Matrix algebra : matrix nilpotent

Question

Suppose that $Tr(A^n) = 0$ for all integer $n>0$. I have to prove that $A$ is nilpotent.

I can see it's true for little $n$ but i do not know how to prove that for all integer $n$. Any idea please?

Answer 1

If $A$ is an $m\times m$ matrix, then $\mathrm{tr}(A)=\lambda_1+\dots+\lambda_m$, where $\lambda_1,\dots,\lambda_m$ are the eigenvalues of $A$. Similarly, $\mathrm{tr}(A^n)=\lambda_1^n+\dots+\lambda_m^n$.

Therefore if $\mathrm{tr}(A^n)=0$ for all $n$, then it follows from Newton's identities that the characteristic polynomial of $A$ is $X^m$.

Therefore the minimal polynomial of $A$ is $X^k$ for some $k\leq m$, i.e. $A$ is nilpotent.

Answer 2

Hint:

As the trace of a matrix is a similarity invariant, prove it first for a Jordan matrix.