線型空間 \(U,\, V\) のテンソル積 \(U \otimes V\) を考える。
\(U,\, V\) の基底
\[
\{ {\bf u}_1, \cdots, {\bf u}_m \} \\
\{ {\bf v}_1, \cdots, {\bf v}_n \}
\]
に対するつぎの集合は，\(U \otimes V\) の基底になる：
\[
\{ {\bf u}_i \otimes {\bf v}_j \,|\, i = 1, \cdots, m; \, j = 1,\cdots, n \}
\]
以下，これの証明：
(1) \(U \otimes V\) の元が，\( {\bf u}_i \otimes {\bf v}_j \) の線型結合で書けること：
\(U \otimes V\) の元は，\( ({\bf x},\, {\bf y}) \in U \times V\) に対する \( {\bf x} \otimes {\bf y} \) であり，
\[
{\bf x} = \sum_i x^i \, {\bf u}_i
\\ {\bf y} = \sum_j y^j \, {\bf v}_j
\]
とするとき，
\[
\begin{align*}
{\bf x} \otimes {\bf y}
&= \left( \sum_i x^i \, {\bf u}_i \right)
\otimes \left( \sum_j y^j \, {\bf v}_j \right)
\\&= \sum_{i, j} ( x^i y^j )\, ( {\bf u}_i \otimes {\bf v}_j )
\end{align*}
\]
(2) \( {\bf u}_i \otimes {\bf v}_j \) が線型独立であること：
先ず，テンソル積の定義から，
\[
{\bf x} \otimes {\bf y} = {\bf x'} \otimes {\bf y'}
\ \Longrightarrow\
(\ \xi\, {\bf x} = {\bf x'} \Longrightarrow \xi\, {\bf y'} = {\bf y} \ )
\]
よって，
\[
{\bf x} \otimes {\bf v}_j = {\bf x'} \otimes {\bf v}_j
\ \Longrightarrow\ {\bf x} = {\bf x'}
\]
特に，
\[
{\bf x} \otimes {\bf v}_j = {\bf 0}
\ \Longrightarrow\ {\bf x} \otimes {\bf v}_j = {\bf 0} \otimes {\bf v}_j
\ \Longrightarrow\ {\bf x} = {\bf 0}
\]
これより，
\[
\sum_j {\bf x}_j \otimes {\bf v}_j = {\bf 0}
\\ \Longrightarrow\ {\bf x}_p \otimes {\bf v}_p
= \sum_{j \ne p} {\bf x}_j \otimes {\bf v}_j
\\ \Longrightarrow\ {\bf x}_p \otimes {\bf v}_p = 0
\\ \Longrightarrow\ {\bf x}_p = 0
\]
最後に，
\[
\sum_{i, j} \xi^{ij} ( {\bf u}_i \otimes {\bf v}_j ) = {\bf 0}
\\ \Longrightarrow\ \sum_j \left( \sum_i \xi^{ij} {\bf u}_i \right) \otimes {\bf v}_j = {\bf 0}
\\ \Longrightarrow\ \sum_i \xi^{ij} {\bf u}_i = {\bf 0} \quad ( j = 1, \cdots, n )
\\ \Longrightarrow\ \left(\ \xi^{ij} = 0 \quad ( i = 1, \cdots, n ) \ \right) \quad ( j = 1, \cdots, n )
\\ \Longrightarrow\ \xi^{ij} = 0 \quad ( i, j = 1, \cdots, n )
\]
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