Splice

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visits member for 1 year, 6 months
seen Dec 8 '11 at 19:28
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Nov
5
 awarded  Yearling
Dec
5
 awarded  Enlightened
Dec
5
 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. :)
Dec
5
 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
Dec
5
 awarded  Nice Answer
Dec
5
 awarded  Editor
Dec
5
 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
Dec
5
 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?
Nov
5
 awarded  Teacher
Nov
5
 comment Residue at $s=1$ for $\zeta$-functions
The link didn't seem to work for some reason - please feel free to edit the answer so it works, I don't know how.
Nov
5
 answered Residue at $s=1$ for $\zeta$-functions