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 one year ago
Best ResponseYou've already chosen the best response.0I would just let \(BC=D\) so that you are simply having to show: \[tr(ABC)=tr (AD)=tr(DA)=tr(BCA) \]

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 one year ago
Best ResponseYou've already chosen the best response.0\[AD = \sum_{j=1}^n A_{ij}D_{jk}\]\[DA = \sum_{j=1}^n D_{ij}A_{jk}\] \[tr(M)=\sum_{i=1}^n M_{ii}\] \[tr(AD) = \sum_{i=1}^n \sum_{j=1}^n A_{ij}D_{ji}\]\[tr(AD) =\sum_{i=1}^n \sum_{j=1}^n D_{ij}A_{ji}\] i and j are dummy indices, so we can rename them, and we can see that we have: \[tr(AD)=tr(DA)\] which shows \[tr(ABC)=tr(BCA)\]

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 one year ago
Best ResponseYou've already chosen the best response.0\[AD = \sum_{j=1}^n A_{ij}D_{jk}\]\[DA = \sum_{j=1}^n D_{ij}A_{jk}\] \[tr(M)=\sum_{i=1}^n M_{ii}\] \[tr(AD) = \sum_{i=1}^n \sum_{j=1}^n A_{ij}D_{ji}\]\[tr(DA) =\sum_{i=1}^n \sum_{j=1}^n D_{ij}A_{ji}\] i and j are dummy indices, so we can rename them, and we can see that we have: \[tr(AD)=tr(DA)\] which shows \[tr(ABC)=tr(BCA)\]

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 one year ago
Best ResponseYou've already chosen the best response.0A is diagonalizeable, then D is the diagonal matrix of eigenvalues: \[tr(A)=tr(PDP^{1})=tr(PP^{1}D) = tr(D) = \text{sum of eigenvalues}\]