Nilpotent matrices condition with traces

Question

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?

Answer 1

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 $0$?

Answer 2

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.