Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $W$.


share|cite|improve this question
Cue someone mentioning the formalism of rigid monoidal categories... :-) – Mariano Suárez-Alvarez Oct 16 '12 at 4:46
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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.