Why is $H = left{ left(begin{array}{rl} z_1&z_2\ -bar z_2&bar z_1 end{array}right) biggvert z_1, z_2...
$begingroup$
Consider the set of matrices $$H = left{ left(begin{array}{rl} z_1&z_2\ -bar z_2&bar z_1 end{array}right) biggvert z_1, z_2 in mathbb C right}.$$ 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:
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 simpler way?
linear-algebra abstract-algebra group-theory complex-numbers vector-spaces
$endgroup$
|
show 1 more comment
$begingroup$
Consider the set of matrices $$H = left{ left(begin{array}{rl} z_1&z_2\ -bar z_2&bar z_1 end{array}right) biggvert z_1, z_2 in mathbb C right}.$$ 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:
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 simpler way?
linear-algebra abstract-algebra group-theory complex-numbers vector-spaces
$endgroup$
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
Mar 31 at 1:10
2
$begingroup$
You multiply elements from H!
$endgroup$
– chhro
Mar 31 at 1:11
4
$begingroup$
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
Mar 31 at 1:12
$begingroup$
@EeveeTrainer ok I got your idea.
$endgroup$
– hopefully
Mar 31 at 1:12
$begingroup$
Notice that the space is one representation of the Quaternions. See the section "Conjugation, the norm, and reciprocal".
$endgroup$
– Somos
Mar 31 at 4:29
|
show 1 more comment
$begingroup$
Consider the set of matrices $$H = left{ left(begin{array}{rl} z_1&z_2\ -bar z_2&bar z_1 end{array}right) biggvert z_1, z_2 in mathbb C right}.$$ 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:
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 simpler way?
linear-algebra abstract-algebra group-theory complex-numbers vector-spaces
$endgroup$
Consider the set of matrices $$H = left{ left(begin{array}{rl} z_1&z_2\ -bar z_2&bar z_1 end{array}right) biggvert z_1, z_2 in mathbb C right}.$$ 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:
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 simpler way?
linear-algebra abstract-algebra group-theory complex-numbers vector-spaces
linear-algebra abstract-algebra group-theory complex-numbers vector-spaces
edited Mar 31 at 6:13
TheSimpliFire
13k62464
13k62464
asked Mar 31 at 1:08
hopefullyhopefully
270215
270215
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
Mar 31 at 1:10
2
$begingroup$
You multiply elements from H!
$endgroup$
– chhro
Mar 31 at 1:11
4
$begingroup$
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
Mar 31 at 1:12
$begingroup$
@EeveeTrainer ok I got your idea.
$endgroup$
– hopefully
Mar 31 at 1:12
$begingroup$
Notice that the space is one representation of the Quaternions. See the section "Conjugation, the norm, and reciprocal".
$endgroup$
– Somos
Mar 31 at 4:29
|
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, $c = -bar{b}$ 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
Mar 31 at 1:10
2
$begingroup$
You multiply elements from H!
$endgroup$
– chhro
Mar 31 at 1:11
4
$begingroup$
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
Mar 31 at 1:12
$begingroup$
@EeveeTrainer ok I got your idea.
$endgroup$
– hopefully
Mar 31 at 1:12
$begingroup$
Notice that the space is one representation of the Quaternions. See the section "Conjugation, the norm, and reciprocal".
$endgroup$
– Somos
Mar 31 at 4:29
3
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
Mar 31 at 1:10
$begingroup$
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
Mar 31 at 1:10
2
2
$begingroup$
You multiply elements from H!
$endgroup$
– chhro
Mar 31 at 1:11
$begingroup$
You multiply elements from H!
$endgroup$
– chhro
Mar 31 at 1:11
4
4
$begingroup$
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
Mar 31 at 1:12
$begingroup$
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
Mar 31 at 1:12
$begingroup$
@EeveeTrainer ok I got your idea.
$endgroup$
– hopefully
Mar 31 at 1:12
$begingroup$
@EeveeTrainer ok I got your idea.
