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How is this set of matrices closed under multiplication?
The Next CEO of Stack OverflowHow to determine if a set is a subspace of the vector space of all complex $2times 2$ matrices?Converting $mathbbC$ linear tranformation with determinant $a+bi$ into an $mathbbR$-linear transformation with determinant $a^2+b^2$.Is this inequality trivial?Showing that a very well-known representation is really a representationWrite out the multiplication table for the following set of matrices over $mathbb Q$Is multiplication an operation in the given set of matrices?Why are (a), (c), (d) true?Let $T:mathbb C^3tomathbb C^3$.Then, adjoint $T^*$ of $T$Prove that set $mathbbS$ forms group under matrix multiplicationAbout subalgebra of Hamilton
$begingroup$
Consider the set of matrices $$H = left left(beginarrayrl z_1&z_2\ -bar z_2&bar z_1 endarrayright) mid z_1, z_2 in mathbb C right.$$ 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:
beginbmatrix
a & b
\
c & d
endbmatrix
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
edited 3 hours ago
Rócherz
3,0013821
asked 4 hours ago
hopefullyhopefully
304214
$endgroup$
|
show 1 more comment
$begingroup$
Consider the set of matrices $$H = left left(beginarrayrl z_1&z_2\ -bar z_2&bar z_1 endarrayright) mid z_1, z_2 in mathbb C right.$$ 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:
beginbmatrix
a & b
\
c & d
endbmatrix
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
edited 3 hours ago
Rócherz
3,0013821
asked 4 hours ago
hopefullyhopefully
304214
$endgroup$
3
$begingroup$
You have to be careful: the matrix you multiply by also has to be of the same form. Thus, $c = -barb$ and $d = bara$. 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
4 hours ago
2
$begingroup$
You multiply elements from H!
$endgroup$
– chhro
4 hours ago
4
$begingroup$
Find $$beginpmatrix z_1 & z_2\ -barz_2& barz_1 endpmatrix beginpmatrix w_1 & w_2\ -barw_2& barw_1 endpmatrix$$ and arrange the entries in the required form!
$endgroup$
– Chinnapparaj R
4 hours ago
$begingroup$
@EeveeTrainer ok I got your idea.
$endgroup$
– hopefully
4 hours ago
$begingroup$
Notice that the space is one representation of the Quaternions. See the section "Conjugation, the norm, and reciprocal".
$endgroup$
– Somos
1 hour ago
|
show 1 more comment
$begingroup$
Consider the set of matrices $$H = left left(beginarrayrl z_1&z_2\ -bar z_2&bar z_1 endarrayright) mid z_1, z_2 in mathbb C right.$$ 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:
beginbmatrix
a & b
\
c & d
endbmatrix
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
edited 3 hours ago
Rócherz
3,0013821
asked 4 hours ago
hopefullyhopefully
304214
$endgroup$
Consider the set of matrices $$H = left left(beginarrayrl z_1&z_2\ -bar z_2&bar z_1 endarrayright) mid z_1, z_2 in mathbb C right.$$ 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:
beginbmatrix
a & b
\
c & d
endbmatrix
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
linear-algebra abstract-algebra group-theory complex-numbers
edited 3 hours ago
Rócherz
3,0013821
asked 4 hours ago
hopefullyhopefully
304214
edited 3 hours ago
Rócherz
3,0013821
asked 4 hours ago
hopefullyhopefully
304214
edited 3 hours ago
Rócherz
3,0013821
edited 3 hours ago
Rócherz
3,0013821
edited 3 hours ago
Rócherz
3,0013821
3,0013821
asked 4 hours ago
hopefullyhopefully
304214
asked 4 hours ago
hopefullyhopefully
304214
asked 4 hours ago
hopefullyhopefully
304214
304214
3
$begingroup$
You have to be careful: the matrix you multiply by also has to be of the same form. Thus, $c = -barb$ and $d = bara$. 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
4 hours ago
2
$begingroup$
You multiply elements from H!
$endgroup$
– chhro
4 hours ago
4
$begingroup$
Find $$beginpmatrix z_1 & z_2\ -barz_2& barz_1 endpmatrix beginpmatrix w_1 & w_2\ -barw_2& barw_1 endpmatrix$$ and arrange the entries in the required form!
$endgroup$
– Chinnapparaj R
4 hours ago
$begingroup$
@EeveeTrainer ok I got your idea.
$endgroup$
– hopefully
4 hours ago
$begingroup$
Notice that the space is one representation of the Quaternions. See the section "Conjugation, the norm, and reciprocal".
