Symplectic equivalent of commuting matrices
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It is well known what happens if two real symmetric matrices commute, i.e. if we have two matrices $A$ and $B$ such that $A=A^T$, $B=B^T$ and $AB=BA$. The answer is given in terms of diagonalization: there is a unitary matrix $M$ such that $A$ and $B$ are transformed into $A'=M^TAM$ and $B'=M^TBM$, and both $A'$ and $B'$ are diagonal.
Here I'm asking if any analogous property holds in the following case.
$A$ and $B$ are symmetric, i.e. $A=A^T$ and $B=B^T$. The following property holds:
$$AOmega B=BOmega A$$ (1)
where $Omega$ is the matrix defining the symplectic bilinear form (skew-symmetric, nonsingular, and hollow), e.g.:
$$Omega = begin{bmatrix}0 & 0 & 1 & 0\0 & 0 & 0 & 1\
-1 & 0 & 0 & 0 \ 0 & -1 & 0 & 0 end{bmatrix}$$
The allowed transformations are the symplectic matrices $M$, i.e. matrices for which the following holds:
$$M^TOmega M=Omega$$
The transformed matrices are $A'=M^TAM$ and $B'=M^TBM$.
My question is if there is a form into which $A'$ and $B'$ can be put, by means of a suitable $M$, provided that Eq.1 holds.
linear-algebra sg.symplectic-geometry
asked Apr 7 at 17:05
Doriano BrogioliDoriano Brogioli
483
New contributor
Doriano Brogioli is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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add a comment |
$begingroup$
It is well known what happens if two real symmetric matrices commute, i.e. if we have two matrices $A$ and $B$ such that $A=A^T$, $B=B^T$ and $AB=BA$. The answer is given in terms of diagonalization: there is a unitary matrix $M$ such that $A$ and $B$ are transformed into $A'=M^TAM$ and $B'=M^TBM$, and both $A'$ and $B'$ are diagonal.
Here I'm asking if any analogous property holds in the following case.
$A$ and $B$ are symmetric, i.e. $A=A^T$ and $B=B^T$. The following property holds:
$$AOmega B=BOmega A$$ (1)
where $Omega$ is the matrix defining the symplectic bilinear form (skew-symmetric, nonsingular, and hollow), e.g.:
$$Omega = begin{bmatrix}0 & 0 & 1 & 0\0 & 0 & 0 & 1\
-1 & 0 & 0 & 0 \ 0 & -1 & 0 & 0 end{bmatrix}$$
The allowed transformations are the symplectic matrices $M$, i.e. matrices for which the following holds:
$$M^TOmega M=Omega$$
The transformed matrices are $A'=M^TAM$ and $B'=M^TBM$.
My question is if there is a form into which $A'$ and $B'$ can be put, by means of a suitable $M$, provided that Eq.1 holds.
linear-algebra sg.symplectic-geometry
asked Apr 7 at 17:05
Doriano BrogioliDoriano Brogioli
483
New contributor
Doriano Brogioli is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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2
$begingroup$
No idea, but a comment, a bit unrelated: for many computations with such things it's better to write $Omega$ as $diag(M,ldots,M)$ where $M=(^0_{-1}{ }^1_0)$. Of course, in practice, it doesn't make that much of a difference.
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– Teo Banica
Apr 7 at 17:24
1
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For $n=2$, there's the formula $Omega AOmega = (det A) A^{-1T}$ ($=(det A) A^{-1}$ here), so if $det A=det B$, then (1) says that $A^{-1}B=B^{-1}A$, and in general, there's an extra constant. Of course, all this is a far cry from the full question, but it might give a hint.
$endgroup$
– Christian Remling
Apr 7 at 18:09
3
$begingroup$
If $A$ and $B$ are complex matrices for which $Omega A$ and $Omega B$ are (anti)symmetric and commute (the latter is equivalent to condition $(1)$), then there's a symplectic $S$ such that $S^{-1}Omega AS=D$ and $S^{-1}Omega BS=E$ are diagonal by Lemma 17 in the paper Carlo posted. Hence $S^top AS=-Omega D$ and $S^top BS=-Omega E$.
