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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 $\sum_{i=1}^{n} a_{i}^2 = \sum_{i=1}^{n} a_{i}^3 = \sum_{i=1}^{n} a_{i}^4$?

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  • $\begingroup$ 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
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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 $k$.

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    $\begingroup$ 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

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