Brian Bi

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.