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– MTyson
Apr 8 at 0:18
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Interesting! Could you please make this an answer, and maybe extend it a bit? For example, I still do not understand if this applies when $A$ is symmetric, i.e. $A=A^T$.
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– Doriano Brogioli
Apr 8 at 9:52
2
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@DorianoBrogioli --- the symmetry of $A$ ensures that $Omega A$ is Hamiltonian (symplectic skew-symmetric), which is the condition needed for Lemma 17 mentioned by MTyson to apply.
$endgroup$
– Carlo Beenakker
Apr 8 at 10:09
add a comment |
$begingroup$
It is well known what happens if two real symmetric matrices commute, i.e. if we have two matrices $A$ and $B$ such that $A=A^T$, $B=B^T$ and $AB=BA$. The answer is given in terms of diagonalization: there is a unitary matrix $M$ such that $A$ and $B$ are transformed into $A'=M^TAM$ and $B'=M^TBM$, and both $A'$ and $B'$ are diagonal.
Here I'm asking if any analogous property holds in the following case.
$A$ and $B$ are symmetric, i.e. $A=A^T$ and $B=B^T$. The following property holds:
$$AOmega B=BOmega A$$ (1)
where $Omega$ is the matrix defining the symplectic bilinear form (skew-symmetric, nonsingular, and hollow), e.g.:
$$Omega = begin{bmatrix}0 & 0 & 1 & 0\0 & 0 & 0 & 1\
-1 & 0 & 0 & 0 \ 0 & -1 & 0 & 0 end{bmatrix}$$
The allowed transformations are the symplectic matrices $M$, i.e. matrices for which the following holds:
$$M^TOmega M=Omega$$
The transformed matrices are $A'=M^TAM$ and $B'=M^TBM$.
My question is if there is a form into which $A'$ and $B'$ can be put, by means of a suitable $M$, provided that Eq.1 holds.
linear-algebra sg.symplectic-geometry
asked Apr 7 at 17:05
Doriano BrogioliDoriano Brogioli
483
New contributor
Doriano Brogioli is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
It is well known what happens if two real symmetric matrices commute, i.e. if we have two matrices $A$ and $B$ such that $A=A^T$, $B=B^T$ and $AB=BA$. The answer is given in terms of diagonalization: there is a unitary matrix $M$ such that $A$ and $B$ are transformed into $A'=M^TAM$ and $B'=M^TBM$, and both $A'$ and $B'$ are diagonal.
Here I'm asking if any analogous property holds in the following case.
$A$ and $B$ are symmetric, i.e. $A=A^T$ and $B=B^T$. The following property holds:
$$AOmega B=BOmega A$$ (1)
where $Omega$ is the matrix defining the symplectic bilinear form (skew-symmetric, nonsingular, and hollow), e.g.:
$$Omega = begin{bmatrix}0 & 0 & 1 & 0\0 & 0 & 0 & 1\
-1 & 0 & 0 & 0 \ 0 & -1 & 0 & 0 end{bmatrix}$$
The allowed transformations are the symplectic matrices $M$, i.e. matrices for which the following holds:
$$M^TOmega M=Omega$$
The transformed matrices are $A'=M^TAM$ and $B'=M^TBM$.
My question is if there is a form into which $A'$ and $B'$ can be put, by means of a suitable $M$, provided that Eq.1 holds.
linear-algebra sg.symplectic-geometry
linear-algebra sg.symplectic-geometry
asked Apr 7 at 17:05
Doriano BrogioliDoriano Brogioli
483
New contributor
Doriano Brogioli is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Apr 7 at 17:05
Doriano BrogioliDoriano Brogioli
483
New contributor
Doriano Brogioli is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Apr 7 at 17:05
Doriano BrogioliDoriano Brogioli
483
New contributor
Doriano Brogioli is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Apr 7 at 17:05
Doriano BrogioliDoriano Brogioli
483
asked Apr 7 at 17:05
Doriano BrogioliDoriano Brogioli
483
483
New contributor
Doriano Brogioli is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Doriano Brogioli is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Doriano Brogioli is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
2
$begingroup$
No idea, but a comment, a bit unrelated: for many computations with such things it's better to write $Omega$ as $diag(M,ldots,M)$ where $M=(^0_{-1}{ }^1_0)$. Of course, in practice, it doesn't make that much of a difference.
