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Let $V$ be vector space $\dim V=N$, and $A\in End(V)$. Denote $$ \wedge^k A^m(\mathbf{v}_1\wedge\dots\wedge\mathbf{v}_k)=\sum_{s_1,\dots,s_k=0,1,\sum_j s_j=m} A^{s_1}\mathbf{v}_1\wedge\dots\wedge A^{s_k}\mathbf{v}_k. $$ ($A^0=I_V$). I want to show that $\wedge^N A^k$ $(k=1,\ldots,N)$ can be expressed as polynomials in $\operatorname{Tr}(A),\operatorname{Tr}(A^2),\ldots,\operatorname{Tr}(A^N)$. (Here $\wedge^N A^k$ can be identified with a number.)
The case $k=1$ is the definition of trace, $k=2,3$ can be calculated directly. I tried to derive a recursion between $\wedge^N A^{k+1}$ and $\wedge^N A^k$ (or with more terms) which using an induction principle would give the statement. The problem here is that (maybe) there are a lot of recursion, but only a few of them can give the right way to the proof.

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up vote 2 down vote accepted

Your $\wedge^k A^m$ is just the polarization:

$\wedge^k A^m = {N \choose m} \wedge^k(\underbrace{A,\dots,A}_m,1,\dots,1)$,

the right-hand side refers to $\wedge^k$ as a $k$-linear form. As we know, $\wedge^N$ is the determinant, so $\wedge^N A^m$ are just the coefficients of the characteristic polynomial. Those are elementary symmetric functions of the eigenvalues, and $\mathrm{tr} \, A^k$ are sums of $k$-th powers, so the polynomials you seek are precisely those appearing in Newton's identities.

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