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Eigenvector kernel problem

  1. Jun 20, 2005 #1
    If you have the following kernel (I think that's what it's called):

    [tex]A-\lambda I=\begin{pmatrix}4 & 1 \\ 4 & 1\end{pmatrix}[/tex]

    You could write the eigenvector as:

    [tex]\operatorname{span}\begin{pmatrix}1 \\ -4\end{pmatrix}[/tex]

    My question is: does it matter how you write the "span" part of it?

    For instance, would [tex]\operatorname{span}\begin{pmatrix}-1 \\ 4\end{pmatrix}[/tex] be preferred (or different) than what I have above?

    Thanks for your help.
     
  2. jcsd
  3. Jun 20, 2005 #2

    HallsofIvy

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    [tex]A-\lambda I=\begin{pmatrix}4 & 1 \\ 4 & 1\end{pmatrix}[/tex]

    doesn't make sense- there's no λ on the right hand side! Did you mean
    [tex]A=\begin{pmatrix}4 & 1 \\ 4 & 1\end{pmatrix}[/tex] and that is the linear operator you want to find eigenvectors for?

    In that case, there are two eigenvalues: 0 and 2.

    Taking 0 as the eigenvalue leads to 4x-y=0 or y= 4x. The eigenvectors are spaned by [1, 4] (not [-1, 4]).
    Taking 2 as the eigenvalue leads to 4x-y= 2x or y= -2x. The eigenvectors are spanned by [1, -2].

    Because the "span" of a single vector is all real multiples of the vector, and -1 is a real number, it doesn't matter whether you use [-1, 4] or [1, -4] or [-1,2] instead of [1,-2].
     
  4. Jun 20, 2005 #3
    Sorry, I should have elaborated. The original matrix (A) was:

    [tex]\begin{pmatrix}3 & 1 \\ 4 & 0\end{pmatrix}[/tex]

    And the eigenvalue I was using was -1. I didn't think that mattered in relation to my question (because my ? dealt with notation), but I don't really know all of the terms, so I appologize.

    Thanks for answering though, the last part of your post answered my question.
     
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