How is this set of matrices closed under multiplication? The Next CEO of Stack OverflowHow to...
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How is this set of matrices closed under multiplication?
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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 $mathbb{C}$ linear tranformation with determinant $a+bi$ into an $mathbb{R}$-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 $mathbb{S}$ forms group under matrix multiplicationAbout subalgebra of Hamilton
$begingroup$
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
edited 1 hour ago
Rócherz
3,0013821
asked 2 hours ago
hopefullyhopefully
294214
$endgroup$
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
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
edited 1 hour ago
Rócherz
3,0013821
asked 2 hours ago
hopefullyhopefully
294214
$endgroup$
Consider the set of matrices $$H = left{ left(begin{array}{rl} z_1&z_2\ -bar z_2&bar z_1 end{array}right) mid z_1, z_2 in mathbb C right}.$$ 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
linear-algebra abstract-algebra group-theory complex-numbers
edited 1 hour ago
Rócherz
3,0013821
asked 2 hours ago
hopefullyhopefully
294214
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
edited 1 hour ago
Rócherz
3,0013821
3,0013821
asked 2 hours ago
hopefullyhopefully
294214
asked 2 hours ago
hopefullyhopefully
294214
asked 2 hours ago
hopefullyhopefully
294214
294214
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 |
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
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
2 hours ago
$begingroup$
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
2
$begingroup$
You multiply elements from H!
$endgroup$
– chhro
2 hours ago
$begingroup$
You multiply elements from H!
$endgroup$
– chhro
2 hours ago
3
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
$begingroup$
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
$begingroup$
@EeveeTrainer ok I got your idea.
$endgroup$
– hopefully
2 hours ago
add a 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.
answered 2 hours 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
20 mins ago
$begingroup$
What about them, exactly?
$endgroup$
– Eevee Trainer
20 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
16 mins 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 $$J := pmatrix{cdot&-1\1&cdot}.$$ 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 24 mins 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
$$begin{bmatrix}
a & b\
-bar{b} & bar{a}
end{bmatrix} begin{bmatrix}
c & d\
-bar{d} & bar{c}
end{bmatrix} =begin{bmatrix}
ac - b bar{d} & ad+bbar{c}\
-bar{a} bar{d} - bar{b}c & bar{a} bar{c}-bar{b}d
end{bmatrix} $$
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.
answered 2 hours 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
20 mins ago
$begingroup$
What about them, exactly?
$endgroup$
– Eevee Trainer
20 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
16 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.
answered 2 hours 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
20 mins ago
$begingroup$
What about them, exactly?
$endgroup$
– Eevee Trainer
20 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
16 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.
answered 2 hours ago
Eevee TrainerEevee Trainer
8,98431640
$endgroup$
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.
answered 2 hours ago
Eevee TrainerEevee Trainer
8,98431640
edited 15 mins ago
answered 2 hours ago
Eevee TrainerEevee Trainer
8,98431640
answered 2 hours ago
Eevee TrainerEevee Trainer
8,98431640
answered 2 hours ago
Eevee TrainerEevee Trainer
8,98431640
8,98431640
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
20 mins ago
$begingroup$
What about them, exactly?
$endgroup$
– Eevee Trainer
20 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
16 mins 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
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
20 mins ago
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What about them, exactly?
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– Eevee Trainer
20 mins 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
16 mins 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
1 hour 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
1 hour ago
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@hopefully Yeah, you're right, I made a typo. Thanks!
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– Eevee Trainer
1 hour ago
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@hopefully Yeah, you're right, I made a typo. Thanks!
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– Eevee Trainer
1 hour 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
20 mins 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
20 mins ago
$begingroup$
What about them, exactly?
$endgroup$
– Eevee Trainer
20 mins ago
$begingroup$
What about them, exactly?
$endgroup$
– Eevee Trainer
20 mins 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
16 mins 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
16 mins 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 $$J := pmatrix{cdot&-1\1&cdot}.$$ 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 24 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 $X, Y in H$, $$(X Y)^dagger J = Y^dagger X^dagger J = Y^dagger JX^top = J Y^top X^top = J (XY)^top .$$
answered 24 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 $X, Y in H$, $$(X Y)^dagger J = Y^dagger X^dagger J = Y^dagger JX^top = J Y^top X^top = J (XY)^top .$$
answered 24 mins 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 $$J := pmatrix{cdot&-1\1&cdot}.$$ 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 24 mins ago
TravisTravis
63.8k769151
answered 24 mins ago
TravisTravis
63.8k769151
answered 24 mins ago
TravisTravis
63.8k769151
answered 24 mins ago
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, $c = -bar{b}$ 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
2 hours ago
2
$begingroup$
You multiply elements from H!
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– 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!
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– Chinnapparaj R
2 hours ago
$begingroup$
@EeveeTrainer ok I got your idea.
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– hopefully
2 hours ago