24
$\begingroup$

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?

$\endgroup$
26
$\begingroup$

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.

$\endgroup$
22
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Out of curiosity, can this be made to work in nonzero characteristic? $\endgroup$ – anon Jun 16 '12 at 19:07
  • $\begingroup$ Ok, but are there any other methods that you know that does not involve the use of symmetric polynomials? $\endgroup$ – yoshi Jun 16 '12 at 19:07
  • 11
    $\begingroup$ @anon, The matrix $\begin{pmatrix}1&1\\0&1\end{pmatrix}$ and all its powers have zero trace in characteristic two. $\endgroup$ – Mariano Suárez-Álvarez Jun 16 '12 at 19:08
  • 2
    $\begingroup$ @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 $\det A$). $\endgroup$ – anon Jun 16 '12 at 19:10
9
$\begingroup$

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.)

$\endgroup$
  • 2
    $\begingroup$ This is how i'd prove the Newton identities, in fact! $\endgroup$ – Mariano Suárez-Álvarez Jun 16 '12 at 20:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.