Let $R$ be a commutative unital ring $R$ with unit element $1$.

For $n\in \mathbb{N}=\{1,2,3,\cdots\}$, let $SL_n(R)$ be the group of all $n\times n$ matrices with entries from $R$ having determinant $1$.

An element $E\in SL_n(R)$ is called an elementary matrix if $E=I+a E_{ij}$, where

• $a\in R$,
• $I$ denotes the identity matrix, and
• $E_{ij}$ the matrix with entry in $i$th row and $j$th column equal to $a$ and all other entries $0$.

The subgroup of $SL_n(R)$ generated by the elementary matrices is denoted by $E_n(R)$.

We know that when $R=\mathbb{C}$, then for all $n$, $SL_n(\mathbb{C})=E_n(\mathbb{C})$.

Now let $R=\ell^\infty$, the algebra of all complex valued bounded sequences with pointwise operations.

Question: Is it true that $SL_n(\ell^\infty)=E_n(\ell^\infty)$ for all $n$?

Let $R$ be a commutative unital ring $R$ with unit element $1$.

For $n\in \mathbb{N}=\{1,2,3,\cdots\}$, let $SL_n(R)$ be the group of all $n\times n$ matrices with entries from $R$ having determinant $1$.

An element $E\in SL_n(R)$ is called an elementary matrix if $E=I+a E_{ij}$, with $i\neq j$, where

• $a\in R$,
• $I$ denotes the identity matrix, and
• $E_{ij}$ the matrix with entry in $i$th row and $j$th column equal to $a$ and all other entries $0$.

The subgroup of $SL_n(R)$ generated by the elementary matrices is denoted by $E_n(R)$.

We know that when $R=\mathbb{C}$, then for all $n$, $SL_n(\mathbb{C})=E_n(\mathbb{C})$.

Now let $R=\ell^\infty$, the algebra of all complex valued bounded sequences with pointwise operations.

Question: Is it true that $SL_n(\ell^\infty)=E_n(\ell^\infty)$ for all $n$?