I was wondering for what matrices (over $\mathbb{C}$) $A$ and $B$ is the equation $\operatorname{tr}(AB) = \operatorname{tr}(A)\operatorname{tr}(B)$ satisfied?
For what matrices $A$ and $B$ is $\operatorname{tr}(AB) = \operatorname{tr}(A)\operatorname{tr}(B)$?
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linear-algebra
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0In the case that $A$ and $B$ are $1 \times 1$ matrices your above formula is satisfied. – 2012-04-24
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0This is not a very natural condition to ask the trace to satisfy. Why do you want to know? – 2012-04-24
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1Even when $B=I$ and dimension $>1$ there is no way – 2012-04-24