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Can a vector be a linear combination of itself?

  1. Sep 1, 2012 #1
    My question is: For a vector x in a vector space V, is x a linear combination of itself?

    I'm not entirely sure since the definition of linear combination in the text I'm working through says:

    A vector x of a set V is a linear combination of vectors of S if there exists vectors in S such that x is the sum of multiples of these vectors. (S is a subset of V)

    Thanks.
     
    Last edited: Sep 1, 2012
  2. jcsd
  3. Sep 1, 2012 #2

    chiro

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    Hey Klungo and welcome to the forums.

    Let x be in S and V. What about x = 1 * x?
     
  4. Sep 1, 2012 #3
    Im asking if x forms a linear combination simply because x=1x.

    For x in any set S, a subset of a vector space V, from x=1x, we have x in spanS.
     
  5. Sep 1, 2012 #4

    chiro

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    I think you know the answer but ultimately the question you need to answer is if the same vector lie in both sets?
     
  6. Sep 1, 2012 #5
    If by both sets, you mean S and V, then yes?

    If by S and spanS, that's where my concern is and I think it's true. My reasoning being as stated above.
     
  7. Sep 1, 2012 #6

    chiro

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    Since you mention subset and not proper subset you can just let S = V and you're done.

    Ultimately though if you can not do this and you want to consider a vector in V but not in S and want to check whether you can use linear combinations of stuff in S to represent something in V\S then you will need to show that perpendicular set is empty.

    What I mean by this is that you have the decomposition v + v_perp = the full space.

    So what this means in your linear combination is that if you have a v_perp term that can't be written in terms of the stuff in S (or linear combinations) then the v_perp term will be non-empty and if this is the case your answer to your question is no.

    So formally you will form a basis for S and for V and compare the two with respect to the rank-nullity theorem. If you have information corresponding to a non-zero vector for v_perp then the answer to your question is no but otherwise, it is yes.
     
  8. Sep 1, 2012 #7
    Sorry, I didn't really understand that. I just began the course in linear algebra and I'm unaware of this for now. To better pin point my the purpose of the thread. I'll elaborate on an example problem.

    Let W be a subset of vector space V. W is a subspace of V if and only if span(W) = W.

    The proof for span(W)=W imples W is a subspace of V is done and ok.
    Now to prove under the assumption that for W a subspace of V, we have span(W) = W.
    I was able to prove W is a subset of span(W).

    Now my goal is to prove span(W) is a subset of W to complete the proof.

    My attempt at this goes: Let x in W be an arbitrary vector. By [Axiom of multiplication identity] we have x=1*x. Thus, x is a linear combination of itself and x must be in span(W). Therefore, W is a subset of span(W).

    Is that part right?


    (Thanks for the help by the way).
     
  9. Sep 1, 2012 #8

    chiro

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    So do you have to prove that span(W) = W or do you assume that to solve your problem?

    I don't know what the axiom of multiplication identity is but since W and span(W) are the same spaces and since x is in both those spaces then your proof should be along the lines of that 1*x is in both spaces, x is in both spaces, 1*x is a linear combination of x and so the statement holds.

    Do you have to show W = span(W) (or not show that this holds)?
     
  10. Sep 1, 2012 #9
    I'm trying to show span(W) = W.

    [Edit: As part of the entire proof, to show W is a subset of span(W)]
    I try to show that if x in W, it is also in spam(W) by showing that
    x is a linear combination of itself by the multiplication identity axiom (which is x=1*x).
    That way, it meets the definition of span(W) which is the set of all linear combinations in W.

    [Edit: Assume I've already showed span(W) is a subset of W.]

    After all, x is in W (by assumption), 1 is a scalar, so im wondering if x is a linear combination of itself by x=x*1 since it looks like that alone meets the definition of a linear combination.

    I apologize if I appear to be repeating some things [many times].
     
    Last edited: Sep 1, 2012
  11. Sep 1, 2012 #10

    chiro

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    Ok I see.

    Well showing equality is the same as showing A is a subset of B and B is a subset of A. So you know W is a subset of span(W) since span(W) must include all vectors in the spanning set corresponding to what they span.

    You've shown span(W) is a subset of W already so using those two you show that W = span(W).

    I think what you are doing is right: just use the definition of spanning where the coeffecients have to be real (and 1 is a real number) where all the vectors have to be in the right set (i.e. W). since x is in W, 1*x is in W and the spanning set has the criteria that z = ax + by + blah blah then you meet the requirement.

    I think you already knew the above and it sounds like this exercise is really a bunch of pointless BS, but I understand that you need marks so I empathize with you.

    If you write out the definition of a spanning set and show that 1 is real, x is in both then I think that should be enough.
     
  12. Sep 1, 2012 #11
    Alright. Thanks for the help. It's a random exercise i did but i used a similar procedure before in other proofs and just wanted to be sure.
     
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