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Same eigenvalues = same jordan form ?

  1. Oct 6, 2011 #1
    Say that for two mxm matrices, they have equivalent eigenvalues

    If this is the case, is it safe to assume that the Jordan forms of both matrices will be the same ?

    My reasoning comes from the fact that a general jordan block is represented
    by the following matrix

    λ 1 0 0
    0 λ 1 0
    0 0 λ 1
    0 0 0 λ

    and so on for an mxm matrix
     
  2. jcsd
  3. Oct 7, 2011 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    No, it is not. The Jordan form depends upon the eigenvectors, not the eigenvalues. If a four by four matrix has [itex]\lambda[/itex] as its only eigenvalue then it can have Jordan forms of
    [tex]\begin{bmatrix}\lambda & 0 & 0 & 0\\ 0 & \lambda & 0 & 0 \\ 0 & 0 & \lambda & 0 \\ 0 & 0 & 0 & \lambda\end{bmatrix}[/tex]
    if it has four independent eigenvectors, or
    [tex]\begin{bmatrix}\lambda & 1 & 0 & 0\\ 0 & \lambda & 0 & 0 \\ 0 & 0 & \lambda & 0 \\ 0 & 0 & 0 & \lambda\end{bmatrix}[/tex]
    if it has three independent eigenvectors, or
    [tex]\begin{bmatrix}\lambda & 1 & 0 & 0\\ 0 & \lambda & 1 & 0 \\ 0 & 0 & \lambda & 0 \\ 0 & 0 & 0 & \lambda\end{bmatrix}[/tex]
    if it has two independent eigenvectors, or
    [tex]\begin{bmatrix}\lambda & 1 & 0 & 0\\ 0 & \lambda & 1 & 0 \\ 0 & 0 & \lambda & 1 \\ 0 & 0 & 0 & \lambda\end{bmatrix}[/tex]
    if it has only one independent eigenvector.
     
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