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Let $A,B \in {M_n}$ and suppose $tr(A^k) = tr(B^k)$ for all $k$=$1,2,...$ . Why do $A$ and $B$ possess the same characteristic polynomial?

Bandar
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  • Welcome to math.se! What have you tried so far? – Travis Willse May 18 '15 at 07:41
  • Also, http://math.stackexchange.com/questions/597743/diagonalizable-matrices-a-and-b-with-mathrmtrak-mathrmtrbk-have-th and http://math.stackexchange.com/questions/923523/showing-that-m-and-n-will-have-same-eigenvalues – Chris Culter May 18 '15 at 07:43
  • I must say I really dislike this "$M_n$" notation, which doesn't mention the field. (I have already seen it in this question: http://math.stackexchange.com/questions/1268459, whose style is somewhat comparable to the one at hand). – PseudoNeo May 18 '15 at 07:45
  • This is duplicate with all the questions linked to in comments above, of which curiously every single one forgets to mention the necessary condition on the characteristic. (The result is not true in characteristic$~p$ with $0<p\leq n$). This is really sad. Closing anyway. – Marc van Leeuwen May 18 '15 at 08:16

1 Answers1

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Because of Newton's identities.

To be more specific, Newton's identities show that the result you quote is true on a field of characteristic $0$. It is false on a positive characteristic field, though: for example, on $\mathbb F_2$, you have $\mathrm{tr}(I_2) = \mathrm{tr}(0) = 0$, but these two $2\times 2$ matrices don't have the same determinant.

PseudoNeo
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