tensor product basis

The following theorem describes a basis of thetensor product (http://planetmath.org/TensorProduct)of two vector spaces, in terms of given bases of thespaces. In passing, it also gives a construction of this tensorproduct. The exact same method can be used also for freemodules over a commutative ring with unit.

tensor product

• Theorem.  Let $U$ and $V$ be vector spaces over a field $𝒦$ with bases

  $${\{{𝐞}_i\}}_{i\in I}\mathit{\hspace{1em}}\text{and}\mathit{\hspace{1em}}{\{{𝐟}_j\}}_{j\in J}$$

  respectively. Then

  $${\{{𝐞}_i\otimes {𝐟}_j\}}_{(i,j)\in I\times J}$$(1)

  is a basis for the tensor product space $U\otimes V$.

Proof.

Let

$$W=\{\psi :I\times J⟶𝒦\text{<span class="ltx\_rule" style="width:1px;background:black;display:inline-block;"> </span>}.f^{-1}\left(𝒦\setminus \{0\}\right)\text{ is finite}\}\text{;}$$

this set is obviously a $𝒦$-vector-space under pointwise additionand multiplication by scalar (see alsothis (http://planetmath.org/FreeVectorSpaceOverASet) article).Let $p:U\times V⟶W$ bethe bilinear map which satisfies

(2)

for all $i,k\in I$ and $j,l\in J$, i.e.,$p({𝐞}_i,{𝐟}_j)\in W$ is the characteristic function of$\left\{(i,j)\right\}$. The reasons (2) uniquelydefines $p$ on the whole of $U\times V$ are that${\{{𝐞}_i\}}_{i\in I}$ is a basis of $U$, ${\{{𝐟}_i\}}_{j\in J}$ is a basis of $V$, and $p$ is bilinear.

Observe that

$${\left\{p({𝐞}_i,{𝐟}_j)\right\}}_{(i,j)\in I\times J}$$

is a basis of $W$. Since one may always define a linear mapby giving its values on the basis elements, this implies that there for every$𝒦$-vector-space $X$ and every map $\gamma :U\times V⟶X$ exists a unique linear map $\widehat{\gamma }:W⟶X$such that

$$\widehat{\gamma }\left(p({𝐞}_i,{𝐟}_j)\right)=\gamma ({𝐞}_i,{𝐟}_j)\mathit{\hspace{1em}}\text{for all }i\in I\text{ and }j\in J\text{.}$$

For $\gamma$ that are bilinear it holds for arbitrary $𝐮={\sum }_{i\in I^{\prime }}{u}_i{𝐞}_i\in U$ and $𝐯={\sum }_{j\in J^{\prime }}{v}_j{𝐟}_j\in V$ that $\gamma (𝐮,𝐯)=(\widehat{\gamma }\circ p)(𝐮,𝐯)$, since

$$\begin{array}{c}\gamma (𝐮,𝐯)=\gamma (\sum _{i\in I^{\prime }}{u}_i{𝐞}_i,\sum _{j\in J^{\prime }}{v}_j{𝐟}_j)=\sum _{i\in I^{\prime }}\sum _{j\in J^{\prime }}{u}_i{v}_j\gamma ({𝐞}_i,{𝐟}_j)=\hfill \\ \hfill =\sum _{i\in I^{\prime }}\sum _{j\in J^{\prime }}{u}_i{v}_j\widehat{\gamma }\left(p({𝐞}_i,{𝐟}_j)\right)=\widehat{\gamma }\left(\sum _{i\in I^{\prime }}\sum _{j\in J^{\prime }}{u}_i{v}_{j}p({𝐞}_i,{𝐟}_j)\right)=\\ \hfill =\widehat{\gamma }\left(p(\sum _{i\in I^{\prime }}{u}_i{𝐞}_i,\sum _{j\in J^{\prime }}{v}_j{𝐟}_j)\right)=\widehat{\gamma }\left(p(𝐮,𝐯)\right)\text{.}\end{array}$$

As this is the defining property of the tensor product $U\otimes V$ however, it follows that $W$ is (an incarnation of) this tensorproduct, with $𝐮\otimes 𝐯:=p(𝐮,𝐯)$.Hence the claim in the theorem is equivalent to the observationabout the basis of $W$.∎