Why are nonsquare matrices not invertible?

I have a theoretical question. Why are non-square matrices not invertible?

I am running into a lot of doubts like this in my introductory study of linear algebra.

I think the simplest way to look at it is considering the dimensions of the Matrices $A$ and $A^{-1 }$ and apply simple multiplication.

So assume, wlog $A$ is $m \times n $, with $n\neq m$ then $A^{-1 }$ has to be $n\times m$ because thats the only way $AA^{-1 }=I_m$

But it must also be true that $A^{-1 } A=I_m$ but now instead of $I_m$ you get $I_n$ wich is not in accordance with the definition of an Inverse ( see ZettaSuro)

Hence $m$ must be equal to $n$

Simple answer: because by definition a matrix is commutative with its inverse on multiplication. That is: $A^{-1}$ is a matrix such that $AA^{-1}=I_n$ and $A^{-1}A=I_n$.

For two matrices to commute on multiplication, both must be square.

More complicated answer: There exists a left inverse and a right inverse that is defined for all matrices including non-square matrices. For a matrix of dimension $m\times{n}$, the left and right inverse are defined as follows:

$$A^L:=\{B|BA=I_n\}$$ $$A^R:=\{B|AB=I_m\}$$

If $A^L=A^R$ , by definition $A^L=A^R=A^{-1}$.

Since this question has just been bumped anyway and I feel like I have something else to add, here are my thoughts:

As pointed out by sigmatau we would have $AA^{-1} = I_m$ and $A^{-1}A = I_n$ and he reasons, that we would have $m=n$ then. I think this is an "unnatural" conclusion: Consider the definition of an inverse of a function $f : A\to B$: It is a function $f^{-1} : B\to A$ with $f^{-1}\circ f = id_A$ and $f\circ f^{-1} = id_B$ and we don't require $id_A = id_B$.

So we can and should ask the question, whether there may be an $A^{-1}$ with $AA^{-1} = I_m$ and $A^{-1}A = I_n$, if $n\neq m$. The answer is no:

The matrix $A$ corresponds to a linear map $f : \mathbb{R}^n \to \mathbb{R}^m, x\mapsto Ax$. By the dimension formula we have:

$n = \dim(\ker f) + \dim(\operatorname{im} f)$.

If $A$ has an inverse, then so does $f$ ($y \mapsto A^{-1}y$). Hence $f$ is injective, so $\dim(\ker f) = 0$ and $f$ is surjective, so $\dim(\operatorname{im} f) = m$. But then $n = 0 + m = m$.