tensor product basisTensorProductBasis2005-07-25 07:37:102007-09-01 05:52:55Theorem% almost certainly you want these
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}The following theorem describes a basis of the
\PMlinkname{tensor product}{TensorProduct}
of two vector spaces, in terms of given bases of the
\PMlinkescapetext{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.
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\begin{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 $U \otimes V$.
\end{theorem}
\begin{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
\PMlinkname{this}{FreeVectorSpaceOverASet} article).
Let \(p\colon U \times V \Fpil W\) be
the bilinear map which satisfies
\begin{equation} \label{Eq:def.p}
p(\vek{e}_i,\vek{f}_j)(k,l) = \begin{cases}
1& \text{if \(i=k\) and \(j=l\),}\\
0& \text{otherwise}
\end{cases}
\end{equation}
for all \(i,k \in I\) and \(j,l \in J\), i.e.,
\(p(\vek{e}_i,\vek{f}_j) \in W\) is the characteristic function of
$\bigl\{(i,j)\bigr\}$. The reasons \eqref{Eq:def.p} 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 \(\gamma\colon U \times V \Fpil
X\) exists a unique linear map \(\widehat{\gamma}\colon W \Fpil X\)
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 \(\vek{u} =
\sum_{i \in I'} u_i\vek{e}_i \in U\) and \(\vek{v} =
\sum_{j \in J'} v_j\vek{f}_j \in V\) that \(\gamma(\vek{u},\vek{v})
= (\widehat{\gamma} \circ\nobreak p)(\vek{u},\vek{v})\), since
\begin{multline*}
\gamma(\vek{u},\vek{v}) =
\gamma\biggl( \sum_{i \in I'} u_i\vek{e}_i,
\sum_{j \in J'} v_j\vek{f}_j \biggr) =
\sum_{i \in I'} \sum_{j \in J'} u_i v_j
\gamma(\vek{e}_i,\vek{f}_j) = \\ =
\sum_{i \in I'} \sum_{j \in J'} u_i v_j
\widehat{\gamma}\bigl( p(\vek{e}_i,\vek{f}_j) \bigr) =
\widehat{\gamma} \biggl(
\sum_{i \in I'} \sum_{j \in J'} u_i v_j p(\vek{e}_i,\vek{f}_j)
\biggr) = \\ =
\widehat{\gamma} \Biggl(
p\biggl( \sum_{i \in I'} u_i \vek{e}_i,
\sum_{j \in J'} v_j \vek{f}_j \biggr)
\Biggr) =
\widehat{\gamma}\bigl( p(\vek{u},\vek{v}) \bigr)
\text{.}
\end{multline*}
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 \(\vek{u} \otimes \vek{v} := p(\vek{u},\vek{v})\).
Hence the claim in the theorem is equivalent to the observation
about the basis of $W$.
\end{proof}