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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 (108k points)

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