Let $A\in M_{n}(\mathbb R)$ and $\mathrm{tr}(A^{m})=0$ for every positive integer $m$. Is $A$ nilpotent? Is $A^{2}-I$ invertible?
A matrix with the property $\mathrm{tr}(A^{m})=0$ for every $m$ is nilpotent?
1 Answers
Write $A$ in its Jordan form: $A=SJS^{-1}$, with $S$ invertible. As $A^m=SJ^mS^{-1}$ for all $m$ and $\mathrm{tr}(SJ^mA^{-1})=\mathrm{tr}(J^m)$, we can work directly with $J$.
Since $\mathrm{tr}(J^m)=0$ for all $m$ and the trace is linear, we conclude that $\mathrm{tr}(p(J))=0$ for every polynomial $p$ such that $p(0)=0$.
Now, since $J$ is upper triangular, the diagonal of $p(J)$ consists of $p(J_{11}),\ldots,p(J_{nn})$. If one or more of the $J_{11},\ldots, J_{nn}$ is nonzero, we can get a polynomial $p$ such that the diagonal of $p(J)$ has positive sum; but this is a contradiction, which shows that the diagonal of $J$ is zero. This implies that $J$ is nilpotent, and so is $A$.
For the last question, $A^2-I=S(J^2-I)S^{-1}$. Since $J^2-I$ is upper triangular with every entry in its diagonal equal to $-1$, its determinant is nonzero, and so it is invertible.
-
0@aliakbar: For the second part, if $A^k=0$ then $(A^2-I)(-A^{2k-2}-A^{2k-4}-A^{2k-6}-\cdots-A^4-A^2-I)=I.$ – 2012-12-08