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?



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












2












$begingroup$



Consider the set of matrices $$H = left{ left(begin{array}{rl} z_1&z_2\ -bar z_2&bar z_1 end{array}right) mid z_1, z_2 in mathbb C right}.$$ 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







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, $c = -bar{b}$ 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
















2












$begingroup$



Consider the set of matrices $$H = left{ left(begin{array}{rl} z_1&z_2\ -bar z_2&bar z_1 end{array}right) mid z_1, z_2 in mathbb C right}.$$ 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







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, $c = -bar{b}$ 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














2












2








2





$begingroup$



Consider the set of matrices $$H = left{ left(begin{array}{rl} z_1&z_2\ -bar z_2&bar z_1 end{array}right) mid z_1, z_2 in mathbb C right}.$$ 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







share|cite|improve this question











$endgroup$





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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 1 hour ago









Rócherz

3,0013821




3,0013821










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














  • 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










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



$$begin{bmatrix}
a & b\
-bar{b} & bar{a}
end{bmatrix} begin{bmatrix}
c & d\
-bar{d} & bar{c}
end{bmatrix} =begin{bmatrix}
ac - b bar{d} & ad+bbar{c}\
-bar{a} bar{d} - bar{b}c & bar{a} bar{c}-bar{b}d
end{bmatrix} $$



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.






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
    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



















1












$begingroup$

Here's an alternative method that, after verification of the simple characterization of this subspace given below, is coordinate-free.



Hint Denote $$J := pmatrix{cdot&-1\1&cdot}.$$ It follows immediately from the definition that $${X in M(2, Bbb C) : textrm{$X$ satisfies $X^dagger J = J X^top$}} .$$




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







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



    $$begin{bmatrix}
    a & b\
    -bar{b} & bar{a}
    end{bmatrix} begin{bmatrix}
    c & d\
    -bar{d} & bar{c}
    end{bmatrix} =begin{bmatrix}
    ac - b bar{d} & ad+bbar{c}\
    -bar{a} bar{d} - bar{b}c & bar{a} bar{c}-bar{b}d
    end{bmatrix} $$



    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.






    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
      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
















    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



    $$begin{bmatrix}
    a & b\
    -bar{b} & bar{a}
    end{bmatrix} begin{bmatrix}
    c & d\
    -bar{d} & bar{c}
    end{bmatrix} =begin{bmatrix}
    ac - b bar{d} & ad+bbar{c}\
    -bar{a} bar{d} - bar{b}c & bar{a} bar{c}-bar{b}d
    end{bmatrix} $$



    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.






    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
      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














    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



    $$begin{bmatrix}
    a & b\
    -bar{b} & bar{a}
    end{bmatrix} begin{bmatrix}
    c & d\
    -bar{d} & bar{c}
    end{bmatrix} =begin{bmatrix}
    ac - b bar{d} & ad+bbar{c}\
    -bar{a} bar{d} - bar{b}c & bar{a} bar{c}-bar{b}d
    end{bmatrix} $$



    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.






    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



    $$begin{bmatrix}
    a & b\
    -bar{b} & bar{a}
    end{bmatrix} begin{bmatrix}
    c & d\
    -bar{d} & bar{c}
    end{bmatrix} =begin{bmatrix}
    ac - b bar{d} & ad+bbar{c}\
    -bar{a} bar{d} - bar{b}c & bar{a} bar{c}-bar{b}d
    end{bmatrix} $$



    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.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited 15 mins ago

























    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














    • 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








    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
    20 mins 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




    $begingroup$
    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.
    $endgroup$
    – Eevee Trainer
    16 mins 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 $$J := pmatrix{cdot&-1\1&cdot}.$$ It follows immediately from the definition that $${X in M(2, Bbb C) : textrm{$X$ satisfies $X^dagger J = J X^top$}} .$$




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







    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 $$J := pmatrix{cdot&-1\1&cdot}.$$ It follows immediately from the definition that $${X in M(2, Bbb C) : textrm{$X$ satisfies $X^dagger J = J X^top$}} .$$




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







      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 $$J := pmatrix{cdot&-1\1&cdot}.$$ It follows immediately from the definition that $${X in M(2, Bbb C) : textrm{$X$ satisfies $X^dagger J = J X^top$}} .$$




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







        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 $$J := pmatrix{cdot&-1\1&cdot}.$$ It follows immediately from the definition that $${X in M(2, Bbb C) : textrm{$X$ satisfies $X^dagger J = J X^top$}} .$$




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








        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 24 mins ago









        TravisTravis

        63.8k769151




        63.8k769151






























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