= Tensor product
{title2=$\otimes$}
{wiki}
A <tensor product> turns a <bilinear map> into a <linear map>. For <vector spaces> $V,W$ over the same <field> $F$, its elements are finite sums of <pure tensors> $v\otimes w$, subject to <bilinearity>:
$$
\begin{aligned}
(v+v')\otimes w&=v\otimes w+v'\otimes w,\\
v\otimes(w+w')&=v\otimes w+v\otimes w',\\
(av)\otimes w&=a(v\otimes w)=v\otimes(aw).
\end{aligned}
$$
The <scalar> $a$ belongs to $F$; writing $V\otimes_F W$ makes this choice explicit. These rules preserve separate <linearity> while allowing arbitrary <linear combinations> of products.<Linear Algebra Done Right by Sheldon Axler (4th edition)>[§9D, 9.71–9.74, printed pp. 372–373: tensor products, bilinearity, and bases]{ref}
For a concrete example, take $V=W=\mathbb R^2$, with <basis vectors> $e_1,e_2$ in the first copy and $f_1,f_2$ in the second. The four <tensor-product basis> elements are
$$
e_1\otimes f_1,\quad e_1\otimes f_2,\quad e_2\otimes f_1,\quad e_2\otimes f_2.
$$
The <pure tensor> made from $v=2e_1+e_2$ and $w=-f_1+3f_2$ expands to
$$
v\otimes w=-2e_1\otimes f_1+6e_1\otimes f_2-e_2\otimes f_1+3e_2\otimes f_2.
$$
Each <basis vector> in one factor can interact with each <basis vector> in the other. This is why <dimensions> multiply; the <tensor-product coefficient matrix> displays these four interactions as a two-by-two grid.
The <tensor product> is different from the <Cartesian product> $V\times W$. With componentwise <vector addition> and <scalar multiplication>, the latter contains pairs and has <dimension> $\dim V+\dim W$. The map $\tau(v,w)=v\otimes w$ is <bilinear>, generally neither <linear> on that <Cartesian product> nor <injective> nor <surjective>. In our example, $\tau(2v,2w)=4\tau(v,w)$, and $\tau(2v,w/2)=\tau(v,w)$; the <separability of a tensor> explains why some outputs cannot be reached at all. Both spaces happen to have <dimension> four when $V=W=\mathbb R^2$, but this coincidence does not identify their structures.
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