Two non-invertible matrices $A$ and $B$ such that $AB = BA= I$ [duplicate]

I know that a square matrix $A$ is called invertible if there exists $B$ such that $AB = BA = I$.

But is it possible to find two matrices $A_{m \times n}$ and $B_{n \times m}$ such that $AB =I_m$ and $BA = I_n$ ?

(where $m \neq n$)

EDIT : I just realized that if $A$ and $B$ are matrices over a field (like $\mathbb{R}$), then $m=n$. But after reading the comments, I am now interested in knowing what conditions will confirm the presence of such an example where $m \neq n$.

Solution 1:

As explained in the comments, this is not possible if the coefficients of your matrices live in a commutative ring with unit.

However, it is possible if the base ring is noncommutative.

As an explicit example, let $R={\rm End}_K(K[X])$, where $K$ is an arbitrary field.

Note that $R$ is a noncommutative ring with unit, where $1_R=Id$, and product is composition of maps (and $0_R$ is the zero map).

Let $x,y,\alpha,\beta:K[X]\to K[X]$ the $K$-linear maps defined by:

$$x(\sum_{n\geq 0}a_nX^n)=\sum_{n\geq 0}a_{2n+1}X^n, \ y(\sum_{n\geq 0}a_nX^n)=\sum_{n\geq 0}a_{2n}X^n$$

$$\alpha(\sum_{n\geq 0}a_nX^n)=\sum_{n\geq 0}a_nX^{2n+1}, \ \beta(\sum_{n\geq 0}a_nX^n)=\sum_{n\geq 0}a_nX^{2n}$$

I let you check that we have the following identities in the ring $R$:

• $\alpha x+\beta y=1$

• $x\alpha=1$ and $x\beta=0$

• $y\alpha=0$ and $y\beta=1$

Now set $A=\begin{pmatrix} \alpha & \beta\end{pmatrix}\in M_{1\times 2}(R)$ and $B=\begin{pmatrix}x \cr y\end{pmatrix}\in M_{2\times 1}(R)$. Then, the relations above imply that $AB=I_1$ and $BA=I_2$.