Condition for invertibility of matrices Let $A \in \mathbb{F}^{n\times n}$ be a $n\times n$-matrix over the field $\mathbb{F}$ and define the function $f: \mathbb{F}^{n\times 1} \rightarrow \mathbb{F}^{n\times 1}, x \mapsto Ax$.
I'm wondering if my following reasoning is correct:
If $f$ is bijective, $A$ is invertible.
Proof.
$f$ is bijective, so there exists an unique inverse $f^{-1}$.
Show that $f^{-1}: x \mapsto A^{-1}x$ is the inverse of $f$.
Let $x \in \mathbb{F}^{n\times 1}$ be chosen freely.
(i) $(f \circ f^{-1})(x)= f(f^{-1}(x))=f(A^{-1}x)=AA^{-1}x=I_{n\times n}\cdot x=x=\text{id}_{\mathbb{F}^{n\times 1}}(x)$.
(ii) $(f^{-1} \circ f)(x)= f^{-1}(f(x))=f^{-1}(Ax)=A^{-1}Ax=I_{n\times n}\cdot x=x=\text{id}_{\mathbb{F}^{n\times 1}}(x)$.
$f^{-1}$ is the inverse of $f$, so $A^{-1}$ exists meaning $A$ is invertible.
What I'm confused about is the general logic of the proof, more specifically the fact that I used $A^{-1}$ (an object whose existence I'm trying to show!) for my definition of $f^{-1}$. Am I allowed to do that? It feels backwards but on the other hand, the inverse if existent is unique, so if I can define a function that fulfills the criteria of an inverse it is the inverse, so $A^{-1}$ should exist.
 A: Indeed there is a problem you don't know that $A^{-1}$ exists so you can't write it down before you proved it is invertible.
You can get around the problem by replacing $A^{-1}$ by another matrix $B$ and deducing that $BA = AB = I$ however to do this you must first prove that $f^{-1}$ is linear or else you cant be sure that there exists a matrix $B$ such that $f^{-1}x = Bx$.
A simpler method is to avoid all this and simply prove that the all linear systems have a unique solution:
$$ Ax = b \iff f(x) = b \iff x = f^{-1}({b)}$$
which proves the invertibility of $A$.
A: Yeah, there is something wrong at logical level in your proof. The fact is that here you can use different approaches depending of what you know exactly about Linear Algebra.
Let me explain better. If you already know that a matrix is invertible if and only if the homogeneous system $Ax=0$ admits only the trivial solution, i.e. the kernel of its linear application (that in your case is $f$) is trivial, then you are done directly. This fact holds simply by using Gauss algorithm.
Let us suppose you just know that a linear map $g\colon \mathbb F^{n}\to \mathbb F^n$ is identified by its associate matrix $B$, i.e. $B$ is the matrix such that $g(x)=Bx$ for any $x\in \mathbb F^n$.
If $f$ is bijective, then $f$ admits an inverse map, let us denote it by $g:=f^{-1}$. Now you can observe that the inverse map of a linear map is linear too:
$$f(g(x+y))=x+y=f(g(x))+f(g(y))=f((g(x)+g(y)) \implies g(x+y)=g(x)+g(y).$$
Therefore $g$ is linear and it has an associate matrix, that we denote by $B$. We claim that $B$ is the inverse of $A$. In fact, you observe that
$BAx=B(f(x))=g(f(x))=x$ for each $x\in \mathbb F^n$.
If yo call by $e_1, \dots e_n$ the canonical basis of $\mathbb F^n$, then you have that $BAe_j=e_j$, and so $BA=I$, that means $B$ is the inverse matrix of $A$ and that $A$ is invertible.
