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How is this set of matrices closed under multiplication?



The Next CEO of Stack OverflowHow to determine if a set is a subspace of the vector space of all complex $2times 2$ matrices?Converting $mathbbC$ linear tranformation with determinant $a+bi$ into an $mathbbR$-linear transformation with determinant $a^2+b^2$.Is this inequality trivial?Showing that a very well-known representation is really a representationWrite out the multiplication table for the following set of matrices over $mathbb Q$Is multiplication an operation in the given set of matrices?Why are (a), (c), (d) true?Let $T:mathbb C^3tomathbb C^3$.Then, adjoint $T^*$ of $T$Prove that set $mathbbS$ forms group under matrix multiplicationAbout subalgebra of Hamilton










2












$begingroup$



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







share|cite|improve this question











$endgroup$







  • 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
    3 hours ago







  • 2




    $begingroup$
    You multiply elements from H!
    $endgroup$
    – chhro
    3 hours ago






  • 4




    $begingroup$
    Find $$beginpmatrix z_1 & z_2\ -barz_2& barz_1 endpmatrix beginpmatrix w_1 & w_2\ -barw_2& barw_1 endpmatrix$$ and arrange the entries in the required form!
    $endgroup$
    – Chinnapparaj R
    3 hours ago










  • $begingroup$
    @EeveeTrainer ok I got your idea.
    $endgroup$
    – hopefully
    3 hours ago










  • $begingroup$
    Notice that the space is one representation of the Quaternions. See the section "Conjugation, the norm, and reciprocal".
    $endgroup$
    – Somos
    5 mins ago















2












$begingroup$



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







share|cite|improve this question











$endgroup$







  • 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
    3 hours ago







  • 2




    $begingroup$
    You multiply elements from H!
    $endgroup$
    – chhro
    3 hours ago






  • 4




    $begingroup$
    Find $$beginpmatrix z_1 & z_2\ -barz_2& barz_1 endpmatrix beginpmatrix w_1 & w_2\ -barw_2& barw_1 endpmatrix$$ and arrange the entries in the required form!
    $endgroup$
    – Chinnapparaj R
    3 hours ago










  • $begingroup$
    @EeveeTrainer ok I got your idea.
    $endgroup$
    – hopefully
    3 hours ago










  • $begingroup$
    Notice that the space is one representation of the Quaternions. See the section "Conjugation, the norm, and reciprocal".
    $endgroup$
    – Somos
    5 mins ago













2












2








2





$begingroup$



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







share|cite|improve this question











$endgroup$





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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 2 hours ago









Rócherz

3,0013821




3,0013821










asked 3 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
    3 hours ago







  • 2




    $begingroup$
    You multiply elements from H!
    $endgroup$
    – chhro
    3 hours ago






  • 4




    $begingroup$
    Find $$beginpmatrix z_1 & z_2\ -barz_2& barz_1 endpmatrix beginpmatrix w_1 & w_2\ -barw_2& barw_1 endpmatrix$$ and arrange the entries in the required form!
    $endgroup$
    – Chinnapparaj R
    3 hours ago










  • $begingroup$
    @EeveeTrainer ok I got your idea.
    $endgroup$
    – hopefully
    3 hours ago










  • $begingroup$
    Notice that the space is one representation of the Quaternions. See the section "Conjugation, the norm, and reciprocal".
    $endgroup$
    – Somos
    5 mins ago












  • 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
    3 hours ago







  • 2




    $begingroup$
    You multiply elements from H!
    $endgroup$
    – chhro
    3 hours ago






  • 4




    $begingroup$
    Find $$beginpmatrix z_1 & z_2\ -barz_2& barz_1 endpmatrix beginpmatrix w_1 & w_2\ -barw_2& barw_1 endpmatrix$$ and arrange the entries in the required form!
    $endgroup$
    – Chinnapparaj R
    3 hours ago










  • $begingroup$
    @EeveeTrainer ok I got your idea.
    $endgroup$
    – hopefully
    3 hours ago










  • $begingroup$
    Notice that the space is one representation of the Quaternions. See the section "Conjugation, the norm, and reciprocal".
    $endgroup$
    – Somos
    5 mins 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
3 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
3 hours ago





2




2




$begingroup$
You multiply elements from H!
$endgroup$
– chhro
3 hours ago




$begingroup$
You multiply elements from H!
$endgroup$
– chhro
3 hours ago




4




4




$begingroup$
Find $$beginpmatrix z_1 & z_2\ -barz_2& barz_1 endpmatrix beginpmatrix w_1 & w_2\ -barw_2& barw_1 endpmatrix$$ and arrange the entries in the required form!
$endgroup$
– Chinnapparaj R
3 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
3 hours ago












$begingroup$
@EeveeTrainer ok I got your idea.
$endgroup$
– hopefully
3 hours ago




$begingroup$
@EeveeTrainer ok I got your idea.
$endgroup$
– hopefully
3 hours ago












$begingroup$
Notice that the space is one representation of the Quaternions. See the section "Conjugation, the norm, and reciprocal".
$endgroup$
– Somos
5 mins ago




$begingroup$
Notice that the space is one representation of the Quaternions. See the section "Conjugation, the norm, and reciprocal".
$endgroup$
– Somos
5 mins ago










2 Answers
2






active

oldest

votes


















4












$begingroup$

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.






share|cite|improve this answer











$endgroup$








  • 2




    $begingroup$
    I think the first term in the second element of the resulting matrix is ad not ab?
    $endgroup$
    – hopefully
    2 hours ago










  • $begingroup$
    @hopefully Yeah, you're right, I made a typo. Thanks!
    $endgroup$
    – Eevee Trainer
    2 hours ago










