Vector space/Tensor product/Bases/Transformation matrix/Fact

Let $K$ be a field, and let $V_{1}, \dots, V_{n}$ denote finite-dimensional vector spaces over $K$ . Let $v_{1 j}, j \in J_{1}, \dots, v_{n j}, j \in J_{n}$ , and $w_{1 j}, j \in J_{1}, \dots, w_{n j}, j \in J_{n}$ , be bases of $V_{1}, \dots, V_{n}$ , with the base change matrices

B_{i} = M_{𝔴_{i}}^{𝔳_{i}} = {(a_{i r s})}_{1 \leq r, s \leq \dim_{K} (V_{i})} .

Then the base change matrix

(with $J = J_{1} \times \dots \times J_{n}$ ) between the bases

v_{1 j_{1}} \otimes_{} \dots \otimes_{} v_{n j_{n}}, (j_{1}, \dots, j_{n}) \in J and w_{1 j_{1}} \otimes_{} \dots \otimes_{} w_{n j_{n}}, (j_{1}, \dots, j_{n}) \in J

of the tensor product is given by the $J \times J$ -matrix with the entries

c_{(j_{1}, \dots, j_{n}), (k_{1}, \dots, k_{n})} = a_{1 j_{1} k_{1}} \dots a_{n j_{n} k_{n}} .

Proof, Write another proof