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Math Help - Distinct Bases for Finite Vector Spaces over Finite Fields

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    Distinct Bases for Finite Vector Spaces over Finite Fields

    Hi, I need help with a linear algebra/algebra question:

    Let V = (F_p)^2 = (Z/pZ)^2 [= GF(p^2)?] and K = F_p = Z/pZ where F_p is the field with p elements for some prime p
    How many distinct bases for V exist?

    A similar question I need help with is:

    Let v_1, v_2 \in (F_p)^3\{0} such that {v_1, v_2} is linearly independent
    Show that there are p^2(p-1) elements v_3 such that {v_1, v_2, v_3} is a basis of (F_p)^3

    Many thanks
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by marcusjh View Post
    Hi, I need help with a linear algebra/algebra question:

    Let V = (F_p)^2 = (Z/pZ)^2 [= GF(p^2)?] and K = F_p = Z/pZ where F_p is the field with p elements for some prime p
    How many distinct bases for V exist?

    A similar question I need help with is:

    Let v_1, v_2 \in (F_p)^3\{0} such that {v_1, v_2} is linearly independent
    Show that there are p^2(p-1) elements v_3 such that {v_1, v_2, v_3} is a basis of (F_p)^3

    Many thanks
    So, it's a simple counting argument, right? So this is really asking how many linearly independent subsets of size 2 are there in \mathbb{F}_p^2. So, say you picked your first element v\in\mathbb{F}_p^2 then you have as many choices as there are elements in \mathbb{F}_p^2-\text{span}(v), how many is that?
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  3. #3
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    i don't think that's quite right. in the first place, your initial choice for v has to be non-zero.

    now, even with that proviso, span(v) is isomorphic to Zp, so it has p elements. that gives us (p^2 - 1)(p^2 - p) choices, according to you.

    so in Z2 x Z2, we should have 6 possible choices of a basis:

    picking (0,1) first, we could form: {(0,1), (1,0)} or {(0,1), (1,1)} as a basis (so far, so good).

    picking (1,0) first, we could form: {(1,0), (0,1)} or {(1,0), (1,1)} as a basis....wait a minute. we already HAVE {(1,0), (0,1)} = {(0,1), (1,0)}

    as a basis....i didn't see anything in the problem stating they should be ORDERED bases.
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Deveno View Post
    i don't think that's quite right. in the first place, your initial choice for v has to be non-zero.

    now, even with that proviso, span(v) is isomorphic to Zp, so it has p elements. that gives us (p^2 - 1)(p^2 - p) choices, according to you.

    so in Z2 x Z2, we should have 6 possible choices of a basis:

    picking (0,1) first, we could form: {(0,1), (1,0)} or {(0,1), (1,1)} as a basis (so far, so good).

    picking (1,0) first, we could form: {(1,0), (0,1)} or {(1,0), (1,1)} as a basis....wait a minute. we already HAVE {(1,0), (0,1)} = {(0,1), (1,0)}

    as a basis....i didn't see anything in the problem stating they should be ORDERED bases.
    Ah indeed, I fdid assume they were ordered. I thought this was leading up to how many different matrices can one form for a linear transformation on \mathbb{F}_p^2.
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  5. #5
    Senior Member Tinyboss's Avatar
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    But that can be fixed trivially by dividing by 2. And I think choosing the initial v nonzero probably went without saying. (I suppose we could consider zero to be in the span of the empty set, and choose each subsequent basis element outside the span of the previous ones, generalizing to n>2, right?)
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    if we generalize to n > 2, we have to remember to keep dividing out the repetitions. we might have started with v1, then chosen v2 not in span(v1), and then v3 not in span(v1,v2),

    but then we could still wind up with {v1,v3,v2} or{v2,v1,v3} or {v2,v3,v1} or {v3,v1,v2} or {v3,v2,v1} as well (it looks like we have to divide by n!).

    however, for the latter problem, if we have already chosen v1 and v2, then span(v1,v2) has p^2 elements, so we have p^3 - p^2 = p^2(p-1) left to choose from.
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