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 Mar 12 asked Notation for “vectorized” function Mar 10 revised Exercising elementary toolkit for quantum computing Fix spacing of outer product bra-ket notation Mar 10 comment Is the sum of the tensor product of a linear operator, the tensor of the sum? I've edited the question. Mar 10 revised Is the sum of the tensor product of a linear operator, the tensor of the sum? Update and clarify question. Mar 10 comment Is the sum of the tensor product of a linear operator, the tensor of the sum? So that gets me to $\sum_{z\in Z}[A(z)\otimes \sum_{y\in Y}B(y,z)]$, and I need to get from there to $[\sum_{z\in Z}A(z)]\otimes C$ for some $C$. That's the crux of the question I'm asking. Mar 10 comment Is the sum of the tensor product of a linear operator, the tensor of the sum? Hmmm. I guess I don't see it them. Can you spell out in an edit to your answer how $\sum_{y\in Y}\sum_{z\in Z}[A(z)\otimes B(y,z)]$ would end up with as $[\sum_{z\in Z}A(z)]\otimes C$. And what would $C$ be (in my case in the question its $I$? Mar 10 revised Is the sum of the tensor product of a linear operator, the tensor of the sum? Remove chat. Mar 10 comment Is the sum of the tensor product of a linear operator, the tensor of the sum? Just to be sure I understand how this applies to my case. The thing I was stuck on was the indices (which are just counters, not arguments). What I've got is $\sum_{y\in Y}\sum_{z\in Z}[A(z)\otimes B(y,z)]$ which is $[\sum_{y\in Y}\sum_{z\in Z}A(z)]\otimes [\sum_{y\in Y}\sum_{z\in Z}B(y,z)]$ or $[\sum_{z\in Z}A(z)]\otimes [\sum_{y\in Y}\sum_{z\in Z}B(y,z)]$. Is that right? Mar 10 comment Is the sum of the tensor product of a linear operator, the tensor of the sum? Yes, I was getting lost (summations are a bane) and confused by the apparent use of two indices in the second term, which is irrelevant. Mar 10 suggested approved edit on Is the sum of the tensor product of a linear operator, the tensor of the sum? Mar 10 comment Is the sum of the tensor product of a linear operator, the tensor of the sum? I see. The key thing is simply that $A \otimes (B+C) = A \otimes B + A \otimes C$, which just follows from the definitions of $A\otimes B$ and $A+B$. I think. Mar 8 comment Is the sum of the tensor product of a linear operator, the tensor of the sum? That helps! What I'm stuck on is that the form of my example is $\sum_{y\in Y}\sum_{z\in Z}A(z)\otimes B(y,z)$, so while that gets me to $\sum_{z\in Z}A(z)\otimes \sum_{y\in Y}B(y,z)$ which for my $A(z)$ is indeed $I\otimes \sum_{y\in Y}B(y,z)$, that leaves the $z$ in the second term orphaned. Mar 8 comment Is the sum of the tensor product of a linear operator, the tensor of the sum? Yes, that gets me started. Can you elaborate a bit (see added example, were I've messed up the subscripts a bit too). Mar 8 revised Is the sum of the tensor product of a linear operator, the tensor of the sum? Question equality Mar 8 asked Is the sum of the tensor product of a linear operator, the tensor of the sum? Mar 5 revised Exercising elementary toolkit for quantum computing added 2 characters in body Mar 5 asked Exercising elementary toolkit for quantum computing Feb 19 revised Using “we have” in maths papers Add author link Feb 18 answered Using “we have” in maths papers Jan 10 awarded Notable Question