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Diagonalization, which eigenvector is found?

  1. Nov 27, 2009 #1
    Hi!

    This might be a silly question, but I can't seem to figure it out and have not found any remarks on it in the literature.

    When diagonalizing an NxN matrix A, we solve the characteristic equation:

    Det(A - mI) = 0

    which gives us the N eigenvalues m. Then, to find the eigenvectors v of A, we solve the eigenvalue problem

    Av = mv

    Now, any scalar multiple of an eigenvector of A is itself an eigenvector with the same eigenvalue. So, which eigenvector do we find when solving the eigenvalue problem? It can't be totally random, can it? Is there a way of determining which of the infinitude of eigenvectors (all with the same "direction") the algorithm chooses?
     
  2. jcsd
  3. Nov 27, 2009 #2
    Usually 3 things are done.

    Normalize the vector so that [itex]\|v\| = 1[/itex]
    Make the smallest entry other than zero equal to 1
    Make the largest equal to one

    [tex]
    \begin{align*}
    (1) \Longrightarrow v_0 &= \frac{1}{\sqrt{v^*v}}v\\
    (2) \Longrightarrow v_0 &= \frac{1}{\min_{v_i\neq 0}(v_i)}v\\
    (3) \Longrightarrow v_0 &= \frac{1}{\max(v_i)}v
    \end{elign*}
    [/tex]

    It is a choice... You can come up yourself with something else anyway
     
  4. Nov 27, 2009 #3
    Thanks trambolin for your answer. However, that was not really my question. I'm aware this is what one usually does when the eigenvectors are found.

    My problem is this: When we solve

    Av = mv (1)

    for the eigenvectors v, we find a set of N eigenvectors. But these are not unique. How does equation (1) "choose" which eigenvector in the 1-d subspace spanned by an eigenvector to "return" since all of them are equally valid?
     
  5. Nov 27, 2009 #4

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Equation (1) doesn't "choose" any specific eigenvector. What typically happens is that you get something like y= 2x, z= 3x- that is, all but one of the components (in the case that there is only one eigenvector corresponding to each eigenvalue) so that the eigenvector is of the form <x, 2x, 3x>= x<1, 2, 3>. You then choose a value of x.

    As far as the diagonalization is concerned, it doesn't matter which you choose- using any of the eigenvectors corresponding to the eigenvalues as columns of "P" will still give you a matrix such that [itex]P^{-1}AP[/itex] is diagonal.
     
  6. Nov 27, 2009 #5
    Ofcourse! It WAS a silly question. This is what happens when you stop doing things by hand and start relying too much on Mathematica to do the thinking for you.

    Thanks guys!
     
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