Motivations for eigenvalues/vectors

  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. 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. HallsofIvy

    HallsofIvy 40,310
    Staff Emeritus
    Science Advisor

    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. Thanks for clearing that up. I wish both of my linear algebra textbooks made it clear which basis we were in.
  6. matt grime

    matt grime 9,396
    Science Advisor
    Homework Helper

    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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