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		+	=Snippets=
		+	Page 47 of notebook, example is {{M|\mathbb{R}^m\otimes\mathbb{R}^n}}
		+	==Definitions==
		+	There are two. First of all is an arbitrary (finite?) operation {{M|\otimes}} where we define:
		+	* <math>v_1\otimes w_1+\cdots+v_k\otimes w_k\in U\otimes W</math> for arbitrary {{M|k\in\mathbb{N} }}
		+	Second is via this property:
		+	* <math>
		+	\begin{xy}
		+	\xymatrix{
		+	{U\times V} \ar@2{->}[r]^t \ar@2{->}[dr]_f & {U\otimes V} \ar@{.>}[d]^{\tilde{f}}\\
		+	& X
		+	}
		+	\end{xy}
		+	</math> (note that the {{M|\implies}} arrows are bilinear, and the single arrows linear maps. THIS IS NON-STANDARD
		+
		+	This diagram says:
		+	* There exists a bilinear map, {{M|t}} such that whenever {{M|f}} is a bilinear map, there exists a linear map, {{M|\tilde{f} }}, such that the diagram commutes
		+
		+	=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.		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.
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	** <math>u\otimes(0v) = u\otimes 0_v</math>		** <math>u\otimes(0v) = u\otimes 0_v</math>
	I am convinced that {{M|1=0_u\otimes v=u\otimes 0_v}} but I am not yet convinced that we must therefor have {{M|1==0_v\otimes 0_u}}		I am convinced that {{M|1=0_u\otimes v=u\otimes 0_v}} but I am not yet convinced that we must therefor have {{M|1==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 <math>u\ne 0\wedge v\ne 0</math> 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.

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Latest revision as of 10:08, 12 June 2015

Snippets

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

Definitions

There are two. First of all is an arbitrary (finite?) operation $\otimes$ where we define:

• $$v_1\otimes w_1+\cdots+v_k\otimes w_k\in U\otimes W$$ for arbitrary $k\in\mathbb{N}$

Second is via this property:

• $$\begin{xy} \xymatrix{ {U\times V} \ar@2{->}[r]^t \ar@2{->}[dr]_f & {U\otimes V} \ar@{.>}[d]^{\tilde{f}}\\ & X } \end{xy}$$ (note that the $\implies$ arrows are bilinear, and the single arrows linear maps. THIS IS NON-STANDARD

This diagram says:

• There exists a bilinear map, $t$ such that whenever $f$ is a bilinear map, there exists a linear map, $\tilde{f}$, such that the diagram commutes

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.