To prove that $T$ is nilpotent under the given conditions, let's proceed with a detailed argument.

Problem Summary

We are tasked with proving that a linear operator $T \in A_F(V)$ is nilpotent (i.e., $T^m = 0$ for some positive integer $m$) over a field $F$ with characteristic 0, under the condition that: $\text{tr}(T^i) = 0, \quad \text{for all } i \geq 1.$

Proof

1. Basic Definitions:

   • Since $F$ has characteristic 0, the trace function satisfies all properties of traces in a field of characteristic 0, particularly cyclic invariance: $\text{tr}(AB) = \text{tr}(BA)$ for any matrices $A, B$.
   • The trace of a linear operator $T$ is the sum of its eigenvalues (counted with multiplicity).

2. Trace Condition:

   • The condition $\text{tr}(T^i) = 0$ for all $i \geq 1$ implies that for every $i$, the sum of the eigenvalues of $T^i$ is 0.

3. Eigenvalues:

   • Let the eigenvalues of $T$ (over some algebraic closure of $F$) be $\lambda_1, \lambda_2, \ldots, \lambda_n$.
   • For any $i$, the eigenvalues of $T^i$ are $\lambda_1^i, \lambda_2^i, \ldots, \lambda_n^i$. Thus, the condition $\text{tr}(T^i) = 0$ implies: $\lambda_1^i + \lambda_2^i + \cdots + \lambda_n^i = 0, \quad \forall i \geq 1.$

4. Vandermonde Argument:

   • Consider the eigenvalues $\lambda_1, \lambda_2, \ldots, \lambda_n$. The conditions $\lambda_1^i + \lambda_2^i + \cdots + \lambda_n^i = 0, \quad \forall i \geq 1,$ imply that all eigenvalues must be 0. This can be shown using the linear independence of power functions $1, x, x^2, \ldots$ over $\mathbb{C}$ (or any algebraic closure of $F$).

   • The trace conditions essentially form a Vandermonde system:

     1 & 1 & \cdots & 1 \\ \lambda_1 & \lambda_2 & \cdots & \lambda_n \\ \lambda_1^2 & \lambda_2^2 & \cdots & \lambda_n^2 \\ \vdots & \vdots & \ddots & \vdots \\ \lambda_1^{n-1} & \lambda_2^{n-1} & \cdots & \lambda_n^{n-1} \end{bmatrix} \begin{bmatrix} 1 \\ 1 \\ \vdots \\ 1 \end{bmatrix} = 0.$$ The only solution is $$\lambda_1 = \lambda_2 = \cdots = \lambda_n = 0$$.

5. Nilpotency:

   • Since all eigenvalues of $T$ are zero, the minimal polynomial of $T$ must be of the form $x^m$, where $m \leq n$ is the size of the vector space $V$.
   • Thus, $T^m = 0$, proving that $T$ is nilpotent.

Conclusion

The operator $T$ is nilpotent because the trace condition forces all eigenvalues of $T$ to be zero.

Let me know if you'd like further clarification or additional details. Here are five related questions for practice:

1. How does the field characteristic affect the trace function?
2. What is the significance of nilpotency in linear algebra and matrix theory?
3. Can you generalize the proof to fields of nonzero characteristic?
4. What role does the Vandermonde matrix play in this proof?
5. How would this proof change if the trace condition were only for finitely many $i$?

Tip: When proving nilpotency, analyze both eigenvalues and minimal polynomials—they often reveal the underlying structure of the operator.