Determine Matrix

Since the matrix $A$ satisfies the equation $x^k-1,$ the minimal polynomial of $A$ divides $x^k-1.$ Since $x^k-1=\prod_{j=0}^{k-1}(x-e^{2\pi ij/k}),$ the roots of $x^k-1$ are all distinct. Hence the roots of the minimal polynomial are also distinct. Therefore the matrix $A$ is diagonalizable.

It suffices to prove that the eigenvalues $\lambda_1,\lambda_2,\dots,\lambda_n$ of $A$ or $p_A(t),$ characteristic polynomial of $A,$ is uniquely determined by $\text{tr}(A),\text{tr}(A^2),\dots,\text{tr}(A^{k-1}).$ We can determine $\text{tr}(A^l)$ for all positive integers $l.$ Since $A$ is always similar to its Jordan canonical form , an upper triangular matrix having $\lambda_1,\lambda_2,\dots,\lambda_n$ on the main diagonal, we get $\text{tr}(A^l)=\sum _{j=0}^n\lambda_j^l.$ Using Newton's identities to express the elementary symmetric polynomials in terms of the power sums, we can find the coefficients of $p_A(t)$ by Vieta's formula .

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Since $A^k = I$ , the minimum polynomial of $A$ is a product of distinct linear factors, and hence $A$ is diagonalisable. If $z_1,z_2,...,z_k$ are the $k$ distinct $k$ th roots of $1$ then (to within similarity) $A$ is determined by the respective multiplicities $u_1,u_2,...,u_k$ of its eigenvalues, and we have $\mathrm{Tr}(a^m) \; = \; \sum_{n=1}^k u_n z_n^m \hspace{2cm} 0 \le m \le k-1$ In other words we have the matrix equation $\left( \begin{array}{ccccc} 1 & 1 & 1 & \cdots & 1 \\ z_1 & z_2 & z_3 & \cdots & z_k \\ z_1^2 & z_2^2 & z_3^2 & \cdots & z_k^3 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ z_1^{k-1} & z_2^{k-1} & z_3^{k-1} & \cdots & z_k^{k-1} \end{array}\right) \left(\begin{array}{c} u_1 \\ u_2 \\ u_3 \\ \vdots \\ u_k \end{array}\right) \; = \; \left(\begin{array}{c} \mathrm{Tr}(I) \\ \mathrm{Tr}(A) \\ \mathrm{Tr}(A^2) \\ \vdots \\ \mathrm{Tr}(a^{k-1}) \end{array} \right)$ It is a standard results that this matrix is nonsingular, with determinant $\pm\prod_{1 \le m < n \le k}(z_m - z_n)$ and so the multiplicities $u_1,u_2,...,u_k$ are determined by the given traces.