Inverse of finite squared matrices. I've usually used that given a square matrix $A$ with determinant $\det(A)\neq0$, then its inverse $A^{-1}$ is the matrix that meets:

$$A^{-1}A=\mathbb{I}$$

and

$$AA^{-1}=\mathbb{I}.$$

However, I've found in several sources, including Wikipedia that if $A$ is finite and square, then:

$$A^{-1}A=\mathbb{I} \Longrightarrow AA^{-1}=\mathbb{I}.$$

I'm struggling to find a proof. Does anybody know one?
If $f, \, g \; : \; \mathbb{R}^{n} \; \longrightarrow \; \mathbb{R}^{n}$ are two linear maps such that : $f \circ g = \mathrm{Id}$ then, it implies that :


$f$ is invertible
$g$ is invertible
$g \circ f = \mathrm{Id}$. 


To prove that $f$ and $g$ are invertible, you only need to prove that $\mathrm{ker}(f) = \lbrace 0 \rbrace$ and $\mathrm{ker}(g)=\lbrace 0 \rbrace$. Let $x \in \mathrm{ker}(g)$. Then, $(f \circ g)(x) = x$ and $(f \circ g)(x) = f(0)=0$. So, $x=0$ and it follows that $\mathrm{ker}(g) = \lbrace 0 \rbrace$ and $g$ is invertible. You can prove as well that $\mathrm{ker}(f) = \lbrace 0 \rbrace$ (so $f$ is invertible).

Let $x \in \mathbb{R}^{n}$. Since $g$ is invertible, there exist $u \in \mathbb{R}^{n}$ such that $x = g(u)$. Therefore, $(g \circ f)(x) = g\big( (f \circ g)(u) \big) = g(u) = x$. As a consequence, $g \circ f = \mathrm{Id}$.
If $A^{-1}A$ is the identity, this means that $A$ must be injective. Since we are in finite dimension and $A$ is square, this means that $A$ is surjective too. So it is invertible: let's call $B$ its inverse. Then, if you have $CA=Id$, it follows that $CAB=B$ and then $C=B$, which is what you wanted: $C$ is actually the inverse af $A$.

Let me remind you that this applies only in finite dimension. In infinite dimension you can very easily have linear operators that are injective but not surjective (or surjective, but not injective).