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!

Proving Subspace

  1. Oct 1, 2012 #1
    The problem has been attached. I am having difficulty expressing myself. The professor said for this problem, it would be best if I use words to answer it.

    1. I must verify the 0 vector is in S+T. Since S and T are subspaces, the 0 vector must exist in both S and T. Thus 0+0=0 and 0 vector is in S+T



    2. I must verify that if X and Y are subspaces in S+T, then I need to check if X+Y is still in the subspace S+T.
    Note: X and Y are of the form S+T, which yields the following:
    So I define X as U1+U2, where U1 is in S and U2 is in T.
    I define Y as V1+V2, where V1 is in S and V2 is in T.
    X+Y=[U1+V1, U2+V2] Since U1 and V1 are both in S, then U1+V1 is in S. Since U2 and V2 are both in T, then U2+V2 are both in T. Thus X+Y is in S.



    3. Lastly I must verify that if Y is in S+T, then I need to check if cY is in S+T, where c is some scalar. Using the same Y=V1+V2 where V1 is in S and V2 is in T. I get cY=c(V1+V2)=cV1+cV2, this satisfies that cV1 is in S and cV2 is in T. Thus cY is in S+T.


    Is this correct?
    Also, if a certain part is incorrect, can you refer clearly to the part or quote just that part? I get confused easily.

    Thanks.
     
  2. jcsd
  3. Oct 2, 2012 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Where? There is no attachment. You seem to be trying to prove "If S and T are vector spaces then S+ T is a vector space but what does the "+" mean here? A S and T assumed to be subspaces of some larger space? If so then S+ T is the "direct sum" (more commonly written S⊕ T), the set of all vectors of the form v= s+ t where s is in S and t is in T.

    Okay, that is good.



    NO, you don't! You must verify that if x and y are vectors in S+ T, then x+ y is still a vector in S+ T.

    This makes sense if X, Y, U1, and V1 are vectors not subspaces.

    As long as you understand that all of your X, Y, U1, and U2 are vectors, not subspaces, then you proof will work.
     
  4. Oct 2, 2012 #3
    Sorry, I always do that, I have attached it now.
     

    Attached Files:

  5. Oct 2, 2012 #4
    Oops, I meant to say "vectors," not subspaces.
     
  6. Oct 2, 2012 #5

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    I'm not sure what the notation in the last line is supposed to mean.
     
  7. Oct 2, 2012 #6
    U1, U2, V1, V2 are vectors.

    When I add X+Y, this yields U1+V1, U2+V2


    Is that what you were asking?
     
  8. Oct 2, 2012 #7

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    I'm saying your notation doesn't make sense. Is it supposed to be an order paired of vectors?
     
  9. Oct 2, 2012 #8

    Uh, I defined X and Y in the same form as S+T. I included a picture in the 3rd post on this thread. I'm not sure what you mean by an ordered pair?
    What I was trying to show with X+Y=[U1+V1, U2+V2], is that since U1 and V1 were in S, then U1+V1 must be in S, and since U2 and V2 are in T, then U2+V2 must be in T, thus X+Y is in S+T.
     
  10. Oct 2, 2012 #9

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    I know what you mean. I'm saying you shouldn't write "X+Y=[U1+V1, U2+V2]". What you wrote in words
    is sufficient. If you want to be a bit more explicit, you can say "Thus X+Y = (U1+V1)+(U2+V2) is in S."

    In other words, I know what mathematical object (U1+V1)+(U2+V2) is. On the other hand, what exactly is [U1+V1, U2+V2] supposed to be? You didn't define what [ , ] means.
     
  11. Oct 2, 2012 #10
    Oh I see, I think I'll stick with the words rather than using that. So my statement would be sufficient to show that X+Y is in S+T?

    I'm also trying to explain the 3rd condition in words. I repasted the part below:
    3. Lastly I must verify that if Y is in S+T, then I need to check if cY is in S+T, where c is some scalar. Using the same Y=V1+V2 where V1 is in S and V2 is in T. I get cY=c(V1+V2)=cV1+cV2, this satisfies that cV1 is in S and cV2 is in T. Thus cY is in S+T.
    Is this worded correctly?
     
  12. Oct 2, 2012 #11

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    Instead of "this satisfies that cV1 is in S and cV2 is in T", I'd say, "Because S and T are subspaces, cV1 and cV2 are elements of S and T, respectively." It's a bit clearer and states explicitly what your logic is. But, yeah, your argument is fine.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Proving Subspace
  1. Proving a subspace (Replies: 22)

  2. Proving a subspace (Replies: 1)

  3. Proving A subspace (Replies: 7)

Loading...