Tensor product

Key ideas: Cartesian product, free vector space, quotient space, equivalence relations

Take the Cartesian product $U\times V$. Consider every element to be a basis element, and form every possible linear combination (over the field $\mathbb{F}$) of $u$'s and $v$'s. This gives us the 'free vector space' over $U\times V$, $F(U\times V)$.

For example, an element of $F(\mathbb{R}\times \mathbb{R})$ can be $2(1,0)+3(2,-2)+4(3,-2)$. We can't simplify this any further.

But we don't actually want this vector space; it's 'too big'. We want to form a quotient space by modding out a subspace, $W$. We're going to specify $W$ by specifying its basis elements, which are all elements of the form: $$(u,v+v')-(u,v)-(u,v')\\(u+u',v) - (u,v) - (u',v)\\(cu,v) - c(u,v)\\(u,cv) - c(u,v).$$
'Modding these out' effectively sets all of these to $0$ in the quotient space $F(U\times V)/W$.

So the vectors of $U\otimes W$ are actually the equivalence classes of $F(U\times V)$ under the following equivalence relations:  $$(u,v+v')\sim (u,v)+(u,v')\\(u+u',v)\sim (u,v)+(u',v)\\(cu,v)\sim c(u,v)\sim (u,cv).$$
An element $(u,v)+W$ is written $u\otimes v$. Then the above rules become: $$u\otimes (v+v')=u\otimes v+u\otimes v'\\(u+u')\otimes v=u\otimes v+u'\otimes v\\c(u\otimes v)=(cu\otimes v)=(u\otimes cv).$$
In other words, $\otimes$ is bilinear on $U\otimes V$.

Let's look at $2(1,0)+3(2,-2)+4(3,-2)$ as it occurs in the tensor product $\mathbb{R}\otimes \mathbb{R}$. It becomes:
$$\begin{align}2(1\otimes 0)+3(2\otimes -2)+4(3\otimes -2)&=2\otimes 0+6\otimes-2+12⊗-2\\&=2\otimes 0+(6+12)\otimes-2\\&=2\otimes 0+18\otimes -2\\&=2\otimes (9*0)+18\otimes -2\\&=18\otimes 0+18\otimes -2\\&=18\otimes -2\\&=-2(18\otimes 1)\\&=-36(1\otimes 1).\end{align}$$
In fact, for any two real numbers $a,b$, we have $(a\otimes b)=ab(1\otimes 1)$. So $\{1\}$ is a basis for $\mathbb{R}$ and $\{1\otimes 1\}$ is a basis for $\mathbb{R}\otimes \mathbb{R}$, which is thus a $1$-dimensional vector space over $\mathbb{R}$, and hence isomorphic to $\mathbb{R}$, namely $\mathbb{R}\otimes \mathbb{R} \cong \mathbb{R}$ (as vector spaces). But note that we can do something in the vector space $\mathbb{R}\otimes \mathbb{R}$ we can't do ordinarily in a vector space: we can 'multiply vectors'.

If $U$ has a basis $\{e_1,\cdots,e_n\}$ and $V$ has a basis $\{e'_1,\cdots,e'_m\}$ then $U\otimes V$ has the basis $\{e_i\otimes e'_j\}$, where $i=1,\cdots,n$ and $j=1,\cdots,m$. Thus $U\otimes V$ has dimension $mn$. In general $U\otimes V$ can be thought of as a generic bilinear map on $U\times V$ (if $U=V=\mathbb{F}$, this becomes ordinary field multiplication). What bilinearity means is that $$u\otimes v=(\sum_{i=1}^n u_ie_i) \otimes (\sum_{j=1}^m v_je'_j)=\sum_{i=1}^n \sum_{j=1}^m u_i v_j(e_i\otimes e'_j).$$ Namely, $u\otimes v$ is a vector in $U\otimes V$, spanned by the basis vectors $e_i\otimes e'_j$, and the coefficients are given by $u_iv_j$ for each basis vector.

Let $n=2,m=3$. The tensor product space is $mn=6$-dimensional. The basis vectors are $e_1\otimes e'_1,e_1\otimes e'_2,e_1\otimes e'_3,e_2\otimes e'_1,e_2\otimes e'_2,e_2\otimes e'_3$. Written as six-component column vectors, they are $$e_1\otimes e'_1=(1,0,0,0,0,0)^T,e_1\otimes e'_2=(0,1,0,0,0,0)^T,e_1\otimes e'_3=(0,0,1,0,0,0)^T\\e_2\otimes e'_1=(0,0,0,1,0,0)^T,e_2\otimes e'_2=(0,0,0,0,1,0)^T,e_2\otimes e'_3=(0,0,0,0,0,1)^T.$$ For general vectors $u$ and $v$, the tensor product is $$u\otimes v=\begin{pmatrix}u_1v_1\\u_1v_2\\u_1v_3\\u_2v_1\\u_2v_2\\u_2v_3\end{pmatrix}.$$