$endgroup$
– Somos
1 hour ago
|
show 1 more comment
3
$begingroup$
You have to be careful: the matrix you multiply by also has to be of the same form. Thus, $c = -barb$ and $d = bara$. 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
4 hours ago
2
$begingroup$
You multiply elements from H!
$endgroup$
– chhro
4 hours ago
4
$begingroup$
Find $$beginpmatrix z_1 & z_2\ -barz_2& barz_1 endpmatrix beginpmatrix w_1 & w_2\ -barw_2& barw_1 endpmatrix$$ and arrange the entries in the required form!
$endgroup$
– Chinnapparaj R
4 hours ago
$begingroup$
@EeveeTrainer ok I got your idea.
$endgroup$
– hopefully
4 hours ago
$begingroup$
Notice that the space is one representation of the Quaternions. See the section "Conjugation, the norm, and reciprocal".
$endgroup$
– Somos
1 hour ago
3
3
$begingroup$
You have to be careful: the matrix you multiply by also has to be of the same form. Thus, $c = -barb$ and $d = bara$. 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
4 hours ago
$begingroup$
You have to be careful: the matrix you multiply by also has to be of the same form. Thus, $c = -barb$ and $d = bara$. 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
4 hours ago
2
2
$begingroup$
You multiply elements from H!
$endgroup$
– chhro
4 hours ago
$begingroup$
You multiply elements from H!
$endgroup$
– chhro
4 hours ago
4
4
$begingroup$
Find $$beginpmatrix z_1 & z_2\ -barz_2& barz_1 endpmatrix beginpmatrix w_1 & w_2\ -barw_2& barw_1 endpmatrix$$ and arrange the entries in the required form!
$endgroup$
– Chinnapparaj R
4 hours ago
$begingroup$
Find $$beginpmatrix z_1 & z_2\ -barz_2& barz_1 endpmatrix beginpmatrix w_1 & w_2\ -barw_2& barw_1 endpmatrix$$ and arrange the entries in the required form!
$endgroup$
– Chinnapparaj R
4 hours ago
$begingroup$
@EeveeTrainer ok I got your idea.
$endgroup$
– hopefully
4 hours ago
$begingroup$
@EeveeTrainer ok I got your idea.
$endgroup$
– hopefully
4 hours ago
$begingroup$
Notice that the space is one representation of the Quaternions. See the section "Conjugation, the norm, and reciprocal".
$endgroup$
– Somos
1 hour ago
$begingroup$
Notice that the space is one representation of the Quaternions. See the section "Conjugation, the norm, and reciprocal".
$endgroup$
– Somos
1 hour ago
|
show 1 more comment
2 Answers
2
active
oldest
votes
$begingroup$
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
$$beginbmatrix
a & b\
-barb & bara
endbmatrix beginbmatrix
c & d\
-bard & barc
endbmatrix =beginbmatrix
ac - b bard & ad+bbarc\
-bara bard - barbc & bara barc-barbd
endbmatrix $$
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:
$$overlinez_1 cdot z_2 = overlinez_1 cdot overlinez_2 ;;;;; textand ;;;;; overlinez_1 + z_2 = overlinez_1 + overlinez_2 ;;;;; textand ;;;;; overlineoverlinez_1 = z_1$$
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.
answered 4 hours ago
Eevee TrainerEevee Trainer
9,01731640
$endgroup$
2
$begingroup$
I think the first term in the second element of the resulting matrix is ad not ab?
$endgroup$
– hopefully
3 hours ago
$begingroup$
@hopefully Yeah, you're right, I made a typo. Thanks!
$endgroup$
– Eevee Trainer
3 hours ago
$begingroup$
what about the terms that contain only one bar, like the second term of the bottom right entry?
$endgroup$
– hopefully
2 hours ago
$begingroup$
What about them, exactly?
$endgroup$
– Eevee Trainer
2 hours 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
2 hours ago
|
show 3 more comments
$begingroup$
Here's an alternative method that, after verification of the simple characterization of this subspace given below, is coordinate-free.
