1. Limited time only! Sign up for a free 30min personal 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!

Equality between subspaces

  1. Nov 23, 2012 #1
    Hi,
    How may I determine whether the subspaces U and W are equal to each other?:

    K is linearly independent wrt V, defined thus:
    K={v1,v2,v3,v4} subset of V
    U and W, subspaces of V, are defined thus:
    U=Sp(K); W=Sp{v1-v2,v2-v3,v3-v4,v4-v1}

    I am not allowed to use equality between dimensions!

    I have tried solving:
    a1v1+a2v2+a3v3+a4v4=b1(v1-v2)+b2(v2-v3)+b3(v3-v4)+b4(v4-v1)
    But I am not sure it got me anywhere :-/.

    I hope one of you could assist. Thanks in advance!
     
  2. jcsd
  3. Nov 23, 2012 #2

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Seems like a reasonable start. Where did that lead?
     
  4. Nov 23, 2012 #3
    It led to a1=b1-b4; a2=-b1+b2; a3=-b2+b3; a4=-b3+b4
    I tried substituting them into a matrix but am not sure I got anything meaningful.
    I also know that when a1v1+a2v2+a3v3+a4v4=0, a1=a2=a3=a4=0.
    Any idea how I should proceed?
     
  5. Nov 23, 2012 #4

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    If you can show that v1-v2, v2-v3, v3-v4, and v4-v1 are independent then you are done. Do you see why?
     
  6. Nov 23, 2012 #5
    I am not sure. If b1(v1-v2)+b2(v2-v3)+b3(v3-v4)+b4(v4-v1)=0 necessarily implies b1=b2=b3=b4=0, why does that mean that U=W? Isn't the homogeneous system merely a private case? Were U and W both linearly independent, why would U be equal to W?
     
  7. Nov 23, 2012 #6

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    If you add those equations up you get a1+a2+a3+a4=0. What would that tell you?
     
  8. Nov 23, 2012 #7
    Does that tell me that {v1,v2,v3,v4} is linearly dependent as a1, a2, a3, a4 are not forcibly all zero?
     
  9. Nov 23, 2012 #8

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    I would say it means that only SOME vectors in Sp{v1,v2,v3,v4} are also in Sp{v1-v2,v2-v3,v3-v4,v4-v1}. Can you tell me why?
     
  10. Nov 23, 2012 #9
    As for the two spans to be equal certain conditions must be met, i.e. a system of equations must be solved which imposes certain limitations on the values of the scalars.
     
  11. Nov 23, 2012 #10

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    That's pretty vague. You wrote a system of equations. One conclusion of that was that if v=a1v1+a2v2+a3v3+a4v4 then v can only be in Sp{v1-v2,v2-v3,v3-v4,v4-v1} if a1+a2+a3+a4=0. Is that condition true for any vectors in Sp{v1,v2,v3,v4}? Is it untrue for any vectors in the span?
     
  12. Nov 23, 2012 #11

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    My point was that since you are given that v1, v2, v3, b4 are independent, the subspace they span has dimension 4. The other 4 vectors, v1-v2,v2-v3,v3-v4,v4-v1, will span that subset if and only if they are independent.
     
  13. Nov 23, 2012 #12
    Okay, so I have therefore tried showing that b1(v1-v2)+b2(v2-v3)+b3(v3-v4)+b4(v4-v1)=0, implies b1=b2=b3=b4=0 (for linear independence). However, based on the fact that a1v1+a2v2+a3v3+a4v4=0 for a1=a2=a3=a4=0, all I got was that b1=b2=b3=b4=any real number (i.e. not necessarily zero)! How shall I proceed?
     
  14. Nov 23, 2012 #13

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Good job. Wouldn't that show that they are NOT linearly independent?
     
  15. Nov 23, 2012 #14
    I was somehow under the impression that U WAS in fact equal to W, so I did expect linear independence in both cases. Is the conclusion then that U is necessarily not equal to W?
     
  16. Nov 23, 2012 #15

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Yes, U isn't equal to W. And not even 'necessarily'. It just plain isn't. (v1-v2)+(v2-v3)+(v3-v4)+(v4-v1)=0. That doesn't sound linearly independent to me.
     
  17. Nov 23, 2012 #16
    Thanks! :-)
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Equality between subspaces
  1. Equal subspaces? (Replies: 3)

Loading...