1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Proof involving subspaces of finite-dimensional vector spaces

  1. Jul 27, 2010 #1
    This is an exercise in a linear algebra textbook that I initially thought was going to be easy, but it took me a while to make the proof convincing.

    Prove: Any subspace of a finite-dimensional vector space is finite-dimensional.

    Here's my attempt. I am not sure about some details and I'm hoping you guys can tell me if it makes sense or suggest improvements, give some feedback, suggest a better method, etc.

    Let V be a finite-dimensional vector space such that dim(V) = n.

    Now let W be a subspace of V and let S be a linearly independent set of vectors {w1, w2, ... , wr} in W.

    If S spans W, then S is a basis for W and W has dimension r. If S does not span W, then add a new vector from W to S such that S is still linearly independent. If W is infinite-dimensional, then we should be able to add as many vectors as we like and S will still not span W.

    However, since all vectors in W are also in V, then S is also a set of vectors in V. Therefore, S can contain at most n vectors while still being linearly independent. And if S contains n vectors, then S spans V and thus W as well.

    This implies that W has dimension of at most n, and thus is finite-dimensional.

    Thanks in advance for any comments.
     
  2. jcsd
  3. Jul 27, 2010 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    That's rather awkwardly stated but I think it is basically correct. One difficulty I see is that you haven't stated explictely what your definition of "finite dimensional" is. Not knowing that, I can't say whether your proof covers it or not. In most books, the definition of "finite dimensional" is NOT that it has a basis containing only a finite number of vectors, because you need finiteness to show that a vector space has a basis (without going into "transfinite induction" or "Zorn's lemma" that is used to prove the existence of a basis for infinite dimensional spaces). Most books say a vector space is "finite dimensional" if and only if it can be spanned by a finite set of vectors. You then show that there exist a basis which is a subset of that spanning set. So I would not worry about a basis. Since V is finite dimensional, it can be spanned by a finite set of vectors. Since W is as subspace of V it can be spanned by that same set and so is finite dimensional.
     
  4. Jul 27, 2010 #3
    Thank you for your reply.

    The textbook gives only the following definition:

    "A nonzero vector space V is called finite-dimensional if it contains a finite set of vectors {v1, v2, ... vn} that forms a basis. If no such set exists, V is called infinite-dimensional. "

    It makes the proof a little more difficult than the definition you stated, where if a vector space can be spanned by a finite set of vectors, then it is finite-dimensional. Using the definition you stated, it seems like the proof immediately follows from the fact that a basis for V spans all of its subspaces as well, so all of its subspaces are finite-dimensional -- like you said, basically.
     
  5. Jul 27, 2010 #4

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Sure, but you can prove if a space is spanned by a finite number of vectors, then it has a basis with a finite number of vectors, right?
     
  6. Jul 27, 2010 #5
    It makes sense that it would be true. Let's see.

    "If a vector space is spanned by a finite number of vectors, then it contains a finite set of vectors which form a basis."

    Let V be a vector space and let S be a linearly independent set of n vectors which spans V.

    While the vectors in S are not necessarily in V, all linearly independent sets of vectors in V can be constructed from linear combinations of the vectors in S.

    This implies that the largest number of vectors that a linearly independent set in V can contain is n. Then, construct from S a set of linearly independent vectors in V to which no other vector from V can be added without making it linearly dependent.

    This new set S' has the following properties:
    a) All vectors in V can be expressed as a linear combination of the vectors in S', otherwise those vectors could be added to S' without making it linearly dependent. Therefore, S' spans V.
    b) It is linearly independent.
    c) It is in V.
    d) It contains at most n vectors.
    And so S' contains a finite number of vectors which form a basis for V.

    Therefore, all vector spaces spanned by a finite number of vectors contain a finite set of vectors which form a basis.

    Now using this result...

    "Any subspace of a finite-dimensional vector space is finite-dimensional."

    If S forms a basis for a finite-dimensional vector space, then it spans its subspaces as well. Then, all subspaces of finite-dimensional vector spaces contain finite sets of vectors which form bases for those subspaces. Therefore, all subspaces of finite-dimensional vector spaces are finite-dimensional.

    How does that look? :)
     
    Last edited: Jul 28, 2010
  7. Jul 28, 2010 #6

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    That's a bit confusing. I'd be a little more explicit about how you get a basis for W. Pick a vector in W, say w1. If {w1} spans W, then you are done. If not, you can find a vector w2 in W that is linearly independent of {w1}, say w2. If {w1,w2} spans W then it's a basis and you are done. If not you can find a third linearly independent vector in W, say w3. You know this has to stop at some point because the maximum number of linearly independent vectors in V is n. Then the w's form a basis for W. Something like that.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Proof involving subspaces of finite-dimensional vector spaces
Loading...