$endgroup$
– Teo Banica
Apr 7 at 17:24
1
$begingroup$
For $n=2$, there's the formula $Omega AOmega = (det A) A^{-1T}$ ($=(det A) A^{-1}$ here), so if $det A=det B$, then (1) says that $A^{-1}B=B^{-1}A$, and in general, there's an extra constant. Of course, all this is a far cry from the full question, but it might give a hint.
$endgroup$
– Christian Remling
Apr 7 at 18:09
3
$begingroup$
If $A$ and $B$ are complex matrices for which $Omega A$ and $Omega B$ are (anti)symmetric and commute (the latter is equivalent to condition $(1)$), then there's a symplectic $S$ such that $S^{-1}Omega AS=D$ and $S^{-1}Omega BS=E$ are diagonal by Lemma 17 in the paper Carlo posted. Hence $S^top AS=-Omega D$ and $S^top BS=-Omega E$.
$endgroup$
– MTyson
Apr 8 at 0:18
$begingroup$
Interesting! Could you please make this an answer, and maybe extend it a bit? For example, I still do not understand if this applies when $A$ is symmetric, i.e. $A=A^T$.
$endgroup$
– Doriano Brogioli
Apr 8 at 9:52
2
$begingroup$
@DorianoBrogioli --- the symmetry of $A$ ensures that $Omega A$ is Hamiltonian (symplectic skew-symmetric), which is the condition needed for Lemma 17 mentioned by MTyson to apply.
$endgroup$
– Carlo Beenakker
Apr 8 at 10:09
add a comment |
2
$begingroup$
No idea, but a comment, a bit unrelated: for many computations with such things it's better to write $Omega$ as $diag(M,ldots,M)$ where $M=(^0_{-1}{ }^1_0)$. Of course, in practice, it doesn't make that much of a difference.
$endgroup$
– Teo Banica
Apr 7 at 17:24
1
$begingroup$
For $n=2$, there's the formula $Omega AOmega = (det A) A^{-1T}$ ($=(det A) A^{-1}$ here), so if $det A=det B$, then (1) says that $A^{-1}B=B^{-1}A$, and in general, there's an extra constant. Of course, all this is a far cry from the full question, but it might give a hint.
$endgroup$
– Christian Remling
Apr 7 at 18:09
3
$begingroup$
If $A$ and $B$ are complex matrices for which $Omega A$ and $Omega B$ are (anti)symmetric and commute (the latter is equivalent to condition $(1)$), then there's a symplectic $S$ such that $S^{-1}Omega AS=D$ and $S^{-1}Omega BS=E$ are diagonal by Lemma 17 in the paper Carlo posted. Hence $S^top AS=-Omega D$ and $S^top BS=-Omega E$.
$endgroup$
– MTyson
Apr 8 at 0:18
$begingroup$
Interesting! Could you please make this an answer, and maybe extend it a bit? For example, I still do not understand if this applies when $A$ is symmetric, i.e. $A=A^T$.
$endgroup$
– Doriano Brogioli
Apr 8 at 9:52
2
$begingroup$
@DorianoBrogioli --- the symmetry of $A$ ensures that $Omega A$ is Hamiltonian (symplectic skew-symmetric), which is the condition needed for Lemma 17 mentioned by MTyson to apply.
$endgroup$
– Carlo Beenakker
Apr 8 at 10:09
2
2
$begingroup$
No idea, but a comment, a bit unrelated: for many computations with such things it's better to write $Omega$ as $diag(M,ldots,M)$ where $M=(^0_{-1}{ }^1_0)$. Of course, in practice, it doesn't make that much of a difference.
