# Finding $\mbox{tr}(A)$ from the condition $\mbox{tr}(A^2) = \mbox{tr}(A^3) = \mbox{tr}(A^4)$?

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

• From where did you get this problem?Thanks. – StammeringMathematician Jul 26 at 17:20
• Sorry As far as I remember I discussed this on this forum or saw a similar but could not understand so asked myself! – 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 $k$.
• 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? – BAYMAX Mar 9 '18 at 7:04