# tensor product basis

The following theorem  describes a basis of the tensor product (http://planetmath.org/TensorProduct) of two vector spaces  , in terms of given bases of the 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.

tensor product

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

 $\{\mathbf{e}_{i}\}_{i\in I}\quad\text{and}\quad\{\mathbf{f}_{j}\}_{j\in J}$

respectively. Then

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

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

###### Proof.

Let

 $W=\left\{\,\psi\colon I\times J\longrightarrow\mathcal{K}\,\,\vrule width 1px% \big{.}\,\,f^{-1}\bigl{(}\mathcal{K}\setminus\{0\}\bigr{)}\text{ is finite}\,% \right\}\text{;}$

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

 $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 $p$ on the whole of $U\times V$ are that $\{\mathbf{e}_{i}\}_{i\in I}$ is a basis of $U$, $\{\mathbf{f}_{i}\}_{j\in J}$ is a basis of $V$, and $p$ is bilinear.

Observe that

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

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

 $\widehat{\gamma}\bigl{(}p(\mathbf{e}_{i},\mathbf{f}_{j})\bigr{)}=\gamma(% \mathbf{e}_{i},\mathbf{f}_{j})\quad\text{for all $$i\in I$$ and $$j\in J$$.}$

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

 $\displaystyle\gamma(\mathbf{u},\mathbf{v})=\gamma\biggl{(}\sum_{i\in I^{\prime% }}u_{i}\mathbf{e}_{i},\sum_{j\in J^{\prime}}v_{j}\mathbf{f}_{j}\biggr{)}=\sum_% {i\in I^{\prime}}\sum_{j\in J^{\prime}}u_{i}v_{j}\gamma(\mathbf{e}_{i},\mathbf% {f}_{j})=\\ \displaystyle=\sum_{i\in I^{\prime}}\sum_{j\in J^{\prime}}u_{i}v_{j}\widehat{% \gamma}\bigl{(}p(\mathbf{e}_{i},\mathbf{f}_{j})\bigr{)}=\widehat{\gamma}\biggl% {(}\sum_{i\in I^{\prime}}\sum_{j\in J^{\prime}}u_{i}v_{j}p(\mathbf{e}_{i},% \mathbf{f}_{j})\biggr{)}=\\ \displaystyle=\widehat{\gamma}\Biggl{(}p\biggl{(}\sum_{i\in I^{\prime}}u_{i}% \mathbf{e}_{i},\sum_{j\in J^{\prime}}v_{j}\mathbf{f}_{j}\biggr{)}\Biggr{)}=% \widehat{\gamma}\bigl{(}p(\mathbf{u},\mathbf{v})\bigr{)}\text{.}$

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 $\mathbf{u}\otimes\mathbf{v}:=p(\mathbf{u},\mathbf{v})$. Hence the claim in the theorem is equivalent     to the observation about the basis of $W$. ∎

Title tensor product basis TensorProductBasis 2013-03-22 15:24:48 2013-03-22 15:24:48 lars_h (9802) lars_h (9802) 11 lars_h (9802) Theorem msc 15A69 basis construction of tensor product TensorProduct FreeVectorSpaceOverASet