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Help with Linear Algebra proof

  1. Jun 1, 2009 #1
    1. The problem statement, all variables and given/known data
    Suppose S, T [tex]\in[/tex] the set of linear transformations
    from V to V. Prove that ST and TS have the same eigenvalues.

    2. Relevant equations
    T=[tex]\lambda[/tex]I


    3. The attempt at a solution
    let v[tex]\in[/tex]V.
    For TS
    T(Sv)=T(I[tex]\lambda[/tex]v)=[tex]\lambda[/tex]T(Iv)
    =[tex]\lambda[/tex]'[tex]\lambda[/tex]Iv.
    For ST
    S(Tv)=S([tex]\lambda[/tex]'v)=[tex]\lambda[/tex]'S(Iv)=
    [tex]\lambda[/tex]'IS(v)=[tex]\lambda[/tex]'I[tex]\lambda[/tex]Iv
    =[tex]\lambda[/tex]'[tex]\lambda[/tex]Iv.
     
  2. jcsd
  3. Jun 1, 2009 #2

    Mark44

    Staff: Mentor

    Your "relevant equation" T = [itex]\lambda[/itex]I doesn't seem at all relevant, as far as I can see.

    If [itex]\lambda[/itex] is an eigenvalue of T, then for some v [itex]\neq[/itex] 0,
    Tv = [itex]\lambda[/itex]v.

    I would approach this by assuming that [itex]\lambda_1[/itex] and [itex]\lambda_2[/itex] are eigenvalues of ST and TS, then showing that the two eigenvalues are equal.
     
  4. Jun 1, 2009 #3
    For ST
    Knowing that Tv=[tex]\lambda[/tex]v, STv=S[tex]\lambda[/tex]v
    =[tex]\lambda[/tex]Sv=[tex]\lambda[/tex] [tex]\lambda[/tex]'v
    [tex]\lambda[/tex] [tex]\lambda[/tex]'=[tex]\lambda[/tex]1

    For TS
    knowing that Sv=[tex]\lambda[/tex]'v, TSv=T[tex]\lambda[/tex]'v
    =[tex]\lambda[/tex]'Tv since we know from ST that Tv=[tex]\lambda[/tex]v
    we get [tex]\lambda[/tex]' Tv=[tex]\lambda[/tex]'[tex]\lambda[/tex]v=[tex]\lambda[/tex] [tex]\lambda[/tex]'v
    [tex]\lambda[/tex][tex]\lambda[/tex]'=[tex]\lambda[/tex]2
     
    Last edited: Jun 1, 2009
  5. Jun 1, 2009 #4

    Mark44

    Staff: Mentor

    Looks OK, if a little rough around the edges.
     
  6. Jun 1, 2009 #5

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    That's really not a very good proof at all. It only works if S and T have the same eigenvectors. The result is true even if they don't. You want to show ST and TS have the same characteristic polynomial. Start with the case where S is invertible. Can you show ST and TS are similar matrices?
     
  7. Jun 1, 2009 #6
    STv=[tex]\lambda[/tex]v
    TSw=[tex]\lambda[/tex]2w
    STv-[tex]\lambda[/tex]v=0
    TSw-[tex]\lambda[/tex]2w=0
    (ST-[tex]\lambda[/tex])v=0
    (TS-[tex]\lambda[/tex]2)w=0
    ST-[tex]\lambda[/tex]=0
    TS-[tex]\lambda[/tex]2=0
    ST=[tex]\lambda[/tex]
    TS=[tex]\lambda[/tex]2
    T=S^-1[tex]\lambda[/tex]2
    T=S^-1[tex]\lambda[/tex]
    plugging in S^-1[tex]\lambda[/tex]2 to ST-[tex]\lambda[/tex]=0 for T
    S(S^-1[tex]\lambda[/tex]2)-[tex]\lambda[/tex]=0
    I[tex]\lambda[/tex]2-[tex]\lambda[/tex]=0
    [tex]\lambda[/tex]2-[tex]\lambda[/tex]=0
    [tex]\lambda[/tex]2=[tex]\lambda[/tex]
    since ST=[tex]\lambda[/tex] and TS=[tex]\lambda[/tex]2
    and [tex]\lambda[/tex]=[tex]\lambda[/tex]2
    ST and TS have the same eigenvalues.
     
