Difference between revisions of "Notes:Tensor product"

Revision as of 07:39, 9 June 2015

Right now (9/6/2015 @ 0827) I have two "definitions" of tensor products. One as detailed on Notes:ToMond and another as the book(s) I have read.

Bilinear function

A function, [ilmath]f:U\times V\rightarrow W[/ilmath] is bilinear if:

• It is linear in both variables, that is:
  • [math]f(\alpha u_1+\beta u_2,v)=\alpha f(u_1,v) + \beta f(u_2,v)[/math] and
  • [math]f(u,\alpha v_1+\beta v_2)=\alpha f(u,v_1) + \beta f(u,v_2)[/math]

As can be seen on Bilinear map (which is a page in need of cleanup!)

Scalar multiplication

Note that:

• [math]f(\lambda u,v)=\lambda f(u,v)[/math] and
• [math]f(u,\lambda v)=\lambda f(u,v)[/math]

So we can conclude that:

• [math]\lambda f(u,v)=f(\lambda u,v)=f(u,\lambda v)[/math]

Tensor product

The tensor product of the vector spaces is [ilmath]U\otimes V[/ilmath] and the elements are [ilmath]u\otimes v[/ilmath] for a bilinear function: [ilmath]\otimes:U\times V\rightarrow W[/ilmath]

Questions

What is the 0 tensor

I have been told that the [ilmath]0[/ilmath] of [ilmath]U\otimes V[/ilmath] is [ilmath]0_U\otimes 0_V[/ilmath] however I am not convinced of this yet. What I do know that the [ilmath]0[/ilmath] vector is given by the [ilmath]0[/ilmath] scalar multiplied by any vector, so I know:

• [math]0(u\otimes v)=[/math]
  • [math](0u)\otimes v=0_u\otimes v[/math]
  • [math]u\otimes(0v) = u\otimes 0_v[/math]

I am convinced that [ilmath]0_u\otimes v=u\otimes 0_v[/ilmath] but I am not yet convinced that we must therefor have [ilmath]=0_v\otimes 0_u[/ilmath]