TITLE: Can we prove that matrix multiplication by its inverse is commutative?
QUESTION [6 upvotes]: We know that $AA^{-1} = I$ and $A^{-1}A = I$, but is there a proof for the commutative property here? Or is this just the definition of invertibility?

REPLY [3 votes]: If $A$ and $B$ are square matrices in $\mathbb R^{n\times n}$ such that $AB=I$, then we can prove that $BA=I$ too.
One way to see this is to consider the $n$ column vectors $B\mathbf e_1, B\mathbf e_2, \ldots, B\mathbf e_n$, where $e_i$s are the standard basis for $\mathbb R^n$. The $B\mathbf e_i$s must be linearly independent (because if we have a linear combination of them, we can multiply that from the left by $A$ and get a linear combination of $\mathbf e_i$s), and any linearly independent set of $n$ vectors is a basis for $\mathbb R^n$.
Now consider an arbitrary column vector $Y\in\mathbb R^n$. We can write $Y$ as a linear combination of the $B\mathbf e_i$s (because they form a basis). Let $X$ be the same linear combination of $\mathbf e_i$s; by linearity we have $BX=Y$. We can now calculate
$$ (BA)Y=(BA)(BX)=B(AB)X=BIX=BX=Y $$
In other words, left multiplication by a $BA$ is the identity, and the only matrix with that property is $I$, so $BA=I$.