# Prove Or Disprove: tr(AB)=tr(A)*tr(B)

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

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

Is the proof valid?

• $\sum a_i\sum b_i\neq \sum a_ib_i$!!! – Pedro Tamaroff Jul 27 '14 at 4:27
• Is there a question? (If you want your proof validated, say so and include the proof-verifcation tag). – Semiclassical Jul 27 '14 at 4:28
• Done the edit, and now? – gbox Jul 27 '14 at 4:29
• (What is true, however, is that ${\rm tr}(AB)={\rm tr}(BA)$ – Pedro Tamaroff Jul 27 '14 at 4:33
• It is true for $n = 1$... – Ivo Terek Jul 27 '14 at 4:45

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.