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Question about showing that a set is a vector space

  1. Feb 15, 2012 #1
    Hi everyone,

    I keep seeing in textbooks and examples online that when someone proves that a set is a vector space, they only use a few of the axioms to prove it.

    Is there a general guideline for when to use all of the axioms, and when you only need to use select ones to prove that a set is a vector space?

    Obviously, you only need one axiom to prove that a set isn't a vector space. But for sets that are vector spaces, I'm not sure.

    Finally, if you have a good example of proving that set is a vector space using all axioms, I would really appreciate it if you could post that as well :-)
     
  2. jcsd
  3. Feb 15, 2012 #2

    Deveno

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    there are 2 different situations, which you are probably thinking "are the same".

    1) to prove an arbitrary set, along with a given addition operation and scalar multiplication is a vector space, one must verify every axiom. there is no short-cut for this situation.

    2) to prove a subset of a given vector space is a subspace, it suffices to verify only 3 conditions:

    a) closure under vector addition
    b) closure under scalar multiplication
    c) being non-empty (or, equivalently, containing the 0-vector of the parent space).

    why does verifying a "subset is a subspace" require only checking these 3 things?

    well, suppose we have a vector space V, and S is some subset we think MIGHT be a vector space in its own right.

    if u and v are 2 elements of S, then u and v are also 2 elements of V, and IN V:

    u+v = v+u, so we don't need to check commutativity of addition again for S.

    the same goes for: associativity of vector addition, and the two "distributive" laws. also, if 1v = v, for every v in V, it certainly still holds for those v that are merely in S (S is, after all, a subset of V). so what does that leave us with?

    well, we must check that "+", restricted to S, is a valid binary operation on S, that for u,v in S, u+v is still in S (so S is "self-contained" as a vector space). a similar consideration holds for a scalar c, and an element v in S: we require that cv also be in S.

    in particular, closure of scalar multiplication tells us that if v is in S, then so is (-1)v. and closure of addition tells us that:

    v + (-1)v = (1 + -1)v = 0v = 0 is also in V....provided that "something" (the "v" we postulated to begin with) is indeed in S to start with (the empty set can NEVER be a vector space, for it has no additive identity).

    from the above discussion, you can see that the 3 criteria i gave earlier automatically give us an identity for S (which is the same 0-vector V has), and additive inverses.

    long story short: it depends on whether you have a parent vector space you can appeal to, or you don't. if the closure rules are satisfied, subspaces inherit many properties from the parent space.
     
  4. Feb 16, 2012 #3

    Bacle2

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    For an easy counter example, show that the set of points in R^2 with first coordinate identically 1 , i.e., the set of points S:={(1,b), b real} is not a vector space. Check against Deveno's axiom #2 .
     
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