# Splice

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bio website location age member for 1 year, 6 months seen Dec 8 '11 at 19:28 profile views 42

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 Nov5 awarded Yearling Dec5 awarded Enlightened Dec5 comment If $\mathrm{Tr}(M^k) = \mathrm{Tr}(N^k)$ for all $1\leq k \leq n$ then how do we show the $M$ and $N$ have the same eigenvalues?um...thanks? I guess I also have tenure at a top 20 math department in the US, but your enthusiasm is appreciated, nonetheless. :) Dec5 revised If $\mathrm{Tr}(M^k) = \mathrm{Tr}(N^k)$ for all $1\leq k \leq n$ then how do we show the $M$ and $N$ have the same eigenvalues?typo - power of (-1) in one expression corrected Dec5 awarded Nice Answer Dec5 awarded Editor Dec5 revised If $\mathrm{Tr}(M^k) = \mathrm{Tr}(N^k)$ for all $1\leq k \leq n$ then how do we show the $M$ and $N$ have the same eigenvalues?extraneous typo "m" in formula for no reason Dec5 answered If $\mathrm{Tr}(M^k) = \mathrm{Tr}(N^k)$ for all $1\leq k \leq n$ then how do we show the $M$ and $N$ have the same eigenvalues? Nov5 awarded Teacher Nov5 comment Residue at $s=1$ for $\zeta$-functionsThe link didn't seem to work for some reason - please feel free to edit the answer so it works, I don't know how. Nov5 answered Residue at $s=1$ for $\zeta$-functions