$M_Λ(A) → A ⊗ M_Λ(C)$

Question

I saw this in here. Let $A$ be a Banach algebra, and let $\Lambda$ be a non-empty set. We denote by $M_\Lambda(A)$ be the set of $\Lambda\times\Lambda$ matrices $(a_{ij})_{i,j\in\Lambda}$ with entries in $A$ such that $\|(a_{ij})\| =\sum_{i,j}\|a_{ij}\|_A < \infty$ . Then $M_\Lambda(A)$ is a Banach algebra with matrix multiplication. The matrix units in $M_\Lambda(\mathbb{C})$ are denoted by $E_{i,j}$, so that $$ E_{i,j}E_{k,l} = \delta_{j,k}E_{i,l} ~~~(i, j, k, l \in\Lambda ) ,$$ where $\delta_{j,k} = 1$ if $j = k$ and $\delta_{j,k }= 0$ if $j \neq k$. The map $$θ : (a_{i,j}) → \sum_{i,j} a_{i,j} \otimes E_{i,j},~~~~ M_\Lambda(A) \to A \hat{\otimes} M_\Lambda(\mathbb{C}) ,$$ is an isometric algebra isomorphism.

Question:How can I show that $θ$ is isometric.

Answer 1

The norm on $M_\Lambda(A)$ is the one obtained by identifying it as a vector space (not an algebra) with $\ell^1(\Lambda\times\Lambda, A)$.

Similarly, the norm on $M_\Lambda({\mathbb C})$ is the one obtained by identifying it as a vector space with $\ell^1(\Lambda\times\Lambda)$.

Fact. Given any index set $I$ and any Banach space $E$, the natural map $\phi: \ell^1(I)\hat\otimes E \to \ell^1(I,E)$ which sends $e_i \otimes x$ to the function $j \mapsto \delta_{ij} x$ (Kronecker delta) is an isometric isomorphism of Banach spaces.

I'm currently too busy/tired to write out the argument but you can find it as a special case of Example 2.19 in Ryan's book Introduction to Tensor Products of Banach Spaces.