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 Post subject: Division Ring
PostPosted: 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 Ring
PostPosted: Sat Jun 25, 2016 11:46 am 
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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 Ring
PostPosted: 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 Ring
PostPosted: 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\,\\
&\quad\quad\quad c_2-b_1\,d_2+c_1\,a_2+d_1\,b_2,d_2\,a_1+b_1\,c_2+d_1\,a_2-c_1\,b_2)
\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\,\\
&\quad\quad\quad\quad\quad c_2-b_1\,d_2+c_1\,a_2+d_1\,b_2,d_2\,a_1+b_1\,c_2+d_1\,a_2-c_1\,b_2\big)
\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 Ring
PostPosted: Sat Jun 25, 2016 11:51 am 
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Joined: Mon Nov 09, 2015 1:52 pm
Posts: 426
Additonal question :

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 Ring
PostPosted: Sat Jun 25, 2016 11:52 am 
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Joined: Mon Nov 09, 2015 1:52 pm
Posts: 426
Additional question

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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