$\operatorname{tr}(A)=\operatorname{tr}(A^{2})= \ldots = \operatorname{tr}(A^{n})=0$ implies $A$ is nilpotent

Solution 1:

Two facts: (1) The trace of a matrix $A$ is the sum of it's eigenvalues and (2) the eigenvalues of $A^k$ are $\lambda^k$ where $\lambda$ is an eigenvalue of $A$ matched with algebraic multiplicities.

This means we can write $$\lambda_1 + \cdots + \lambda_n = 0$$ $$\lambda_1^2 + \cdots + \lambda_n^2 = 0$$ $$\cdots$$ $$\lambda_1^n + \cdots + \lambda_n^n = 0$$

It is also a known fact that all symmetric polynomials can be expressed as a polynomial in the above symmetric polynomial forms. The characteristic equation of $A$,

$$\det(A-\lambda I) = (\lambda - \lambda_1) \cdots (\lambda - \lambda_n) =\\ \lambda^n + f_{n-1}(\lambda_1, \dots, \lambda_n) \lambda^{n-1} + \cdots + f_0(\lambda_1, \cdots, \lambda_n),$$

has coefficients symmetric in $\lambda_1, \cdots, \lambda_n$. This means they must be constant as the expression in terms of the powers of eigenvalues can be evaluated with those expressions equal to zero. I.e., the coefficients are in fact not dependent on the eigenvalues at all! But since each of the coefficients will go to zero as $|\lambda_1 \cdots \lambda_n| \to 0$ the constant terms must be zero and hence the characteristic equation is $\lambda^n$ identically.

Solution 2:

Use the Cayley-Hamilton theorem in combination with Newton's identities.

Solution 3:

To complete your first attempt of proof, you need the following lemma for Vandermonde matrices

Lemma: Let $V=\begin{bmatrix} 1 & \alpha_1 & \alpha_1^2 & \dots & \alpha_1^{n-1}\\ 1 & \alpha_2 & \alpha_2^2 & \dots & \alpha_2^{n-1}\\ 1 & \alpha_3 & \alpha_3^2 & \dots & \alpha_3^{n-1}\\ \vdots & \vdots & \vdots & \ddots &\vdots \\ 1 & \alpha_n & \alpha_n^2 & \dots & \alpha_n^{n-1} \end{bmatrix}$

Then $\det V=\prod\limits_{1\leq i<j\leq n}(\alpha_i-\alpha_j)$.

Corollary: $\det V=0\Leftrightarrow\exists i\neq j\ \alpha_i=\alpha_j$

Now, take the Jordan form as you suggested. Note that, for the purpose of calculating $tr A$ you need only non-zero eigenvalues.

Let $\alpha_1,\ldots,\alpha_m$ these eigenvalues $(m\leq n)$, $r_1,\ldots,r_m$ the dimensions of their generalized eigenspaces. You have the same relation as before

$r_{1}\alpha_{1}+\ldots+r_{m}\alpha_{m}=0 \\ r_{1}\alpha_{1}^{2}+\ldots+r_{m} \alpha_{m}^{2}=0\\ \ldots \\ r_{1}\alpha_{1}^{n}+\ldots +r_{m}\alpha_{m}^{n}=0$

That can be written $$\begin{bmatrix} \alpha_1 & \alpha_2 & \dots & \alpha_m\\ \alpha^2_1 & \alpha_2^2 & \dots & \alpha_m^2\\ \alpha_1^3 & \alpha_2^3 & \dots & \alpha_m^3\\ \vdots & \vdots & \ddots &\vdots \\ \alpha_1^n & \alpha_2^n & \dots & \alpha_m^n \end{bmatrix}\begin{bmatrix} r_1\\r_2\\\vdots\\r_m\end{bmatrix}=0$$

Now, by hypothesis $\alpha_i$ are non-zero and differ one from another. Now, take the first $m\times m$ minor of the system's coefficients matrix.

It's the transponse of a Vandermonde matrix V, each column of which has been multiplied by a non-zero element. The first lemma ensures that $\det V\neq 0$.

Therefore, the whole coefficient matrix has rank $m$, i.e. it represents an injective map $\mathbb{C}^m\rightarrow\mathbb{C}^n$. This implies $\forall i,\ r_i=0$, which tells you that the inital $A$ is indeed nilpotent.

Solution 4:

I think you can show that the characteristic polynomial $\det(\lambda I - A)$ of $A$ is in fact $\lambda^n$ using newtons formula on symmetric polynomials.