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One property of nilpotent matrices is that a matrix $N$ is nilpotent if and only if $\operatorname{tr}(N^k)=0$ for all $k>0$. How can this property be proved?

JJ Beck
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  • look at this http://math.stackexchange.com/questions/159167/trace-of-powers-of-a-nilpotent-matrix – jim Jan 30 '14 at 02:41
  • What can you say about the eigenvalues of a nilpotent matrix, using the definition of eigenvalue? – user25004 Jan 30 '14 at 02:42

2 Answers2

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Let $\lambda_1,\ldots,\lambda_n$ the eigenvalues of $N$ repeated with their multiplicities and notice that if $\lambda_k$ is an eigenvalue of $N^k$ so since $\operatorname{tr}(N^k)=0$ we find the system of equations $$\sum_{i=1}^n\lambda_i^k=0\quad k=1,\ldots,n$$

Now we solve this system by induction:

  • the case $n=1$ is immediate
  • assume we have the result for $n-1$
  • Let $$P(x)=(x-\lambda_1)\cdots(x-\lambda_n)=x^n+a_{n-1}x^{n-1}+\cdots+a_0$$ so $$0=\sum_{i=1}^nP(\lambda_i)=\sum_{i=1}^n\lambda_i^n+a_{n-1}\sum_{i=1}^n\lambda_i^{n-1}+\cdots+na_0=na_0$$ hence $a_0=0$ and then $0$ is eigenvalue of $N$ and WLOG assume $\lambda_n=0$ and finaly the induction's hypothesis gives $\lambda_1=\cdots=\lambda_{n-1}=0$. What we can say about a matrix which all its eigenvalues $0$?
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This follows from Engel's theorem: a nilpotent Matrix can ( by a suitable Base change) be brought in strictly triangular form (with 0's on the diagonal) from which the claim is immediate.

user39082
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