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Interesting proof

  1. Jun 15, 2009 #1
    1. The problem statement, all variables and given/known data
    Let U be a fixed nxn matrix, and consider the operator T:Msub(n,n)---->Msub(n,n)
    given by T(A)=UA (look familiar?:biggrin:)
    Show that if dim[Esub(c)(U)]=d then dim[Esub(c)(T)]=nd.

    2. Relevant equations



    3. The attempt at a solution
    The author provided a small hint. He suggested that we define a basis
    B={e1...ed} where each matrix in the eigenspace of U has one column
    from B and the rest are 0. What gets me is this:

    Suppose d=3 and n=3
    M1=[tex]\left\begin{bmatrix}1 & 0&0 \\ 0 &0&0\\0&0&0\end{bmatrix}\right[/tex]
    M2=[tex]\left\begin{bmatrix}0 & 0&0 \\ 0 &1&0\\0&0&0\end{bmatrix}\right[/tex]
    M3=[tex]\left\begin{bmatrix}0 & 0&0 \\ 0 &0&0\\0&0&1\end{bmatrix}\right[/tex]
    How can this be 9 dimensional?
    Wait, unless he means this: That the matrices count as eigenvectors themselves
    and should be treated as such:
    [tex]\left\begin{bmatrix}M1\\ &M2&\\&&M3\end{bmatrix}\right[/tex]
     
    Last edited: Jun 15, 2009
  2. jcsd
  3. Jun 15, 2009 #2

    Dick

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    It's like your second picture. Msub(n,n) has dimension n^2, right?
     
  4. Jun 15, 2009 #3
    Hey Dick, whats up?
    Okay good, that makes sense! I'll be back later, have an errand to run.
     
  5. Jun 15, 2009 #4
    Alright lets try it.
    Define a basis B={e1...ed} with ei a column in B.
    ei is a column matrix with 1 at position i and 0 at position=/=i.
    Set d=3 and n=3.
    now let U=[tex]\left\begin{bmatrix}a1,1 & a1,2&a1,3\\ a2,1&a2,2&a2,3\\a3,1&a3,2&a3,3\end{bmatrix}\right [/tex]
    a basis for [Esub(c)(U)] is [tex]\left\begin{bmatrix}e1&&\\ &e2&\\&&e3\end{bmatrix}\right [/tex]
    with ei a 1x3 matrix. We cannot add each matrix directly, since we would end up with
    [tex]\left\begin{bmatrix}e1\\ 0\\e3\end{bmatrix}\right [/tex] after adding say, e1 and e3
    since the sum is not an eigenvector, so the direct sum was taken to get the basis for
    [Esub(c)(U)] not done yet, just want to know if its right so far.
     
    Last edited: Jun 15, 2009
  6. Jun 15, 2009 #5

    Dick

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    I don't get it. I haven't followed enough of your posts yet to know how to magically guess what you mean, so I can't help along that line. If you mean B={e1...ed} is a basis for the subspace with eigenvalue c, then consider a full basis {v1,...,vn} and consider the number of linearly independent ways to define X(vi). Where T(X)=UX=cX.
     
    Last edited: Jun 15, 2009
  7. Jun 15, 2009 #6
    still a bit lost, theres three columns, isn't that three dimensions?
     
  8. Jun 15, 2009 #7

    Dick

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    Ok, take v1 in a completely arbitrary basis. UX(v1)=c*X(v1), right? That means X(v1) MUST be an eigenvector of U with eigenvalue c, yes? That means X(v1) must be in the span of {e1...ed}. That gives me d linearly independent choices for defining X(v1). How many choices for X(v2)?
     
  9. Jun 15, 2009 #8

    Dick

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    Yes, there are three dimensions for X(vi), but there are three vi's, v1, v2 and v3. Doesn't that make 3x3=9 dimensions?
     
  10. Jun 16, 2009 #9
    there are d choices for X(v2).
    So what you've done is take the basis {e1...ed} and the shared characteristic
    between those vectors is the fact that they are eigenvectors belonging to c. Since X(v1) is an eigenvector with eigenvalue c, it is a member of the eigenspace with basis {e1...ed}.
    One question though. How can linearly independent vectors share the same
    eigenvalue? For e2, if X(e2) gets multiplied by U, wouldn't it get a different eigenvalue?
     
  11. Jun 16, 2009 #10

    Dick

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    {e1...ed} are supposed to be linearly independent eigenvectors sharing the same eigenvalue "c", right? What's wrong with that? The identity matrix has a full basis of eigenvectors all sharing the same eigenvalue, "1".
     
  12. Jun 16, 2009 #11
    Wait, I'm onto something....
    Should I take this into consideration?
    [tex]
    \left\begin{bmatrix}M1\\ &M2&\\&&M3\end{bmatrix}\right
    [/tex]
    I don't want to sound too crazy, but do you mean that
    we have three v1's in that there are three matrices here,
    each with a [1 0 0] (rotate this row so that 1 is at the top)
    Each [1 0 0] are in different dimensions, so they are linearly
    independent, but multiplying a [1 0 0] by U should give c1,1*[1 0 0]
     
  13. Jun 16, 2009 #12

    Dick

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    If you are going to do it by drawing matrices, pick a special basis <v1,v2,...vn> where v1=e1, v2=e2, ... vd=ed. The case d=n is really too easy. In that case it will turn out that X can be ANY nonzero matrix, right? Since if d=n any vector is an eigenvector.
     
  14. Jun 16, 2009 #13
    That makes sense. I think, but why are we using X? Aren't we dealing with U?
     
  15. Jun 16, 2009 #14

    Dick

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    Have you forgotten the problem? You want to show dim[Esub(c)(T)]=nd. An element of Esub(c)(T) is an n^2xn^2 matrix. I'm calling it X. So T(X)=UX=cX.
     
  16. Jun 16, 2009 #15
    Oh that's right so X is the eigenvector.
     
  17. Jun 16, 2009 #16
    Oh that's right so X is the eigenvector. But the whole d elements to choose from
    doesn't make much sense to me if we have a linearly independent set {e1...ed}
    I can't really see why they would have the same eigenvalues if they are linear independent (e1...ed)
    if you were to choose e1 to be your v1 and multiply X(v1)*U wouldn't you be getting c1,1*v1
    and if you were to choose e2 to be your v1 and multiply X(v1)*U wouldn't you be getting
    c2,2*v1? I'm not trying to argue, I'm just concerned that I don't understand this.
     
    Last edited: Jun 16, 2009
  18. Jun 16, 2009 #17

    Dick

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    We are trying to define a basis for the subspace Esub(c)(T). So we can count the number of elements in the basis. You were trying to do it by describing the subspace in matrix form.
     
  19. Jun 16, 2009 #18
    Oh, sorry I edited it.
    But the whole d elements to choose from
    doesn't make much sense to me if we have a linearly independent set {e1...ed}
    I can't really see why they would have the same eigenvalues if they are linear independent (e1...ed)
    if you were to choose e1 to be your v1 and multiply X(v1)*U wouldn't you be getting c1,1*v1
    and if you were to choose e2 to be your v1 and multiply X(v1)*U wouldn't you be getting
    c2,2*v1? I'm not trying to argue, I'm just concerned that I don't understand this.
     
  20. Jun 16, 2009 #19

    Dick

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    I'm concerned you don't understand it too. What's the definition of Esubc(U)?
     
  21. Jun 16, 2009 #20
    That is the eigenspace of U which contains the eigenvectors that have elements of U
    as eigenvalues.
     
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