$endgroup$
– Teo Banica
Apr 7 at 17:24
$begingroup$
No idea, but a comment, a bit unrelated: for many computations with such things it's better to write $Omega$ as $diag(M,ldots,M)$ where $M=(^0_{-1}{ }^1_0)$. Of course, in practice, it doesn't make that much of a difference.
$endgroup$
– Teo Banica
Apr 7 at 17:24
1
1
$begingroup$
For $n=2$, there's the formula $Omega AOmega = (det A) A^{-1T}$ ($=(det A) A^{-1}$ here), so if $det A=det B$, then (1) says that $A^{-1}B=B^{-1}A$, and in general, there's an extra constant. Of course, all this is a far cry from the full question, but it might give a hint.
$endgroup$
– Christian Remling
Apr 7 at 18:09
$begingroup$
For $n=2$, there's the formula $Omega AOmega = (det A) A^{-1T}$ ($=(det A) A^{-1}$ here), so if $det A=det B$, then (1) says that $A^{-1}B=B^{-1}A$, and in general, there's an extra constant. Of course, all this is a far cry from the full question, but it might give a hint.
$endgroup$
– Christian Remling
Apr 7 at 18:09
3
3
$begingroup$
If $A$ and $B$ are complex matrices for which $Omega A$ and $Omega B$ are (anti)symmetric and commute (the latter is equivalent to condition $(1)$), then there's a symplectic $S$ such that $S^{-1}Omega AS=D$ and $S^{-1}Omega BS=E$ are diagonal by Lemma 17 in the paper Carlo posted. Hence $S^top AS=-Omega D$ and $S^top BS=-Omega E$.
$endgroup$
– MTyson
Apr 8 at 0:18
$begingroup$
If $A$ and $B$ are complex matrices for which $Omega A$ and $Omega B$ are (anti)symmetric and commute (the latter is equivalent to condition $(1)$), then there's a symplectic $S$ such that $S^{-1}Omega AS=D$ and $S^{-1}Omega BS=E$ are diagonal by Lemma 17 in the paper Carlo posted. Hence $S^top AS=-Omega D$ and $S^top BS=-Omega E$.
$endgroup$
– MTyson
Apr 8 at 0:18
$begingroup$
Interesting! Could you please make this an answer, and maybe extend it a bit? For example, I still do not understand if this applies when $A$ is symmetric, i.e. $A=A^T$.
$endgroup$
– Doriano Brogioli
Apr 8 at 9:52
$begingroup$
Interesting! Could you please make this an answer, and maybe extend it a bit? For example, I still do not understand if this applies when $A$ is symmetric, i.e. $A=A^T$.
$endgroup$
– Doriano Brogioli
Apr 8 at 9:52
2
2
$begingroup$
@DorianoBrogioli --- the symmetry of $A$ ensures that $Omega A$ is Hamiltonian (symplectic skew-symmetric), which is the condition needed for Lemma 17 mentioned by MTyson to apply.
$endgroup$
– Carlo Beenakker
Apr 8 at 10:09
$begingroup$
@DorianoBrogioli --- the symmetry of $A$ ensures that $Omega A$ is Hamiltonian (symplectic skew-symmetric), which is the condition needed for Lemma 17 mentioned by MTyson to apply.
$endgroup$
– Carlo Beenakker
Apr 8 at 10:09
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Let $A$ and $B$ be complex symmetric matrices of even dimension which satisfy condition $(1)$ and for which $Omega A$ and $Omega B$ are diagonalizable. Then $Omega A$ and $Omega B$ are Hamiltonian (not (anti)symmetric as my comment said):
$$Omega^top (Omega A)^top Omega=-Omega A^topOmega^topOmega=-Omega A.$$
Condition $(1)$ means that $Omega A$ and $Omega B$ commute. By Lemma 17 in the paper in Carlo's answer, there's a symplectic $S$ such that $S^{-1}Omega AS=D$ and $S^{-1}Omega BS=E$ are diagonal. Hence $S^top AS=-Omega D$ and $S^top BS=-Omega E$ are of the same form.
answered Apr 8 at 13:04
MTysonMTyson
1,4061611
$endgroup$
$begingroup$
Very useful and complete. Could you please comment on the requisite that $Omega A$ and $Omega B$ are diagonalizable? Can be expressed as a more explicit condition on $A$ and $B$? And what happens if they are not diagonalizable?
