Undergrad The tensor product of tensors confusion

[wrobel]

The standard definition of tensor product of two vector spaces (perhaps infinite dimensional) is as follows. let $E,F$ be vector spaces (say over $\mathbb{R}$) and let $B(E,F)$ be a space of bilinear functions $f:E\times F\to \mathbb{R}$.

Define a mapping $u_{xy}:B(E,F)\to \mathbb{R}$ as follows
$u_{xy}(f)=f(x,y)$ so that $u_{xy}\in (B(E,F))^*.$ We have also got a bilinear mapping
$$\chi:E\times F\to (B(E,F))^*,\quad \chi(x,y)=u_{xy}.$$
By definition the tensor product $E\otimes F$ is the linear span of $$\chi(E\times F);\quad x\otimes y:=u_{xy}.$$

The main feature is as follows. Any bilinear function $A:E\times F\to W$ (W is some other vector space) can be presented as follows $A=\tilde A\chi$
here $\tilde A:E\otimes F\to W$ is a linear mapping.

 

[GR191511]

The standard definition of tensor product of two vector spaces (perhaps infinite dimensional) is as follows. let $E,F$ be vector spaces (say over $\mathbb{R}$) and let $B(E,F)$ be a space of bilinear functions $f:E\times F\to \mathbb{R}$.

Define a mapping $u_{xy}:B(E,F)\to \mathbb{R}$ as follows
$u_{xy}(f)=f(x,y)$ so that $u_{xy}\in (B(E,F))^*.$ We have also got a bilinear mapping
$$\chi:E\times F\to (B(E,F))^*,\quad \chi(x,y)=u_{xy}.$$
By definition the tensor product $E\otimes F$ is the linear span of $$\chi(E\times F);\quad x\otimes y:=u_{xy}.$$

The main feature is as follows. Any bilinear function $A:E\times F\to W$ (W is some other vector space) can be presented as follows $A=\tilde A\chi$
here $\tilde A:E\otimes F\to W$ is a linear mapping.

Thanks

 

[WWGD]

In a general sense, the tensor product of two vector spaces $V, W$ over the same field is a third vector space $V \otimes W$, whose dimension is the product of those of $V, W$and so that every bilinear map from $V \times W \rightarrow Z$ , becomes a linear map from $V\otimes W \rightarrow Z$, for $Z$ any vector space.
Idea is to transform multilinear maps into linear ones, as the latter are simpler and easier to deal with.