For a commutative ring $R$, a matrix $A \in M\_n(R)$ is invertible iff $\det (A)$ is a unit in $R$. Is there a similar criterion to determine invertibility (having two-sided inverse) of a matrix over a noncommutative ring ? I am particularly interested in determining invertibility of matrices in $M\_n(F \langle x, y \rangle )$, the free associative algebra in $2$ indeterminates over a field $F$.


Passing onto the polynomial ring using the relation $xy - yx = 0$, we may conclude that the `determinant' should be in $F$ which serves as a necessary condition to filter out certain matrices as being non-invertible. Based on this, I have a question as stated below.


Let $f \in F \langle x, y \rangle$ and $a \in F - \{ 0 \}$. Clearly matrices of the form 
$$\begin{bmatrix}
1 & f\\
0 & 1
\end{bmatrix},
\begin{bmatrix}
0 & 1\\
1 & 0
\end{bmatrix},
\begin{bmatrix}
a & 0\\
0 & 1
\end{bmatrix}
$$ are invertible in $M\_2(F \langle x, y \rangle)$. By composing such matrices, one gets other examples of invertible matrices in $M\_2(F \langle x, y \rangle)$.


**Question :** Can any invertible matrix in $M\_2(F \langle x, y \rangle)$ be decomposed as a product of matrices of the above $3$ types ?



invertibility of matrix over free associative algebra
You are asking whether the following holds.



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> > **Claim.** Let $F$ be a field. Then the [free associative algebra](https://en.wikipedia.org/wiki/Free_algebra) $F\langle x,y\rangle$ is a $GE\_2$-ring in the sense of P. M. Cohn [2].
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The answer is **yes**. Indeed, free associative algebras over a field are more generally $GE$-rings [Theorem 3.4, 2]. This fact has been used by Bass [1] to prove that a finitely generated projective module over a free algebra with coefficients in a [PID](https://en.wikipedia.org/wiki/Principal_ideal_ring) is free.
By contrast, free commutative algebras over a field, i.e., polynomial rings over a field **aren't** $GE$-rings if the number of indeterminates exceeds $1$ [Proposition 7.3, 2]. (I have made this remark in another [MO post](https://mathoverflow.net/questions/240078/is-the-class-of-commutative-generalized-euclidean-rings-stable-under-quotient-an)).


We recall below the definition of a $GE$-ring in the sense of P. M. Cohn and we give subsequently a proof of the claim. 


Let $R$ be an associative ring with identity and let $E\_n(R)$ be the subgroup of $GL\_n(R)$ generated by matrices obtained from the identity matrix by replacing an off-diagonal entry by some $r∈R$. Let $D\_n(R)$ be the group of invertible diagonal matrices; note that it normalizes $E\_n(R)$. 


A ring $R$ is termed a $GE\_n$-ring if $GL\_n(R)=E\_n(R)D\_n(R)$. 
The ring $R$ is a $GE$-ring in the sense of P. M. Cohn if it is a $GE\_n$-ring for every $n > 1$. 


Let us turn to the proof of the claim. Let $S$ be the free monoid on $\{x, y\}$ endowed with the [shortlex order](https://en.wikipedia.org/wiki/Shortlex_order), i.e., words are ordered according to word length and words of the same length are ordered lexicographically. Then $S$ is a bi-ordered monoid and it is not difficult to show that $F\langle x,y \rangle = F[S]$ is an integral domain with unit group $F^{\times}$. Let $\varphi: S \rightarrow \omega$ be an order-preserving bijection. The following lemma is instrumental.



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> > **Lemma.** Let $R = F\langle x,y \rangle$ and let $a\_1, a\_2 \in R$. Assume there are $b\_1, b\_2 \in F\langle x,y \rangle$, not both zero, such that $a\_1b\_1 + a\_2 b\_2 = 0$. Then we can find $E \in E\_2(R)$ and $a \in R$ such that $(a\_1, a\_2)E = (a, 0)$.
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> > *Proof*. For $r \in R$, write $r = \sum\_{w \in S} r\_w w$, where each $r\_w \in F$ and $\text{supp}(r) \Doteq \{w \in S \,\vert\, r\_w \neq 0\}$ is a finite subset of $S$. Since $R$ is an integral domain, we can assume, without loss of generality, that none of $a\_1, a\_2, b\_1$ and $b\_2$ is zero. Setting $n(r) \Doteq \varphi(\max(\text{supp}(r))) \in \omega$ for $r \neq 0$, we argue by induction on $n(a\_1) + n(a\_2)$. Let $w\_i = \max(\text{supp}(a\_i)) \in S$, for $i = 1,2$.
> >  Since $a\_1b\_1 + a\_2 b\_2 = 0$, we can find $v\_1, v\_2 \in S$ such that $w\_1 v\_1 = w\_2 v\_2$. Swapping $a\_1$ and $a\_2$ if needed, we can assume that $w\_1 \le w\_2$, so that $w\_1$ is an initial subword of $w\_2$. Therefore we can find $s \in S$ and $\lambda \in F$ such that either $a\_2' = 0$ or
> >  $n(a\_2') < n(a\_2)$ where $a\_2' = a\_2 - \lambda a\_1 s$. We set $a\_1' = a\_1$, 
> >  $b\_1' = b\_1 + \lambda sb\_2$ and $b\_2' = b\_2$. Then we have $a\_1' b\_1' + a\_2'b\_2' = 0$ and $(a\_1', a\_2')$ is obtained from $(a\_1, a\_2)$ by right-multiplication of a matrix in $E\_2(R)$. If $a\_2' = 0$, we are done. Otherwise the induction hypothesis applies to $(a\_1', a\_2')$, which yields the result.
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We are now in position to prove the claim.



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> > *Proof of the claim.* Let $A = (a\_{ij}) \in GL\_2(R)$ with $R = F\langle x,y \rangle$. As $A$ has a right inverse, we can find $b\_1, b\_2 \in R$, not both zero, such that $a\_{11} b\_1 + a\_{12}b\_2 = 0$. By the above lemma, we can assume that $a\_{12} = 0$. It follows that $a\_{11} \in F^{\times}$, so that $A$ can be reduced to an invertible diagonal matrix through an elementary row operation. Therefore $A \in E\_2(R)D\_2(R)$. 
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[1] H. Bass, "Projective modules over free groups are free", 1964.  

[2] P. M. Cohn, "On the structure of the $GL\_2$ of a ring", 1966. 