    Last edited: Jun 1, 2009
  8. Jun 1, 2009 #7
    (ST-[itex]\lambda[/itex])v=0
    (TS-[itex]\lambda[/itex]2)w=0
    do not imply
    ST-[itex]\lambda[/itex]=0
    TS-[itex]\lambda[/itex]2=0
     
  9. Jun 1, 2009 #8
    Dick's suggestion is good, and if you start with a singular matrix S,you can substitute S-[itex]\epsilon[/itex]I instead of S, and let [itex]\epsilon[/itex] tend to 0. It's a standard homotopy trick frequently used in linear algebra
     
  10. Jun 1, 2009 #9
    still dont get it
     
  11. Jun 1, 2009 #10

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Do you know similar matrices have the same eigenvalues? Can you show if S is nonsingular then ST and TS are similar? Do you know that nonsingular matrices are dense in the space of all matrices? Do you know that the coefficients of the characteristic polynomial are continuous functions of the elements of the matrices? You have to put all of these things together to get it. Big hint: TS=S^(-1)*(ST)*(S).
     
  12. Jun 1, 2009 #11
    Oh, my bad, I should've explained that I'm taking this off of Axler's book, which is my text, he says nothing of similar matrices,
    so I'm not too familiar with those. Though I see your point in that there
    can be an eigenvector (1,0) that undergoes [tex]\lambda[/tex](1,0)= (2*1, 2*0)=(2,0)
    and another that undergoes [tex]\lambda[/tex](0, 1)=(2*0, 2*1)=(0,2)
    the same transformations.
     
  13. Jun 2, 2009 #12
    TS=[tex]\lambda[/tex]2
    ST=[tex]\lambda[/tex]1
    TS=S^(-1)[tex]\lambda[/tex]1S
    TS=[tex]\lambda[/tex]1I
    TS=[tex]\lambda[/tex]1
    Since [tex]\lambda[/tex]1=TS=[tex]\lambda[/tex]2,
    [tex]\lambda[/tex]1=[tex]\lambda[/tex]2
    ST=T^(-1)TST
    ST=T^(-1)[tex]\lambda[/tex]2T
    ST=I[tex]\lambda[/tex]2

    I still don't know what I just did. Does the one below work?
    If not, then I'll stop trying to prove this in this manner.
    w and v are either different or the same, as opposed to just using v.
    I'm not trying to be the my proofs better than yours guy,
    I just need convincing.

    For ST
    Knowing that Tv=[tex]\lambda[/tex]v, STv=S[tex]\lambda[/tex]v
    =[tex]\lambda[/tex]Sv=[tex]\lambda[/tex] [tex]\lambda[/tex]'v
    [tex]\lambda[/tex] [tex]\lambda[/tex]'=[tex]\lambda[/tex]1


    For TS
    knowing that Sv=[tex]\lambda[/tex]'w, TSv=T[tex]\lambda[/tex]'w
    =[tex]\lambda[/tex]'Tw since we know from ST that Tv=[tex]\lambda[/tex]v
    we get [tex]\lambda[/tex]' Tw=[tex]\lambda[/tex]'[tex]\lambda[/tex]v=[tex]\lambda[/tex] [tex]\lambda[/tex]'w
    [tex]\lambda[/tex][tex]\lambda[/tex]'=[tex]\lambda[/tex]2
     
    Last edited: Jun 2, 2009
  14. Jun 2, 2009 #13
    Check your latex codes,
    and I think Dick has given the proof:TS=S^(-1)*(ST)*(S), that's all of it, plus the homotopy trick I mentioned,then the proof will be finished.
     
  15. Jun 2, 2009 #14

    Dick

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    Science Advisor
    Homework Helper

    I really don't think you can prove the general case that way. The eigenvalues and eigenvectors of S and T don't have to be the same as those of ST and TS. Nor is it generally true that the product of the eigenvalues of S and T are the eigenvalues of ST and TS. Try some nontrivial examples. Say S=[[2,1],[1,2]] (eigenvalues 3 and 1) and T=[[1,1],[8,-1]] (eigenvalues -3 and 3). The eigenvalues of ST or TS are (9+/-3*sqrt(21))/2. Now do you believe me?
     
    Last edited: Jun 2, 2009
  16. Jun 2, 2009 #15
    Oh, ok I see that. Yeah, thats my problem with these proofs,
    I tend to neglect some cases. Thanks for your efforts.
     
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