4\right\}: a_{i}=0}$

$\displaystyle{\implies \forall\,i\in\left\{1,2,3,4\right\}: \left(a_{i},0,0,0\right)=\left(0,0,0,0\right)=0_{S}}$

In conclusion, the vectors $\displaystyle{e_{i}\,,1\leq i\leq 4}$ are linear-indepedent and thus the set

$\displaystyle{A=\left\{e_{i}\in\mathbb{R}^4: 1\leq i\leq 4\right\}}$ is a basis of $\displaystyle{\left(\mathbb{R}^4,+,\star\right)}$

and $\displaystyle{\dim_{S}\,\mathbb{R}^4=4\neq 2}$ .

The basis in $\displaystyle{\mathbb{H}=\left\{\begin{pmatrix} z&w \\ -\overline{w}&\overline{z} \end{pmatrix}: z\,,w\in\mathbb{C}\right\}}$

is (according to the isomorphism $\displaystyle{f}$ above) the set

$\displaystyle{A'=\left\{I_{2},\begin{pmatrix} i&0 \\ 0&-i \end{pmatrix}\,,\begin{pmatrix} 0&1 \\ -1 & 0 \end{pmatrix}\,,\begin{pmatrix} 0&i \\ i& 0 \end{pmatrix}\right\}}$

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