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How do I find a Jordan Canonical form of a 3x3 matrix with eigenvalue 0 which has an algebric multiplicity of 3?

\begin{bmatrix} {5} & -9 & -4 \\ {6} & {-11} & {-5} \\ {-7} & {13} & {6} \\ \end{bmatrix}

I get to the eigenvalue 0 with a characteristic polynomial of \lambda^{3}

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If the eigenvalue zero has algebraic multiplicity of $\;3\;$ , then the characteristic polynomial must be $\;x^3\;$, thus the matrix is nilpotent, and thus it is one of the following:

$$\begin{pmatrix}0&0&0\\0&0&0\\0&0&0\end{pmatrix}\;,\;\;\begin{pmatrix}0&1&0\\0&0&0\\0&0&0\end{pmatrix}\;,\;\;\begin{pmatrix}0&1&0\\0&0&1\\0&0&0\end{pmatrix}$$

In the first case, the geometric multiplicity of $\;0\;$ is three, in the second one it is two, and it is one in the last one.

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  • $\begingroup$ I got tothe third matrix with a geometric multiplicity of 1, but then I am not able to calculate \begin{bmatrix} {5} & -9 & -4 \\ {6} & {-11} & {-5} \\ {-7} & {13} & {6} \\ \end{bmatrix}\{pmatrix}^{2} as a result of \J^2 $\endgroup$ – Maria Apr 4 '17 at 15:53
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    $\begingroup$ You changed completely your original question... In your original post there was no matrix at all and you only asked what you wrote in your first two lines! Don't do that: upvote/accept this question's answers and ask a new question ! $\endgroup$ – DonAntonio Apr 4 '17 at 17:50

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