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Let $A$ be an $n\times n$ complex nilpotent matrix. Then we know that because all eigenvalues of $A$ must be 0, it follows that $\text{tr}(A^n)=0$ for all positive integers $n$.

What I would like to show is the converse, that is, if $\text{tr}(A^n)=0$ for all positive integers $n$, then $A$ is nilpotent. I tried to show that 0 must be an eigenvalue of $A$, then try to show that all other eigenvalues must be equal to 0. However, I am stuck at the point where I need to show that $\det(A)=0$.

May I know of the approach to show that $A$ is nilpotent?

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What are the eigenvalues of $A^2, A^3 ...$? [as values, and in terms of the eigenvalues of $A$ in general, whether or not nilpotent]. Cayley Hamilton might help. –  Mark Bennet Jun 16 '12 at 19:01
For the record, here is a related more general question. –  Marc van Leeuwen May 18 at 8:21

3 Answers 3

If the eigenvalues of $A$ are $\lambda_1$, $\dots$, $\lambda_n$, then the eigenvalues of $A^k$ are $\lambda_1^k$, $\dots$, $\lambda_n^k$. It follows that if all powers of $A$ have zero trace, then $$\lambda_1^k+\dots+\lambda_n^k=0\qquad\text{for all $k\geq1$.}$$ Using Newton's identities to express the elementary symmetric functions of the $\lambda_i$'s in terms of their power sums, we see that all the coefficients of the characteristic polynomial of $A$ (except that of greatest degree, of course) are zero. This means that $A$ is nilpotent.

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Out of curiosity, can this be made to work in nonzero characteristic? –  anon Jun 16 '12 at 19:07
Ok, but are there any other methods that you know that does not involve the use of symmetric polynomials? –  yoshi Jun 16 '12 at 19:07
@anon, The matrix $\begin{pmatrix}1&1\\0&1\end{pmatrix}$ and all its powers has zero trace in characteristic two. –  Mariano Suárez-Alvarez Jun 16 '12 at 19:08
@yoshi: Your question is basically "can $p_k=0$ for all $k$ be used to prove $e_n=0$?" It is literally about two types of symmetric functions, just with different words ($\mathrm{tr}(A^n)$ and $\det A$). –  anon Jun 16 '12 at 19:10

Assume that for all $k=1,\ldots,n$, $\mathrm{tr}(A^k) = 0$ where $A$ is a $n\times n$ matrix.
We consider the eigenvalues in $\mathbb C$.

Suppose $A$ is not nilpotent, so $A$ has some non-zero eigenvalues $\lambda_1,\ldots,\lambda_r$.
Let $n_i$ the multiplicity of $\lambda_i$ then $$\left\{\begin{array}{ccc}n_1\lambda_1+\cdots+n_r\lambda_r&=&0 \\ \vdots & & \vdots \\ n_1\lambda_1^r+\cdots+n_r\lambda_r^r&=&0\end{array}\right.$$ So we have $$\left(\begin{array}{cccc}\lambda_1&\lambda_2&\cdots&\lambda_r\\\lambda_1^2 & \lambda_2^2 & \cdots & \lambda_r^2 \\ \vdots & \vdots & \vdots & \vdots \\ \lambda_1^r & \lambda_2^r & \cdots & \lambda_r^r\end{array}\right)\left(\begin{array}{c}n_1 \\ n_2 \\ \vdots \\ n_r \end{array}\right)=\left(\begin{array}{c}0 \\ 0\\ \vdots \\ 0\end{array}\right)$$ But $$\mathrm{det}\left(\begin{array}{cccc}\lambda_1&\lambda_2&\cdots&\lambda_r\\\lambda_1^2 & \lambda_2^2 & \cdots & \lambda_r^2 \\ \vdots & \vdots & \vdots & \vdots \\ \lambda_1^r & \lambda_2^r & \cdots & \lambda_r^r\end{array}\right)=\lambda_1\cdots\lambda_r\,\mathrm{det}\left(\begin{array}{cccc} 1 & 1 & \cdots & 1 \\ \lambda_1&\lambda_2&\cdots&\lambda_r\\\lambda_1^2 & \lambda_2^2 & \cdots & \lambda_r^2 \\ \vdots & \vdots & \vdots & \vdots \\ \lambda_1^{r-1} & \lambda_2^{r-1} & \cdots & \lambda_r^{r-1}\end{array}\right)\neq 0$$ (Vandermonde)

So the system has a unique solution which is $n_1=\ldots=n_r=0$. Contradiction.

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Here is an argument that does not involve Newton's identities, although it is still closely related to symmetric functions. Write $$f(z) = \sum_{k\ge 0} z^k \text{tr}(A^k) = \sum_{i=1}^n \frac{1}{1 - z \lambda_i}$$

where $\lambda_i$ are the eigenvalues of $A$. As a meromorphic function, $f(z)$ has poles at the reciprocals of all of the nonzero eigenvalues of $A$. Hence if $f(z) = n$ identically, then there are no such nonzero eigenvalues.

The argument using Newton's identities, however, proves the stronger statement that we only need to require $\text{tr}(A^k) = 0$ for $1 \le k \le n$. Newton's identities are in fact equivalent to the identity $$f(z) = n - \frac{z p'(z)}{p(z)}$$

where $p(z) = \prod_{i=1}^n (1 - z \lambda_i)$. To prove this identity it suffices to observe that $$\log p(z) = \sum_{i=1}^n \log (1 - z \lambda_i)$$

and differentiating both sides gives $$\frac{p'(z)}{p(z)} = \sum_{i=1}^n \frac{- \lambda_i}{1 - z \lambda_i}.$$

(The argument using Newton's identities is also valid over any field of characteristic zero.)

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This is how i'd prove the Newton identities, in fact! –  Mariano Suárez-Alvarez Jun 16 '12 at 20:33

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