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$\mathrm{tr}(AB)=\sum\limits_{i=1}^n \sum\limits_{j=1}^n a_{ij}*b_{ji}$
$\mathrm{tr}(A)*\mathrm{tr}(B)=\sum\limits_{i=1}^n a_{ii}*\sum\limits_{i=1}^n b_{ii}$

Therefore $\mathrm{tr}(AB) \neq \mathrm{tr}(A)*\mathrm{tr}(B)$

Is the proof valid?

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    $\begingroup$ $\sum a_i\sum b_i\neq \sum a_ib_i$!!! $\endgroup$ – Pedro Tamaroff Jul 27 '14 at 4:27
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    $\begingroup$ Is there a question? (If you want your proof validated, say so and include the proof-verifcation tag). $\endgroup$ – Semiclassical Jul 27 '14 at 4:28
  • $\begingroup$ Done the edit, and now? $\endgroup$ – gbox Jul 27 '14 at 4:29
  • $\begingroup$ @Semiclassical thanks $\endgroup$ – gbox Jul 27 '14 at 4:30
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    $\begingroup$ (What is true, however, is that ${\rm tr}(AB)={\rm tr}(BA)$ $\endgroup$ – Pedro Tamaroff Jul 27 '14 at 4:33
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It is false. Let's think small. Consider the identity matrix, of order $n > 1$. Then: $$n = \mathrm{tr}(I) = \mathrm{tr}(I \cdot I) \neq \mathrm{tr}(I)~ \mathrm{tr}(I) = n^2$$ It is important to try some silly cases and gain intuition about the affirmation before tackling summations, etc.

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You have not given a reason why those expressions are not identically equal. They will be equal in some special cases. The easiest way to prove that such an identity doesn't hold is to give a counterexample. Try 2-by-2 matrices.

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