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Let $A\otimes B$ denote the tensor product of two matrices, $A$ and $B$. I can show the trace of it is the same as the product of the traces of $A$ and $B$, which follows from computation.

Is there some conceptual explanation for this? I believe there should be one related to the fact that $V\otimes W$ is the space of functions over $\{1,2,\dots n\}\times \{1,2,3,\dots m\}$ if $V$ is of dimension $n$ and $W$ of dimension $m$.And $A\otimes B$ acts on this space if $A$ acts on $V$ and $B$ acts on $W$.

Thanks!

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Cue someone mentioning the formalism of rigid monoidal categories... :-) –  Mariano Suárez-Alvarez Oct 16 '12 at 4:46
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up vote 9 down vote accepted

Let $V$ be a finite dimensional vector space over some fixed field of coefficients $k$. If we interpret $End(V)$ as $V\otimes V^*$, then the trace map is just the natural pairing $V\otimes V^* \to k$.

The multiplicativity of traces then comes from the fact that the trace map $(V\otimes W)\otimes (V\otimes W)^* \to k$ can be factored as the tensor product of the individual pairing $V\otimes V^* \to k$ and $W\otimes W^* \to k,$ and the map $k\otimes k \cong k.$ The latter pairing is just given by multiplication of elements of $k$, and this gives the multiplicative nature of traces.

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