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How is this set of matrices closed under multiplication?

2

$begingroup$

Consider the set of matrices $$H = left{ left(begin{array}{rl} z_1&z_2\ -bar z_2&bar z_1 end{array}right) mid z_1, z_2 in mathbb C right}.$$ It is a four-dimensional real subspace of the vector space $L_2(mathbb C)$, and enjoys the following remarkable properties:

$1)$ $H$ is closed under multiplicacion, i.e., it is a real subalgebra of the algebra $L_2(mathbb C)$;

I have tried to multiply it with this matrix:
begin{bmatrix}
a & b
\
c & d
end{bmatrix}

where $a$, $b$, $c$, and $d$ are complex numbers but I got a very big formula that I do not know how this formula still is in $H$. Is there any suggestions for proving this in a simplier way?

linear-algebra abstract-algebra group-theory complex-numbers

share|cite|improve this question

edited 1 hour ago

Rócherz

3,0013821

asked 2 hours ago

hopefullyhopefully

294214

$endgroup$

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  You have to be careful: the matrix you multiply by also has to be of the same form. Thus, $c = -bar{b}$ and $d = bar{a}$. I might write up a fuller explanation of how this holds in a second (if it does, I gotta check) - I just wanted to point out that since it seemed like the first likely place where you might have tripped up.
  $endgroup$
  – Eevee Trainer
  2 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  You multiply elements from H!
  $endgroup$
  – chhro
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  Find $$begin{pmatrix} z_1 & z_2\ -bar{z_2}& bar{z_1} end{pmatrix} begin{pmatrix} w_1 & w_2\ -bar{w_2}& bar{w_1} end{pmatrix}$$ and arrange the entries in the required form!
  $endgroup$
  – Chinnapparaj R
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @EeveeTrainer ok I got your idea.
  $endgroup$
  – hopefully
  2 hours ago
  

  
  
  
  

add a comment | 

2

$begingroup$

Consider the set of matrices $$H = left{ left(begin{array}{rl} z_1&z_2\ -bar z_2&bar z_1 end{array}right) mid z_1, z_2 in mathbb C right}.$$ It is a four-dimensional real subspace of the vector space $L_2(mathbb C)$, and enjoys the following remarkable properties:

$1)$ $H$ is closed under multiplicacion, i.e., it is a real subalgebra of the algebra $L_2(mathbb C)$;

I have tried to multiply it with this matrix:
begin{bmatrix}
a & b
\
c & d
end{bmatrix}

where $a$, $b$, $c$, and $d$ are complex numbers but I got a very big formula that I do not know how this formula still is in $H$. Is there any suggestions for proving this in a simplier way?

linear-algebra abstract-algebra group-theory complex-numbers

share|cite|improve this question

edited 1 hour ago

Rócherz

3,0013821

asked 2 hours ago

hopefullyhopefully

294214

$endgroup$

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  You have to be careful: the matrix you multiply by also has to be of the same form. Thus, $c = -bar{b}$ and $d = bar{a}$. I might write up a fuller explanation of how this holds in a second (if it does, I gotta check) - I just wanted to point out that since it seemed like the first likely place where you might have tripped up.
  $endgroup$
  – Eevee Trainer
  2 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  You multiply elements from H!
  $endgroup$
  – chhro
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  Find $$begin{pmatrix} z_1 & z_2\ -bar{z_2}& bar{z_1} end{pmatrix} begin{pmatrix} w_1 & w_2\ -bar{w_2}& bar{w_1} end{pmatrix}$$ and arrange the entries in the required form!
  $endgroup$
  – Chinnapparaj R
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @EeveeTrainer ok I got your idea.
  $endgroup$
  – hopefully
  2 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

Consider the set of matrices $$H = left{ left(begin{array}{rl} z_1&z_2\ -bar z_2&bar z_1 end{array}right) mid z_1, z_2 in mathbb C right}.$$ It is a four-dimensional real subspace of the vector space $L_2(mathbb C)$, and enjoys the following remarkable properties:

$1)$ $H$ is closed under multiplicacion, i.e., it is a real subalgebra of the algebra $L_2(mathbb C)$;

I have tried to multiply it with this matrix:
begin{bmatrix}
a & b
\
c & d
end{bmatrix}

where $a$, $b$, $c$, and $d$ are complex numbers but I got a very big formula that I do not know how this formula still is in $H$. Is there any suggestions for proving this in a simplier way?

