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About subalgebra of Hamilton

Matrix representation of $mathbbC$Subfields of Matrix Ringscomposition series of moduleHow to prove the tensor product of two copies of $mathbbH$ is isomorphic to $M_4 (mathbbR)$?Calculate C*-subalgebra generated by $A$, $A^*$ and $mathbb 1$Logic behind cubic resolution of $x^4+px^2+qx+r=0$Finding a composition series for the Frobenius group $F_20$Compute the matrix with the standard basis.Show that $D^n$ is simple and uniqueIs this method to solve a system of equations in modular arithmetic correct?

0

$begingroup$

could someone help me with this question?

I have to prove that $$H=leftleft(beginarraylcr a&-barb\b&baraendarrayright):;a,binmathbbCright$$
is a real subalgebra of $M_2(mathbbC)$.

My attempt

I am trying to use the following facts, but I don't know if they can help:

• $H$ is an algebra isomorphic to $leftleft(beginarraylcrc_1&-c_2&-c_3&-c_4\c_2&c_1&-c_4&c_3\c_3&c_4&c_1&-c_2\c_4&-c_3&c_2&c_1endarrayright):;c_1,c_2,c_3,c_4inmathbbRright$. This is a real subalgebra of $M_4(mathbbC)$




• $mathbbC$ is an algebra isomorphic to $leftleft(beginarraylcrp&-q\q&pendarrayright):;p,qinmathbbRright$. This a subalgebra of $M_2(mathbbR)$

Does someone know if it possible prove it by using this?
Is there a better (shorter) way?

Thanks.

abstract-algebra

share|cite|improve this question

edited yesterday

J. W. Tanner

3,0121320

asked yesterday

prosepprosep

1

New contributor

prosep is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

add a comment | 

0

$begingroup$

could someone help me with this question?

I have to prove that $$H=leftleft(beginarraylcr a&-barb\b&baraendarrayright):;a,binmathbbCright$$
is a real subalgebra of $M_2(mathbbC)$.

My attempt

I am trying to use the following facts, but I don't know if they can help:

• $H$ is an algebra isomorphic to $leftleft(beginarraylcrc_1&-c_2&-c_3&-c_4\c_2&c_1&-c_4&c_3\c_3&c_4&c_1&-c_2\c_4&-c_3&c_2&c_1endarrayright):;c_1,c_2,c_3,c_4inmathbbRright$. This is a real subalgebra of $M_4(mathbbC)$




• $mathbbC$ is an algebra isomorphic to $leftleft(beginarraylcrp&-q\q&pendarrayright):;p,qinmathbbRright$. This a subalgebra of $M_2(mathbbR)$

Does someone know if it possible prove it by using this?
Is there a better (shorter) way?

Thanks.

abstract-algebra

share|cite|improve this question

edited yesterday

J. W. Tanner

3,0121320

asked yesterday

prosepprosep

1

New contributor

prosep is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

add a comment | 

$begingroup$

could someone help me with this question?

I have to prove that $$H=leftleft(beginarraylcr a&-barb\b&baraendarrayright):;a,binmathbbCright$$
is a real subalgebra of $M_2(mathbbC)$.

My attempt

I am trying to use the following facts, but I don't know if they can help:

• $H$ is an algebra isomorphic to $leftleft(beginarraylcrc_1&-c_2&-c_3&-c_4\c_2&c_1&-c_4&c_3\c_3&c_4&c_1&-c_2\c_4&-c_3&c_2&c_1endarrayright):;c_1,c_2,c_3,c_4inmathbbRright$. This is a real subalgebra of $M_4(mathbbC)$




• $mathbbC$ is an algebra isomorphic to $leftleft(beginarraylcrp&-q\q&pendarrayright):;p,qinmathbbRright$. This a subalgebra of $M_2(mathbbR)$

Does someone know if it possible prove it by using this?
Is there a better (shorter) way?

Thanks.

abstract-algebra

share|cite|improve this question

edited yesterday

J. W. Tanner

3,0121320

asked yesterday

prosepprosep

1

New contributor

prosep is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this question

edited yesterday

J. W. Tanner

3,0121320

asked yesterday

prosepprosep

1

New contributor

prosep is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|cite|improve this question

edited yesterday

J. W. Tanner

3,0121320

asked yesterday

prosepprosep

1

New contributor

prosep is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

edited yesterday

J. W. Tanner

3,0121320

edited yesterday

J. W. Tanner

3,0121320

asked yesterday

prosepprosep

1

New contributor

prosep is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked yesterday

prosepprosep

1

1 Answer
1

active

oldest

votes

0

$begingroup$

Yes, there is a better way. Prove that:

• the sum of two elements of $H$ is again an element of $H$;


• the product of an element of $H$ by a real number is an element of $H$;


• the product of two elements of $H$ is again an element of $H$.

share|cite|improve this answer

answered yesterday

José Carlos SantosJosé Carlos Santos

166k22132235

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks. I have thought of this way but I wondered if there is a shorter proof.
  $endgroup$
  – prosep
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I seriously doubt that.
  $endgroup$
  – José Carlos Santos
  yesterday
  

  
  
  
  

add a comment | 

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0

$begingroup$

Yes, there is a better way. Prove that:

• the sum of two elements of $H$ is again an element of $H$;


• the product of an element of $H$ by a real number is an element of $H$;


• the product of two elements of $H$ is again an element of $H$.

share|cite|improve this answer

answered yesterday

José Carlos SantosJosé Carlos Santos

166k22132235

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks. I have thought of this way but I wondered if there is a shorter proof.
  $endgroup$
  – prosep
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I seriously doubt that.
  $endgroup$
  – José Carlos Santos
  yesterday
  

  
  
  
  

add a comment | 

0

$begingroup$

Yes, there is a better way. Prove that:

• the sum of two elements of $H$ is again an element of $H$;


• the product of an element of $H$ by a real number is an element of $H$;


• the product of two elements of $H$ is again an element of $H$.

share|cite|improve this answer

answered yesterday

José Carlos SantosJosé Carlos Santos

166k22132235

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks. I have thought of this way but I wondered if there is a shorter proof.
  $endgroup$
  – prosep
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I seriously doubt that.
  $endgroup$
  – José Carlos Santos
  yesterday
  

  
  
  
  

add a comment | 

$begingroup$

Yes, there is a better way. Prove that:

• the sum of two elements of $H$ is again an element of $H$;


• the product of an element of $H$ by a real number is an element of $H$;


• the product of two elements of $H$ is again an element of $H$.

share|cite|improve this answer

answered yesterday

José Carlos SantosJosé Carlos Santos

166k22132235

$endgroup$

share|cite|improve this answer

answered yesterday

José Carlos SantosJosé Carlos Santos

166k22132235

answered yesterday

José Carlos SantosJosé Carlos Santos

166k22132235

answered yesterday

José Carlos SantosJosé Carlos Santos

166k22132235