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Invertibility of Symplectic Matrices

  1. Jan 25, 2012 #1

    WWGD

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    Hi, All:

    The Wikipedia page on symplectic matrices:

    http://en.wikipedia.org/wiki/Symplectic_vector_space ,

    claims that symplectic matrices are invertible

    , i.e., skew-symmetric nxn-

    matrix with entries w(b_i,b_j) , satisfying the properties:

    i)w(b_i,b_i)=0

    ii)w(b_i,b_j)=-w(b_j,b_i)

    iii)w(b_i,.)=0 , i.e., w(b_i,b_j)=0 for all b_j

    are invertible.

    Even for small n , calculating the determinant seems to get out of hand;

    Is there an easy way of seeing this?

    TIA

    Thanks.

    iii)w(bi,.)=0 , then bi=0
     
  2. jcsd
  3. Jan 25, 2012 #2

    morphism

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    Call the matrix of your symplectic form J. Then a matrix A is symplectic (wrt your form) iff A^t J A = J. From this you get (detA)^2 = 1, since det J != 0 (which is because J is coming from nondegenerate bilinear form). In particular, detA=+/-1 is nonzero.

    In fact, you can show that detA=1, but this much more difficult...
     
  4. Jan 25, 2012 #3

    WWGD

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    A couple of followup questions:

    i)AFAIK, symplectic forms and symmetric/quadratic forms represent a generalization of

    , I think it was inner-product and maybe the norm or length, maybe even a metric;

    anyone know?

    TIA.
     
  5. Jan 25, 2012 #4

    WWGD

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    Thanks, morphism, but I was actually trying to show that J itself is nonsingular; I don't see

    how the nondegeneracy condition helps show invertibility. I guess one could choose a basis

    {b1,..,bn}, and then set up something like:

    w(bi,bj)=biTJbj , where T is the matrix (w_ij:=w(bi,bj)) , for the above choice

    of basis {b1,b2,..,bn} and then w(bi,bj)=0 for all bj only if bi=0 . Maybe we can

    also set up a symplectic basis {b1',b2',..,bn'} , i.e., a basis so that w(bi',bj')=

    δi'j' . But then the result may be basis-dependent.

    Any Ideas?
     
  6. Jan 25, 2012 #5

    morphism

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    A symplectic form is a nondegenerate skew-symmetric bilinear form.
    A (real) inner product is a nondegenerate symmetric bilinear form. (Well, it's not just nondegenerate - it's positive-definite.)

    The invertibility is basis-independent, because there's an invertible change of basis matrix involved. I.e. if you write A for the matrix of your nondegenerate bilinear form in one basis, and write B for it in another basis, then there will be an invertible matrix S such that A=SBS^{-1}.
     
    Last edited: Jan 25, 2012
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