Trace matters!

Given $Tr(A^2)=Tr(A^3)=Tr(A^4)$ . Suppose $(\lambda_1,\lambda_2\ldots\lambda_n)$ be the eigenvalues of $A$ .

Condition implies $\sum\limits_{k=1}^{n}\lambda_k^2=\sum\limits_{k=1}^{n}\lambda_k^3=\sum\limits_{k=1}^{n}\lambda_k^4$ .

Consider the vectors $x_1=(\lambda_1,\lambda_2\ldots\lambda_n)$ and $x_2=( \lambda_1^2,\lambda_2^2\ldots\lambda_n^2)$ .

We have $|\langle x_1, x_2\rangle|^2=\left(\sum\limits_{k=1}^{n}\lambda_k^3\right)^2$

also, $\langle x_1, x_1\rangle=||x_1||^2=\sum\limits_{k=1}^{n}\lambda_k^2=\sum\limits_{k=1}^{n}\lambda_k^3$

and $\langle x_2, x_2\rangle=||x_2||^2=\sum\limits_{k=1}^{n}\lambda_k^4=\sum\limits_{k=1}^{n}\lambda_k^3$

Now we have $|\langle x_1, x_2\rangle|^2=\langle x_1, x_1\rangle\cdot\langle x_2, x_2\rangle$

This is the case of equality in Cauchy-Schwarz inequality and hence we can conclude $x_2=k\cdot x_1$ for some scalar $k$ .

i.e., $(\lambda_1^2,\lambda_2^2\ldots\lambda_n^2)=k\cdot (\lambda_1,\lambda_2\ldots\lambda_n)$ .

for each $\lambda_j$ , we have $\lambda_j^2=k\lambda_j$ . Since $\lambda_j\neq 0 \,\forall \, 0\leq j\leq n$ , $\lambda_j=k$ for all $0\leq j\leq n$

Again since we have $\sum\limits_{k=1}^{n}\lambda_k^2=\sum\limits_{k=1}^{n}\lambda_k^3=\sum\limits_{k=1}^{n}\lambda_k^4\implies nk^2=nk^3=nk^4\implies k=1$ .

Hence $\sum\limits_{k=1}^{n}\lambda_k=n$