  • $begingroup$
    what about the terms that contain only one bar, like the second term of the bottom right entry?
    $endgroup$
    – hopefully
    1 hour ago










  • $begingroup$
    What about them, exactly?
    $endgroup$
    – Eevee Trainer
    1 hour 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
    1 hour ago


















1












$begingroup$

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 $X, Y in H$, $$(X Y)^dagger J = Y^dagger X^dagger J = Y^dagger JX^top = J Y^top X^top = J (XY)^top .$$







share|cite|improve this answer









$endgroup$













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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    4












    $begingroup$

    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.






    share|cite|improve this answer











    $endgroup$








    • 2




      $begingroup$
      I think the first term in the second element of the resulting matrix is ad not ab?
      $endgroup$
      – hopefully
      2 hours ago










    • $begingroup$
      @hopefully Yeah, you're right, I made a typo. Thanks!
      $endgroup$
      – Eevee Trainer
      2 hours ago










    • $begingroup$
      what about the terms that contain only one bar, like the second term of the bottom right entry?
      $endgroup$
      – hopefully
      1 hour ago










    • $begingroup$
      What about them, exactly?
      $endgroup$
      – Eevee Trainer
      1 hour 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
      1 hour ago















    4












    $begingroup$

    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.






    share|cite|improve this answer











    $endgroup$








    • 2




      $begingroup$
      I think the first term in the second element of the resulting matrix is ad not ab?
      $endgroup$
      – hopefully
      2 hours ago










    • $begingroup$
      @hopefully Yeah, you're right, I made a typo. Thanks!
      $endgroup$
      – Eevee Trainer
      2 hours ago










    • $begingroup$
      what about the terms that contain only one bar, like the second term of the bottom right entry?
      $endgroup$
      – hopefully
      1 hour ago










    • $begingroup$
      What about them, exactly?
      $endgroup$
      – Eevee Trainer
      1 hour 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
      1 hour ago













    4












    4








    4





    $begingroup$

    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.






    share|cite|improve this answer











    $endgroup$



    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.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited 1 hour ago

























    answered 3 hours ago









    Eevee TrainerEevee Trainer

    8,99631640




    8,99631640







    • 2




      $begingroup$
      I think the first term in the second element of the resulting matrix is ad not ab?
      $endgroup$
      – hopefully
      2 hours ago










    • $begingroup$
      @hopefully Yeah, you're right, I made a typo. Thanks!
      $endgroup$
      – Eevee Trainer
      2 hours ago










    • $begingroup$
      what about the terms that contain only one bar, like the second term of the bottom right entry?
      $endgroup$
      – hopefully
      1 hour ago










    • $begingroup$
      What about them, exactly?
      $endgroup$
      – Eevee Trainer
      1 hour 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
      1 hour ago












    • 2




      $begingroup$
      I think the first term in the second element of the resulting matrix is ad not ab?
      $endgroup$
      – hopefully
      2 hours ago










    • $begingroup$
      @hopefully Yeah, you're right, I made a typo. Thanks!
      $endgroup$
      – Eevee Trainer
      2 hours ago










    • $begingroup$
      what about the terms that contain only one bar, like the second term of the bottom right entry?
      $endgroup$
      – hopefully
      1 hour ago










    • $begingroup$
      What about them, exactly?
      $endgroup$
      – Eevee Trainer
      1 hour 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
      1 hour ago







    2




    2




    $begingroup$
    I think the first term in the second element of the resulting matrix is ad not ab?
    $endgroup$
    – hopefully
    2 hours ago




    $begingroup$
    I think the first term in the second element of the resulting matrix is ad not ab?
    $endgroup$
    – hopefully
    2 hours ago












    $begingroup$
    @hopefully Yeah, you're right, I made a typo. Thanks!
    $endgroup$
    – Eevee Trainer
    2 hours ago




    $begingroup$
    @hopefully Yeah, you're right, I made a typo. Thanks!
    $endgroup$
    – Eevee Trainer
    2 hours ago












    $begingroup$
    what about the terms that contain only one bar, like the second term of the bottom right entry?
    $endgroup$
    – hopefully
    1 hour ago




    $begingroup$
    what about the terms that contain only one bar, like the second term of the bottom right entry?
    $endgroup$
    – hopefully
    1 hour ago












    $begingroup$
    What about them, exactly?
    $endgroup$
    – Eevee Trainer
    1 hour ago




    $begingroup$
    What about them, exactly?
    $endgroup$
    – Eevee Trainer
    1 hour 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
    1 hour 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
    1 hour ago











    1












    $begingroup$

    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 $X, Y in H$, $$(X Y)^dagger J = Y^dagger X^dagger J = Y^dagger JX^top = J Y^top X^top = J (XY)^top .$$







    share|cite|improve this answer









    $endgroup$

















      1












      $begingroup$

      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 $X, Y in H$, $$(X Y)^dagger J = Y^dagger X^dagger J = Y^dagger JX^top = J Y^top X^top = J (XY)^top .$$







      share|cite|improve this answer









      $endgroup$















        1












        1








        1





        $begingroup$

        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 $X, Y in H$, $$(X Y)^dagger J = Y^dagger X^dagger J = Y^dagger JX^top = J Y^top X^top = J (XY)^top .$$







        share|cite|improve this answer









        $endgroup$



        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 $X, Y in H$, $$(X Y)^dagger J = Y^dagger X^dagger J = Y^dagger JX^top = J Y^top X^top = J (XY)^top .$$








        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 1 hour ago









        TravisTravis

        63.8k769151




        63.8k769151



























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