Why is inverse multiplication shown as commutative?

Question

Can someone explain why this is shown as commutative when matrix multiplication is not? Is this a specific case with inverses?

"A square $(n\times n)$ matrix $A$ is said to have an inverse $A^{-1}$ iff $(A\times A^{-1})=(A^{-1}\times A)=I$. In this case, the matrix $A$ is called invertible."

I thought at first it may be because it is a square, but I cannot render why square would be different.
If it is because it is square, then all inverses are square matrices, so then, are all inverse relations commutative?

Yes, for square matrices if you know that $AB=I$ then you immediately know that $BA=I$. — Artem, May 20 '23 at 20:39
Not all matrix multiplications commute. Also, we have the notion of "left inverse" and "right inverse" when we look at group structure. — Sean Roberson, May 20 '23 at 20:39
Such commutativity is by hypothesis, i.e. that's the definition of a two-sided inverse (vs. one-sided, i.e. left or right inverse). Often "two sided" is ommitted, i.e. "inverse" means "two-sided inverse". — Bill Dubuque, May 20 '23 at 22:40
It's easy to prove that if both a left and right inverse exist then they must be equal. But generally $,AB=1\Rightarrow BA=1,$ fails, for example let $B$ be a shift map $,(v_1,v_2,v_3,\cdots,)\to(0,v_1,v_2,v_3,\cdots,),$ on a infinite dimension vector space. — Bill Dubuque, May 21 '23 at 00:07

GEdgar · Answer 1 · 2023-05-20T21:26:39.403

The definition quoted

$(*)\quad$if $A B=B A=I$, then $B$ is called an inverse of $A$

In that definition, we postulate that multiplication in both orders yields $I$. So (as far as the definition is concerned) if $BA=I$ it could happen that $AB \ne I$, and then $B$ would not be "an inverse of $A$".

As some of the comments have indicated, in fact for square matrices, from $AB=I$ we may deduce $BA=I$. So definition $(*)$ is stronger than needed. But in some other similar settings (but not square matrices) we can, indeed have $BA=I$ but $AB \ne I$. So it is perhaps educational to include the extra condition of $(*)$ even in this case.

Why is inverse multiplication shown as commutative?

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