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Assuming that ring morphisms have to take identity elements to identity elements, I think that one we can show even more the following for the first $n$:th Weyl algebra(the proof for higher order Weyl algebras is analogous), with very basic methods;.
Choose an arbitrary positive integer $n$ and put $A_n := \mathbb{C}\langle x_1,\ldots,x_n,y_1,\ldots,y_n \rangle / I$ where $I$ is the ideal generated by the elements $y_1x_1-x_1y_1-1,\ldots,y_nx_n-x_ny_n-1$ and $x_i x_j-x_jx_i, y_i y_j - y_j y_i$ for $i,j \in 1,\ldots,n$.
There does not exist a positive integer $n$ m$and any an associative, commutative and unital ring$R$it such that there is impossible to have a ring morphism $$\phi : \mathbb{C}\langle x,y\rangle/(yx-xy-1) A_n \to M_n(R).$$M_m(R).$$Proof: Seeking for a contradiction, suppose that there is some n m and some associative, commutative and unital ring R such that \phi exists. Denote the images in M_n(R) M_m(R) of x x_1 respectively y, y_1, under \phi, by A:=\phi(x) A:=\phi(x_1) respectively B:=\phi(y). B:=\phi(y_1). The image of 1 will be the identity matrix I. Consider the element yx-xy=1, y_1x_1-x_1y_1=1, the image of which, under \phi, is equal to$$ BA-AB=\phi(yx-xy)=\phi(1)=IBA-AB=\phi(y_1x_1-x_1y_1)=\phi(1)=I. $$Hence the matrices A and B have to satisfy BA-AB=I. Taking the trace of the left hand side of this equality yields$$ tr(BA-AB)=tr(BA)-tr(AB)=tr(AB)-tr(AB)=0 $$whereas the trace of the right hand side is equal to tr(I)=n. tr(I)=m. This is a contradiction. Corollary of the above proof: The same claim holds if we replace M_n(R) M_m(R) by any unital Banach algebra. This is easily seen by using the following well-known fact: The identity element of a unital Banach algebra can not be a commutator, i.e. ab-ba\neq 1 for any elements a,b of the Banach algebra. This applies to the case M_n(R) M_m(R) with R=\mathbb{C}, because M_n(\mathbb{C}) M_m(\mathbb{C}) is a unital C*-algebra and in particular a unital Banach algebra. So in this particular case, this gives us an alternative way of making the desired conclusion without using the trace. 2 Remarked that the proof holds for higher order Weyl algebras. Assuming that ring morphisms have to take identity elements to identity elements, I think that one can show even more for the first Weyl algebra (the proof for higher order Weyl algebras is analogous), with very basic methods; Claim: For any positive integer n and any unital ring R it is impossible to have a ring morphism$$ \phi : \mathbb{C}\langle x,y\rangle/(yx-xy-1) \to M_n(R).$$Proof: Seeking for a contradiction, suppose that there is some n and some unital ring R such that \phi exists. Denote the images in M_n(R) of x respectively y, under \phi, by A:=\phi(x) respectively B:=\phi(y). The image of 1 will be the identity matrix I. Consider the element yx-xy=1, the image of which, under \phi, is equal to$$ BA-AB=\phi(yx-xy)=\phi(1)=I. $$Hence the matrices A and B have to satisfy BA-AB=I. Taking the trace of the left hand side of this equality yields$$ tr(BA-AB)=tr(BA)-tr(AB)=tr(AB)-tr(AB)=0 $$whereas the trace of the right hand side is equal to tr(I)=n. This is a contradiction. Corollary of the above proof: The same claim holds if we replace M_n(R) by any unital Banach algebra. This is easily seen by using the following well-known fact: The identity element of a unital Banach algebra can not be a commutator, i.e. ab-ba\neq 1 for any elements a,b of the Banach algebra. This applies to the case M_n(R) with R=\mathbb{C}, because M_n(\mathbb{C}) is a unital C*-algebra and in particular a unital Banach algebra. So in this particular case, this gives us an alternative way of making the desired conclusion without using the trace. 1 Assuming that ring morphisms have to take identity elements to identity elements, I think that one can show even more for the first Weyl algebra, with very basic methods; Claim: For any positive integer n and any unital ring R it is impossible to have a ring morphism$$ \phi : \mathbb{C}\langle x,y\rangle/(yx-xy-1) \to M_n(R).$$Proof: Seeking for a contradiction, suppose that there is some n and some unital ring R such that \phi exists. Denote the images in M_n(R) of x respectively y, under \phi, by A:=\phi(x) respectively B:=\phi(y). The image of 1 will be the identity matrix I. Consider the element yx-xy=1, the image of which, under \phi, is equal to$$ BA-AB=\phi(yx-xy)=\phi(1)=I. $$Hence the matrices A and B have to satisfy BA-AB=I. Taking the trace of the left hand side of this equality yields$$ tr(BA-AB)=tr(BA)-tr(AB)=tr(AB)-tr(AB)=0$$whereas the trace of the right hand side is equal to$tr(I)=n$. This is a contradiction. Corollary of the above proof: The same claim holds if we replace$M_n(R)$by any unital Banach algebra. This is easily seen by using the following well-known fact: The identity element of a unital Banach algebra can not be a commutator, i.e.$ab-ba\neq 1$for any elements$a,b$of the Banach algebra. This applies to the case$M_n(R)$with$R=\mathbb{C}$, because$M_n(\mathbb{C})\$ is a unital C*-algebra and in particular a unital Banach algebra. So in this particular case, this gives us an alternative way of making the desired conclusion without using the trace.