If $Tr(A^k)=0$ for $1\leq k\leq n$ is $A^n=0_n$, where $A\in M_n(\mathbb R)$?

Question

I know that the relation $Tr(A^k)=0$ is equivalent to $\sum_1^n(\alpha_i^k)=0$ where $\alpha_i$ is an eigenvalue of A and that $\alpha_i=0$ for any i is a solution to the system, but is it unique?

Edit: I forgot to specify the size of the matrix the first time around

Answer 1

I take, as per edit, that we are talking $n \times n$ matrices.

Yes, because of Newton's identities.

They imply that all the elementary symmetric functions in the eigenvalues of $A$ are $0$, so the characteristic polynomial of $A$ is $x^n$.