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Math Help - Trace of a tensor product

  1. #1
    MHF Contributor Swlabr's Avatar
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    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.
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  2. #2
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    Quote Originally Posted by Swlabr View Post
    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 nn 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 mm 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 mnmn matrices.
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  3. #3
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    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).
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  4. #4
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by Opalg View Post
    The trace on the nn 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 mm 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 mnmn matrices.
    Should condition (2) not be tr(Id) = n where Id is an n \times n matrix?...
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  5. #5
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    Quote Originally Posted by Swlabr View Post
    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.
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