$endgroup$
– Doriano Brogioli
Apr 8 at 20:59
1
$begingroup$
@DorianoBrogioli If $S^top AS=-Omega D$ then $S^{-1}Omega AS=D$, so this is a necessary condition. I don't know any equivalent conditions on $A$ nor if there's a Jordan-like form that the matrices can always be put into.
$endgroup$
– MTyson
Apr 8 at 21:58
1
$begingroup$
There is a symplectic version of the Jordan normal form (of a single matrix). Have a look at the paper Conjugacy classes in linear groups.
$endgroup$
– Tobias Diez
Apr 10 at 9:49
add a comment |
$begingroup$
The symplectic counterpart of the fact that a family of commuting diagonalizable matrices is simultaneously diagonalizable is discussed in section 3.1 of On the diagonalizability of a matrix by a symplectic equivalence, similarity or congruence transformation
answered Apr 7 at 19:41
Carlo BeenakkerCarlo Beenakker
80.5k9193295
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2
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I don't think this paper ever considers condition (1) of the OP; "commuting" there just means commuting.
$endgroup$
– Christian Remling
Apr 7 at 20:27
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I agree. As far as I can understand, the paper has to do with the topic but does not directly contain the answer, and further elaboration and work is needed. Maybe the comment of MTyson (to the question) can help.
$endgroup$
– Doriano Brogioli
Apr 8 at 9:54
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
$begingroup$
Let $A$ and $B$ be complex symmetric matrices of even dimension which satisfy condition $(1)$ and for which $Omega A$ and $Omega B$ are diagonalizable. Then $Omega A$ and $Omega B$ are Hamiltonian (not (anti)symmetric as my comment said):
$$Omega^top (Omega A)^top Omega=-Omega A^topOmega^topOmega=-Omega A.$$
Condition $(1)$ means that $Omega A$ and $Omega B$ commute. By Lemma 17 in the paper in Carlo's answer, there's a symplectic $S$ such that $S^{-1}Omega AS=D$ and $S^{-1}Omega BS=E$ are diagonal. Hence $S^top AS=-Omega D$ and $S^top BS=-Omega E$ are of the same form.
answered Apr 8 at 13:04
MTysonMTyson
1,4061611
$endgroup$
$begingroup$
Very useful and complete. Could you please comment on the requisite that $Omega A$ and $Omega B$ are diagonalizable? Can be expressed as a more explicit condition on $A$ and $B$? And what happens if they are not diagonalizable?
$endgroup$
– Doriano Brogioli
Apr 8 at 20:59
1
$begingroup$
@DorianoBrogioli If $S^top AS=-Omega D$ then $S^{-1}Omega AS=D$, so this is a necessary condition. I don't know any equivalent conditions on $A$ nor if there's a Jordan-like form that the matrices can always be put into.
$endgroup$
– MTyson
Apr 8 at 21:58
1
$begingroup$
There is a symplectic version of the Jordan normal form (of a single matrix). Have a look at the paper Conjugacy classes in linear groups.
$endgroup$
– Tobias Diez
Apr 10 at 9:49
add a comment |
$begingroup$
Let $A$ and $B$ be complex symmetric matrices of even dimension which satisfy condition $(1)$ and for which $Omega A$ and $Omega B$ are diagonalizable. Then $Omega A$ and $Omega B$ are Hamiltonian (not (anti)symmetric as my comment said):
$$Omega^top (Omega A)^top Omega=-Omega A^topOmega^topOmega=-Omega A.$$
Condition $(1)$ means that $Omega A$ and $Omega B$ commute. By Lemma 17 in the paper in Carlo's answer, there's a symplectic $S$ such that $S^{-1}Omega AS=D$ and $S^{-1}Omega BS=E$ are diagonal. Hence $S^top AS=-Omega D$ and $S^top BS=-Omega E$ are of the same form.
answered Apr 8 at 13:04
MTysonMTyson
1,4061611
$endgroup$
$begingroup$
Very useful and complete. Could you please comment on the requisite that $Omega A$ and $Omega B$ are diagonalizable? Can be expressed as a more explicit condition on $A$ and $B$? And what happens if they are not diagonalizable?
