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Prove that the matrix $A = \begin{pmatrix} 1&1&1 \\ -1&-1&-1 \\ 1&1&0 \end{pmatrix}$ is nilpotent, and find its invariants and Jordan form.

So far, I've verified directly that $A^3=0$, so $A$ is clearly nilpotent. We also know that the eigenvalues of any nilpotent matrix are $0$. Using the characteristic polynomial $\lambda^3=0$, I've confirmed this. We have an eigenvalue $0$ of algebraic multiplicity 3.

This is where I get stuck. I'm not sure how to find the invariants or Jordan form. I think you could take $J_i - \lambda_i I$ for all $i=1,2,3$, where $J_i$ are Jordan blocks, but I'm not sure how to find those. Any hints would be greatly appreciated.

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    $\begingroup$ First note that if $A$ is nilpotent, then all its eigenvalues are $0$, not just some of them. Next, use the fact that $A^2 \ne 0$ to show it must contain at least one Jordon block whose size it at least $3$. $\endgroup$ – Stephen Montgomery-Smith Apr 30 '14 at 3:34
  • $\begingroup$ @StephenMontgomery-Smith Since the index of nilpotence is $3=\dim V$, it follows that the invariant is 3, correct? In which case the Jordan form is given by a standard $3\times 3$ nilpotent matrix? $\endgroup$ – Hildegarde Apr 30 '14 at 3:38
  • $\begingroup$ Yes, that is correct. $\endgroup$ – Stephen Montgomery-Smith Apr 30 '14 at 3:39
  • $\begingroup$ @StephenMontgomery-Smith Ah, I didn't think it would be so simple! Thanks for your help. $\endgroup$ – Hildegarde Apr 30 '14 at 3:40
  • $\begingroup$ A lot of these kinds of problems depend upon the matrix being small. If it was a $100 \times 100$ matrix satisfying $A^3 = 0$, you would be stuck! $\endgroup$ – Stephen Montgomery-Smith Apr 30 '14 at 3:42

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