Hint Denote $$J := pmatrixcdot&-1\1&cdot.$$ It follows immediately from the definition that $$X in M(2, Bbb C) : textrm$X$ satisfies $X^dagger J = J X^top$ .$$
So, for $X, Y in H$, $$(X Y)^dagger J = Y^dagger X^dagger J = Y^dagger JX^top = J Y^top X^top = J (XY)^top .$$
answered 2 hours ago
TravisTravis
63.8k769151
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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
$$beginbmatrix
a & b\
-barb & bara
endbmatrix beginbmatrix
c & d\
-bard & barc
endbmatrix =beginbmatrix
ac - b bard & ad+bbarc\
-bara bard - barbc & bara barc-barbd
endbmatrix $$
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:
$$overlinez_1 cdot z_2 = overlinez_1 cdot overlinez_2 ;;;;; textand ;;;;; overlinez_1 + z_2 = overlinez_1 + overlinez_2 ;;;;; textand ;;;;; overlineoverlinez_1 = z_1$$
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.
answered 4 hours ago
Eevee TrainerEevee Trainer
9,01731640
$endgroup$
2
$begingroup$
I think the first term in the second element of the resulting matrix is ad not ab?
$endgroup$
– hopefully
3 hours ago
$begingroup$
@hopefully Yeah, you're right, I made a typo. Thanks!
$endgroup$
– Eevee Trainer
3 hours ago
$begingroup$
what about the terms that contain only one bar, like the second term of the bottom right entry?
$endgroup$
– hopefully
2 hours ago
$begingroup$
What about them, exactly?
$endgroup$
– Eevee Trainer
2 hours 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
2 hours 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
$$beginbmatrix
a & b\
-barb & bara
endbmatrix beginbmatrix
c & d\
-bard & barc
endbmatrix =beginbmatrix
ac - b bard & ad+bbarc\
-bara bard - barbc & bara barc-barbd
endbmatrix $$
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:
$$overlinez_1 cdot z_2 = overlinez_1 cdot overlinez_2 ;;;;; textand ;;;;; overlinez_1 + z_2 = overlinez_1 + overlinez_2 ;;;;; textand ;;;;; overlineoverlinez_1 = z_1$$
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.
answered 4 hours ago
Eevee TrainerEevee Trainer
9,01731640
$endgroup$
2
$begingroup$
I think the first term in the second element of the resulting matrix is ad not ab?
$endgroup$
– hopefully
3 hours ago
$begingroup$
@hopefully Yeah, you're right, I made a typo. Thanks!
$endgroup$
– Eevee Trainer
3 hours ago
$begingroup$
what about the terms that contain only one bar, like the second term of the bottom right entry?
$endgroup$
– hopefully
2 hours ago
$begingroup$
What about them, exactly?
$endgroup$
– Eevee Trainer
2 hours 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
2 hours 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
$$beginbmatrix
a & b\
-barb & bara
endbmatrix beginbmatrix
c & d\
-bard & barc
endbmatrix =beginbmatrix
ac - b bard & ad+bbarc\
-bara bard - barbc & bara barc-barbd
endbmatrix $$
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:
$$overlinez_1 cdot z_2 = overlinez_1 cdot overlinez_2 ;;;;; textand ;;;;; overlinez_1 + z_2 = overlinez_1 + overlinez_2 ;;;;; textand ;;;;; overlineoverlinez_1 = z_1$$
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.
answered 4 hours ago
Eevee TrainerEevee Trainer
9,01731640
$endgroup$
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
$$beginbmatrix
a & b\
-barb & bara
endbmatrix beginbmatrix
c & d\
-bard & barc
endbmatrix =beginbmatrix
ac - b bard & ad+bbarc\
-bara bard - barbc & bara barc-barbd
endbmatrix $$
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:
$$overlinez_1 cdot z_2 = overlinez_1 cdot overlinez_2 ;;;;; textand ;;;;; overlinez_1 + z_2 = overlinez_1 + overlinez_2 ;;;;; textand ;;;;; overlineoverlinez_1 = z_1$$
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.
answered 4 hours ago
Eevee TrainerEevee Trainer
9,01731640
edited 2 hours ago
answered 4 hours ago
Eevee TrainerEevee Trainer
9,01731640
answered 4 hours ago
Eevee TrainerEevee Trainer
9,01731640
answered 4 hours ago
Eevee TrainerEevee Trainer
9,01731640
9,01731640
2
$begingroup$
I think the first term in the second element of the resulting matrix is ad not ab?
$endgroup$
– hopefully
3 hours ago
$begingroup$
@hopefully Yeah, you're right, I made a typo. Thanks!
$endgroup$
– Eevee Trainer
3 hours ago
$begingroup$
what about the terms that contain only one bar, like the second term of the bottom right entry?
$endgroup$
– hopefully
2 hours ago
$begingroup$
What about them, exactly?
