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Eigenvalue proof

  1. Jun 13, 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.
    Show that c is an eigenvalue of T if and only if it is an eigenvalue of U.

    2. Relevant equations



    3. The attempt at a solution
    If T(A)=UA then T(A)-UA=0 (T-U)A=0.
    Let v be an eigenvector of T so Tv=cv.
    If v is an eigenvector[tex]\in[/tex]A then A
    is not a zero matrix so for (T-U)A=0 we
    have Tv-Uv=0 so cv-Uv=0
    Uv=cv so c must be an eigenvalue of U for
    Uv=Tv=cv and for Uv-Tv=0 for v[tex]\in[/tex]A.
     
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  3. Jun 13, 2009 #2

    Dick

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    You are going to have to rethink that whole thing. You have a concept wrong. T is an operator, not a nxn matrix. So writing something like (T-U) makes no sense at all. Furthermore an eigenvalue of T is number c such that for some nxn matrix T(X)=cX. Get it? EIgenvectors of T are actually MATRICIES.

    Start it this way. If c is an eigenvalue of T, then there is a matrix X such that T(X)=cX=UX. Now you want to show c is an eigenvalue of U.
     
    Last edited: Jun 13, 2009
  4. Jun 13, 2009 #3
    Knowing that T(X)-cX=0, for X is an eigenvector of T and c is an eigenvalue of T,
    T(X)=cX but since T(X)=UX UX=cX therefore
    UX-cX=0=T(X)-cX (this equation shows that for T(X) to be=UX, they must have the
    same eigenvalues).

    Yeah for that first attempt, I got T mixed up with a matrix representation of T.
     
    Last edited: Jun 13, 2009
  5. Jun 13, 2009 #4

    Dick

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    UX=cX doesn't really tell you much directly. To say c is an eigenvalue of U, you want to show there is a VECTOR v such that Uv=cv. X is a matrix. Hint: X is a nonzero matrix (why nonzero?).
     
  6. Jun 13, 2009 #5
    X is nonzero because if T(X)=cX, and c is an eigenvalue, then X must
    not be a zero matrix since eigenvectors are nonzero.

    Hang on, this will get a bit weird:
    Consider V of dimension n.
    The basis of V is <v1...vn> with each vi a 1xn column with a single nonzero
    element (vi ) at row i. If vi is the eigenvector for c, then we let all other vectors
    be multiplied by 0 and add them up (direct sum) to get the nxn matrix with vi as the only nonzero,
    which equates to vi. So X=vi and so T(X)=cX=T(vi)=cvi since UX=cX and cX=cvi,
    UX=cvi, and since X=vi, Uvi=cvi. Therefore, Uvi-cvi=0=T(vi)-cvi
    and for this to be true, Uvi must share its eigenvalue with T(vi), which it does.

    Why does X=vi? Because vi is single column matrix and X is an nxn matrix
    with one nonzero element.
     
    Last edited: Jun 13, 2009
  7. Jun 13, 2009 #6

    Dick

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    I think you've got the right idea but you did express it oddly. If X is a nonzero linear transformation, then there must a vector u such that Xu is nonzero. That's about all there is to it. Then v=Xu must be an eigenvector of U, right? Don't forget there is still a converse to prove.
     
  8. Jun 13, 2009 #7
    I know, it was weird :biggrin:. I'll be back later on to tackle that part.
     
  9. Jun 13, 2009 #8
    Ok, I'm back. The converse is: if c is an eigenvalue of U, then it is also an eigenvalue of
    T for UX=T(X)

    Knowing that UX=cX, c is an eigenvalue so X is a nonzero nxn matrix. Proving that
    vi exists in a similar manner as before, we get Uvi=cvi and since vi=X,
    and T(X)=UX=cX, T(vi)=Uvi=cvi. Thus T(vi)=cvi so c is an eigenvalue of T
    so T(vi)-cvi=0=Uvi-cvi.
     
  10. Jun 13, 2009 #9

    Dick

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    Stop with the first statement! You DON'T know UX=cX. You don't even know what X is. You are given that U has a eigenvalue c, not T. What you do know is that there is a vector v such that Uv=cv. Pay attention to the premise, ok? Now try and find an X such that UX=cX.
     
  11. Jun 14, 2009 #10
    Since c is an eigenvalue, v must be a nonzero vector. Since Uv is
    an nxn matrix of one nonzero element (cv). Now, factoring out the scalar
    c gives the scalar *a matrix containing v which we call X so we have cX.
    Since U*v produced the nxn matrix, and cX is the result of "pullng the c out",
    multiplying c and X should give the same product as Uv so Uv=cX.
    Since Uv-cX=0 and Uv-cv=0 Uv-cX=0=Uv-cv and
    -cX-cv=Uv-Uv so cX-cv=0 so X must=v therefore since
    Uv=cv, UX=cX since v=X.

    To visualize what I meant by "pulling the c out" take an nxn matrix
    and put cv in an entry. Then factor out the c and we are left with
    c*the matrix with v instead of cv.

    That is one ugly proof :bugeye:
     
  12. Jun 14, 2009 #11

    Dick

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    It's not even a proof, it's gibberish. Let's focus on one part. You said v=X. You know that's nonsense, right? v is a vector and X is a matrix. Pay attention to what things are and if you say a thing of one type equals a thing of another type, recoil in horror. Don't even write it down. In what sense can a linear transformation "equal" a vector? Equal in super quotes. There is a good answer. Think really hard.
     
    Last edited: Jun 14, 2009
  13. Jun 14, 2009 #12
    well in this case when Tv=jv with jv being the nxn matrix with only jv as
    the nonzero which is "equivalent" to the vector jv (chose j because
    we dont know if c is an eigenvalue of Tv).
     
