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dan815

  • one year ago

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  1. dan815
    • one year ago
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    @Empty

  2. dan815
    • one year ago
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    |dw:1443248047236:dw|

  3. Empty
    • one year ago
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    http://planetmath.org/proofofpropertiesoftraceofamatrix 3

  4. Empty
    • one year ago
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    I would just let \(BC=D\) so that you are simply having to show: \[tr(ABC)=tr (AD)=tr(DA)=tr(BCA) \]

  5. dan815
    • one year ago
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  6. dan815
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  7. dan815
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    |dw:1443248682017:dw|

  8. Empty
    • one year ago
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    \[AD = \sum_{j=1}^n A_{ij}D_{jk}\]\[DA = \sum_{j=1}^n D_{ij}A_{jk}\] \[tr(M)=\sum_{i=1}^n M_{ii}\] \[tr(AD) = \sum_{i=1}^n \sum_{j=1}^n A_{ij}D_{ji}\]\[tr(AD) =\sum_{i=1}^n \sum_{j=1}^n D_{ij}A_{ji}\] i and j are dummy indices, so we can rename them, and we can see that we have: \[tr(AD)=tr(DA)\] which shows \[tr(ABC)=tr(BCA)\]

  9. dan815
    • one year ago
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    |dw:1443248823878:dw|

  10. Empty
    • one year ago
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    \[AD = \sum_{j=1}^n A_{ij}D_{jk}\]\[DA = \sum_{j=1}^n D_{ij}A_{jk}\] \[tr(M)=\sum_{i=1}^n M_{ii}\] \[tr(AD) = \sum_{i=1}^n \sum_{j=1}^n A_{ij}D_{ji}\]\[tr(DA) =\sum_{i=1}^n \sum_{j=1}^n D_{ij}A_{ji}\] i and j are dummy indices, so we can rename them, and we can see that we have: \[tr(AD)=tr(DA)\] which shows \[tr(ABC)=tr(BCA)\]

  11. Empty
    • one year ago
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    A is diagonalizeable, then D is the diagonal matrix of eigenvalues: \[tr(A)=tr(PDP^{-1})=tr(PP^{-1}D) = tr(D) = \text{sum of eigenvalues}\]

  12. Empty
    • one year ago
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  13. dan815
    • one year ago
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  14. dan815
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  15. Empty
    • one year ago
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  17. Empty
    • one year ago
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  18. dan815
    • one year ago
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  21. dan815
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