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Consider the set of $n\times n$ complex matrices $X\subseteq \mathbb{C}^{n^2}$ such that $X^n=0$. What is a minimal set of at most $n^2-1$ polynomials $f_1,\ldots,f_m$ such that the set of matrices is the variety of $f_1,\ldots,f_m$?

I don't really know where to start here. What polynomials would all those matrices be zero on (beside the obvious zero polynomial)?

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I assume that the question is asking about polynomials on the entries of this matrix $X$, rather than polynomials on the matrix itself. If this is not the case, let me know. – Omnomnomnom Jan 30 '14 at 1:15
@Omnomnomnom My bad, sorry. You're right. – JJ Beck Jan 30 '14 at 1:21
up vote 3 down vote accepted

Hint: $X$ is a nilpotent matrix if and only if $$ \operatorname{trace}(X^i) = 0 $$ For $i = 1,\dots,n$.

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Thanks.. sure, so we get $n$ polynomials from those relations. How can we know it's a minimal set though (i.e. we can't take out some polynomial)? – JJ Beck Jan 30 '14 at 6:09
The key is to think in terms of the eigenvalues of the matrix. We need enough equations to guarantee that all $n$ eigenvalues of $X$ (up to algebraic multiplicity) are zero, which is equivalent to the statement that $X^n = 0$ for some $n$. – Omnomnomnom Jan 30 '14 at 13:58
The first $n$ moments of the zeros of a monic polynomial of degree $n$ uniquely determine its coefficients. The trace of $X^i$ is the same as $\sum_{k=1}^n {\lambda_k}^i$, where each $\lambda_k$ is a root of the characteristic polynomial of $X$. $X$ is nilpotent if and only if its characteristic polynomial is $p(t) = t^n$. – Omnomnomnom Jan 30 '14 at 14:04
Is there anything unclear about my explanations? I'd be happy to elaborate further. Regarding the determination of coefficients: Newton's identities are useful to this effect. – Omnomnomnom Jan 31 '14 at 16:40
Thank you, Omnomnomnom. I think it is clear now. :) – JJ Beck Jan 31 '14 at 23:22

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