  14. Jun 14, 2009 #13

    Dick

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    I'm not sure I understand what that means. But here's another hint. Suppose you could find a linear transformation X such that X acting on ANY vector u would give you a multiple of v (where Uv=cv). Is that what you are looking for? That's sort of like "X=v". If that's what you want, can you define one? Remember you can define a linear transformation by specifying it's action on a basis.
     
  15. Jun 14, 2009 #14
    Actually that was kind of what I meant by X=v. And remember when I said
    at the way beginning (first post) that T-U deal? I imagined the T to be
    your X or a representation as a matrix, but I didn't really specify that.
    And in the v in the proof where I said v=X
    v was supposed to be a single column matrix representation of v,
    my fault for not pointing that out.
     
  16. Jun 14, 2009 #15

    Dick

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    I THOUGHT that was what you meant. But I can't read through the "Uv-cv=0 Uv-cX=0=Uv-cv " stuff you are posting as a posting as a proof. Here's one clear way to do it. "Pick a basis {v1,v2,...,vn}. Define X as a linear transformation such that X(vi)=v for all i". Do you see how that makes X an eigenvector of T with eigenvalue c? You could also say " X(v1)=v and X(vi)=0 for i>1" which I think is your single column way. I think you are thinking the right things, but your proof statements just don't convey that.
     
  17. Jun 14, 2009 #16
    I did a similar method to that in this proof:

    Hang on, this will get a bit weird:
    Consider V of dimension n.
    The basis of V is <v1...vn> with each vi a 1xn column with a single nonzero
    element (vi ) at row i. If vi is the eigenvector for c, then we let all other vectors
    be multiplied by 0 and add them up (direct sum) to get the nxn matrix with vi as the only nonzero,
    which equates to vi. So X=vi and so T(X)=cX=T(vi)=cvi since UX=cX and cX=cvi,
    UX=cvi, and since X=vi, Uvi=cvi. Therefore, Uvi-cvi=0=T(vi)-cvi
    and for this to be true, Uvi must share its eigenvalue with T(vi), which it does.

    Why does X=vi? Because vi is single column matrix and X is an nxn matrix
    with one nonzero element.


    But for the second part, I didn't explicitly define a basis for that vector
    represented by a column matrix. But, I will modify it a bit.
    Okay using Dick's basis <v1....vn> we represent a vector with a column matrix
    (lets use v1 as the lonely nonzero). This will be called I(v1). Now we multiply
    to get UI(v1). Factoring out c1,1 gives c1,1*a nxn matrix with v1 as the
    lonely nonzero which we will call X. Now UI(v1)=cX. Since I(v1) is
    a matrix with one nonzero element (v1) as is the case with X,
    and multiplying X by U will produce the same product as multiplying I(v1)
    by U, X=I(v1) so UX=cX.
     
    Last edited: Jun 14, 2009
  18. Jun 14, 2009 #17

    Dick

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    Why are you making this so complicated? If you want to talk in terms of matrices, to get an 'eigenmatrix' X with the same eigenvalue as Uv=cv, just define each column of X to be proportional to v. So the column space of X is just span{v}. I.e.X(u)=k(u)*v where k(u) is a constant depending on u. You can just say that, I think it's the same thing you are trying to say in a more complicated, not necessarily grammatical way. But that STILL doesn't mean you get to say X=v, or X=v1. They aren't equal.
     
    Last edited: Jun 14, 2009
  19. Jun 14, 2009 #18
    Whoops, I edited it. And yes it makes no sense to compare a line with a representation of a line. They are different things, its like saying an apple is a tree.

    But for the second part, I didn't explicitly define a basis for that vector
    represented by a column matrix. But, I will modify it a bit.
    Okay using Dick's basis <v1....vn> we represent a vector with a column matrix
    (lets use v1 as the lonely nonzero). This will be called I(v1). Now we multiply
    to get UI(v1). Factoring out c1,1 gives c1,1*a nxn matrix with v1 as the
    lonely nonzero which we will call X. Now UI(v1)=cX. Since I(v1) is
    a matrix with one nonzero element (v1) as is the case with X,
    and multiplying X by U will produce the same product as multiplying I(v1)
    by U, X=I(v1) so UX=cX.
     
  20. Jun 14, 2009 #19

    Dick

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    Where v1 is the eigenvector of U, right? Say so. You mean I(v1) to be the matrix such that c1,1=1 and all of the rest of the elements are zero in the basis {v1,...,vn}, right? See? It's still confusing me. If so that's fine. I've been trying to get you to state the same thing using fewer words and symbols, but that doesn't seem to be happening. But the proof is essentially correct and that's the important part.
     
  21. Jun 14, 2009 #20
    You are absolutely right. Now I realize how delicate proofs can be in that
    I can make ONE mistake (ie saying that a matrix equals a vector) and that
    puts an arrow through the proof's heart, even if the proof is otherwise correct.
    Had I mentioned the fact that I was representing the vector in matrix form,
    this thread would've been done a long time ago. And I need to use
    fewer words and get to the point. This will only get better with practice,
    as is any skill.

    Now, what I meant is that I(v1) is a column matrix with I a column
    with a 1 at 1,1 and 0 everywhere else.
    This column matrix gets multiplied to U to get UI(v1).
    Now, at position 1,1 of U is the eigenvalue c1,1 and after
    multiplying we have c1,1*v1 at position 1,1 and 0 everywhere else
    on the matrix. Now we have c1,1*I(v1) with I(v1) a nxn matrix
    (it turned into one) with v1 at 1,1 and 0 everywhere else. So UI(v1)=c1,1*I(v1)
     
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