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Check my reasoning (proof)

  1. Oct 26, 2009 #1
    Ok I am doing some additional problems becuase my professor assigns evens and i do odd and evens so i can at least check my work in backs of book.


    Prove that if W is a subspace of vector space V, dim(W) is less than or equal to dim(V).

    Proof: We know W is a subspace of V and therefore has a basis. The basis in W is in V and we also know V has a basis. If the basis for W has same dimension as a basis for V then W is V and we are done. Since V contains W that means the basis for W must be a linear combination of vectors in basis of V. So since every w in W can be generated by a basis in V then then dimV cannot be smaller than dimW. there for dimW<=dimV.
     
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  3. Oct 26, 2009 #2
    Linear Algebra Proof, (please check my reasoning)

    Here is my questionable attempt at a proof for my LA class
    my professor assigns evens so we cant even check our homework so i do both odds and evens but even then I can't check my proofs so easily.

    Prove that if W is a subspace of vector space V, dim(W) is less than or equal to dim(V).

    Proof: We know W is a subspace of V and therefore has a basis. The basis in W is in V and we also know V has a basis. If the basis for W has same dimension as a basis for V then W is V and we are done. Since V contains W that means the basis for W must be a linear combination of vectors in basis of V. So since every w in W can be generated by a basis in V then then dim(V) cannot be smaller than dim(W). there for dim(W)<=dim(V).
     
  4. Oct 26, 2009 #3
    Re: Linear Algebra Proof, (please check my reasoning)

    bump please help! don't be shy
     
  5. Oct 26, 2009 #4
    more rigourously, try wording it in terms of the replacement theorem. I think you may have to prove that the subspace is finite-dimensional as well, if V is finite dimensional
     
  6. Oct 26, 2009 #5

    HallsofIvy

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    I think it would be a little clearer to use the fact that, for V of dimension n, a set of more than n vectors in V cannot be independent.
     
  7. Oct 26, 2009 #6
    Re: Linear Algebra Proof, (please check my reasoning)

    It can be challenging to give advice on such problems because I don't know what theorems/corollaries you've already proved, or even if your vector spaces are all finite dimensional. I'll assume that all spaces are finite dimensional.

    You should consider the case W = {0}, although it's trivial.

    Remark: Bases don't have dimensions, only spaces and subspaces do. You probably meant to say something like: If the basis for W has the same number of elements as a basis for V, then W = V and we are done.

    Again, your terminology is inexact. A basis can't be a linear combination of vectors, but the elements of the basis (which are vectors) can be.

    I find your statements in the last quote to be confusing. Have you already proved that, if dim (V) = n, then no set of linearly independent vectors in V can contain more than n elements? If so, I suggest that you try to base a proof on that.

    Please post again if my comments aren't clear or if you have any questions.

    Petek
     
  8. Oct 26, 2009 #7
    Please note that the original poster also posed this question here in the homework thread.

    @Dosmascerveza: Please post a question only in one thread to avoid duplication of efforts by those who are trying to help you. Thanks.

    Petek
     
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