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Matrix rep. of form, before and after tensoring.

  1. Mar 17, 2009 #1


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    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)


    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)

  2. jcsd
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