I know the usual definition of Tensor product (cartesian product of basis of $U$ and $V$, with bilinear property) but I started confusing myself as I was reading this new definition.
The tensor product $u\otimes v$ of two finite-dimensional vector
spaces $u$ and $v$(over the same field) is the dual of the vector
space of all bilinear forms on $u\oplus v$. For each pair of $x$ and
$y$, with $x$ in $u$ and $y$ in $v$, the tensor product $z=x\otimes y$
of $x$ and $y$ is the element of $u\otimes v$ defined by $z(w)=w(x.y)$
for every bilinear form $w$.
What does this function $z$ exactly look like? Does this suggest that the tensor product $z=x\otimes y$ is a function that sends all bilinear forms (which form a vector space) to something that looks like $w(x,y)$, which is a scalar value? How can I understand this definition?
I'd really appreciate any help.
Best Answer
Let $k$ be a field. Recall that if $Z$ is a $k$-vector space, the dual space of $Z$ is the $k$-vector space $Z^*$ of all the linear maps $f : Z \to k$.
Now, if $U$ and $V$ are two $k$-vector spaces, consider the $k$-vector space $\operatorname{Bil}_k(U,V;k)$ of all the bilinear maps$^\color{blue}1$ $b : U \times V \to k$. So, here $U \otimes V$ denotes the dual space of $\operatorname{Bil}_k(U,V;k)$, that is, $$U \otimes V := \operatorname{Bil}_k(U,V;k)^*.$$ Thus, an element $z$ of $U \otimes V$ is a linear map $z : \operatorname{Bil}_k(U,V;k) \to k$. Hence, given $u \in U$ and $v \in V$, the tensor product $u \otimes v \in U \otimes V$ is the linear map $u \otimes v : \operatorname{Bil}_k(U,V;k) \to k$ such that $$\forall b \in \operatorname{Bil}_k(U,V;k) : \quad (u \otimes v)(b) = b(u,v).$$ The linearity of $u \otimes v$ follows from the definition of the operations in $\operatorname{Bil}_k(U,V;k)$: if $b_1,b_2 \in \operatorname{Bil}_k(U,V;k)$ and $\lambda \in k$, then $(\lambda b_1+b_2)(u,v) := \lambda b_1(u,v)+b_2(u,v)$.
$^\color{blue}1$ Which is not the same as $(U \times V)^*$.