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I'm trying to find the Jordan normal form and the corresponding Jordan basis for the following matrix:

$$\begin{pmatrix} 0 & 3 & 3 \\ -1 & 8 & 6 \\2 & -14 & -10\end{pmatrix}$$

I've found the eigenvalues to be $0,-1$, with eigenvectors $\begin{pmatrix} -2 \\ -2 \\ 1 \end{pmatrix}$ and $\begin{pmatrix} -3 \\ -3 \\4 \end{pmatrix}$

However, when I try to compute the third vector to complete my Jordan basis I get a probelm. I know I need to solve $v_2=(A+I)v_3$, which I have done (and checked) and found the third one to be $\begin{pmatrix} -3/2 \\ -1/2 \\ 0\end{pmatrix}$. But this doesn't seem to give me that $J=C^{-1}AC$.

What have I done wrong?

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    @Moo I'm still not sure what's wrong with what I've done though? Why does solving that system not work? – 2017-01-09
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    @Moo So have I just made a mistake in solving the system? – 2017-01-09
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    Actually, your eigenvector should be $(-2,-1,1)$ for the zero eigenvalue, so you likely wrote it wrong or simple algebra mistake. The other eigenvectors are correct. – 2017-01-09
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    send http://m.wolframalpha.com/input/?i=jordandecomposition+%7B%7B0%2C3%2C3%7D%2C%7B-1%2C8%2C6%7D%2C%7B2%2C-14%2C-10%7D%7D&x=0&y=0 – 2017-01-09
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    @Moo I knew it would be something stupid! You're right, it was just a numerical error in computing the eigenvector corresponding to the zero eigenvalue. Thank you. – 2017-01-09

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