$endgroup$
– Eevee Trainer
2 hours 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
2 hours ago
|
show 3 more comments
2
$begingroup$
I think the first term in the second element of the resulting matrix is ad not ab?
$endgroup$
– hopefully
3 hours ago
$begingroup$
@hopefully Yeah, you're right, I made a typo. Thanks!
$endgroup$
– Eevee Trainer
3 hours ago
$begingroup$
what about the terms that contain only one bar, like the second term of the bottom right entry?
$endgroup$
– hopefully
2 hours ago
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What about them, exactly?
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– Eevee Trainer
2 hours ago
1
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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.
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– Eevee Trainer
2 hours ago
2
2
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I think the first term in the second element of the resulting matrix is ad not ab?
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– hopefully
3 hours ago
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I think the first term in the second element of the resulting matrix is ad not ab?
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– hopefully
3 hours ago
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@hopefully Yeah, you're right, I made a typo. Thanks!
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– Eevee Trainer
3 hours ago
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@hopefully Yeah, you're right, I made a typo. Thanks!
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– Eevee Trainer
3 hours ago
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what about the terms that contain only one bar, like the second term of the bottom right entry?
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– hopefully
2 hours ago
$begingroup$
what about the terms that contain only one bar, like the second term of the bottom right entry?
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– hopefully
2 hours ago
$begingroup$
What about them, exactly?
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– Eevee Trainer
2 hours ago
$begingroup$
What about them, exactly?
$endgroup$
– Eevee Trainer
2 hours ago
1
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
2 hours ago
$begingroup$
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.
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– Eevee Trainer
2 hours ago
|
show 3 more comments
$begingroup$
Here's an alternative method that, after verification of the simple characterization of this subspace given below, is coordinate-free.
Hint Denote $$J := pmatrixcdot&-1\1&cdot.$$ It follows immediately from the definition that $$X in M(2, Bbb C) : textrm$X$ satisfies $X^dagger J = J X^top$ .$$
So, for $X, Y in H$, $$(X Y)^dagger J = Y^dagger X^dagger J = Y^dagger JX^top = J Y^top X^top = J (XY)^top .$$
answered 2 hours 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 $$J := pmatrixcdot&-1\1&cdot.$$ It follows immediately from the definition that $$X in M(2, Bbb C) : textrm$X$ satisfies $X^dagger J = J X^top$ .$$
So, for $X, Y in H$, $$(X Y)^dagger J = Y^dagger X^dagger J = Y^dagger JX^top = J Y^top X^top = J (XY)^top .$$
answered 2 hours 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 $$J := pmatrixcdot&-1\1&cdot.$$ It follows immediately from the definition that $$X in M(2, Bbb C) : textrm$X$ satisfies $X^dagger J = J X^top$ .$$
So, for $X, Y in H$, $$(X Y)^dagger J = Y^dagger X^dagger J = Y^dagger JX^top = J Y^top X^top = J (XY)^top .$$
answered 2 hours ago
TravisTravis
63.8k769151
$endgroup$
Here's an alternative method that, after verification of the simple characterization of this subspace given below, is coordinate-free.
Hint Denote $$J := pmatrixcdot&-1\1&cdot.$$ It follows immediately from the definition that $$X in M(2, Bbb C) : textrm$X$ satisfies $X^dagger J = J X^top$ .$$
So, for $X, Y in H$, $$(X Y)^dagger J = Y^dagger X^dagger J = Y^dagger JX^top = J Y^top X^top = J (XY)^top .$$
answered 2 hours ago
TravisTravis
63.8k769151
answered 2 hours ago
TravisTravis
63.8k769151
answered 2 hours ago
TravisTravis
63.8k769151
answered 2 hours ago
TravisTravis
63.8k769151
63.8k769151
add a comment |
add a comment |
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3
$begingroup$
You have to be careful: the matrix you multiply by also has to be of the same form. Thus, $c = -barb$ and $d = bara$. 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.
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– Eevee Trainer
4 hours ago
2
$begingroup$
You multiply elements from H!
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– chhro
4 hours ago
4
$begingroup$
Find $$beginpmatrix z_1 & z_2\ -barz_2& barz_1 endpmatrix beginpmatrix w_1 & w_2\ -barw_2& barw_1 endpmatrix$$ and arrange the entries in the required form!
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– Chinnapparaj R
4 hours ago
$begingroup$
@EeveeTrainer ok I got your idea.
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– hopefully
4 hours ago
$begingroup$
Notice that the space is one representation of the Quaternions. See the section "Conjugation, the norm, and reciprocal".
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– Somos
1 hour ago