Brian Bi
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Exercise 3.10.1 Let \(A_{ij}\) denote the element of \(\mathrm{Mat}_m(k)\) that has a 1 in the \(i\)th row and \(j\)th column and 0 for all other entries; evidently the \(A_{ij}\)'s form a basis of \(\mathrm{Mat}_m(k)\). Let \(B_{ij}\) denote the similar element of \(\mathrm{Mat}_n(k)\). Let \(C_{ij}\) denote the similar element of \(\mathrm{Mat}_{mn}(k)\).

The tensor product \(\mathrm{Mat}_m(k) \otimes \mathrm{Mat}_n(k)\) has a basis given by elements of the form \(A_{ij} \otimes B_{pq}\) where \(1 \le i, j \le m, 1 \le p, q \le n\). Define \(\varphi : \mathrm{Mat}_m(k) \otimes \mathrm{Mat}_n(k) \to \mathrm{Mat}_{mn}(k)\) to be the unique linear map such that \[ \varphi(A_{ij} \otimes B_{pq}) = C_{i+(m-1)p, j+(m-1)q} \] For each \(r \in \{1, \ldots, mn\}\) there is exactly one pair \((i, p) \in \{1, \ldots, m\} \times \{1, \ldots, n\}\) with \(r = i+(m-1)p\), therefore \(\varphi\) maps a basis of its domain to a basis of its codomain; it is an isomorphism of vector spaces.

Furthermore \begin{align*} \varphi((A_{ij} \otimes B_{pq})(A_{i'j'} \otimes B_{p'q'})) &= \varphi(A_{ij}A_{i'j'} \otimes B_{pq}B_{p'q'}) \\ &= \varphi(\delta_{ji'} A_{ij'} \otimes \delta_{qp'} B_{pq'}) \\ &= \delta_{ji'} \delta_{qp'} C_{i+(m-1)p, j'+(n-1)q'} \\ &= \delta_{j+(n-1)q, i'+(m-1)p'} C_{i+(m-1)p, j'+(n-1)q'} \\ &= C_{i+(m-1)p, j+(n-1)q} C_{i'+(m-1)p', j'+(n-1)q'} \\ &= \varphi(A_{ij} \otimes B_{pq}) \varphi(A_{i'j'} \otimes B_{p'q'}) \\ \end{align*} so the identity \(\varphi(ab) = \varphi(a)\varphi(b)\) holds whenever \(a, b\) belong to the previously described basis of \(\mathrm{Mat}_m(k) \otimes \mathrm{Mat}_n(k)\). By linearity, \(\varphi(ab) = \varphi(a)\varphi(b)\) for all \(a, b \in \mathrm{Mat}_m(k) \otimes \mathrm{Mat}_n(k)\).

The identity of \(\mathrm{Mat}_m(k) \otimes \mathrm{Mat}_n(k)\) is \[ I = \left[\sum_{i=1}^m A_{ii}\right] \otimes \left[\sum_{j=1}^n B_{jj}\right] = \sum_{i=1}^m \sum_{j=1}^n A_{ii} \otimes B_{jj} \] so \begin{align*} \varphi(I) &= \sum_{i=1}^m \sum_{j=1}^n \varphi(A_{ii} \otimes B_{jj}) \\ &= \sum_{i=1}^m \sum_{j=1}^n C_{i+(m-1)j,i+(m-1)j} \\ &= \sum_{r=1}^{mn} C_{rr} \end{align*} and the last expression is the identity of \(\mathrm{Mat}_{mn}(k)\). So \(\varphi\) is an isomorphism of algebras.