I was just wondering if anyone knew of a neat proof of the fact that $\displaystyle Tr(A \otimes B) = Tr(A)Tr(B)$ for $\displaystyle A$ and $\displaystyle B$ matrices.

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- Mar 25th 2010, 07:25 AMSwlabrTrace of a tensor product
I was just wondering if anyone knew of a neat proof of the fact that $\displaystyle Tr(A \otimes B) = Tr(A)Tr(B)$ for $\displaystyle A$ and $\displaystyle B$ matrices.

- Mar 25th 2010, 11:30 AMOpalg
The trace on the n×n matrices is the unique linear functional satisfying the conditions (1) $\displaystyle \text{Tr}(XY) = \text{Tr}(YX) $ and (2) $\displaystyle \text{Tr}(I) =1$.

For matrices $\displaystyle A\in M_m$ (the set of m×m matrices) and $\displaystyle B\in M_n$, the functional $\displaystyle f(A\otimes B) = \text{Tr}(AB)$ is bilinear, and therefore extends to a linear functional on $\displaystyle M_m\otimes M_n\cong M_{mn}$. So we just have to check that f satisfies (1) and (2). Condition (1) is bilinear, so you only need to check it on simple tensors:

$\displaystyle \begin{aligned}f\bigl((A_1\otimes B_1)(A_2\otimes B_2)\bigr) &= f(A_1A_2\otimes B_1B_2) \\ &= \text{Tr}(A_1A_2)\text{Tr}(B_1B_2) \\ &= \text{Tr}(A_2A_1)\text{Tr}(B_2B_1) \\ &= f(A_2A_1\otimes B_2B_1) = f\bigl((A_2\otimes B_2)(A_1\otimes B_1)\bigr).\end{aligned}$

Condition (2) is trivially satisfied, so f is the functional Tr on the mn×mn matrices. - Mar 25th 2010, 07:36 PMNonCommAlg
if $\displaystyle A=[a_{ij}],$ then using the definition of $\displaystyle A \otimes B$ we have: $\displaystyle \text{Tr}(A \otimes B)=\sum_{i} \text{Tr}(a_{ii}B)= \sum_i a_{ii} \text{Tr}(B)=\left( \sum_i a_{ii} \right) \text{Tr}(B)=\text{Tr}(A) \text{Tr}(B).$

- Mar 26th 2010, 02:42 AMSwlabr
- Mar 26th 2010, 03:15 AMOpalg
Yes it should. (I'm used to dealing with the normalised trace, where you divide by the dimension so that the identity has trace 1.) For the usual trace, condition (2) in my previous comment is equally obvious, because both sides become mn when you put A and B equal to the appropriately-sized identity matrices.