Let $A$ be an $n \times n$ matrix with real eigenvalues such that $$\mbox{tr}(A^2) = \mbox{tr}(A^3) = \mbox{tr}(A^4)$$ Then what would be $\mbox{tr}(A)$?

I thought of finding $\sum_{i=1}^{n} \lambda_{i}$ from $$\sum_{i=1}^{n} \lambda_{i}^2 = \sum_{i=1}^{n} \lambda_{i}^3 = \sum_{i=1}^{n} \lambda_{i}^4$$

after this, I could try $\sum_{i=1}^{n} \lambda_{i}^2 - \lambda_{i}^3 = 0$ and $\sum_{i=1}^{n} \lambda_{i}^3 - \lambda_{i}^4 = 0$, how can I proceed with this?

Also in another way it can also be put like this - finding $\sum_{i=1}^{n} a_{i}$ where $a_{i} \in \Bbb{R}$ given that $\sum_{i=1}^{n} a_{i}^2 = \sum_{i=1}^{n} a_{i}^3 = \sum_{i=1}^{n} a_{i}^4$?

  • $\begingroup$ From where did you get this problem?Thanks. $\endgroup$ – StammeringMathematician Jul 26 at 17:20
  • $\begingroup$ Sorry As far as I remember I discussed this on this forum or saw a similar but could not understand so asked myself! $\endgroup$ – BAYMAX Jul 26 at 23:01

With the vectors $u=(\newcommand{\la}{\lambda}\la_1,\ldots,\la_n)$ and $v=(\la_1^2,\ldots,\la_n^2)$ you have equality in Cauchy-Schwarz. That is $u\cdot v=|u||v|$. This means that $u$ and $v$ are scalar multiples of each other. Assume they are nonzero, then there is a $t$ such that $\la_i^2=t\la_i$ for each $i$. So $\la_i\in\{0,t\}$. Let there be $m$ nonzero $\la_i$. Then the trace of $A^2$ is $mt^2$ and that of $A^3$ is $mt^3$. Therefore $t=1$. Then the trace of $A^k$ is $m1^k=m$ for all $k$.

  • 1
    $\begingroup$ Very nice idea!!!, but how do we know the number of non-zero eigenvalues ? like the answer must be in terms of $n$ or some known variables? $\endgroup$ – BAYMAX Mar 9 '18 at 7:04
  • $\begingroup$ you can change your basis so all the zero eigenvalues dont factor into it without changing the trace, trace is base invariant. $\endgroup$ – shai horowitz Mar 9 '18 at 7:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.