Prove that two $n \times n$ matrices $A$ and $B$ have the same eigen values if and only if $\operatorname{trace}(A^{k}) = \operatorname{trace}(B^{k})$.
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The trace is the sum of the eigenvalues (counted with multiplicity). The eigenvalues of $X^k$ are the $k$th powers of the eigenvalues of $X$ (counted with multiplicity) if $X$ is a matrix.
Therefore, we are reduced to the following problem: if $\lambda_1,\dots,\lambda_n$ and $\gamma_1,\dots,\gamma_n$ are real numbers such that $\sum_{i=1}^{n} \lambda_i^k=\sum_{i=1}^{n} \gamma_i^k$ for all $k\geq 1$, then prove that $\{\lambda_1,\dots,\lambda_n\}=\{\gamma_1,\dots,\gamma_n\}$.
Now, refer to Omnomnomnom's (have I got the right number of o's, n's and m's? :)) excellent answer to conclude.
Hope this helps!
Amitesh Datta
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It seems like Newton's Identities are really what you need as they determine the characteristic polynomial uniquely and therefore its roots. On the other hand, is it possible to see directly (easily?) that the system of equations given by the traces have a unique solution? In other words, it seems like setting up that system of equations is unnecessary if you just refer to Newton's identities. – Alex R. Apr 23 '14 at 15:12