$endgroup$
– hopefully
Mar 31 at 1:12
$begingroup$
Notice that the space is one representation of the Quaternions. See the section "Conjugation, the norm, and reciprocal".
$endgroup$
– Somos
Mar 31 at 4:29
$begingroup$
Notice that the space is one representation of the Quaternions. See the section "Conjugation, the norm, and reciprocal".
$endgroup$
– Somos
Mar 31 at 4:29
|
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
$$begin{bmatrix}
a & b\
-bar{b} & bar{a}
end{bmatrix} begin{bmatrix}
c & d\
-bar{d} & bar{c}
end{bmatrix} =begin{bmatrix}
ac - b bar{d} & ad+bbar{c}\
-bar{a} bar{d} - bar{b}c & bar{a} bar{c}-bar{b}d
end{bmatrix} $$
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.
$endgroup$
2
$begingroup$
I think the first term in the second element of the resulting matrix is ad not ab?
$endgroup$
– hopefully
Mar 31 at 2:14
$begingroup$
@hopefully Yeah, you're right, I made a typo. Thanks!
$endgroup$
– Eevee Trainer
Mar 31 at 2:20
$begingroup$
what about the terms that contain only one bar, like the second term of the bottom right entry?
$endgroup$
– hopefully
Mar 31 at 3:04
$begingroup$
What about them, exactly?
$endgroup$
– Eevee Trainer
Mar 31 at 3:05
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
Mar 31 at 3:09
|
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 $$J := pmatrix{cdot&-1\1&cdot}.$$ It follows immediately from the definition that $${X in M(2, Bbb C) : textrm{$X$ satisfies $X^dagger J = J X^top$}} .$$
So, for $X, Y in H$, $$(X Y)^dagger J = Y^dagger X^dagger J = Y^dagger JX^top = J Y^top X^top = J (XY)^top .$$
$endgroup$
add a comment |
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2 Answers
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2 Answers
2
active
oldest
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active
oldest
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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
$$begin{bmatrix}
a & b\
-bar{b} & bar{a}
end{bmatrix} begin{bmatrix}
c & d\
-bar{d} & bar{c}
end{bmatrix} =begin{bmatrix}
ac - b bar{d} & ad+bbar{c}\
-bar{a} bar{d} - bar{b}c & bar{a} bar{c}-bar{b}d
end{bmatrix} $$
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.
$endgroup$
2
$begingroup$
I think the first term in the second element of the resulting matrix is ad not ab?
$endgroup$
– hopefully
Mar 31 at 2:14
$begingroup$
@hopefully Yeah, you're right, I made a typo. Thanks!
$endgroup$
– Eevee Trainer
Mar 31 at 2:20
$begingroup$
what about the terms that contain only one bar, like the second term of the bottom right entry?
$endgroup$
– hopefully
Mar 31 at 3:04
$begingroup$
What about them, exactly?
$endgroup$
– Eevee Trainer
Mar 31 at 3:05
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
Mar 31 at 3:09
|
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.
$endgroup$
2
$begingroup$
I think the first term in the second element of the resulting matrix is ad not ab?
$endgroup$
– hopefully
Mar 31 at 2:14
$begingroup$
@hopefully Yeah, you're right, I made a typo. Thanks!
$endgroup$
– Eevee Trainer
Mar 31 at 2:20
$begingroup$
what about the terms that contain only one bar, like the second term of the bottom right entry?
$endgroup$
– hopefully
Mar 31 at 3:04
$begingroup$
What about them, exactly?
$endgroup$
– Eevee Trainer
Mar 31 at 3:05
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
Mar 31 at 3:09
|
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.
