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Cardinality vs. Dimension, Solution of homogeneous equations

  1. Jan 30, 2012 #1
    1. The problem statement, all variables and given/known data

    Show that the number of distinct solutions of a system of linear equations (in any number of equations, and unknowns) over the field Zp is either 0, or a power of p.


    3. The attempt at a solution

    First off, I was wondering whether there is any difference between "cardinality" of a vector space and "dimension". Aren't both just the size of the basis? (the cardinality of V = dim(V)??)
    This is just because my prof keeps switching between both and confusing the rest of us.

    for the question, suppose the system is m equations in n unknowns.
    The case of 0 is trivial, so if we take the subset W of Zpn of solutions over Zp, W is a vector space, and dim(W) = n.
    I'm not sure how to put this technically, but for each unknown more than the number of equations (say we parametrize them) we have p possible choices, and thus pt solutions.

    There's something missing, but am I on the right track?
     
  2. jcsd
  3. Jan 30, 2012 #2

    morphism

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    No. The cardinality of V isn't equal to the dimension of V. Easy counterexample: an n-dimensional vector space over Z/pZ has p^n elements (which is essentially what you've noticed with your W). What is true is that the cardinality of a basis for V is equal to the dimension of V (and this is the definition of dimension).

    Anyway, you've basically solved the problem, although you've said some questionable things. You've noticed W is a finite-dimensional vector space over Z/pZ, which is correct. If the dimension of W is t (it's not necessarily n), then you can choose a basis w_1,...,w_t for W. Any element of W will thus look like a_1w_1+...+a_tw_t with a_i in Z/pZ. Now count these suckers.
     
  4. Jan 30, 2012 #3
    Oh ok thank you. So what is cardinality then (talking about a vector space)?

    So then the dimension is pt! It makes sense to say dim(W) = t, but I don't understand why it isn't n:
    The system is in n unknowns, and as you said, you can form a basis {w1, w2,...,wt}. But in the standard basis, you would have {e1,...,en}, thus dim(W) = n since all bases of W have the same cardinality. Why not n?
     
  5. Jan 30, 2012 #4

    morphism

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    The cardinality of a vector space is just its cardinality as a set.
    Try solving
    x_1 + x_2 = 0
    x_1 + x_2 + x_3 = 0
    in Z/2Z, for example. Is dimW=3?
     
  6. Jan 30, 2012 #5
    Thank you, makes perfect sense.
    Oh alright. Essentially saying dimW = t is just saying it is finite dimensional.
     
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