linear-algebra abstract-algebra group-theory complex-numbers

share|cite|improve this question

edited 1 hour ago

Rócherz

3,0013821

asked 2 hours ago

hopefullyhopefully

294214

$endgroup$

share|cite|improve this question

edited 1 hour ago

Rócherz

3,0013821

asked 2 hours ago

hopefullyhopefully

294214

share|cite|improve this question

edited 1 hour ago

Rócherz

3,0013821

asked 2 hours ago

hopefullyhopefully

294214

edited 1 hour ago

Rócherz

3,0013821

edited 1 hour ago

Rócherz

3,0013821

asked 2 hours ago

hopefullyhopefully

294214

asked 2 hours ago

hopefullyhopefully

294214

2 Answers
2

active

oldest

votes

4

$begingroup$

So, for a set $S$ of matrices (or any sort of element) to be closed under an operation $ast$ on it, we require that, for all $a,b in S, a ast b in S$.

As I noted in the comments, your issue lied in multiplying a matrix of $H$ by a generic matrix of complex elements, which is too general to have closure. You have to take two generic matrices of the set. So, let $a,b,c,d in Bbb C$ and then consider the multiplication

$$begin{bmatrix}
a & b\
-bar{b} & bar{a}
end{bmatrix} begin{bmatrix}
c & d\
-bar{d} & bar{c}
end{bmatrix} =begin{bmatrix}
ac - b bar{d} & ad+bbar{c}\
-bar{a} bar{d} - bar{b}c & bar{a} bar{c}-bar{b}d
end{bmatrix} $$

You can see immediately the left two matrices are of the form of matrices in $H$; on the right is their product. You can verify that it, too, matches by noting a couple of properties of the complex conjugate:

$$overline{z_1 cdot z_2} = overline{z_1} cdot overline{z_2} ;;;;; text{and} ;;;;; overline{z_1 + z_2} = overline{z_1} + overline{z_2} ;;;;; text{and} ;;;;; overline{overline{z_1}} = z_1$$

where $z_1,z_2 in Bbb C$. So if...

• ...the bottom-left entry is the negative of the conjugate of the top-right


• ...the bottom-right entry is the conjugate of the top-left

...then the product is in the form for a matrix in $H$. It does happen to hold, and thus $H$ is closed under matrix multiplication.

share|cite|improve this answer

edited 10 mins ago

answered 1 hour ago

Eevee TrainerEevee Trainer

8,98431640

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  I think the first term in the second element of the resulting matrix is ad not ab?
  $endgroup$
  – hopefully
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @hopefully Yeah, you're right, I made a typo. Thanks!
  $endgroup$
  – Eevee Trainer
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  what about the terms that contain only one bar, like the second term of the bottom right entry?
  $endgroup$
  – hopefully
  15 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What about them, exactly?
  $endgroup$
  – Eevee Trainer
  15 mins ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Take the conjugate of the top-left entry and you see they are equal if you use the properties in my post (the conjugate of a sum/product is the sum/product of the conjugates). One property I did leave out was that the conjugate of a conjugate is the original number, so I will add that. But the core idea is effectively that each factor of a number, when you take the conjugate, becomes its conjugate.
  $endgroup$
  – Eevee Trainer
  11 mins ago
  

  
  
  
  

 | 
show 3 more comments

1

$begingroup$

Here's an alternative method that, after verification of the simple characterization of this subspace given below, is coordinate-free.

Hint Denote $$J := pmatrix{cdot&-1\1&cdot}.$$ It follows immediately from the definition that $${X in M(2, Bbb C) : textrm{$X$ satisfies $X^dagger J = J X^top$}} .$$

So, for $X, Y in H$, $$(X Y)^dagger J = Y^dagger X^dagger J = Y^dagger JX^top = J Y^top X^top = J (XY)^top .$$

share|cite|improve this answer

answered 19 mins ago

TravisTravis

63.8k769151

$endgroup$

add a comment | 

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4

$begingroup$

So, for a set $S$ of matrices (or any sort of element) to be closed under an operation $ast$ on it, we require that, for all $a,b in S, a ast b in S$.

