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Number of jordan blocks in Jordan decomposition

  1. Dec 12, 2013 #1
    Given a matrix $$A$$. Is it possible to have a Jordan block form like:

    $$\begin{pmatrix}
    \lambda & 1 & 0 & 0\\
    0 & \lambda & 0 & 0 \\
    0 & 0 & \lambda & 1\\
    0 & 0 & 0 & \lambda\\
    \end{pmatrix}$$

    ?
     
    Last edited: Dec 12, 2013
  2. jcsd
  3. Dec 12, 2013 #2

    jgens

    User Avatar
    Gold Member

    The matrix you named above has exactly that jordan canonical form. Sooooooo...yes.
     
  4. Dec 12, 2013 #3
    Then I have another question.

    The dimension of ker(A-λI) is the number of Jordan blocks associated to λ. And n, where n is given by dim(ker(A-λI)^n)=dim(ker(A-λI)^(n+1)) is the size of the largest Jordan block. The total size is dim(ker(A-λI).

    So, how do we know which form is the correct one in the following case?

    $$\begin{pmatrix}
    \lambda & 1 & 0 & 0 & 0 & 0 & 0\\
    0 & \lambda & 0 & 0 & 0 & 0 & 0\\
    0 & 0 & \lambda & 1 & 0 & 0 & 0\\
    0 & 0 & 0 &\lambda & 0 & 0 & 0\\
    0 & 0 & 0 & 0 &\lambda & 1 & 0\\
    0 & 0 & 0 & 0 &0& \lambda & 1\\
    0 & 0 & 0 & 0 & 0 & 0 & \lambda\\
    \end{pmatrix}$$

    or

    $$\begin{pmatrix}
    \lambda & 0 & 0 & 0 & 0 & 0 & 0\\
    0 & \lambda & 1 & 0 & 0 & 0 & 0\\
    0 & 0 & \lambda & 1 & 0 & 0 & 0\\
    0 & 0 & 0 &\lambda & 0 & 0 & 0\\
    0 & 0 & 0 & 0 &\lambda & 1 & 0\\
    0 & 0 & 0 & 0 &0& \lambda & 1\\
    0 & 0 & 0 & 0 & 0 & 0 & \lambda\\
    \end{pmatrix}$$

    If there isn't enough information, what else do I need?
     
  5. Dec 12, 2013 #4

    jgens

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    Gold Member

    I mean there are algorithms to turn any given matrix into its Jordan canonical form, so if you have a specific matrix, then there is always a way to tell the difference. To distinguish between the above cases you can always check that the matrices obtained by restricting the domain to a certain subspace have the desired block form. I have no idea if there are better ways of checking beyond these basic tricks, however, since I rarely use canonical forms.
     
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