$$A_1^k + A_2^k + \cdots + A_m^k = 0, \qquad \forall k \in \mathbb N^+$$


$$\mbox{Tr}(A_1^k) + \mbox{Tr}(A_2^k) + \cdots + \mbox{Tr}(A_m^k) = 0$$

where $A_1, A_2, \dots, A_m$ are $n\times n $ matrices. Would that mean that all the eigenvalues of these matrices are $0$?

The book where this is stated says it could be proved by using Newton's formulae but I haven't heard of those before.

  • $\begingroup$ Do you mean this holds $\forall k$? $\endgroup$ – lisyarus Apr 15 at 12:31
  • $\begingroup$ Yes, if k>=1. I forgot to write that the sum of all these matrices at the k-th power is also On : A1^k+A2^k+...+Am^k=On $\endgroup$ – Raluca Pelin Apr 15 at 12:33
  • $\begingroup$ Are you trying to prove the desired result using both of the known facts or just the one involving the trace? $\endgroup$ – Ian Apr 15 at 12:48
  • $\begingroup$ Both. The one involving the trace comes from the fact that the sum is 0, and the said Newtow's formulae are said to have sth to do with the sum of the eigenvalues. $\endgroup$ – Raluca Pelin Apr 15 at 12:51
  • 1
    $\begingroup$ To begin with, can you handle the case of $m=1$? If you have not found them by yourself already, here are Newton's identities. $\endgroup$ – Bill O'Haran Apr 15 at 13:01

Write the equalities linking the traces for $k=1,\cdots ,mn$ and define the $(\lambda_i)_{i\leq mn}$ as follows

$spectrum(A_1)=(\lambda_1,\cdots ,\lambda_n),spectrum(A_2)=(\lambda_{n+1},\cdots,\lambda_{2n})$ and so on.

Use the fact that $tr(A_1^k)=\sum_{i\leq n}\lambda_i^k$. We obtain a system of $mn$ equations with $mn$ complex unknowns. The sole solution is the trivial one and we are done. This can be proved using the Newton's identities. cf on this website

"Traces of all positive powers of a matrix are 0 implies it is nilpotent".

Remark. The above result is still valid over a field of characteristic $> mn$.

A question to the OP. In your book, does the author study the possible simultaneous triangularization of the $(A_i)$ ?


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