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Center of $A=leftleft(beginsmallmatrix a & b \ -overlineb & overlineaendsmallmatrixright)bigvert a,binmathbbCright$

Proving $left| fracvecvright| =1$, $vecvne vec0$Finding diagonal and unitary matricesProof, wheather a subset of a Group is a SubgroupProblem with spanning set and matricesWhat is the angle between those two matrices over $mathbbC$?Calculate the following seminorm of $A$Find the nullspace of a function complex vectorspace of polynomials of degree 2 or lessDetermine if the three matrices span the vector space of $2times 2$ matricesFind base in a $4times 4$ matrix.General form of elements of $SU(2)$

0

$begingroup$

Let $A=leftbeginpmatrix a & b \ -overlineb & overlineaendpmatrix Bigvert,, a,binmathbbCright$. I want to determine $Z(A)$.

I've seen that $A$ is actually the matrix representation of the qauternions, so what I've found by searching the web is that $Z(A)=alpha I $. However I'm trying to calculate $A$ explicitely.

So if $xin Z(A)$ then $xy=yx$ for all $yin A$. So by taking $y=beginpmatrix i & 0 \ 0 & -iendpmatrixin A$ then for $x=beginpmatrix a & b \ -overlineb & overlineaendpmatrix$ we should have

$$beginpmatrix ia & -ib \ -ioverlineb & -ioverlineaendpmatrix=beginpmatrix ia & ib \ -ioverlineb & -ioverlineaendpmatrix$$

So we may conclude $b=0$ as $-ib$ should equal $ib$. Likewise if we take $y=beginpmatrix 0 & 1 \ -1 & 0endpmatrixin A$ we should have

$$beginpmatrix b & a \ -overlinea & -overlinebendpmatrix=beginpmatrix -overlineb & overlinea \ -a & -bendpmatrix$$

So $a$ should be real as $a=overlinea$.

Does this proof work?

linear-algebra proof-verification

share|cite|improve this question

edited Mar 13 at 10:49

StubbornAtom

6,21811339

asked Mar 13 at 10:36

JtaJta

340211

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You can translate the proof of the duplicate by inserting your matrix representation for quaternions. Then it is exactly the same explicit calculation.
  $endgroup$
  – Dietrich Burde
  Mar 13 at 10:38
  

  
  
  
  

add a comment | 

0

$begingroup$

Let $A=leftbeginpmatrix a & b \ -overlineb & overlineaendpmatrix Bigvert,, a,binmathbbCright$. I want to determine $Z(A)$.

I've seen that $A$ is actually the matrix representation of the qauternions, so what I've found by searching the web is that $Z(A)=alpha I $. However I'm trying to calculate $A$ explicitely.

So if $xin Z(A)$ then $xy=yx$ for all $yin A$. So by taking $y=beginpmatrix i & 0 \ 0 & -iendpmatrixin A$ then for $x=beginpmatrix a & b \ -overlineb & overlineaendpmatrix$ we should have

$$beginpmatrix ia & -ib \ -ioverlineb & -ioverlineaendpmatrix=beginpmatrix ia & ib \ -ioverlineb & -ioverlineaendpmatrix$$

So we may conclude $b=0$ as $-ib$ should equal $ib$. Likewise if we take $y=beginpmatrix 0 & 1 \ -1 & 0endpmatrixin A$ we should have

$$beginpmatrix b & a \ -overlinea & -overlinebendpmatrix=beginpmatrix -overlineb & overlinea \ -a & -bendpmatrix$$

So $a$ should be real as $a=overlinea$.

Does this proof work?

linear-algebra proof-verification

share|cite|improve this question

edited Mar 13 at 10:49

StubbornAtom

6,21811339

asked Mar 13 at 10:36

JtaJta

340211

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You can translate the proof of the duplicate by inserting your matrix representation for quaternions. Then it is exactly the same explicit calculation.
  $endgroup$
  – Dietrich Burde
  Mar 13 at 10:38
  

  
  
  
  

add a comment | 

$begingroup$

Let $A=leftbeginpmatrix a & b \ -overlineb & overlineaendpmatrix Bigvert,, a,binmathbbCright$. I want to determine $Z(A)$.

