According to Specht's theorem, a necessary and sufficient condition for unitary equivalent of two matrices $A$ and $B$ is that $\text{Tr} W(A,A^{\dagger})=\text{Tr} W(B,B^{\dagger})$ for all words $W$. In case $A$ and $B$ are both Hermitian, this reduces to $\text{Tr} A^n = \text{Tr} B^n$ for all $n$ (considering $n \leq N$ where $N$ is the matrix size should be sufficient). My question is if there is a way to compute the eigenvalues of an $N \times N$ Hermitian matrix A if the matrix itself is not given but only $\text{Tr} A^n$ for all $n \leq N$.
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$\begingroup$ I think you want $n\leq N$, not $n < N$. $\endgroup$– darij grinbergCommented Jan 29, 2012 at 17:22
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6$\begingroup$ Your question is to compute $N$ reals $x_1$, $x_2$, ..., $x_N$ given the sums $x_1^n+x_2^n+...+x_N^n$ for all $1\leq n\leq N$. This can be done in two steps: First, get the elementary symmetric polynomials $e_1$, $e_2$, ..., $e_N$ in the $x_1$, $x_2$, ..., $x_N$ from these sums (see en.wikipedia.org/wiki/… for how to do this), and then compute the $x_1$, $x_2$, ..., $x_N$ as the zeroes of the polynomial $X^n - e_1 X^{n-1} + e_2 X^{n-2} \pm ... + \left(-1\right)^n e_n$. $\endgroup$– darij grinbergCommented Jan 29, 2012 at 17:25
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1$\begingroup$ Voting to close, this is too elementary for MO. $\endgroup$– Dmitri PavlovCommented Jan 29, 2012 at 17:35
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1$\begingroup$ Computing eigenvalues exactly is usually impossible in practice, though there are good approximative numerical methods in many cases. As darij's theoretical remarks show, you ultimately have to solve equations for which there is no effective method. It's far easier in general when working with matrix entries to decide whether or not two big matrices are similar, without knowing their eigenvalues explicitly. $\endgroup$– Jim HumphreysCommented Jan 29, 2012 at 18:17
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