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!

Linear algebra proof

  1. Jun 10, 2009 #1
    1. The problem statement, all variables and given/known data
    Prove or give a counterexample: If U is a subspace of V that is invariant
    under every operator on V, then U={0} or U=V.


    2. Relevant equations



    3. The attempt at a solution
    Counterexample: Suppose dimV=n and dimU=m with m<n.
    let u=u1+...+um in U. Since u1+...+um=u1+...+um+...+0n
    we can apply any T and get T(u1)+...+T(um)+..+T(0n)
    =c1u1+...+cmum+cm+1*0+...+cn*0=c1u1+...+cmum in U.
     
  2. jcsd
  3. Jun 10, 2009 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    It looks like you are claiming to prove that ANY proper subspace of V is invariant under any ANY operator T. Does that really seem likely?? Do you really think for ANY operator T, that T(u1)=c1*u1?? Show me an operator T and a subspace U where that is NOT true. Be concrete. Work in say R^3 with basis {e1,e2,e3}. Pick say, U=span{e1}. Define an operator T such that U is not invariant.
     
    Last edited: Jun 10, 2009
  4. Jun 10, 2009 #3
    T(e1)=T(e1)+T(0)+T(0)=0*e1+ c1*0+c2*0=0 So T maps from U to {0}.
     
  5. Jun 10, 2009 #4

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    But what's T(e1)?. Look, you can define a linear transformation by giving the values of T(e1), T(e2) and T(e3). Here's an example. Define S by S(e1)=e2, S(e2)=0 and S(e3)=0. Is U=span{e1} invariant under S?
     
  6. Jun 10, 2009 #5
    Oh okay e2 is not in U so you didn't map from U to U.
    Now I will prove that U={0} or V.
    For T to map from U to U under any transformation T,
    let u=0 so T(u)=0 so all members in the rangeT are
    0 and for the domain to match the range, U={0}.
    So we mapped from {0} to {0}. And U can also be V
    to be invariant under T since we map from V to V if
    we let U=V.
     
  7. Jun 10, 2009 #6

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    I think you got the point of the example S, but your 'proof' doesn't convey that. It looks like you just defined T to be the zero map, T(v)=0 for any v. ANY subspace is invariant under that map. Why? The point to the example was that given a subspace U, if I can find a vector v that is not in U, I can define a linear transformation T that maps a nonzero vector in U to v. Try and use the example to model a proof for a general U. You may want to pick a specially chosen basis for V. Can you start?
     
  8. Jun 10, 2009 #7
    Basis for V <v1....vn> basis for U <v1....vm>.
    Let m<n.
    Now, a possible T maps vi to vm+1 1[tex]\leq[/tex]i[tex]\leq[/tex]m. Since vm+1
    is not within U's basis, U is not invariant under T.
     
  9. Jun 10, 2009 #8

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    That's it! Simple, isn't it? Now for what choices of U is this impossible? I.e. what happens if m=n or m=0?
     
    Last edited: Jun 10, 2009
  10. Jun 10, 2009 #9
    let m=n. Basis for U=<v1...vm>. Since U is a subspace of V, <v1...vm> is in the basis
    of V. But since dimU=dimV, U=V. Now we can map T(vi) to vn or "lower" without falling out
    of the basis for U.
    let m=0. Basis for U=<0>. T(0)=c*0=0. Since 0 is in {0}, we mapped from {0} to {0}.
     
  11. Jun 10, 2009 #10

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    I guess. But it's not so much that you "CAN map T(vi) to vn or "lower" without falling out
    of the basis", you can always do that, it's that you CAN'T map T(vi) outside of the basis.
     
  12. Jun 10, 2009 #11

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    That's what I was thinking you were thinking. Expressing yourself clearly takes some practice. Keep practicing.
     
  13. Jun 10, 2009 #12
    I deleted that message, sorry lol.
    That's the thing with abstract concepts, I can understand, but can't express.
    Alright, were done here.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Linear algebra proof
  1. Linear Algebra Proof (Replies: 8)

  2. Linear Algebra proof (Replies: 21)

Loading...