$endgroup$
– Doriano Brogioli
Apr 8 at 20:59
1
$begingroup$
@DorianoBrogioli If $S^top AS=-Omega D$ then $S^{-1}Omega AS=D$, so this is a necessary condition. I don't know any equivalent conditions on $A$ nor if there's a Jordan-like form that the matrices can always be put into.
$endgroup$
– MTyson
Apr 8 at 21:58
1
$begingroup$
There is a symplectic version of the Jordan normal form (of a single matrix). Have a look at the paper Conjugacy classes in linear groups.
$endgroup$
– Tobias Diez
Apr 10 at 9:49
add a comment |
$begingroup$
Let $A$ and $B$ be complex symmetric matrices of even dimension which satisfy condition $(1)$ and for which $Omega A$ and $Omega B$ are diagonalizable. Then $Omega A$ and $Omega B$ are Hamiltonian (not (anti)symmetric as my comment said):
$$Omega^top (Omega A)^top Omega=-Omega A^topOmega^topOmega=-Omega A.$$
Condition $(1)$ means that $Omega A$ and $Omega B$ commute. By Lemma 17 in the paper in Carlo's answer, there's a symplectic $S$ such that $S^{-1}Omega AS=D$ and $S^{-1}Omega BS=E$ are diagonal. Hence $S^top AS=-Omega D$ and $S^top BS=-Omega E$ are of the same form.
answered Apr 8 at 13:04
MTysonMTyson
1,4061611
$endgroup$
Let $A$ and $B$ be complex symmetric matrices of even dimension which satisfy condition $(1)$ and for which $Omega A$ and $Omega B$ are diagonalizable. Then $Omega A$ and $Omega B$ are Hamiltonian (not (anti)symmetric as my comment said):
$$Omega^top (Omega A)^top Omega=-Omega A^topOmega^topOmega=-Omega A.$$
Condition $(1)$ means that $Omega A$ and $Omega B$ commute. By Lemma 17 in the paper in Carlo's answer, there's a symplectic $S$ such that $S^{-1}Omega AS=D$ and $S^{-1}Omega BS=E$ are diagonal. Hence $S^top AS=-Omega D$ and $S^top BS=-Omega E$ are of the same form.
answered Apr 8 at 13:04
MTysonMTyson
1,4061611
edited Apr 8 at 13:24
answered Apr 8 at 13:04
MTysonMTyson
1,4061611
answered Apr 8 at 13:04
MTysonMTyson
1,4061611
answered Apr 8 at 13:04
MTysonMTyson
1,4061611
1,4061611
$begingroup$
Very useful and complete. Could you please comment on the requisite that $Omega A$ and $Omega B$ are diagonalizable? Can be expressed as a more explicit condition on $A$ and $B$? And what happens if they are not diagonalizable?
$endgroup$
– Doriano Brogioli
Apr 8 at 20:59
1
$begingroup$
@DorianoBrogioli If $S^top AS=-Omega D$ then $S^{-1}Omega AS=D$, so this is a necessary condition. I don't know any equivalent conditions on $A$ nor if there's a Jordan-like form that the matrices can always be put into.
$endgroup$
– MTyson
Apr 8 at 21:58
1
$begingroup$
There is a symplectic version of the Jordan normal form (of a single matrix). Have a look at the paper Conjugacy classes in linear groups.
$endgroup$
– Tobias Diez
Apr 10 at 9:49
add a comment |
$begingroup$
Very useful and complete. Could you please comment on the requisite that $Omega A$ and $Omega B$ are diagonalizable? Can be expressed as a more explicit condition on $A$ and $B$? And what happens if they are not diagonalizable?
$endgroup$
– Doriano Brogioli
Apr 8 at 20:59
1
$begingroup$
@DorianoBrogioli If $S^top AS=-Omega D$ then $S^{-1}Omega AS=D$, so this is a necessary condition. I don't know any equivalent conditions on $A$ nor if there's a Jordan-like form that the matrices can always be put into.
