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Eigenvectors of a matrix in Jordan Normal Form

  1. Dec 19, 2011 #1
    Let [itex]A =\begin{bmatrix} -14 & 5 & -2 & 1 \\ -38 & 13 & -4 & 5 \\ 25 & -10 & 7 & 5 \\ -17 & 5 & -2 & 4 \end{bmatrix}[/itex] .

    The Jordan Normal Form of A is [itex]J =\begin{bmatrix} 2 & 1 & 0 & 0 \\ 0 & 2 & 0 & 0 \\ 0 & 0 & 3 & 1 \\ 0 & 0 & 0 & 3 \end{bmatrix}[/itex] .

    Now, I've been told that eigenvectors can be read off as the columns with no 1's on the superdiagonal.

    So for the eigenvalue 2, an eigenvector is [itex]\begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \end{bmatrix}[/itex] and for the eigenvalue 3, an eigenvector is [itex]\begin{bmatrix} 0 \\ 0 \\ 1 \\ 0 \end{bmatrix}[/itex] .

    But [itex]\text{Ker} (A-2I) = \left\{ \begin{bmatrix} w \\ 3w \\ 0 \\ w \end{bmatrix} : w\in\mathbb{R} \right\} = \text{span} \left\{ \begin{bmatrix} 1 \\ 3 \\ 0 \\ 1 \end{bmatrix} \right\}[/itex]

    but clearly [itex]\begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \end{bmatrix} \notin \text{Ker} (A-2I)[/itex]

    All eigenvectors should be in the kernel so why isn't this - isn't it an eigenvector?

    Similarly [itex]\text{Ker} (A-3I) = \left\{ \begin{bmatrix} 0 \\ \frac{2}{5}z \\ z \\ 0 \end{bmatrix} : z\in\mathbb{R} \right\} = \text{span} \left\{ \begin{bmatrix} 0 \\ 2 \\ 5 \\ 0 \end{bmatrix} \right\}[/itex]

    but clearly [itex]\begin{bmatrix} 0 \\ 0 \\ 1 \\ 0 \end{bmatrix} \notin \text{Ker} (A-3I)[/itex]
     
  2. jcsd
  3. Dec 19, 2011 #2

    Ray Vickson

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    The vectors (1,0,0,0)^T and (0,0,1,0)^T are eigenvectors of J, not of A.

    RGV
     
  4. Dec 19, 2011 #3
    Of course, [itex]J=P^{-1}AP \neq A[/itex] for some change of basis matrix [itex]P[/itex].
     
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