I've seen that $A$ is actually the matrix representation of the qauternions, so what I've found by searching the web is that $Z(A)=alpha I $. However I'm trying to calculate $A$ explicitely.

So if $xin Z(A)$ then $xy=yx$ for all $yin A$. So by taking $y=beginpmatrix i & 0 \ 0 & -iendpmatrixin A$ then for $x=beginpmatrix a & b \ -overlineb & overlineaendpmatrix$ we should have

$$beginpmatrix ia & -ib \ -ioverlineb & -ioverlineaendpmatrix=beginpmatrix ia & ib \ -ioverlineb & -ioverlineaendpmatrix$$

So we may conclude $b=0$ as $-ib$ should equal $ib$. Likewise if we take $y=beginpmatrix 0 & 1 \ -1 & 0endpmatrixin A$ we should have

$$beginpmatrix b & a \ -overlinea & -overlinebendpmatrix=beginpmatrix -overlineb & overlinea \ -a & -bendpmatrix$$

So $a$ should be real as $a=overlinea$.

Does this proof work?

linear-algebra proof-verification

share|cite|improve this question

edited Mar 13 at 10:49

StubbornAtom

6,21811339

asked Mar 13 at 10:36

JtaJta

340211

$endgroup$

share|cite|improve this question

edited Mar 13 at 10:49

StubbornAtom

6,21811339

asked Mar 13 at 10:36

JtaJta

340211

share|cite|improve this question

edited Mar 13 at 10:49

StubbornAtom

6,21811339

asked Mar 13 at 10:36

JtaJta

340211

edited Mar 13 at 10:49

StubbornAtom

6,21811339

edited Mar 13 at 10:49

StubbornAtom

6,21811339

asked Mar 13 at 10:36

JtaJta

340211

asked Mar 13 at 10:36

JtaJta

340211

1 Answer
1

active

oldest

votes

0

$begingroup$

Yes, your strategy is correct. Technically you should add a comment at the end that all such matrices are obviously elements of the centre because they're proportional to the identity matrix, but yes, one can constrain centre elements by identifying specific matrices that are hard to commute with.

share|cite|improve this answer

answered Mar 13 at 10:57

J.G.J.G.

30.6k23149

$endgroup$

add a comment | 

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0

$begingroup$

Yes, your strategy is correct. Technically you should add a comment at the end that all such matrices are obviously elements of the centre because they're proportional to the identity matrix, but yes, one can constrain centre elements by identifying specific matrices that are hard to commute with.

share|cite|improve this answer

answered Mar 13 at 10:57

J.G.J.G.

30.6k23149

$endgroup$

add a comment | 

0

$begingroup$

Yes, your strategy is correct. Technically you should add a comment at the end that all such matrices are obviously elements of the centre because they're proportional to the identity matrix, but yes, one can constrain centre elements by identifying specific matrices that are hard to commute with.

share|cite|improve this answer

answered Mar 13 at 10:57

J.G.J.G.

30.6k23149

$endgroup$

add a comment | 

$begingroup$

Yes, your strategy is correct. Technically you should add a comment at the end that all such matrices are obviously elements of the centre because they're proportional to the identity matrix, but yes, one can constrain centre elements by identifying specific matrices that are hard to commute with.

share|cite|improve this answer

answered Mar 13 at 10:57

J.G.J.G.

30.6k23149

$endgroup$

share|cite|improve this answer

answered Mar 13 at 10:57

J.G.J.G.

30.6k23149

answered Mar 13 at 10:57

J.G.J.G.

30.6k23149

answered Mar 13 at 10:57

J.G.J.G.

30.6k23149