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 6 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 6 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
6 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
5 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 6 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 6 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 6 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 6 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 6 hours ago
Doriano BrogioliDoriano Brogioli
361
asked 6 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
6 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
5 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
6 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
5 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
6 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
6 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
5 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
5 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 4 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
3 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 4 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
3 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 4 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
3 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 4 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 4 hours ago
Carlo BeenakkerCarlo Beenakker
79.7k9190292
answered 4 hours ago
Carlo BeenakkerCarlo Beenakker
79.7k9190292
answered 4 hours ago
Carlo BeenakkerCarlo Beenakker
79.7k9190292
answered 4 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
3 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
3 hours ago
$begingroup$
I don't think this paper ever considers condition (1) of the OP; "commuting" there just means commuting.
$endgroup$
– Christian Remling
3 hours ago
$begingroup$
I don't think this paper ever considers condition (1) of the OP; "commuting" there just means commuting.
$endgroup$
– Christian Remling
3 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
6 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
5 hours ago