I would like to do an operation on a matrix acting on a tensor product vector space that is a bit unusual. It is similar to a partial trace, but not quite that. Say I have a tensor product vector space $V \otimes V$. What I would like to be able to do, is a linear map from $L(V \otimes V)$ to $L(V)$ defined through:

$$ A \otimes B \to AB. $$

In other words I want to turn a tensor product into a normal matrix product.

Assuming the usual Kronecker product convention for the tensor products, how can I compute this operation for an arbitrary matrix $T \in L(V \otimes V)$ (so $T$ can not necessarily be written as a simple tensor product)?

I know that in "tensor terminology" this is a contraction of two indices, if $T$ was viewed as a rank four tensor, I'm just not sure how to actually do it given a representation of $T$ as a matrix.

  • $\begingroup$ By the universal property, for every bilinear function $g:L(V)\times L(V)\rightarrow L(V)$, we have its linearization $G:L(V)\otimes L(V)\rightarrow L(V)$, defined by the formula $G(\sum_{i=1}^n A_i\otimes B_i)=\sum_{i=1}^n g(A_i,B_i)$. Now, let $g(A,B)=AB$. Remember that $L(V\otimes V)\simeq L(V)\otimes L(V)$. $\endgroup$ – Daniel Oct 3 '14 at 19:25

This isn't too hard if you already know about the partial trace. Recall that for $A \otimes B \in V \otimes V$, we have $$ \DeclareMathOperator{\tr}{tr} \tr_1(A \otimes B) = \tr(A)B\\ \tr_2(A \otimes B) = \tr(B) A\\ \tr(A \otimes B) = \tr(A) \tr(B) $$ So, given an arbitrary $T \in V \otimes V$, define $$ p(T) = \frac{\tr_2(T)\tr_1(T)}{\tr(T)} $$ And verify that, whenever $T = A \otimes B$, $p(T) = AB$.

  • $\begingroup$ Oh, I see now that my question wasn't well defined. I actually ment for the operation to be linear, meaning $\sum_n A_n \otimes B_n$ goes to $\sum_n A_n B_n$. This should uniquely define a linear map, I hope. But I don't think the one you gave is the one I'm after. I have edited my question to emphasize that the map should be linear. $\endgroup$ – arne Oct 3 '14 at 18:51
  • $\begingroup$ My mistake. I think you're right about all that. $\endgroup$ – Omnomnomnom Oct 3 '14 at 19:05

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