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When constructing the jordan normal form of the matrix $A$ one has to construct the jordan blocks corresponding to the eigenvalues $\lambda_i$ of the matrix. If the eigenvalue is simple or semi-simple the block is trivial to construct but in the last case, when $\text{mult}(\lambda_i) \not= \text{kern}(A - \lambda I)$ i feel there is some abiugisity in how the block should be created.

I will use the examlpe on wikipedia to illustrate. In it $\lambda_3 = \lambda_4 = 4$ and we find $x = (1, 0, 0, 0)^T$ and $y = (1, 0, -1, 1)^T$

QUESTION: In which order should $x$ and $y$ appear as column vectors in the matrix $P$ such that $A = PJP^{-1}$ Is there any rule I can apply to determine the order of the vectors?

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    $\begingroup$ It depends on whether you like your Jordan blocks to have the 1s above the diagonal or below the diagonal. The two forms are similar, though. They should appear in $P$ in whatever order you choose for your ordered basis. In the example in Wikipedia, $x$ is a generalized eigenvector but not an eigenvector, and $y=(A-4I)x$ is an eigenvector. Since in the Jordan form the eigenvector occurs before the generalized eigenvector (the third column corresponds to an eigenvector, but the fourth column does not), then the eigenvector occurs first in $P^{-1}$ (following your ntoation, not Wikipedia's) $\endgroup$
    – Arturo Magidin
    Commented Dec 3, 2012 at 19:56
  • $\begingroup$ Thanks. This means that the $lambdas$ in the $J$ matrix corresponding to a generalized eigenvector should have a 1 over them? (if i follow wikipedias example.) What if there are several general eigenvectors in the block? Does the order of them matter or is the only important thing that the eigenvector appear before the generalized eigenvectors. $\endgroup$
    – lysgaard
    Commented Dec 4, 2012 at 9:53

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