Let's now consider the map $A:U\to U, u\mapsto Au$ or $$\begin{pmatrix} u_1\\u_2\\ \vdots\\ u_n\end{pmatrix}\mapsto \begin{pmatrix} a_{21}&a_{22}&\cdots&a_{2n}\\ \vdots&\vdots&\ddots&\vdots\\a_{n1}&a_{n2}&\cdots&a_{nn}\end{pmatrix}\begin{pmatrix} u_1\\u_2\\ \vdots\\ u_n\end{pmatrix}.$$ On the tensor product space $U\otimes V$, the same matrix can still act on the vectors, so that $u\mapsto Au$, but $v\mapsto v$ untouched. The matrix is $A\otimes I$. $$A\otimes I=\begin{pmatrix}a_{11}&0&0&a_{12}&0&0\\0&a_{11}&0&0&a_{12}&0\\0&0&a_{11}&0&0&a_{12}\\a_{21}&0&0&a_{22}&0&0\\0&a_{21}&0&0&a_{22}&0\\0&0&a_{21}&0&0&a_{22}\\\end{pmatrix}$$ If we act it on $u\otimes v$, $$\begin{align}(A\otimes I)(u\otimes v)&=\begin{pmatrix}a_{11}&0&0&a_{12}&0&0\\0&a_{11}&0&0&a_{12}&0\\0&0&a_{11}&0&0&a_{12}\\a_{21}&0&0&a_{22}&0&0\\0&a_{21}&0&0&a_{22}&0\\0&0&a_{21}&0&0&a_{22}\\\end{pmatrix} \begin{pmatrix}u_1v_1\\u_1v_2\\u_1v_3\\u_2v_1\\u_2v_2\\u_2v_3\end{pmatrix}\\&=\begin{pmatrix}(a_{11}u_1+a_{12}u_2)v_1\\(a_{11}u_1+a_{12}u_2)v_2\\(a_{11}u_1+a_{12}u_2)v_3\\(a_{21}u_1+a_{22}u_2)v_1\\(a_{21}u_1+a_{22}u_2)v_2\\(a_{21}u_1+a_{22}u_2)v_3\end{pmatrix}\\&=(Au)\otimes v \end{align}$$  We can see that the matrix $A$ indeed acts only on $u\in U$, and leaves $v\in V$ untouched.

Similarly, the matrix $B:V\to V, w\mapsto Bw$ can also act on $U\otimes V$ as $I\otimes B$: $$I\otimes B=\begin{pmatrix}b_{11}&b_{12}&b_{13}&0&0&0\\b_{21}&b_{22}&b_{23}&0&0&0\\b_{31}&b_{32}&b_{33}&0&0&0\\0&0&0&b_{11}&b_{12}&b_{13}\\0&0&0&b_{21}&b_{22}&b_{23}\\0&0&0&b_{31}&b_{32}&b_{33}\end{pmatrix},$$ which acts on $u\otimes v$ as $$\begin{align}(I\otimes b)(u\otimes v)&=\begin{pmatrix}b_{11}&b_{12}&b_{13}&0&0&0\\b_{21}&b_{22}&b_{23}&0&0&0\\b_{31}&b_{32}&b_{33}&0&0&0\\0&0&0&b_{11}&b_{12}&b_{13}\\0&0&0&b_{21}&b_{22}&b_{23}\\0&0&0&b_{31}&b_{32}&b_{33}\end{pmatrix}\begin{pmatrix}u_1v_1\\u_1v_2\\u_1v_3\\u_2v_1\\u_2v_2\\u_2v_3\end{pmatrix}\\ &=\begin{pmatrix}u_1(b_{11}v_1+b_{12}v_2+b_{13}v_3)\\ u_1(b_{21}v_1+b_{22}v_2+b_{23}v_3)\\u_1(b_{31}v_1+b_{32}v_2+b_{33}v_3)\\ u_2(b_{11}v_1+b_{12}v_2+b_{13}v_3)\\u_2(b_{21}v_1+b_{22}v_2+b_{23}v_3)\\ u_2(b_{31}v_1+b_{32}v_2+b_{33}v_3)\end{pmatrix}\\ &=u\otimes (Bv)\end{align}.$$  If you have two matrices, their multiplications are done on each vector space separately, $$(A_1\otimes I)(A_2\otimes I)=(A_1A_2)\otimes I\\ (I\otimes B_1)(I\otimes B_2)=I\otimes(B_1B_2)\\(A\otimes I)(I\otimes B)=(I\otimes B)(A\otimes I)=(A\otimes B).$$
We can write out $A\otimes B$ explicitly: $$A\otimes B=\begin{pmatrix} a_{11}b_{11}&a_{11}b_{12}&a_{11}b_{13}&a_{12}b_{11}&a_{12}b_{12}&a_{12}b_{13}\\ a_{11}b_{21}&a_{11}b_{22}&a_{11}b_{23}&a_{12}b_{21}&a_{12}b_{22}&a_{12}b_{23}\\ a_{11}b_{31}&a_{11}b_{32}&a_{11}b_{33}&a_{12}b_{31}&a_{12}b_{32}&a_{12}b_{33}\\ a_{21}b_{11}&a_{21}b_{12}&a_{21}b_{13}&a_{22}b_{11}&a_{22}b_{12}&a_{22}b_{13}\\ a_{21}b_{21}&a_{21}b_{22}&a_{21}b_{23}&a_{22}b_{21}&a_{22}b_{22}&a_{22}b_{23}\\ a_{21}b_{31}&a_{21}b_{32}&a_{21}b_{33}&a_{22}b_{31}&a_{22}b_{32}&a_{22}b_{33} \end{pmatrix}.$$ One can verify that $(A\otimes B)(u\otimes v)=(Au)\otimes (Bv)$.
Other useful formulae are $\text{det}(A\otimes B)=(\text{det}A)^m(\text{det}B)^n$ and $\text{Tr}(A\otimes B)=(\text{Tr}A)(\text{Tr}B)$.

