I saw this in [here](http://www.researchgate.net/publication/225152216_Biflatness_of_semigroup_algebras). 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.



$M_Λ(A) → A ⊗ M_Λ(C)$
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*.


