Symplectic equivalent of commuting matricesProving “almost all matrices over C are diagonalizable”.Polar...
Symplectic equivalent of commuting matrices
Proving “almost all matrices over C are diagonalizable”.Polar decomposition for quaternionic matrices?Characterizing symplectic matrices relative to a partial Iwasawa decompositionApproximating commuting matrices by commuting diagonalizable matricesSymplectic block-diagonalization of a real symmetric Hamiltonian matrixDo skew symmetric matrices ever naturally represent linear transformations?Constant symplectic structureCenter of matricesDiagonalization of real symmetric matrices with symplectic matricesIs every stable matrix orthogonally similar to a $D$-skew-symmetric matrix?
$begingroup$
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 yesterday
Doriano BrogioliDoriano Brogioli
462
New contributor
Doriano Brogioli is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$
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 yesterday
Doriano BrogioliDoriano Brogioli
462
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.
$endgroup$
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
yesterday
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
yesterday
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
yesterday
$begingroup$
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
18 hours ago
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
18 hours ago
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 yesterday
Doriano BrogioliDoriano Brogioli
462
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 yesterday
Doriano BrogioliDoriano Brogioli
462
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 yesterday
Doriano BrogioliDoriano Brogioli
462
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 yesterday
Doriano BrogioliDoriano Brogioli
462
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 yesterday
Doriano BrogioliDoriano Brogioli
462
asked yesterday
Doriano BrogioliDoriano Brogioli
462
462
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
yesterday
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
yesterday
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
yesterday
$begingroup$
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
18 hours ago
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
18 hours ago
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
yesterday
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
yesterday
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
yesterday
$begingroup$
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
18 hours ago
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
18 hours ago
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
yesterday
$begingroup$
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
yesterday
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
yesterday
$begingroup$
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
yesterday
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
yesterday
$begingroup$
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
yesterday
$begingroup$
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
18 hours ago
$begingroup$
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
18 hours ago
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
18 hours ago
$begingroup$
@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
18 hours ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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 yesterday
Carlo BeenakkerCarlo Beenakker
79.8k9190293
$endgroup$
2
$begingroup$
I don't think this paper ever considers condition (1) of the OP; "commuting" there just means commuting.
$endgroup$
– Christian Remling
yesterday
$begingroup$
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
18 hours ago
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 15 hours ago
MTysonMTyson
1,3811611
$endgroup$
$begingroup$
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
7 hours ago
$begingroup$
@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
6 hours ago
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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 yesterday
Carlo BeenakkerCarlo Beenakker
79.8k9190293
$endgroup$
2
$begingroup$
I don't think this paper ever considers condition (1) of the OP; "commuting" there just means commuting.
$endgroup$
– Christian Remling
yesterday
$begingroup$
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
18 hours ago
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 yesterday
Carlo BeenakkerCarlo Beenakker
79.8k9190293
$endgroup$
2
$begingroup$
I don't think this paper ever considers condition (1) of the OP; "commuting" there just means commuting.
$endgroup$
– Christian Remling
yesterday
$begingroup$
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
18 hours ago
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 yesterday
Carlo BeenakkerCarlo Beenakker
79.8k9190293
$endgroup$
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 yesterday
Carlo BeenakkerCarlo Beenakker
79.8k9190293
answered yesterday
Carlo BeenakkerCarlo Beenakker
79.8k9190293
answered yesterday
Carlo BeenakkerCarlo Beenakker
79.8k9190293
answered yesterday
Carlo BeenakkerCarlo Beenakker
79.8k9190293
79.8k9190293
2
$begingroup$
I don't think this paper ever considers condition (1) of the OP; "commuting" there just means commuting.
$endgroup$
– Christian Remling
yesterday
$begingroup$
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
18 hours ago
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
yesterday
$begingroup$
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
18 hours ago
2
2
$begingroup$
I don't think this paper ever considers condition (1) of the OP; "commuting" there just means commuting.
$endgroup$
– Christian Remling
yesterday
$begingroup$
I don't think this paper ever considers condition (1) of the OP; "commuting" there just means commuting.
$endgroup$
– Christian Remling
yesterday
$begingroup$
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
18 hours ago
$begingroup$
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
18 hours ago
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 15 hours ago
MTysonMTyson
1,3811611
$endgroup$
$begingroup$
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
7 hours ago
$begingroup$
@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
6 hours ago
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 15 hours ago
MTysonMTyson
1,3811611
$endgroup$
$begingroup$
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
7 hours ago
$begingroup$
@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
6 hours ago
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 15 hours ago
MTysonMTyson
1,3811611
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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 15 hours ago
MTysonMTyson
1,3811611
edited 14 hours ago
answered 15 hours ago
MTysonMTyson
1,3811611
answered 15 hours ago
MTysonMTyson
1,3811611
answered 15 hours ago
MTysonMTyson
1,3811611
1,3811611
$begingroup$
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?
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– Doriano Brogioli
7 hours ago
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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
6 hours ago
add a comment |
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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
7 hours ago
$begingroup$
@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
6 hours ago
$begingroup$
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
7 hours ago
$begingroup$
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
7 hours ago
$begingroup$
@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
6 hours ago
$begingroup$
@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
6 hours ago
add a comment |
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
yesterday
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
yesterday
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
yesterday
$begingroup$
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. $A=A^T$.
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– Doriano Brogioli
18 hours ago
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
18 hours ago