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A question about a new way to find eigenvectors that i noticed

  1. Mar 2, 2008 #1
    i got this question in which we are given the matrix T
    and we need to find the eigenvalues and the independent spaces (i dont know what is independent space) of T^2 +2*T

    the problem is that he started to solve the question as i would have solved it
    but then he puts a big X on it and does something else
    i cant understand it??(and he gets all the point for it)

    it looks as if he skips the finding the roots of polinomial step
    why???

    http://img253.imageshack.us/my.php?image=img86091xg4.jpg
     
    Last edited: Mar 2, 2008
  2. jcsd
  3. Mar 2, 2008 #2

    Dick

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    The solver appears to have realized that he didn't have to compute the eigenvalues of T^2+2*T since he already knew that the eigenvalues of T were +1 and -1, apparently from a previous problem. This let him immediately conclude the eigenvalues of T^2+2*T are 3 and -1. Once he knew the eigenvalues he substituted them in for lambda and seems to have read off the eigenvectors more or less by inspection. It's not a new way of computing eigenvalues.
     
  4. Mar 2, 2008 #3

    HallsofIvy

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    If [itex]\lambda[/itex] is an eigenvalue of T, with eigenvector v, then Tv= [itex]\lambda[/itex]v. From that, [itex](T^2+ 2T)v= T(T(v))+ 2T(v)= T(\lambda v)- 2\lambda= \lambda T(v)- 2\lambda= \lambda(\lambda v)- 2\lambda v= (\lambda^2- 2\lambda) v[/itex].

    In other words if [itex]\lambda[/itex] is an eigenvalue of T with eigenvector v, then [itex]\lambda^2- 2\lambda[/itex] is an eigenvalue of T2- 2T with eigenvector v.

    It is easier to find the eigenvalues of T and then use that formula than to find the eigenvalues of T2- 2T directly.
     
    Last edited: Mar 4, 2008
  5. Mar 4, 2008 #4
    ok i understood how you got the formula from the T^2 + 2T epression

    what now??
    how do i mix up the eigenvalues of T with this formula in order to get the new values
    ??
     
  6. Mar 4, 2008 #5

    HallsofIvy

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    What do you mean by "mix up the eigenvalues of T" and what "new values" are you talking about?

    If you mean "How do I go from the eigenvalues of T to the eigenvalues of T2- 2T?", that's exactly what I told you before.:
     
  7. Mar 4, 2008 #6
    correct me if i am wrong

    x-eigenvalue of T
    y-eigen value of the expression
    y(x)=x^2-2*x

    so if x=-1 then for that "old" eigen value we get y=3
    and we do that proccess for every eigenvalue
     
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