$endgroup$
– MTyson
Apr 8 at 21:58
1
$begingroup$
There is a symplectic version of the Jordan normal form (of a single matrix). Have a look at the paper Conjugacy classes in linear groups.
$endgroup$
– Tobias Diez
Apr 10 at 9:49
$begingroup$
Very useful and complete. Could you please comment on the requisite that $Omega A$ and $Omega B$ are diagonalizable? Can be expressed as a more explicit condition on $A$ and $B$? And what happens if they are not diagonalizable?
$endgroup$
– Doriano Brogioli
Apr 8 at 20:59
$begingroup$
Very useful and complete. Could you please comment on the requisite that $Omega A$ and $Omega B$ are diagonalizable? Can be expressed as a more explicit condition on $A$ and $B$? And what happens if they are not diagonalizable?
$endgroup$
– Doriano Brogioli
Apr 8 at 20:59
1
1
$begingroup$
@DorianoBrogioli If $S^top AS=-Omega D$ then $S^{-1}Omega AS=D$, so this is a necessary condition. I don't know any equivalent conditions on $A$ nor if there's a Jordan-like form that the matrices can always be put into.
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– MTyson
Apr 8 at 21:58
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@DorianoBrogioli If $S^top AS=-Omega D$ then $S^{-1}Omega AS=D$, so this is a necessary condition. I don't know any equivalent conditions on $A$ nor if there's a Jordan-like form that the matrices can always be put into.
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– MTyson
Apr 8 at 21:58
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There is a symplectic version of the Jordan normal form (of a single matrix). Have a look at the paper Conjugacy classes in linear groups.
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– Tobias Diez
Apr 10 at 9:49
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There is a symplectic version of the Jordan normal form (of a single matrix). Have a look at the paper Conjugacy classes in linear groups.
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– Tobias Diez
Apr 10 at 9:49
add a comment |
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The symplectic counterpart of the fact that a family of commuting diagonalizable matrices is simultaneously diagonalizable is discussed in section 3.1 of On the diagonalizability of a matrix by a symplectic equivalence, similarity or congruence transformation
answered Apr 7 at 19:41
Carlo BeenakkerCarlo Beenakker
80.5k9193295
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2
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I don't think this paper ever considers condition (1) of the OP; "commuting" there just means commuting.
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– Christian Remling
Apr 7 at 20:27
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I agree. As far as I can understand, the paper has to do with the topic but does not directly contain the answer, and further elaboration and work is needed. Maybe the comment of MTyson (to the question) can help.
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– Doriano Brogioli
Apr 8 at 9:54
add a comment |
$begingroup$
The symplectic counterpart of the fact that a family of commuting diagonalizable matrices is simultaneously diagonalizable is discussed in section 3.1 of On the diagonalizability of a matrix by a symplectic equivalence, similarity or congruence transformation
answered Apr 7 at 19:41
Carlo BeenakkerCarlo Beenakker
80.5k9193295
$endgroup$
2
$begingroup$
I don't think this paper ever considers condition (1) of the OP; "commuting" there just means commuting.
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– Christian Remling
Apr 7 at 20:27
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I agree. As far as I can understand, the paper has to do with the topic but does not directly contain the answer, and further elaboration and work is needed. Maybe the comment of MTyson (to the question) can help.
$endgroup$
– Doriano Brogioli
Apr 8 at 9:54
add a comment |
$begingroup$
The symplectic counterpart of the fact that a family of commuting diagonalizable matrices is simultaneously diagonalizable is discussed in section 3.1 of On the diagonalizability of a matrix by a symplectic equivalence, similarity or congruence transformation
answered Apr 7 at 19:41
Carlo BeenakkerCarlo Beenakker
80.5k9193295
$endgroup$
The symplectic counterpart of the fact that a family of commuting diagonalizable matrices is simultaneously diagonalizable is discussed in section 3.1 of On the diagonalizability of a matrix by a symplectic equivalence, similarity or congruence transformation
answered Apr 7 at 19:41
Carlo BeenakkerCarlo Beenakker
80.5k9193295
answered Apr 7 at 19:41
Carlo BeenakkerCarlo Beenakker
80.5k9193295
answered Apr 7 at 19:41
Carlo BeenakkerCarlo Beenakker
80.5k9193295
answered Apr 7 at 19:41
Carlo BeenakkerCarlo Beenakker
80.5k9193295
80.5k9193295
2
$begingroup$
I don't think this paper ever considers condition (1) of the OP; "commuting" there just means commuting.
