Revision as of 17:14, 9 June 2015

Snippets

Page 47 of notebook, example is $\mathbb{R}^m\otimes\mathbb{R}^n$

Notes

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, $f:U\times V\rightarrow W$ is bilinear if:

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

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

Scalar multiplication

Note that:

• $$f(\lambda u,v)=\lambda f(u,v)$$ and
• $$f(u,\lambda v)=\lambda f(u,v)$$

So we can conclude that:

• $$\lambda f(u,v)=f(\lambda u,v)=f(u,\lambda v)$$

Tensor product

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

Questions

What is the 0 tensor

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

• $$0(u\otimes v)=$$
  • $$(0u)\otimes v=0_u\otimes v$$
  • $$u\otimes(0v) = u\otimes 0_v$$

I am convinced that $0_u\otimes v=u\otimes 0_v$ but I am not yet convinced that we must therefor have $=0_v\otimes 0_u$

Solved

Page 46 of notebook, gist is to show that (u,v) is in the kernel when u=0 or v=0 (done on bilinear page) then suppose that $$u\ne 0\wedge v\ne 0$$ and construct bilinear map with $f(u,v)\ne 0$ then by the characteristic property of the tensor product the tensor product is non zero.