PlanetMath: tensor product basis

The following theorem describes a basis of the tensor product of two vector spaces, in terms of given bases of the component spaces. In passing, it also gives a construction of this tensor product. The exact same method can be used also for free modules over a commutative ring with unit.

Proof. Let

$\displaystyle W = \left\{ \, \psi\colon I \times J \longrightarrow \mathcal{K} ... ...\bigl( \mathcal{K}\setminus \{0\} \bigr) \text{ is finite} \, \right\}\text{;}$

this set is obviously a $\mathcal{K}$ -vector-space under pointwise addition and multiplication by scalar (see also this article). Let $p\colon U \times V \longrightarrow W$ be the bilinear map which satisfies

$\displaystyle p(\mathbf{e}_i,\mathbf{f}_j)(k,l) = \begin{cases}1& \text{if $i=k$ and $j=l$,}\\ 0& \text{otherwise} \end{cases}$

(2)

for all $i,k \in I$ and $j,l \in J$ , i.e., $p(\mathbf{e}_i,\mathbf{f}_j) \in W$ is the characteristic function of $\bigl\{(i,j)\bigr\}$ . The reasons (2) uniquely defines

on the whole of $U \times V$ are that $\{\mathbf{e}_i\}_{i \in I}$ is a basis of

, $\{\mathbf{f}_i\}_{j \in J}$ is a basis of

, and

is bilinear.

Observe that

$\displaystyle \bigl\{ p(\mathbf{e}_i,\mathbf{f}_j) \bigr\}_{(i,j) \in I \times J}$

is a basis of

. Since one may always define a linear map by giving its values on the basis elements, this implies that there for every $\mathcal{K}$ -vector-space

and every map $\gamma\colon U \times V \longrightarrow X$ exists a unique linear map $\widehat{\gamma}\colon W \longrightarrow X$ such that

$\displaystyle \widehat{\gamma}\bigl( p(\mathbf{e}_i,\mathbf{f}_j) \bigr) = \gamma(\mathbf{e}_i,\mathbf{f}_j)$ for all $i \in I$ and $j \in J$ .

For $\gamma$ that are bilinear it holds for arbitrary $\mathbf{u} = \sum_{i \in I'} u_i\mathbf{e}_i \in U$ and $\mathbf{v} = \sum_{j \in J'} v_j\mathbf{f}_j \in V$ that $\gamma(\mathbf{u},\mathbf{v}) = (\widehat{\gamma} \circ\nobreak p)(\mathbf{u},\mathbf{v})$ , since

$\begin{multline*} \gamma(\mathbf{u},\mathbf{v}) = \gamma\biggl( \sum_{i \in I'} ... ... \widehat{\gamma}\bigl( p(\mathbf{u},\mathbf{v}) \bigr) \text{.} \end{multline*}$

As this is the defining property of the tensor product $U \otimes V$ however, it follows that

is (an incarnation of) this tensor product, with $\mathbf{u} \otimes \mathbf{v} := p(\mathbf{u},\mathbf{v})$ . Hence the claim in the theorem is equivalent to the observation about the basis of

. $\qedsymbol$