Is there any upper bound $m$ of $n$ such that $A=0$ is equivalent to $Tr(A^n)=0$ for $n=1...m$?

Question

Suppose I have a matrix $A=0$, hence I know $Tr(A^n)=0\forall n\ge1$, but is there some bound for $n$ such that $Tr(A^n)=0$ for $n=1...m$ is equivalent to $A=0$? For example, when $A$ is positive semidefinite matrix, $m=1$. When $A$ is hermitian matrix, $m=2$. Hence I guess $m$ should be a finite number?

Answer 1

No, even $\text{tr}(A^k) = 0$ for all $k \ge 1$ doesn't imply $A = 0$. Over a field of characteristic zero it implies that all the eigenvalues of $A$ are zero, hence that $A$ is nilpotent, so e.g. we could have $A = \left[ \begin{array}{cc} 0 & 1 \\ 0 & 0 \end{array} \right]$. And in positive characteristic it doesn't even imply that.

This already occurs if $A$ is an $n \times n$ matrix and $\text{tr}(A^k) = 0$ for $1 \le k \le n$ so we never need to check past $k = n$.

Answer 2

In general from $\operatorname{tr}(A^k)=0$, $k=1, \ldots, n$ you (only) get $n! A^n = 0$.

This is because you can express the coefficients of the characteristic polynomial $P_A(\cdot)$ of $A$ in terms of $\operatorname{tr}(A^k)=0$, $k=1, \ldots, n$. Now use Hamilton Cayley $P_A(A) = 0$ and get $n! A^n = 0$ ( $n!$ because we have some denominators in the above expressions).

Example: $A$ is $4\times 4$. Eliminate $x_i$'s from the system $$A - x_1 = 0 \\ t_k - \sum_{i=1}^4 x^k_i, \ \ k=1, \ldots, 4$$

and get

$$A^4 - t_1 A^3 + \frac{1}{2} ( t_1^2 - 2 t_2) A^2 + \frac{1}{6}( 3 t_1 t_2 - 2 t_3 - t_1^3) A + \frac{1}{24}(8 t_1 t_3 +3 t_2^2- 6 t_1^2 t_2 - 6 t_4)=0 $$

For a general formula, consider the equality

$$(I- A X) = \exp ( \log (I - A X))$$ Take determinants and get

$$\det(I - A X) = \exp( \operatorname{tr}( I- A X))$$

On RHS, use

$$\log (I - A X) = -A X- \frac{A^2 X^2}{2} - \frac{A^3 X^3}{3} - \cdots$$

Take the traces, then $\exp$, truncate at level $n$ and get $\det(I - A X)$. Now use the fact that $X= \frac{1}{A}$ is a root.

Answer 3

Another simple example is that if $A=\left( \begin{matrix} a& 0\\ b& 0\\ \end{matrix} \right) $, then $Tr\left( A^n \right) =a^n$ which has no restriction on $b$.