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I thought a bit more about this. Strictly speaking the construction of the tensor product that you link to does use a basis. This basis is the Product of the or
by mathgenius 9y ago
I thought a bit more about this. Strictly speaking the construction of the tensor product that you link to does use a basis. This basis is the Product of the original two vector spaces. Also, the relations that you quotient by, this is a basis for a subspace. We didn't need a basis for the original spaces, but ended up using bases elsewhere.
- Sniffnoy 9y agoThat's certainly true, but that's also clearly not what was being talked about. The problem was (easier) avoiding choosing bases to perform a construction, and (harder) avoiding assuming all vector spaces have bases. Using bases that you directly contruct isn't something anyone ever really has reason to avoid. (And really, neither is the second in the finite-dimensional case, but hey, may as well if you can, right?)
- zodiac 9y agoYou can define tensor products and quotient spaces without bases! (It's a bit of work though)
- jacobolus 9y agohttp://geocalc.clas.asu.edu/pdf/MathViruses.pdf http://geocalc.clas.asu.edu/pdf/MathViruses.pdf