# Jordan Canonical form with 3 eigenvalues =0?

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}

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.