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Eigenvalues, linear transformations

  1. Feb 13, 2010 #1
    1. The problem statement, all variables and given/known data

    T: V-> V, dimV = n, satisfies the condition that T2 = T

    1. Show that if v [tex]\in[/tex] V \ {0} then v [tex]\in[/tex] kerT or Tv is an eigenvector for eigenvalue 1.

    2. Show that T is diagonalisable.

    2. Relevant equations



    3. The attempt at a solution

    I have shown in an earlier part of the question that the eigenvalues of T are 0 and 1.

    1. Considering eigenvalue 1, we get T(v) = v, T(T(v)) = T(v), T(v) = T(v), this is true so T(v) is an eigenvector.

    If v [tex]\in[/tex]kerT then we have T(v) = v, but then surely we get 0 = v which it isn't?

    Also, I think I have worked back from the answer to show that these work, rather than showing that these are the only possible eigenvectors satisfying T(v) = v.

    2. Suppose I have shown the 1st part correctly.
    dim(kerT) +dim(Tv) = dim(kerT) + dim (imT) = dimV
    So we have n linearly independent eigenvectors which means T is diagonal. I'm not so sure about this claim.
     
  2. jcsd
  3. Feb 13, 2010 #2

    Dick

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    If v is in ker(T) the T(v)=0*v. Yes, it's an eigenvector. The means every vector in ker(T) is an eigenvector and you've just shown every vector in im(T) is also an eigenvector. As you've said the sum of their dimensions is n. So, right, you wouldn't have any problems finding a basis of eigenvectors.
     
  4. Feb 14, 2010 #3

    HallsofIvy

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    Better is the other way: if u is in the image of T, then u= T(v) for some v so T(u)= T(T(v))= T2(v)= T(v)= u. Therefore any vector in Im(T) is an eigenvector with eigenvalue 1.

    No, T(v) is not equal to v. Saying that v is in the kernel of T means that T(v)= 0 which, in turn, gives T2(v)= T(T(v))= T(0)= 0. Therefore, any vector in ker(T) is an eigenvector with eigenvalue 0.

    Yes, that is correct. Every vector in V is an eigenvector of V with eigenvalue 0 or 1.
     
  5. Feb 14, 2010 #4

    Dick

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    Ooops. Now that's not correct. If u is in ker(T) and v is in im(T) and u and v are nonzero, u+v is not an eigenvector of T.
     
  6. Feb 14, 2010 #5
    So the way I have shown T is diagonalisable is incorrect?
     
  7. Feb 14, 2010 #6

    Dick

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    No, it's not incorrect. All you need to do is find a basis of eigenvectors. My only point was that doesn't prove ALL vectors are eigenvectors.
     
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