Can the tensor product be visualized as a machine for processing vectors?

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The discussion explores the visualization of tensor products, emphasizing the challenge of intuitively understanding them. The author suggests using lower-dimensional spaces to build intuition for higher-rank objects, proposing that visualizing the tensor product as a machine processing vectors can aid comprehension. They illustrate this by explaining how the tensor product maps one vector into another, providing a clear mathematical relationship. The author, drawing from an engineering background, shares practical applications of this concept, particularly in relation to stress tensors. Overall, the conversation highlights the importance of visualization techniques in grasping complex mathematical concepts.
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"Seeing" Tensor Products

Is there a way to "visualize" the tensor product of two (or ##n##) vectors/tensors/algebras/etc.?

I'm having a lot of trouble making the tensor product feel intuitive. I know its properties, and I can usually apply it without too much of a problem, but it does not feel "easy." Any ideas?
 
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For tensor actions in general, I always started by thinking about the properties in lower dimensional spaces, where they can make more physical sense or can be mapped out/drawn explicitly on paper. It tends to bolster your intuition for what happens with higher rank objects.

Then, you can just think of a general action as something that you can already visualize, but in a projected sense (ie, you are imagining only a part of the true behavior, as more exists "behind the scenes," but it is enough to know what's going on).
 
I'm probably coming at this from a different perspective from you, but maybe this can help. My background is in engineering. I like to visualize that tensor product as a machine for processing vectors. You feed a vector to the machine, and the tensor product maps the vector into a new vector. If \vec{L}\otimes \vec{R} is the vector product of the vectors \vec{L} and \vec{R}, then if I dot the vector product on the right by a vector \vec{V}, I get:

(\vec{L}\otimes \vec{R})\centerdot\vec{V}=\vec{L}(\vec{R}\centerdot\vec{V})
This is a vector in the direction of \vec{L}, with a magnitude equal to the magnitude of \vec{L} times (\vec{R}\centerdot\vec{V}). If I dot the vector product on the left by a vector \vec{V}, I get an analogous result.

As a physicist or mathematician, I don't know if this makes any sense to you. But to me as an engineer it makes perfect sense, and I have used it many time in practice when working with the engineering stress tensor to map a unit normal vector to a surface into the traction vector.

Chet
 

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