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Math Help - Integral Basis...

  1. #1
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    Integral Basis...

    Suppose I have a basis A=(a1,a2,...,an) for the nullspace of a matrix with integer coefficients (i.e. Q-linear combinations of A will give me the span of A). I want to find the integral basis for this (perhaps my terminology isn't correct... but what I mean is I want to find a basis B=(b1,b2,...,bn) where the bi's are integers, such that Z-linear combinations of B will give me the span of the A.)

    I'm not sure how to do this. In algebra class, given a field Q(sqrt(d)) I remember finding integral basis' for this... but I can't figure out whether what I'm doing here is at all similar... or much easier... any suggestions would be appreciated!
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  2. #2
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    Re: Integral Basis...

    Hey gummy_ratz.

    Will the basis of A span A and thus give you a basis such that all linear combinations of those vector span the set A? If you have a full-rank system, then the basis should provide you with your basis b.
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  3. #3
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    Re: Integral Basis...

    suppose A = {a1,...,an} is a basis for a vector space V.

    suppose further that k ≠ 0 in the field F that V is a vector space over.

    is B = {ka1,...,kan} also a basis? let's see:

    suppose c1(ka1) +...+ cn(kan) = 0.

    then k(c1a1 +...+ cnan) = 0.

    since k ≠ 0, c1a1 +...+ cnan = 0.

    by the linear independence of the aj, c1 =...= cn = 0.

    thus the kaj are also linearly independent.

    we know that the aj span V. this means given ANY v in V we can write:

    v = c1a1 +...+ cnan for some cj in F.

    thus v = (c1 /k)(ka1) +....+ (cn/k)(kan), so the kaj span V as well.

    now, suppose that each aj = pj/qj, where the p's and q's are relatively prime integers for each j.

    what about k = lcm(q1,..,qn)? this certainly gives a basis that is all integers.

    however, i don't believe you can realize the span of this basis as Z-linear combinations, only as Q-linear combinations.

    let me give an example:

    suppose we have the matrix:

    A = \begin{bmatrix}2&1&3\\4&-5&1\\6&-4&4 \end{bmatrix}

    which has a null space with basis {(8,5,-7)}. now (1,5/8,-7/8) is also in this null space, but NO Z-multiple of (8,5,-7) will ever give us (1,5/8,-7/8).
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  4. #4
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    Re: Integral Basis...

    Hi Deveno,

    What I'm looking for is not a basis with integer coefficients that spans the same space as the null space of A... But I want a basis with integer coefficients that hits every vector with integer coefficients in the nullspace of A, i.e. null(A)nZ.

    I know that given a basis for my nullspace, I can easily find a basis with integer coefficients by multiplying by a scalar k (e.g. the least common multiple of the the denominators of my generators), but is there a way to make sure that I'm hitting everything in null(A)nZ (the elements in the null space of A which have integer coefficients)?

    So in your example, I'm not hoping to hit (1,5/8,-7/8) because it has rational coefficients... but I want to make sure I hit every vector with integer coefficients... (For example, suppose I chose (16, 10, -14) for my basis. Q-linear combinations will give me the span of null(A), but Z-linear combinations will not give me (8,5,-7) ).
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