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MathsGee Android Q&A

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Are the anti-symmetrized sum of traces of products of \(k\) matrices is always 0?
in Mathematics by Platinum (147,718 points) | 80 views

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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 (147,718 points)

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MathsGee Android Q&A

MathsGee Android Q&A