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