For sake of brevity let $A$ denote the Banach algebra formed by equipping $\ell^1({\mathbb N})$ with *pointwise* multiplication. This algebra is clearly not isomorphic as a Banach algebra to any uniform algebra (since it has idempotents of arbitrarily large norm).


It has been known since the 1970s that there exists a Hilbert space $H$ and a continuous injective algebra homomorphism $\theta:A\to B(H)$ which has (norm-)closed range. That is, $A$ is isomorphic to a non-self-adjoint operator algebra. However, the proof is non-explicit; the only argument I have seen uses the fact that multiplication $A \otimes A \to A$ extends continuously to the injective tensor product of $A$ with itself, and then uses abstract GNS-flavoured techniques to build the required space $H$. See



> 
> Varopoulos, N. Th. [Some remarks on $Q$-algebras](http://www.ams.org/mathscinet-getitem?mr=338780). Ann. Inst. Fourier (Grenoble) 22 (1972), no. 4, 1–11.
> 
> 
> Davie, A. M. [Quotient algebras of uniform algebras](http://www.ams.org/mathscinet-getitem?mr=324427). J. London Math. Soc. (2) 7 (1973), 31–40.
> 
> 
> 


For each $n$ let $A\_n$ be the subalgebra of $A$ generated by the first $n$ minimal idempotents (so $A\_n$ is just ${\mathbb C}^n$ with the $\ell^1$-norm and pointwise product).


**UPDATE** I originally mis-stated part of the question, as Andreas, Bill and Gideon were all quick to point out; my apologies for the confusion.
**UPDATE, ENCORE** My apologies for having omitted yet another condition in my haste to get the original edits done - thanks to fedja for catching this


*Taking the existence of $\theta$ as read*, it is straightforward to show that given $a\in A\_n$ there exists a subspace $H\_n\subset H$ of dimension $n+1$ and a homomorphism $\theta\_n A\_n \to B(H\_n)$ such that


$$ c\Vert a\Vert\_1 < \Vert\theta\_n(a)\Vert \leq C \Vert a\Vert\_1 $$
where the constants $c$ and $C$ are strictly positive and independent of $n$ and $a$.


By a crude compactness argument, this implies the following:



> 
> There exist constants $0 < c < C$ such that, for each $n\geq 1$, we can find $m(n)$ and idempotents $E\_1,\dots, E\_n$ in $M\_{m(n)}({\mathbb C})$, which satisfy
> $$ E\_jE\_k=0 \quad\hbox{for $j\neq k$} $$
> and
> $$ c\sum\_{j=1}^n |a\_j| \leq \left\Vert \sum\_{j=1}^n a\_j E\_j \right\Vert \leq C \sum\_{j=1}^n |a\_j| $$
> for all $a\_1,\dots, a\_n\in {\mathbb C}$.
> 
> 
> 


My question is this: **does anybody know of an *explicit* construction of such $E\_1,\dots, E\_n$?** I would be interested just to know of a construction which works for a sequence $n\_1 < n\_2< \dots$ Also, how small can we make the containing matrix algebra? (The compactness argument referred to above produces an embedding into the block-diagonal matrices with block size $n+1$ -- I haven't tried to work out just how `long' the diagonal has to be.)


These seem to me like questions that must have been looked at before, so perhaps the answer would be as simple as supplying an appropriate reference.
(The choice to work over complex scalars here is not important; a solution just for real scalars would be just as welcome.)


**Remark:** perhaps it might be useful to use the fairly well-known equivalence between the $\ell^1$-norm and the norm
$$ \Vert x \Vert\_S := \sup\_{F \subseteq {\mathbb N}} \left\Vert \sum\_{i\in F} x\_i \right\Vert $$



"Explicit" embedding of $\ell^1$ as a closed subalgebra of a direct sum of matrix algebras
You can manage to get $m(n)=(n+2)2^{n}$ (this can certainly be made a bit better). By your remark, it is enough to construct projections $E\_1,\dots,E\_n$ in $M\_{n+2}(\mathbb C)$ such that:


1. $E\_i E\_j=0$ for $i \ne j$
2. $\|E\_i\|\leq 3$
3. $\| \sum s\_i E\_i\| \geq |\sum s\_i|$ for any complex numbers $s\_1,\dots s\_n$.


(Then consider $\oplus\_{F \subset (1,\dots,n)} M\_{|F|+2}$ inside $M\_{m(n)}(\mathbb C)$).


We use the following elementary fact: there exists $\Omega$ a unit vector in the euclidean space $\mathbb C^{n+2}$ and $x\_1,\dots,x\_n$ and $y\_1,\dots,y\_n$ of norm $\sqrt 3$ in $\mathbb C^{n+2}$ such that $\langle x\_i,y\_j\rangle=1$ if $i=j$ and $0$ if $i\ne j$, and such that $\langle x\_i,\Omega\rangle=\langle y\_i,\Omega\rangle=1$.
(if $\Omega,e\_0,\dots,e\_n$ is an orthonormal basis, take $x\_i=\Omega+e\_0+e\_i$ and $y\_i=\Omega-e\_0+e\_i$).


Then the matrices $E\_i=x\_i^\* \otimes y\_i$ (the rank one operator mapping $x$ to $\langle x\_i,x\rangle y\_i$ satisfy the conditions above. The only point perhaps not completely evident is the third point, but this is because $\langle \Omega,\sum\_i s\_i E\_i \Omega \rangle = \sum\_i s\_i$.