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Now we present the formal definitions related to tensor product.

As with most of linear algebra, one has the usual choice between a concrete, base-oriented definition and an abstract, base-free definition. The latter definition has clear advantages since we can use it for important, more general applications. Nevertheless, to build intuition, we first provide the concrete (base-oriented) definition of the tensor product of vector spaces over a field.

Definition:
Suppose $A$ and $B$ are vector spaces over a field $\mathbb{F}$, with bases $\{a_1,\cdots,a_m\}$ and $\{b_1,\cdots,b_n\}$ respectively. The tensor product $A\otimes B$ is defined as the vector space of dimension $mn$ over $\mathbb{F}$ with base $\{a_i \otimes b_j: 1\leq i \leq m, 1\leq j \leq n\}$. For any $a=\sum_i \alpha_i a_i \in A$ and $b=\sum_j \beta_j b_j \in B$ with $\alpha_i, \beta_i$ in $\mathbb{F}$, we define the simple tensor $a\otimes b$ to be $\sum_{i,j} \alpha_i \beta_j a_i\otimes b_j$. Thus $\alpha(a\otimes b)=\alpha a\otimes b=a\otimes \alpha b$, $\forall \alpha \in \mathbb{F}$.

Despite its intuitive immediacy, this definition hampers development of the theory because of its reliance on the choice of base.

We now try an abstract approach, which provides direct, computation-free proofs of all the important properties. An extra benefit is the ability to work with modules and algebras over arbitrary commutative rings, since the existence of a base no longer is required to define the tensor product.

A balanced map is a function $\psi:M\times N \to G$, where $G=(G,+)$ is an Abelian group, satisfying the following properties:
$$\phi(a_1+a_2,b)=\phi(a_1,b)+\phi(a_2,b)\\ \phi(a,b_1+b_2)=\phi(a,b_1)+\phi(a,b_2)\\ \phi(ar,b)=\phi(a,rb)$$
for all $a_i\in M,b_i\in N,r\in R$. The tensor product is a group $M\otimes_R N$, endowed with a balanced map $\phi_{M,N}:M\times N\to M\otimes_R N$, such that for any balanced map $\psi:M\times N \to G$ there is a unique group homomorphism $\overline{\psi}:M\otimes_R N\to G$ such that $$\overline{\psi}\phi_{M,N}=\psi,$$
namely, the diagram
$$M\times N \xrightarrow{\qquad \psi \qquad} G\\ \; \phi_{M,N} \searrow \quad \qquad \nearrow \overline{\psi}\\ \qquad\;\; M\otimes_R N$$
is commutative. Our universal construction is achieved by taking the free Abelian group, that is, the free $\mathbb{Z}$-module, generated by what are to become the simple tensors, and factoring out all the desired relations.

Definition: Given a right $R$-module $M$ and an $R$-module $N$ over an arbitrary ring $R$, $M\otimes_R N$ is defined to be $F/K$, where $F$ is the free $\mathbb{Z}$-module with basis $M\times N$, the set of ordered pairs $\{(a,b):a\in M, b\in N\}$, and where $K$ is the submodule generated by the elements
$$(a_1+a_2,b)-(a_1,b)-(a_2,b);\\ (a,b_1+b_2)-(a,b_1)-(a,b_2);\\ (ar,b)-(a,rb)$$
for all $a_i\in M,b_i\in N,r\in R$.

However abstract the construction of $M\otimes_R N$, it must be "correct", being unique up to isomorphism, by "abstract nonsense".

References
What are tensor products?
Algebra by Ernest Shult
Graduate Algebra: Noncommutative View by Louis Halle Rowen [p137-140]

More to read
How to lose your fear of tensor products