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.

Theorem. Let $U$ and $V$ be vector spaces over a field $\K$ with bases $$ \{\vek{e}_i\}_{i \in I} \quad\text{and}\quad \{\vek{f}_j\}_{j \in J} $$ respectively. Then \begin{equation} \label{Eq:ProdBas} \{ \vek{e}_i \otimes \vek{f}_j\}_{(i,j) \in I \times J} \end{equation}is a basis for the tensor product space $U \otimes V$ .

Proof. Let $$ W = \setOfBig{ \psi\colon I \times J \Fpil \K }{ f^{-1}\bigl( \K \setminus \{0\} \bigr) \text{ is finite} }\text{;} $$ this set is obviously a $\K$ -vector-space under pointwise addition and multiplication by scalar (see also this article). Let be the bilinear map which satisfies

(1)

for all and , i.e., is the characteristic function of $\bigl\{(i,j)\bigr\}$ . The reasons (1) uniquely defines $p$ on the whole of $U \times V$ are that $\{\vek{e}_i\}_{i \in I}$ is a basis of $U$ , $\{\vek{f}_i\}_{j \in J}$ is a basis of $V$ , and $p$ is bilinear.

Observe that $$ \bigl\{ p(\vek{e}_i,\vek{f}_j) \bigr\}_{(i,j) \in I \times J} $$ is a basis of $W$ . Since one may always define a linear map by giving its values on the basis elements, this implies that there for every $\K$ -vector-space $X$ and every map exists a unique linear map such that $$ \widehat{\gamma}\bigl( p(\vek{e}_i,\vek{f}_j) \bigr) = \gamma(\vek{e}_i,\vek{f}_j) \quad\text{for all \(i \in I\) and \(j \in J\).} $$ For $\gamma$ that are bilinear it holds for arbitrary and that , since

As this is the defining property of the tensor product $U \otimes V$ however, it follows that $W$ is (an incarnation of) this tensor product, with . Hence the claim in the theorem is equivalent to the observation about the basis of $W$ .