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After playing around for a bit I found one:

$$ \begin{bmatrix} 0&1&0\\ 0&0&1\\ 0&0&0 \end{bmatrix} $$

but I couldn't find a good systematic way.

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It is not hard to come up with non-zero nilpotent matrices. – André Nicolas Mar 5 '13 at 6:12
I get lots of junk mail about solving nilpotency problems... – copper.hat Mar 5 '13 at 6:13
@AndréNicolas thank you for telling me what I was looking for! That+Wikipedia was a great help. – crf Mar 6 '13 at 4:31
up vote 1 down vote accepted

All strictly triangular matrices are nilpotent (see here) & hence can have $0$ as the only eigenvalue.

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$$A = \begin{bmatrix} 0&a&c\\ 0&0&b\\ 0&0&0 \end{bmatrix} $$

with $ab \ne 0$ and $c$ arbitrary.

So the only eigenvalue is $\lambda_{1,2,3} = 0$, with multiplicity three.

Update: from CH's comment, the restriction on $a, b$ is not needed.

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Why do you care about the values of $a,b$? – copper.hat Mar 5 '13 at 6:15
@copper.hat: You are correct, I don't need that restriction. Since $ab \ne 0$, the first two rows are not proportional and their cross-product is $(ab, 0, 0)$, so the eigenspace, $E(0) = ker A$ is the line through $e_1 = (1, 0, 0)$. However, we don't need anything quite so rigid for this problem. I tried to do something to help match the example from the OP. – Amzoti Mar 5 '13 at 6:21
Nice post, and update, too...I've done my fare share of edits/updates based on feedback! – amWhy Apr 25 '13 at 0:25

Using the Schur decomposition, it is exactly the set of matrices $U T U^*$, where $U$ is unitary, and $T$ is strictly upper triangular.

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