Today I came accross to . And after some study, I have derived the following solution. Request to the experts, kindly let me know if I have made any mistakes.  The question is: if $M, N$ be $n$ square matrices such that tr$(M^k)=$ tr$(N^k)$ for all $k=1, 2, \cdots, n$ then prove that $M$ and $N$ will have same eigenvalues. What about this one?  We know that if $\psi_A(x)=x^n+c_{n-1}x^{n-1}+c_{n-2}x^{n-2}+\cdots+c_1x+(-1)^n|A|I_n$ be characteristic polynomial of $A$, then the coefficients are given by  $$c_{n-m}=\frac{(-1)^m}{m}\left|                               \begin{array}{ccccc}                                 t_1 &amp; m-1 &amp; 0 &amp; \cdots &amp;  0 \\                                 t_2 &amp; t_1 &amp; m-2 &amp; \cdots &amp;  0  \\                                 \vdots  &amp; \vdots &amp;   &amp;   &amp;  \vdots \\                                 t_{m-1} &amp; t_{m-2} &amp; t_2 &amp; t_1 &amp; 1 \\                                 t_m &amp; t_{m-1} &amp; t_3 &amp; t_2 &amp; t_1 \\                               \end{array}                             \right|$$  where $t_r:=$ tr$(A^r)$.  Assume that $\psi_{M}(x)=x^n+a_{n-1}x^{n-1}+a_{n-2}x^{n-2}+\cdots+a_1x+(-1)^n|M|I_n$ and $\psi_N(x)=x^n+b_{n-1}x^{n-1}+b_{n-2}x^{n-2}+\cdots+b_1x+(-1)^n|N|I_n$. Then $$a_r=b_r$$ (by using tr$(M^k)=$ tr$(N^k)$ and the above determinant).   Hence $\psi_M(x)=\psi_N(x)$ which means $M$ and $N$ will have same eigenvalues.
Let $M,N$ be $n \times n$ square matrices over an algebraically closed field with the properties that the trace of both matrices coincides along with all powers of the matrix. More specifically, suppose that $\mathrm{Tr}(M^k) = \mathrm{Tr}(N^k)$ for all $1\leq k \leq n$. The following questions about eigenvalues is then natural and I was thinking it would be an application of Cayley-Hamilton but I am having trouble writing out a proof.  How do we show that $M$ and $N$ have the same eigenvalues?  Added (because this question is now target of many duplicates, it should state its hypotheses properly). Assume that all the mentioned values of $k$ are nonzero in the field considered; in other words either the field is of characteristic $0$, or else its prime characteristic $p$ satisfies $p&gt;n$.
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