Symplectic equivalent of commuting matricesProving “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?
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.:
$$Omega = beginbmatrix0 & 0 & 1 & 0\0 & 0 & 0 & 1\
-1 & 0 & 0 & 0 \ 0 & -1 & 0 & 0 endbmatrix$$
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 11 hours ago
Doriano BrogioliDoriano Brogioli
361
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.
$endgroup$
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.:
$$Omega = beginbmatrix0 & 0 & 1 & 0\0 & 0 & 0 & 1\
-1 & 0 & 0 & 0 \ 0 & -1 & 0 & 0 endbmatrix$$
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 11 hours ago
Doriano BrogioliDoriano Brogioli
361
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
10 hours ago
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 $A^-1B=B^-1A$, 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
10 hours ago
$begingroup$
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 $S^-1Omega AS=D$ and $S^-1Omega BS=E$ are diagonal by Lemma 17 in the paper Carlo posted. Hence $S^top AS=-Omega D$ and $S^top BS=-Omega E$.
$endgroup$
– MTyson
4 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.:
$$Omega = beginbmatrix0 & 0 & 1 & 0\0 & 0 & 0 & 1\
-1 & 0 & 0 & 0 \ 0 & -1 & 0 & 0 endbmatrix$$
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 11 hours ago
Doriano BrogioliDoriano Brogioli
361
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.:
$$Omega = beginbmatrix0 & 0 & 1 & 0\0 & 0 & 0 & 1\
-1 & 0 & 0 & 0 \ 0 & -1 & 0 & 0 endbmatrix$$
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 11 hours ago
Doriano BrogioliDoriano Brogioli
361
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 11 hours ago
Doriano BrogioliDoriano Brogioli
361
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 11 hours ago
Doriano BrogioliDoriano Brogioli
361
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 11 hours ago
Doriano BrogioliDoriano Brogioli
361
asked 11 hours ago
Doriano BrogioliDoriano Brogioli
361
361
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
10 hours ago
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 $A^-1B=B^-1A$, 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
10 hours ago
$begingroup$
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 $S^-1Omega AS=D$ and $S^-1Omega BS=E$ are diagonal by Lemma 17 in the paper Carlo posted. Hence $S^top AS=-Omega D$ and $S^top BS=-Omega E$.
$endgroup$
– MTyson
4 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
10 hours ago
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 $A^-1B=B^-1A$, 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
10 hours ago
$begingroup$
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 $S^-1Omega AS=D$ and $S^-1Omega BS=E$ are diagonal by Lemma 17 in the paper Carlo posted. Hence $S^top AS=-Omega D$ and $S^top BS=-Omega E$.
$endgroup$
– MTyson
4 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
10 hours ago
$begingroup$
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
10 hours ago
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 $A^-1B=B^-1A$, 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
10 hours ago
$begingroup$
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 $A^-1B=B^-1A$, 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
10 hours ago
$begingroup$
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 $S^-1Omega AS=D$ and $S^-1Omega BS=E$ are diagonal by Lemma 17 in the paper Carlo posted. Hence $S^top AS=-Omega D$ and $S^top BS=-Omega E$.
$endgroup$
– MTyson
4 hours ago
$begingroup$
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 $S^-1Omega AS=D$ and $S^-1Omega BS=E$ are diagonal by Lemma 17 in the paper Carlo posted. Hence $S^top AS=-Omega D$ and $S^top BS=-Omega E$.
$endgroup$
– MTyson
4 hours ago
add a comment |
1 Answer
1
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 8 hours ago
Carlo BeenakkerCarlo Beenakker
79.7k9190292
$endgroup$
$begingroup$
I don't think this paper ever considers condition (1) of the OP; "commuting" there just means commuting.
$endgroup$
– Christian Remling
7 hours ago
add a comment |
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$begingroup$
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 8 hours ago
Carlo BeenakkerCarlo Beenakker
79.7k9190292
$endgroup$
$begingroup$
I don't think this paper ever considers condition (1) of the OP; "commuting" there just means commuting.
$endgroup$
– Christian Remling
7 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 8 hours ago
Carlo BeenakkerCarlo Beenakker
79.7k9190292
$endgroup$
$begingroup$
I don't think this paper ever considers condition (1) of the OP; "commuting" there just means commuting.
$endgroup$
– Christian Remling
7 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 8 hours ago
Carlo BeenakkerCarlo Beenakker
79.7k9190292
$endgroup$
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 8 hours ago
Carlo BeenakkerCarlo Beenakker
79.7k9190292
answered 8 hours ago
Carlo BeenakkerCarlo Beenakker
79.7k9190292
answered 8 hours ago
Carlo BeenakkerCarlo Beenakker
79.7k9190292
answered 8 hours ago
Carlo BeenakkerCarlo Beenakker
79.7k9190292
79.7k9190292
$begingroup$
I don't think this paper ever considers condition (1) of the OP; "commuting" there just means commuting.
$endgroup$
– Christian Remling
7 hours ago
add a comment |
$begingroup$
I don't think this paper ever considers condition (1) of the OP; "commuting" there just means commuting.
$endgroup$
– Christian Remling
7 hours ago
$begingroup$
I don't think this paper ever considers condition (1) of the OP; "commuting" there just means commuting.
$endgroup$
– Christian Remling
7 hours ago
$begingroup$
I don't think this paper ever considers condition (1) of the OP; "commuting" there just means commuting.
$endgroup$
– Christian Remling
7 hours ago
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.
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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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
10 hours ago
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 $A^-1B=B^-1A$, 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
10 hours ago
$begingroup$
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 $S^-1Omega AS=D$ and $S^-1Omega BS=E$ are diagonal by Lemma 17 in the paper Carlo posted. Hence $S^top AS=-Omega D$ and $S^top BS=-Omega E$.
$endgroup$
– MTyson
4 hours ago