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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