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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.

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    $\begingroup$ 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 $\endgroup$
    – Svyatoslav
    Jan 9, 2021 at 21:34
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    $\begingroup$ 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. $\endgroup$ Jan 9, 2021 at 21:51

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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 $(U \times V)^*$.

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  • $\begingroup$ 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 $h$? $\endgroup$
    – able20
    Jan 9, 2021 at 22:36
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    $\begingroup$ $h$ 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. $\endgroup$
    – azif00
    Jan 10, 2021 at 1:25

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