# Trace of a tensor product

• March 25th 2010, 08:25 AM
Swlabr
Trace of a tensor product
I was just wondering if anyone knew of a neat proof of the fact that $Tr(A \otimes B) = Tr(A)Tr(B)$ for $A$ and $B$ matrices.
• March 25th 2010, 12:30 PM
Opalg
Quote:

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

The trace on the n×n matrices is the unique linear functional satisfying the conditions (1) $\text{Tr}(XY) = \text{Tr}(YX)$ and (2) $\text{Tr}(I) =1$.

For matrices $A\in M_m$ (the set of m×m matrices) and $B\in M_n$, the functional $f(A\otimes B) = \text{Tr}(AB)$ is bilinear, and therefore extends to a linear functional on $M_m\otimes M_n\cong M_{mn}$. 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:

\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.
• March 25th 2010, 08:36 PM
NonCommAlg
if $A=[a_{ij}],$ then using the definition of $A \otimes B$ we have: $\text{Tr}(A \otimes B)=\sum_{i} \text{Tr}(a_{ii}B)= \sum_i a_{ii} \text{Tr}(B)=\left( \sum_i a_{ii} \right) \text{Tr}(B)=\text{Tr}(A) \text{Tr}(B).$
• March 26th 2010, 03:42 AM
Swlabr
Quote:

Originally Posted by Opalg
The trace on the n×n matrices is the unique linear functional satisfying the conditions (1) $\text{Tr}(XY) = \text{Tr}(YX)$ and (2) $\text{Tr}(I) =1$.

For matrices $A\in M_m$ (the set of m×m matrices) and $B\in M_n$, the functional $f(A\otimes B) = \text{Tr}(AB)$ is bilinear, and therefore extends to a linear functional on $M_m\otimes M_n\cong M_{mn}$. 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:

\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.

Should condition (2) not be $tr(Id) = n$ where $Id$ is an $n \times n$ matrix?...
• March 26th 2010, 04:15 AM
Opalg
Quote:

Originally Posted by Swlabr
Should condition (2) not be $tr(Id) = n$ where $Id$ is an $n \times n$ matrix?...

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.