# Confusion on the definition of Tensor product

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.

• Tensor is not a function - roughly speaking, this is a two-dimensional (in your case) matrix. Every element of a tensor has two numbers, designating its place in the matrix (number of a row and number of a column), and could (as in your case) be composed from elements of two vectors (by means of its multiplication). For instance, element Z34 (row 3 column 4) = X3*Y4 (third element of vector X multiplied by fourth element of vector Y. This is not a definition, just illustration. quora.com/Whats-the-basic-definition-of-tensor Jan 9, 2021 at 21:34
• It is quite confusing to use lower-case letters to denote vector spaces; a common convention in this context is to reserve lower-case letters for vectors and upper-case letters for vector spaces. I also do not recommend thinking about the tensor product this way; it is both unnecessarily complicated and only works in the finite-dimensional case, whereas we routinely use and need infinite-dimensional tensor products in many contexts. Jan 9, 2021 at 21:51

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 $$(\lambda b_1+b_2)(u,v) := \lambda b_1(u,v)+b_2(u,v)$$.
Exercise: If $$Z$$ is a $$k$$-vector space, and $$U,V$$ are finite dimensional, prove that for any bilinear map $$h : U \times V \to Z$$ there exists a unique linear map $$\tilde h : U \otimes V \to Z$$ sending $$u \otimes v$$ to $$h(u,v)$$, for any $$u \in U$$ and $$v \in V$$.
$$^\color{blue}1$$ Which is not the same as $$(U \times V)^*$$.
• Thanks a lot! This clarifies the point where I was confused about. Does the exercise follow from the fact that $\tilde h$ is a double dual space of $h$? Jan 9, 2021 at 22:36
• $h$ is a bilinear map, not a vector space. Pick bases $u_1,\dots,u_n$ and $v_1,\dots,v_m$ for $U$ and $V$, respectively, and note that the $nm$ elements $u_i \otimes v_j$ form a basis for $U \otimes V$. Then define $\tilde{h} : U \otimes V \to Z$ in the basis elements. Jan 10, 2021 at 1:25