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Homework Help: Linear Algebra, proof

  1. Jun 8, 2010 #1
    Hi all,
    I need some proofs for my assignment, the question is like below:

    Let L be a linear map from the vector space V to the vector space W.

    • ker(L) is a subset of V which consists of vectors u such that Lu = 0. Is ker(L) a
    vector subspace of V ? Give a proof.
    • Let S be a subset of V which consists of vectors which are not contained in ker(L).
    Is S a vector subspace of V ? Give a proof.
    • Let T be a subset of W which consists of vectors which are contained in Range L.
    Is T a vector subspace of W? Give a proof.
    • Let T be a subset of W which consists of vectors which are not contained in
    Range L. Is T a vector subspace of W? Give a proof.

    Thanks in advance
     
  2. jcsd
  3. Jun 9, 2010 #2
    You'll need to show some of your work; what you've tried, concepts you think might be involved, etc. Then we can help you!
     
  4. Jun 9, 2010 #3
    you are right, but unfortunately I don't have many things to show. We have done with entire linear algebra in two weeks and now I am supposed to finish this assignment. It just does not settle this fast in my mind,I am trying though. That's why I need some help. Thanks again
     
  5. Jun 9, 2010 #4

    lanedance

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    so people are happy to help lead you through the problemif you try, but won;t just do it for you ;)

    If you get lost the definitions are a good place to start... so for 1) teh question is:

    Is Ker(L) a subspace of V, where L is a linear map from V to V?

    so what is the definition of a subspace?
     
  6. Jun 9, 2010 #5
    well, (it says here) a subspace of a vector space is a nonempty subset that satisfies the requirements for a vector space.
     
  7. Jun 9, 2010 #6
    ok,from the definition ker(L) is a vector subspace of V (I guess). but is this definition enough for the proof?
     
  8. Jun 9, 2010 #7

    lanedance

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    NO - you need to show it. There are quite a few properties required to show a set, equipped with addition & scalar multiplication is a vector space (~10 if i remember correctly)

    To show a subset of a vector space is a "subspace" is a little less onerous, you need to show 3 things:
    - it has the zero vector
    - it si "closed" under addition
    - it is "closed" under scalar multiplication

    see this for subspace
    http://en.wikipedia.org/wiki/Linear_subspace

    and for a general vector space
    http://en.wikipedia.org/wiki/Vector_space
     
    Last edited: Jun 9, 2010
  9. Jun 10, 2010 #8
    Thank you very much, here is my final proof :)
    - ker(L) is nonempty since Lu=0, the zero vector of V, is in ker(L)
    - if u belongs ker(L) and a is a scalar, then
    L(au)=aL(U)=a.0=0, therefore au belongs ker(L)
    -if u1, u2 belong ker(L), then
    L(u1+u2)=Lu1+Lu2=0+0=0, so u1+u2 belongs ker(L)
    hence ker(L) is a subspace of V. Am I right?

    Proof for the second statement:
    S is a subset of V which consists of vectors which are not contained in ker(L).
    therefore S does not have a zero vector, 0v belongs ker(L), and S is not a subspace of V.

    The last two are similar to first two.
    I guess I makes sense but do you agree?
    Thanks
     
  10. Jun 10, 2010 #9

    lanedance

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    subspace proof looks good

    2nd statement is valid as well

    another way to do the 2nd one without using the zero vector is, pick u & v such that:
    [tex] L(u) = 0, \ u \ \in Ker(L) [/tex]
    [tex] L(v) \neq 0, \ v\ \in V [/tex]
    [tex] L(u+v) = L(u) + L(v) = 0 + L(v) \neq 0, \ \to (u+v) \in V [/tex]
    [tex] (u+v) + (-v) = u \notin V [/tex]
    so its not closed
     
  11. Jun 10, 2010 #10
    Thank you.
     
  12. Jun 11, 2010 #11

    HallsofIvy

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    Well done, lanedance and jwhite2531.
     
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