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Why is $H = left left(beginarrayrl z_1&z_2\ -bar z_2&bar z_1 endarrayright) biggvert z_1, z_2 in mathbb C right$ closed?
The Next CEO of Stack OverflowRREF of $left[beginarrayc A & B endarrayright]$ is $left[beginarrayc I & Xendarrayright]$. Prove that $X=A^-1B$.How to find the coordinate vector of $left[beginarrayrx\yendarrayright]$ with respect to some non-standard basis $mathcalB$Converting $mathbbC$ linear tranformation with determinant $a+bi$ into an $mathbbR$-linear transformation with determinant $a^2+b^2$.Is it possible to express $left(beginarraycca & -a \ a-1 & 1-a \ endarray right)$ as a certain product of two matrices?Matrices made of gluing $beginpmatrix z & iw \ i bar w & bar z endpmatrix$ blocks have a real determinantIs this inequality trivial?Showing that a very well-known representation is really a representationWhy are (a), (c), (d) true?Let $T:mathbb C^3tomathbb C^3$.Then, adjoint $T^*$ of $T$Need help to understand the proof of collinearity of 3 complex numbers
$begingroup$
Consider the set of matrices $$H = left left(beginarrayrl z_1&z_2\ -bar z_2&bar z_1 endarrayright) biggvert z_1, z_2 in mathbb C right.$$ 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 multiplication, 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 simpler way?
linear-algebra abstract-algebra group-theory complex-numbers vector-spaces
edited yesterday
TheSimpliFire
13k62464
asked yesterday
hopefullyhopefully
301214
$endgroup$
|
show 1 more comment
$begingroup$
Consider the set of matrices $$H = left left(beginarrayrl z_1&z_2\ -bar z_2&bar z_1 endarrayright) biggvert z_1, z_2 in mathbb C right.$$ 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 multiplication, 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 simpler way?
linear-algebra abstract-algebra group-theory complex-numbers vector-spaces
edited yesterday
TheSimpliFire
13k62464
asked yesterday
hopefullyhopefully
301214
$endgroup$
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
yesterday
2
$begingroup$
You multiply elements from H!
$endgroup$
– chhro
yesterday
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
yesterday
$begingroup$
@EeveeTrainer ok I got your idea.
$endgroup$
– hopefully
yesterday
$begingroup$
Notice that the space is one representation of the Quaternions. See the section "Conjugation, the norm, and reciprocal".
$endgroup$
– Somos
yesterday
|
show 1 more comment
$begingroup$
Consider the set of matrices $$H = left left(beginarrayrl z_1&z_2\ -bar z_2&bar z_1 endarrayright) biggvert z_1, z_2 in mathbb C right.$$ 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 multiplication, 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 simpler way?
linear-algebra abstract-algebra group-theory complex-numbers vector-spaces
edited yesterday
TheSimpliFire
13k62464
asked yesterday
hopefullyhopefully
301214
$endgroup$
Consider the set of matrices $$H = left left(beginarrayrl z_1&z_2\ -bar z_2&bar z_1 endarrayright) biggvert z_1, z_2 in mathbb C right.$$ 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 multiplication, 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 simpler way?
linear-algebra abstract-algebra group-theory complex-numbers vector-spaces
linear-algebra abstract-algebra group-theory complex-numbers vector-spaces
edited yesterday
TheSimpliFire
13k62464
asked yesterday
hopefullyhopefully
301214
edited yesterday
TheSimpliFire
13k62464
asked yesterday
hopefullyhopefully
301214
edited yesterday
TheSimpliFire
13k62464
edited yesterday
TheSimpliFire
13k62464
edited yesterday
TheSimpliFire
13k62464
13k62464
asked yesterday
hopefullyhopefully
301214
asked yesterday
hopefullyhopefully
301214
asked yesterday
hopefullyhopefully
301214
301214
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
yesterday
2
$begingroup$
You multiply elements from H!
$endgroup$
– chhro
yesterday
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
yesterday
$begingroup$
@EeveeTrainer ok I got your idea.
$endgroup$
– hopefully
yesterday
$begingroup$
Notice that the space is one representation of the Quaternions. See the section "Conjugation, the norm, and reciprocal".
$endgroup$
– Somos
yesterday
|
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
yesterday
2
$begingroup$
You multiply elements from H!
$endgroup$
– chhro
yesterday
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
yesterday
$begingroup$
@EeveeTrainer ok I got your idea.
$endgroup$
– hopefully
yesterday
$begingroup$
Notice that the space is one representation of the Quaternions. See the section "Conjugation, the norm, and reciprocal".
$endgroup$
– Somos
yesterday
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
yesterday
$begingroup$
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
yesterday
2
2
$begingroup$
You multiply elements from H!
$endgroup$
– chhro
yesterday
$begingroup$
You multiply elements from H!
$endgroup$
– chhro
yesterday
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
yesterday
$begingroup$
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
yesterday
$begingroup$
@EeveeTrainer ok I got your idea.
$endgroup$
– hopefully
yesterday
$begingroup$
@EeveeTrainer ok I got your idea.
