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Motivations for eigenvalues/vectors

  1. May 1, 2005 #1
    Hi,

    I understand one of the motivations for eigenvalues/vectors is when you need
    to compute A^k * x. So we like to write,

    A = C*D*C^-1 and then A^k = C * D^k * C^-1, and D^k is trivial to compute.

    My professor said C^-1 and C can be though of as change of coordinate
    matrices. But from which basis? For example, C^-1 would take me from
    *some* basis to the basis of eigenvectors. But what is this *some* basis?

    Is it assumed that everything is coordinitized relative to some basis B in
    R^n. And then I want to change to the basis of eigenvectors B'?
     
  2. jcsd
  3. May 1, 2005 #2
    Short answer: Yes.

    A is the matrix of a transformation wrt some basis B. D is the matrix of the same transformation wrt the eigenbasis B'. [itex]C^{-1}[/itex] takes vectors from B to B', and C takes vectors back from B' to B.
     
  4. May 1, 2005 #3

    HallsofIvy

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    A general linear transformation can be written as a matrix in a given basis. If all you are given is the matrix, then the corresponding "given basis" is <1, 0,...>, <0,1,...> , etc. The basis you are changing to is the basis consisting of the eigenvectors for the matrix A.

    (A matrix can be diagonalized if and only if there exist a basis consisting entirely of eigenvectors of the matrix.)
     
  5. May 1, 2005 #4
    Thanks for clearing that up. I wish both of my linear algebra textbooks made it clear which basis we were in.
     
  6. May 2, 2005 #5

    matt grime

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    But it obviously goes from whatever basis A is written with respect to, and it doesn't matter what that basis is, which is why the book didn't state it.
     
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