As I noted in the comments, your issue lied in multiplying a matrix of $H$ by a generic matrix of complex elements, which is too general to have closure. You have to take two generic matrices of the set. So, let $a,b,c,d in Bbb C$ and then consider the multiplication

$$begin{bmatrix}
a & b\
-bar{b} & bar{a}
end{bmatrix} begin{bmatrix}
c & d\
-bar{d} & bar{c}
end{bmatrix} =begin{bmatrix}
ac - b bar{d} & ad+bbar{c}\
-bar{a} bar{d} - bar{b}c & bar{a} bar{c}-bar{b}d
end{bmatrix} $$

You can see immediately the left two matrices are of the form of matrices in $H$; on the right is their product. You can verify that it, too, matches by noting a couple of properties of the complex conjugate:

$$overline{z_1 cdot z_2} = overline{z_1} cdot overline{z_2} ;;;;; text{and} ;;;;; overline{z_1 + z_2} = overline{z_1} + overline{z_2} ;;;;; text{and} ;;;;; overline{overline{z_1}} = z_1$$

where $z_1,z_2 in Bbb C$. So if...

• ...the bottom-left entry is the negative of the conjugate of the top-right


• ...the bottom-right entry is the conjugate of the top-left

...then the product is in the form for a matrix in $H$. It does happen to hold, and thus $H$ is closed under matrix multiplication.

share|cite|improve this answer

edited 10 mins ago

answered 1 hour ago

Eevee TrainerEevee Trainer

8,98431640

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  I think the first term in the second element of the resulting matrix is ad not ab?
  $endgroup$
  – hopefully
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @hopefully Yeah, you're right, I made a typo. Thanks!
  $endgroup$
  – Eevee Trainer
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  what about the terms that contain only one bar, like the second term of the bottom right entry?
  $endgroup$
  – hopefully
  15 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What about them, exactly?
  $endgroup$
  – Eevee Trainer
  15 mins ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Take the conjugate of the top-left entry and you see they are equal if you use the properties in my post (the conjugate of a sum/product is the sum/product of the conjugates). One property I did leave out was that the conjugate of a conjugate is the original number, so I will add that. But the core idea is effectively that each factor of a number, when you take the conjugate, becomes its conjugate.
  $endgroup$
  – Eevee Trainer
  11 mins ago
  

  
  
  
  

 | 
show 3 more comments

4

$begingroup$

So, for a set $S$ of matrices (or any sort of element) to be closed under an operation $ast$ on it, we require that, for all $a,b in S, a ast b in S$.

As I noted in the comments, your issue lied in multiplying a matrix of $H$ by a generic matrix of complex elements, which is too general to have closure. You have to take two generic matrices of the set. So, let $a,b,c,d in Bbb C$ and then consider the multiplication

$$begin{bmatrix}
a & b\
-bar{b} & bar{a}
end{bmatrix} begin{bmatrix}
c & d\
-bar{d} & bar{c}
end{bmatrix} =begin{bmatrix}
ac - b bar{d} & ad+bbar{c}\
-bar{a} bar{d} - bar{b}c & bar{a} bar{c}-bar{b}d
end{bmatrix} $$

You can see immediately the left two matrices are of the form of matrices in $H$; on the right is their product. You can verify that it, too, matches by noting a couple of properties of the complex conjugate:

$$overline{z_1 cdot z_2} = overline{z_1} cdot overline{z_2} ;;;;; text{and} ;;;;; overline{z_1 + z_2} = overline{z_1} + overline{z_2} ;;;;; text{and} ;;;;; overline{overline{z_1}} = z_1$$

where $z_1,z_2 in Bbb C$. So if...

• ...the bottom-left entry is the negative of the conjugate of the top-right


• ...the bottom-right entry is the conjugate of the top-left

...then the product is in the form for a matrix in $H$. It does happen to hold, and thus $H$ is closed under matrix multiplication.

share|cite|improve this answer

edited 10 mins ago

answered 1 hour ago

Eevee TrainerEevee Trainer

8,98431640

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  I think the first term in the second element of the resulting matrix is ad not ab?
  $endgroup$
  – hopefully
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @hopefully Yeah, you're right, I made a typo. Thanks!
  $endgroup$
  – Eevee Trainer
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  what about the terms that contain only one bar, like the second term of the bottom right entry?
  $endgroup$
  – hopefully
  15 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What about them, exactly?
  $endgroup$
  – Eevee Trainer
  15 mins ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Take the conjugate of the top-left entry and you see they are equal if you use the properties in my post (the conjugate of a sum/product is the sum/product of the conjugates). One property I did leave out was that the conjugate of a conjugate is the original number, so I will add that. But the core idea is effectively that each factor of a number, when you take the conjugate, becomes its conjugate.
  $endgroup$
  – Eevee Trainer
  11 mins ago
  

  
  
  
  

 | 
show 3 more comments

$begingroup$

So, for a set $S$ of matrices (or any sort of element) to be closed under an operation $ast$ on it, we require that, for all $a,b in S, a ast b in S$.

