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
The exact meaning of 'Mom made me a sandwich'
Is there a difference between "Fahrstuhl" and "Aufzug"
Why is my new battery behaving weirdly?
Why don't programming languages automatically manage the synchronous/asynchronous problem?
Is French Guiana a (hard) EU border?
Would a grinding machine be a simple and workable propulsion system for an interplanetary spacecraft?
Solving system of ODEs with extra parameter
How did people program for Consoles with multiple CPUs?
If Nick Fury and Coulson already knew about aliens (Kree and Skrull) why did they wait until Thor's appearance to start making weapons?
Is there always a complete, orthogonal set of unitary matrices?
Is it my responsibility to learn a new technology in my own time my employer wants to implement?
0 rank tensor vs 1D vector
"misplaced omit" error when >centering columns
Bartok - Syncopation (1): Meaning of notes in between Grand Staff
Is it professional to write unrelated content in an almost-empty email?
Can I use the load factor to estimate the lift?
How to count occurrences of text in a file?
How to get from Geneva Airport to Metabief, Doubs, France by public transport?
Newlines in BSD sed vs gsed
Why, when going from special to general relativity, do we just replace partial derivatives with covariant derivatives?
Is it possible to replace duplicates of a character with one character using tr
Is it convenient to ask the journal's editor for two additional days to complete a review?
When you upcast Blindness/Deafness, do all targets suffer the same effect?
Why does standard notation not preserve intervals (visually)
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 2 hours ago
Rócherz
3,0013821
asked 2 hours ago
hopefullyhopefully
294214
$endgroup$
add a 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 2 hours 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, $c = -barb$ 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
2 hours ago
2
$begingroup$
You multiply elements from H!
$endgroup$
– chhro
2 hours ago
3
$begingroup$
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
2 hours ago
$begingroup$
@EeveeTrainer ok I got your idea.
$endgroup$
– hopefully
2 hours ago
add a comment |
$begingroup$
Consider the set of matrices $$H = left left(beginarrayrl z_1&z_2\ -bar z_2&bar z_1 endarrayright) mid z_1, z_2 in mathbb C right.$$ 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 2 hours ago
Rócherz
3,0013821
asked 2 hours ago
hopefullyhopefully
294214
$endgroup$
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 2 hours ago
Rócherz
3,0013821
asked 2 hours ago
hopefullyhopefully
294214
edited 2 hours ago
Rócherz
3,0013821
asked 2 hours ago
hopefullyhopefully
294214
edited 2 hours ago
Rócherz
3,0013821
edited 2 hours ago
Rócherz
3,0013821
edited 2 hours 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, $c = -barb$ 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
2 hours ago
2
$begingroup$
You multiply elements from H!
$endgroup$
– chhro
2 hours ago
3
$begingroup$
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
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, $c = -barb$ 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
2 hours ago
2
$begingroup$
You multiply elements from H!
$endgroup$
– chhro
2 hours ago
3
$begingroup$
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
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, $c = -barb$ 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
2 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
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 $$beginpmatrix z_1 & z_2\ -barz_2& barz_1 endpmatrix beginpmatrix w_1 & w_2\ -barw_2& barw_1 endpmatrix$$ and arrange the entries in the required form!
$endgroup$
– Chinnapparaj R
2 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
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
$$beginbmatrix
a & b\
-barb & bara
endbmatrix beginbmatrix
c & d\
-bard & barc
endbmatrix =beginbmatrix
ac - b bard & ad+bbarc\
-bara bard - barbc & bara barc-barbd
endbmatrix $$
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 2 hours ago
Eevee TrainerEevee Trainer
8,98631640
$endgroup$
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
35 mins ago
$begingroup$
What about them, exactly?
$endgroup$
– Eevee Trainer
35 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
31 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 $$J := pmatrixcdot&-1\1&cdot.$$ 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 39 mins ago
TravisTravis
63.8k769151
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
);
);
, "mathjax-editing");
StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);
else
createEditor();
);
function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);
);
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3168941%2fhow-is-this-set-of-matrices-closed-under-multiplication%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
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 2 hours ago
Eevee TrainerEevee Trainer
8,98631640
$endgroup$
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
35 mins ago
$begingroup$
What about them, exactly?
$endgroup$
– Eevee Trainer
35 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
31 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
$$beginbmatrix
a & b\
-barb & bara
endbmatrix beginbmatrix
c & d\
-bard & barc
endbmatrix =beginbmatrix
ac - b bard & ad+bbarc\
-bara bard - barbc & bara barc-barbd
endbmatrix $$
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 2 hours ago
Eevee TrainerEevee Trainer
8,98631640
$endgroup$
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
35 mins ago
$begingroup$
What about them, exactly?
$endgroup$
– Eevee Trainer
35 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
31 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
$$beginbmatrix
a & b\
-barb & bara
endbmatrix beginbmatrix
c & d\
-bard & barc
endbmatrix =beginbmatrix
ac - b bard & ad+bbarc\
-bara bard - barbc & bara barc-barbd
endbmatrix $$
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 2 hours ago
Eevee TrainerEevee Trainer
8,98631640
$endgroup$
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 2 hours ago
Eevee TrainerEevee Trainer
8,98631640
edited 30 mins ago
answered 2 hours ago
Eevee TrainerEevee Trainer
8,98631640
answered 2 hours ago
Eevee TrainerEevee Trainer
8,98631640
answered 2 hours ago
Eevee TrainerEevee Trainer
8,98631640
8,98631640
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
35 mins ago
$begingroup$
What about them, exactly?
$endgroup$
– Eevee Trainer
35 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
31 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
35 mins ago
$begingroup$
What about them, exactly?
$endgroup$
– Eevee Trainer
35 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
31 mins ago
2
2
$begingroup$
I think the first term in the second element of the resulting matrix is ad not ab?
$endgroup$
– hopefully
1 hour ago
$begingroup$
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
$begingroup$
@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
35 mins ago
$begingroup$
what about the terms that contain only one bar, like the second term of the bottom right entry?
$endgroup$
– hopefully
35 mins ago
$begingroup$
What about them, exactly?
$endgroup$
– Eevee Trainer
35 mins ago
$begingroup$
What about them, exactly?
$endgroup$
– Eevee Trainer
35 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
31 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.
$endgroup$
– Eevee Trainer
31 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 $$J := pmatrixcdot&-1\1&cdot.$$ 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 39 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 $$J := pmatrixcdot&-1\1&cdot.$$ 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 39 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 $$J := pmatrixcdot&-1\1&cdot.$$ 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 39 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 $$J := pmatrixcdot&-1\1&cdot.$$ 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 39 mins ago
TravisTravis
63.8k769151
answered 39 mins ago
TravisTravis
63.8k769151
answered 39 mins ago
TravisTravis
63.8k769151
answered 39 mins ago
TravisTravis
63.8k769151
63.8k769151
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3168941%2fhow-is-this-set-of-matrices-closed-under-multiplication%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
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
2 hours ago
2
$begingroup$
You multiply elements from H!
$endgroup$
– chhro
2 hours ago
3
$begingroup$
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
2 hours ago
$begingroup$
@EeveeTrainer ok I got your idea.
$endgroup$
– hopefully
2 hours ago