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Are the anti-symmetrized sum of traces of products of $k$ matrices is always 0?
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The anti-symmetrized sum of traces of products of $k$ matrices is always 0 for even $\mathrm{k}$. But it's generally nonzero for odd $\mathrm{k}$, which turns out to be a useful quantity in certain applications. #math #algebra #matrices

Let $A_{1}, A_{2}, \ldots, A_{k}$ be $n \times n$ matrices. Show that the antisymmetrized sum of traces
$B=\sum_{\pi \in S_{k}} \operatorname{sgn}(\pi) \operatorname{Tr}\left[A_{\pi(1)} A_{\pi(2)} \cdots A_{\pi(k)}\right]$
satisfies
$B=(-1)^{k-1} B$
It then follows that $B=0$ for even $k$ (for all $A_{i}$ 's!). Show, however, that $B$ can be nonzero for odd $k$.

The sum runs over all permutations $\pi$ of the indices $\{1,2, \ldots, k\}$. The sign of $\pi, \operatorname{sgn}(\pi)$, is $+1$ if $\pi$ is a product of an even number of transpositions, and is $-1$ if $\pi$ is a product of an odd number.
by Platinum (93,241 points)

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