Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Matrix rep. of form, before and after tensoring.

  1. Mar 17, 2009 #1

    WWGD

    User Avatar
    Science Advisor
    Gold Member

    Hi, everyone ,I would appreciate your help:

    Say A is a torsion-free Abelian group, with basis {v_1,..,v_n} ; AFAIK , a torsion-free modules over a PID are free (after $9.99 S&H :smile:)

    and we have a symmetric quadratic bilinear form

    Q(a,b) : AxA-->Z , with a matrix representation

    Q_ij = Q(v_i,v_j); i,j=1,..,n.

    Now, we tensor A with IR over Z , i.e., A(X)IR , tensor over Z -integers, where IR

    is the reals (with IR as an Abelian group , aka, Z-module, I assume;

    not clear in the paper, but I assume it is tensoring of Z-modules, which would seem obvious)


    Question:

    How does the matrix rep. of Q(a,b) change after
    the tensoring above (given that the map on A(x)IR is
    now (Z-)linear and not bilinear anymore.)?

    Do we just use the basis rep. of A(x)R for the
    matrix, using basis {v_i(x)1; i=1,..,n} ?.
    Is the matrix rep. still diagonalizable? ( the
    original matrix is symmetric, so diagonalizable)

    Thanks.
     
  2. jcsd
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Can you offer guidance or do you also need help?
Draft saved Draft deleted



Similar Discussions: Matrix rep. of form, before and after tensoring.
Loading...