## dan815 one year ago http://prntscr.com/8kl0ln

1. dan815

@Empty

2. dan815

|dw:1443248047236:dw|

3. Empty
4. Empty

I would just let $$BC=D$$ so that you are simply having to show: $tr(ABC)=tr (AD)=tr(DA)=tr(BCA)$

5. dan815

|dw:1443248064461:dw|

6. dan815

|dw:1443248515776:dw|

7. dan815

|dw:1443248682017:dw|

8. Empty

$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

|dw:1443248823878:dw|

10. Empty

$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

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

|dw:1443251868992:dw|

13. dan815

|dw:1443252078878:dw|

14. dan815

|dw:1443252173838:dw|

15. Empty

|dw:1443252224172:dw|

16. Empty

|dw:1443252363232:dw|

17. Empty

|dw:1443252428676:dw|

18. dan815

|dw:1443252614256:dw|

19. dan815

|dw:1443252653544:dw|

20. Empty

|dw:1443252656505:dw|

21. dan815

|dw:1443252836312:dw|

22. dan815

|dw:1443252908436:dw|

23. dan815

|dw:1443253010127:dw|