$endgroup$
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
$$begin{bmatrix}
a & b\
-bar{b} & bar{a}
end{bmatrix} begin{bmatrix}
c & d\
-bar{d} & bar{c}
end{bmatrix} =begin{bmatrix}
ac - b bar{d} & ad+bbar{c}\
-bar{a} bar{d} - bar{b}c & bar{a} bar{c}-bar{b}d
end{bmatrix} $$
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.
edited Mar 31 at 3:10
answered Mar 31 at 1:22
Eevee TrainerEevee Trainer
9,91631740
9,91631740
2
$begingroup$
I think the first term in the second element of the resulting matrix is ad not ab?
$endgroup$
– hopefully
Mar 31 at 2:14
$begingroup$
@hopefully Yeah, you're right, I made a typo. Thanks!
$endgroup$
– Eevee Trainer
Mar 31 at 2:20
$begingroup$
what about the terms that contain only one bar, like the second term of the bottom right entry?
$endgroup$
– hopefully
Mar 31 at 3:04
$begingroup$
What about them, exactly?
$endgroup$
– Eevee Trainer
Mar 31 at 3:05
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
Mar 31 at 3:09
|
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
Mar 31 at 2:14
$begingroup$
@hopefully Yeah, you're right, I made a typo. Thanks!
$endgroup$
– Eevee Trainer
Mar 31 at 2:20
$begingroup$
what about the terms that contain only one bar, like the second term of the bottom right entry?
$endgroup$
– hopefully
Mar 31 at 3:04
$begingroup$
What about them, exactly?
$endgroup$
– Eevee Trainer
Mar 31 at 3:05
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
Mar 31 at 3:09
2
2
$begingroup$
I think the first term in the second element of the resulting matrix is ad not ab?
$endgroup$
– hopefully
Mar 31 at 2:14
$begingroup$
I think the first term in the second element of the resulting matrix is ad not ab?
$endgroup$
– hopefully
Mar 31 at 2:14
$begingroup$
@hopefully Yeah, you're right, I made a typo. Thanks!
$endgroup$
– Eevee Trainer
Mar 31 at 2:20
$begingroup$
@hopefully Yeah, you're right, I made a typo. Thanks!
$endgroup$
– Eevee Trainer
Mar 31 at 2:20
$begingroup$
what about the terms that contain only one bar, like the second term of the bottom right entry?
$endgroup$
– hopefully
Mar 31 at 3:04
$begingroup$
what about the terms that contain only one bar, like the second term of the bottom right entry?
$endgroup$
– hopefully
Mar 31 at 3:04
$begingroup$
What about them, exactly?
$endgroup$
– Eevee Trainer
Mar 31 at 3:05
$begingroup$
What about them, exactly?
$endgroup$
– Eevee Trainer
Mar 31 at 3:05
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.
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– Eevee Trainer
Mar 31 at 3:09
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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
Mar 31 at 3:09
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show 3 more comments
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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 $X, Y in H$, $$(X Y)^dagger J = Y^dagger X^dagger J = Y^dagger JX^top = J Y^top X^top = J (XY)^top .$$
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add a comment |
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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 $X, Y in H$, $$(X Y)^dagger J = Y^dagger X^dagger J = Y^dagger JX^top = J Y^top X^top = J (XY)^top .$$
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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 $X, Y in H$, $$(X Y)^dagger J = Y^dagger X^dagger J = Y^dagger JX^top = J Y^top X^top = J (XY)^top .$$
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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 $X, Y in H$, $$(X Y)^dagger J = Y^dagger X^dagger J = Y^dagger JX^top = J Y^top X^top = J (XY)^top .$$
answered Mar 31 at 3:01
TravisTravis
64k769151
64k769151
add a comment |
add a comment |
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3
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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.
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– Eevee Trainer
Mar 31 at 1:10
2
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You multiply elements from H!
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– chhro
Mar 31 at 1:11
4
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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!
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– Chinnapparaj R
Mar 31 at 1:12
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@EeveeTrainer ok I got your idea.
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– hopefully
Mar 31 at 1:12
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Notice that the space is one representation of the Quaternions. See the section "Conjugation, the norm, and reciprocal".
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– Somos
Mar 31 at 4:29