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Find the Jordan canonical form of

A=$\left[ {\begin{array}{c} 1 & 1 & -1 \\ 0 & 0 & 2 \\ 0 & -1 & 3 \end{array} } \right]$

I wang to solve its eigenvalues first and then get their corresponding eigenvectors and generalized eigenvectors to find the Jordan norm form. But it's hard to solve eigenvalues from characteristic polynomials, can you give me some other approaches? Thank you!

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    $\begingroup$ Well, $\begin{pmatrix}1\\0\\0\end{pmatrix}$ is clearly an eigenvector. And, with a bit more luck, $\begin{pmatrix}0\\1\\1\end{pmatrix}$ is too. $\endgroup$ – user228113 Jul 28 '16 at 15:34
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$$\det(A-\lambda I)=\det(\left(\left[ {\begin{array}{c} 1-\lambda & 1 & -1 \\ 0 & -\lambda & 2 \\ 0 & -1 & 3-\lambda \end{array} } \right]\right)=(1-\lambda)\det(\left(\left[ {\begin{array}{c} -\lambda & 2 \\ -1 & 3-\lambda \end{array} } \right]\right)$$

One of the eigenvalue is $1$, the remaining part is a quadratic equation.

Furthermore, for the remaining part, we can visually inspect $\lambda=1,2$ are the remaining root by making the rows to be multiple of each other.

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