$endgroup$
– Christian Remling
Apr 7 at 20:27
$begingroup$
I agree. As far as I can understand, the paper has to do with the topic but does not directly contain the answer, and further elaboration and work is needed. Maybe the comment of MTyson (to the question) can help.
$endgroup$
– Doriano Brogioli
Apr 8 at 9:54
add a comment |
2
$begingroup$
I don't think this paper ever considers condition (1) of the OP; "commuting" there just means commuting.
$endgroup$
– Christian Remling
Apr 7 at 20:27
$begingroup$
I agree. As far as I can understand, the paper has to do with the topic but does not directly contain the answer, and further elaboration and work is needed. Maybe the comment of MTyson (to the question) can help.
$endgroup$
– Doriano Brogioli
Apr 8 at 9:54
2
2
$begingroup$
I don't think this paper ever considers condition (1) of the OP; "commuting" there just means commuting.
$endgroup$
– Christian Remling
Apr 7 at 20:27
$begingroup$
I don't think this paper ever considers condition (1) of the OP; "commuting" there just means commuting.
$endgroup$
– Christian Remling
Apr 7 at 20:27
$begingroup$
I agree. As far as I can understand, the paper has to do with the topic but does not directly contain the answer, and further elaboration and work is needed. Maybe the comment of MTyson (to the question) can help.
$endgroup$
– Doriano Brogioli
Apr 8 at 9:54
$begingroup$
I agree. As far as I can understand, the paper has to do with the topic but does not directly contain the answer, and further elaboration and work is needed. Maybe the comment of MTyson (to the question) can help.
$endgroup$
– Doriano Brogioli
Apr 8 at 9:54
add a comment |
Doriano Brogioli is a new contributor. Be nice, and check out our Code of Conduct.
Doriano Brogioli is a new contributor. Be nice, and check out our Code of Conduct.
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No idea, but a comment, a bit unrelated: for many computations with such things it's better to write $Omega$ as $diag(M,ldots,M)$ where $M=(^0_{-1}{ }^1_0)$. Of course, in practice, it doesn't make that much of a difference.
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– Teo Banica
Apr 7 at 17:24
1
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For $n=2$, there's the formula $Omega AOmega = (det A) A^{-1T}$ ($=(det A) A^{-1}$ here), so if $det A=det B$, then (1) says that $A^{-1}B=B^{-1}A$, and in general, there's an extra constant. Of course, all this is a far cry from the full question, but it might give a hint.
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– Christian Remling
Apr 7 at 18:09
3
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If $A$ and $B$ are complex matrices for which $Omega A$ and $Omega B$ are (anti)symmetric and commute (the latter is equivalent to condition $(1)$), then there's a symplectic $S$ such that $S^{-1}Omega AS=D$ and $S^{-1}Omega BS=E$ are diagonal by Lemma 17 in the paper Carlo posted. Hence $S^top AS=-Omega D$ and $S^top BS=-Omega E$.
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– MTyson
Apr 8 at 0:18
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Interesting! Could you please make this an answer, and maybe extend it a bit? For example, I still do not understand if this applies when $A$ is symmetric, i.e. $A=A^T$.
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– Doriano Brogioli
Apr 8 at 9:52
2
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@DorianoBrogioli --- the symmetry of $A$ ensures that $Omega A$ is Hamiltonian (symplectic skew-symmetric), which is the condition needed for Lemma 17 mentioned by MTyson to apply.
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– Carlo Beenakker
Apr 8 at 10:09