The defining matrix of a symplectic matrix
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Just a beginner in symplectic geometry, and the definition of symplectic matrix bothers me.
A $2ntimes 2n$ real matrix $M$ is said to be symplectic if it satisfies the following condition:
$$M^TOmega M=Omega$$
where $Omega$ is a fixed $2ntimes 2n$ real, invertible and skew-symmetric matrix.
My question is: since $Omega$ can be arbitrary, so if $Omega,Delta$ are both satisfy the condition, then the following statement must be true:
$$M^TOmega M=Omega Rightarrow M^TDelta M=Delta.$$
But I don't know how to prove this. Can anyone help me? Thanks.
linear-algebra symplectic-geometry symplectic-linear-algebra
asked Nov 15 at 9:30
Arc Walker
11
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Arc Walker is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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add a comment |
up vote
0
down vote
favorite
Just a beginner in symplectic geometry, and the definition of symplectic matrix bothers me.
A $2ntimes 2n$ real matrix $M$ is said to be symplectic if it satisfies the following condition:
$$M^TOmega M=Omega$$
where $Omega$ is a fixed $2ntimes 2n$ real, invertible and skew-symmetric matrix.
My question is: since $Omega$ can be arbitrary, so if $Omega,Delta$ are both satisfy the condition, then the following statement must be true:
$$M^TOmega M=Omega Rightarrow M^TDelta M=Delta.$$
But I don't know how to prove this. Can anyone help me? Thanks.
linear-algebra symplectic-geometry symplectic-linear-algebra
asked Nov 15 at 9:30
Arc Walker
11
New contributor
Arc Walker is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Just a beginner in symplectic geometry, and the definition of symplectic matrix bothers me.
A $2ntimes 2n$ real matrix $M$ is said to be symplectic if it satisfies the following condition:
$$M^TOmega M=Omega$$
where $Omega$ is a fixed $2ntimes 2n$ real, invertible and skew-symmetric matrix.
My question is: since $Omega$ can be arbitrary, so if $Omega,Delta$ are both satisfy the condition, then the following statement must be true:
$$M^TOmega M=Omega Rightarrow M^TDelta M=Delta.$$
But I don't know how to prove this. Can anyone help me? Thanks.
linear-algebra symplectic-geometry symplectic-linear-algebra
asked Nov 15 at 9:30
Arc Walker
11
New contributor
Arc Walker is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Just a beginner in symplectic geometry, and the definition of symplectic matrix bothers me.
A $2ntimes 2n$ real matrix $M$ is said to be symplectic if it satisfies the following condition:
$$M^TOmega M=Omega$$
where $Omega$ is a fixed $2ntimes 2n$ real, invertible and skew-symmetric matrix.
My question is: since $Omega$ can be arbitrary, so if $Omega,Delta$ are both satisfy the condition, then the following statement must be true:
$$M^TOmega M=Omega Rightarrow M^TDelta M=Delta.$$
But I don't know how to prove this. Can anyone help me? Thanks.
linear-algebra symplectic-geometry symplectic-linear-algebra
linear-algebra symplectic-geometry symplectic-linear-algebra
asked Nov 15 at 9:30
Arc Walker
11
New contributor
Arc Walker is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Nov 15 at 9:30
Arc Walker
11
New contributor
Arc Walker is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Nov 15 at 9:30
Arc Walker
11
New contributor
Arc Walker is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Nov 15 at 9:30
Arc Walker
11
asked Nov 15 at 9:30
Arc Walker
11
11
New contributor
Arc Walker is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Arc Walker is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Arc Walker is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
1 Answer
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There are more than one symplectic structure oon a vector space, but they are isomorphic not equal, there exists a linear invertible map such that $fcircDelta =Omegacirc f$ where $Omega$ and $Delta$ are the linear map associated to the corresponding matrices.
answered Nov 15 at 9:33
Tsemo Aristide
54.3k11344
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
There are more than one symplectic structure oon a vector space, but they are isomorphic not equal, there exists a linear invertible map such that $fcircDelta =Omegacirc f$ where $Omega$ and $Delta$ are the linear map associated to the corresponding matrices.
answered Nov 15 at 9:33
Tsemo Aristide
54.3k11344
add a comment |
up vote
2
down vote
There are more than one symplectic structure oon a vector space, but they are isomorphic not equal, there exists a linear invertible map such that $fcircDelta =Omegacirc f$ where $Omega$ and $Delta$ are the linear map associated to the corresponding matrices.
answered Nov 15 at 9:33
Tsemo Aristide
54.3k11344
add a comment |
up vote
2
down vote
up vote
2
down vote
There are more than one symplectic structure oon a vector space, but they are isomorphic not equal, there exists a linear invertible map such that $fcircDelta =Omegacirc f$ where $Omega$ and $Delta$ are the linear map associated to the corresponding matrices.
answered Nov 15 at 9:33
Tsemo Aristide
54.3k11344
There are more than one symplectic structure oon a vector space, but they are isomorphic not equal, there exists a linear invertible map such that $fcircDelta =Omegacirc f$ where $Omega$ and $Delta$ are the linear map associated to the corresponding matrices.
answered Nov 15 at 9:33
Tsemo Aristide
54.3k11344
answered Nov 15 at 9:33
Tsemo Aristide
54.3k11344
answered Nov 15 at 9:33
Tsemo Aristide
54.3k11344
answered Nov 15 at 9:33
Tsemo Aristide
54.3k11344
54.3k11344
add a comment |
add a comment |
Arc Walker is a new contributor. Be nice, and check out our Code of Conduct.
Arc Walker is a new contributor. Be nice, and check out our Code of Conduct.
Arc Walker is a new contributor. Be nice, and check out our Code of Conduct.
Arc Walker is a new contributor. Be nice, and check out our Code of Conduct.
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