$Tr(A^2)=Tr(A^3)=Tr(A^4)$ then find $Tr(A)$

Question

Let $A$ be a non singular $n\times n$ matrix with all eigenvalues real and $$Tr(A^2)=Tr(A^3)=Tr(A^4).$$Find $Tr(A)$.

I considered $2\times 2$ matrix $\begin{bmatrix}a&b\\c&d\end{bmatrix}$ and tried computing traces of $A^2,A^3,A^4$ and ended up with following

1. $Tr(A^2)=Tr(A)^2-2\det(A)$
2. $Tr(A^3)=Tr(A)^3-3Tr(A)\det(A)$
3. $Tr(A^4)=Tr(A)^4-4Tr(A)^2\det(A)+2\det(A)$

I have no idea how to proceed from here...

Answer 1

Let's denote the eigenvalues by $t_j$. Then, by Cauchy-Schwarz $$ \sum t_j^3 \le \left( \sum t_j^2 \sum t_j^4\right)^{1/2} , $$ which equals $\sum t_j^3$ by assumption. Equality in the CSI means that the vectors are linearly dependent, so $t_j=ct_j^2$. This says that there is only one eigenvalue ($=1/c$), and clearly it must be $1$ then. So $\textrm{tr}\, A=n$.

Answer 2

A slightly different approach: we have $\operatorname{trace}((A-A^2)^2)=0$. As $A$ has a real spectrum, so does $A-A^2$. So, the previous trace condition implies that $A-A^2$ is nilpotent. Hence every eigenvalue of $A$ is equal to its square. As $A$ is nonsingular, every eigenvalue of $A$ must be equal to $1$. Hence the trace of $A$ is $n$.