As I noted in the comments, your issue lied in multiplying a matrix of $H$ by a generic matrix of complex elements, which is too general to have closure. You have to take two generic matrices of the set. So, let $a,b,c,d in Bbb C$ and then consider the multiplication

$$begin{bmatrix}
a & b\
-bar{b} & bar{a}
end{bmatrix} begin{bmatrix}
c & d\
-bar{d} & bar{c}
end{bmatrix} =begin{bmatrix}
ac - b bar{d} & ad+bbar{c}\
-bar{a} bar{d} - bar{b}c & bar{a} bar{c}-bar{b}d
end{bmatrix} $$

You can see immediately the left two matrices are of the form of matrices in $H$; on the right is their product. You can verify that it, too, matches by noting a couple of properties of the complex conjugate:

$$overline{z_1 cdot z_2} = overline{z_1} cdot overline{z_2} ;;;;; text{and} ;;;;; overline{z_1 + z_2} = overline{z_1} + overline{z_2} ;;;;; text{and} ;;;;; overline{overline{z_1}} = z_1$$

where $z_1,z_2 in Bbb C$. So if...

• ...the bottom-left entry is the negative of the conjugate of the top-right


• ...the bottom-right entry is the conjugate of the top-left

...then the product is in the form for a matrix in $H$. It does happen to hold, and thus $H$ is closed under matrix multiplication.

share|cite|improve this answer

edited 10 mins ago

answered 1 hour ago

Eevee TrainerEevee Trainer

8,98431640

$endgroup$

share|cite|improve this answer

edited 10 mins ago

answered 1 hour ago

Eevee TrainerEevee Trainer

8,98431640

answered 1 hour ago

Eevee TrainerEevee Trainer

8,98431640

answered 1 hour ago

Eevee TrainerEevee Trainer

8,98431640

1

$begingroup$

Here's an alternative method that, after verification of the simple characterization of this subspace given below, is coordinate-free.

Hint Denote $$J := pmatrix{cdot&-1\1&cdot}.$$ It follows immediately from the definition that $${X in M(2, Bbb C) : textrm{$X$ satisfies $X^dagger J = J X^top$}} .$$

So, for $X, Y in H$, $$(X Y)^dagger J = Y^dagger X^dagger J = Y^dagger JX^top = J Y^top X^top = J (XY)^top .$$

share|cite|improve this answer

answered 19 mins ago

TravisTravis

63.8k769151

$endgroup$

add a comment | 

1

$begingroup$

Here's an alternative method that, after verification of the simple characterization of this subspace given below, is coordinate-free.

Hint Denote $$J := pmatrix{cdot&-1\1&cdot}.$$ It follows immediately from the definition that $${X in M(2, Bbb C) : textrm{$X$ satisfies $X^dagger J = J X^top$}} .$$

So, for $X, Y in H$, $$(X Y)^dagger J = Y^dagger X^dagger J = Y^dagger JX^top = J Y^top X^top = J (XY)^top .$$

share|cite|improve this answer

answered 19 mins ago

TravisTravis

63.8k769151

$endgroup$

add a comment | 

$begingroup$

Here's an alternative method that, after verification of the simple characterization of this subspace given below, is coordinate-free.

Hint Denote $$J := pmatrix{cdot&-1\1&cdot}.$$ It follows immediately from the definition that $${X in M(2, Bbb C) : textrm{$X$ satisfies $X^dagger J = J X^top$}} .$$

So, for $X, Y in H$, $$(X Y)^dagger J = Y^dagger X^dagger J = Y^dagger JX^top = J Y^top X^top = J (XY)^top .$$

share|cite|improve this answer

answered 19 mins ago

TravisTravis

63.8k769151

$endgroup$

share|cite|improve this answer

answered 19 mins ago

TravisTravis

63.8k769151

answered 19 mins ago

TravisTravis

63.8k769151

answered 19 mins ago

TravisTravis

63.8k769151