$endgroup$
– hopefully
yesterday
$begingroup$
Notice that the space is one representation of the Quaternions. See the section "Conjugation, the norm, and reciprocal".
$endgroup$
– Somos
yesterday
$begingroup$
Notice that the space is one representation of the Quaternions. See the section "Conjugation, the norm, and reciprocal".
$endgroup$
– Somos
yesterday
|
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 yesterday
Eevee TrainerEevee Trainer
9,23631640
$endgroup$
2
$begingroup$
I think the first term in the second element of the resulting matrix is ad not ab?
$endgroup$
– hopefully
yesterday
$begingroup$
@hopefully Yeah, you're right, I made a typo. Thanks!
$endgroup$
– Eevee Trainer
yesterday
$begingroup$
what about the terms that contain only one bar, like the second term of the bottom right entry?
$endgroup$
– hopefully
yesterday
$begingroup$
What about them, exactly?
$endgroup$
– Eevee Trainer
yesterday
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
yesterday
|
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 yesterday
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 yesterday
Eevee TrainerEevee Trainer
9,23631640
$endgroup$
2
$begingroup$
I think the first term in the second element of the resulting matrix is ad not ab?
$endgroup$
– hopefully
yesterday
$begingroup$
@hopefully Yeah, you're right, I made a typo. Thanks!
$endgroup$
– Eevee Trainer
yesterday
$begingroup$
what about the terms that contain only one bar, like the second term of the bottom right entry?
$endgroup$
– hopefully
yesterday
$begingroup$
What about them, exactly?
$endgroup$
– Eevee Trainer
yesterday
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
yesterday
|
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 yesterday
Eevee TrainerEevee Trainer
9,23631640
$endgroup$
2
$begingroup$
I think the first term in the second element of the resulting matrix is ad not ab?
$endgroup$
– hopefully
yesterday
$begingroup$
@hopefully Yeah, you're right, I made a typo. Thanks!
$endgroup$
– Eevee Trainer
yesterday
$begingroup$
what about the terms that contain only one bar, like the second term of the bottom right entry?
$endgroup$
– hopefully
yesterday
$begingroup$
What about them, exactly?
$endgroup$
– Eevee Trainer
yesterday
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
yesterday
|
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 yesterday
Eevee TrainerEevee Trainer
9,23631640
$endgroup$
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 yesterday
Eevee TrainerEevee Trainer
9,23631640
edited yesterday
answered yesterday
Eevee TrainerEevee Trainer
9,23631640
answered yesterday
Eevee TrainerEevee Trainer
9,23631640
answered yesterday
Eevee TrainerEevee Trainer
9,23631640
9,23631640
2
$begingroup$
I think the first term in the second element of the resulting matrix is ad not ab?
$endgroup$
– hopefully
yesterday
$begingroup$
@hopefully Yeah, you're right, I made a typo. Thanks!
$endgroup$
– Eevee Trainer
yesterday
$begingroup$
what about the terms that contain only one bar, like the second term of the bottom right entry?
$endgroup$
– hopefully
yesterday
$begingroup$
What about them, exactly?
$endgroup$
– Eevee Trainer
yesterday
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.
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– Eevee Trainer
yesterday
|
show 3 more comments
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
yesterday
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@hopefully Yeah, you're right, I made a typo. Thanks!
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– Eevee Trainer
yesterday
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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
yesterday
$begingroup$
What about them, exactly?
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– Eevee Trainer
yesterday
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.
$endgroup$
– Eevee Trainer
yesterday
2
2
$begingroup$
I think the first term in the second element of the resulting matrix is ad not ab?
$endgroup$
– hopefully
yesterday
$begingroup$
I think the first term in the second element of the resulting matrix is ad not ab?
$endgroup$
– hopefully
yesterday
$begingroup$
@hopefully Yeah, you're right, I made a typo. Thanks!
$endgroup$
– Eevee Trainer
yesterday
$begingroup$
@hopefully Yeah, you're right, I made a typo. Thanks!
$endgroup$
– Eevee Trainer
yesterday
$begingroup$
what about the terms that contain only one bar, like the second term of the bottom right entry?
$endgroup$
– hopefully
yesterday
$begingroup$
what about the terms that contain only one bar, like the second term of the bottom right entry?
$endgroup$
– hopefully
yesterday
$begingroup$
What about them, exactly?
$endgroup$
– Eevee Trainer
yesterday
$begingroup$
What about them, exactly?
$endgroup$
– Eevee Trainer
yesterday
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
yesterday
$begingroup$
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
yesterday
|
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 yesterday
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 yesterday
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 yesterday
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 yesterday
TravisTravis
63.8k769151
answered yesterday
TravisTravis
63.8k769151
answered yesterday
TravisTravis
63.8k769151
answered yesterday
TravisTravis
63.8k769151
63.8k769151
add a comment |
add a comment |
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$begingroup$
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
yesterday
2
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You multiply elements from H!
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– chhro
yesterday
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
yesterday
$begingroup$
@EeveeTrainer ok I got your idea.
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– hopefully
yesterday
$begingroup$
Notice that the space is one representation of the Quaternions. See the section "Conjugation, the norm, and reciprocal".
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– Somos
yesterday