2:19 AM
@MadSpaces Here's how I think about tensor products. Let $f$ be a bilinear map on $V$ and $W$ to some other vector space $X$. What things must be in this codomain vector space $X$?
Well, for any $v\in V,w\in W$, $X$ must contain $f(v,w)$, more-or-less by definition.
It must also contain $f(v,w)+f(v',w')$ (for any $v,v'\in V,w,w'\in W$) because it's a vector space.
Now, here's a question. Say I pick one element $v\in V$ and two elements $w,w'\in W$. Must $f(v,w)+f(v,w')$ be in the image of $f$?
Yes - because $f$ is bilinear, $f(v+w)+f(v,w')$ equals $f(v,w+w')$.
(On the other hand, there's no guarantee $f(v,w)+f(v',w')$, with two vectors chosen from each of $V$ and $W$, is in the image of $f$.)
These expressions are all things we can write down that live in $X$, without having actually ever seen an element of $X$ directly.
Let's thing about these expressions themselves, rather than the elements of $X$ they represent. Say two expressions are the same if we can prove using the bilinearity laws that they must refer to the same element of $X$.
(So I'm considering $f(v+w)+f(v,w')$ and $f(v,w+w')$ to be the same expression. I'm not considering $f(v,w)$ and $f(v',w)$ to be the same expression even if they coincidentally refer to the same element of $X$, because I can't prove it using the bilinearity laws.)
Then we can say the set of expressions is its own vector space.
This set of expressions is the vector space $V\otimes W$, just in different notation.
Instead of talking about $v\otimes w$, I was talking about the expression $f(v,w)$.
But they behave exactly the same. $v\otimes w+v\otimes w'=v\otimes(w+w')$, exactly like how $f(v,w)+f(v,w')=f(v,w+w')$ for a mystery function $f$.
When is $v\otimes w+v'\otimes w'$ not a pure tensor? Exactly when you can't prove $f(v,w)+f(v',w')$ is in the image of $f$ for a mystery function $f$.
I probably should have led with that phrasing. Tensor products describe the behavior of a mystery bilinear function.
