So, here's the result that I'm trying to prove:
Let $A \in M(n \times n,F)$. Then, $rank(A) = n$ if and only if $A$ is invertible.
Proof Attempt:
We will first prove the backwards direction. Let $A$ be an invertible matrix and consider the linear map $f:F^n \to F^n$ associated with it. It is going to be an isomorphism, since $A$ is invertible.
Now, consider $Ker(f)$. Let $x \in Ker(f)$. Then:
$$f(x) = 0 = f(0)$$
By injectivity, that means that $Ker(f) = \{0\}$. So, we have:
$$\dim(Ker(f)) + rank(f) = n$$
$$\implies rank(f) = n$$
Now, suppose that $rank(A) = n$. Then, all we have to do is to show that the associated map $f$ is bijective. This will prove invertibility.
$$rank(f) = n \implies dim(Ker(f)) = 0$$
So, $Ker(f) = \{0\}$. Now, let $f(x) = f(y)$ for some $x,y \in F^n$. Then:
$$f(x)-f(y) = 0$$
$$f(x-y) = 0 \implies x-y \in Ker(f)$$
$$\implies x = y$$
This shows injectivity. To prove surjectivity, we note that the columns of the matrix $A$ are the images of the basis vectors of $F^n$. Let $(v_1,v_2,\ldots,v_n)$ be a basis of $F^n$.
Since $rank(A) = n$, it holds that $(f(v_1),f(v_2),\ldots,f(v_n))$ is a linearly independent list of vectors. I also claim that this is a basis for $F^n$. If it were not so, then $n < \dim(F^n) = n$. That would be a contradiction.
Let $w \in F^n$. Then, we have:
$$w = \sum_{k = 1}^{n} a_k f(v_k)$$
$$\implies w = f[\sum_{k=1}^{n} a_k v_k]$$
$$ \implies w \in Im(f)$$
This proves surjectivity. Since $f$ is surjective and injective, it follows that it is an isomorphism and the corresponding matrix $A$ is invertible.
This proves the desired result.
Does my proof above work? If it doesn't, how could i improve it?
Best Answer
Your proof sounds good. I will provide here an alternative way to approach the implication ($\Leftarrow$).
Let $A = [a^{T}_{1},a^{T}_{2},\ldots,a^{T}_{n}]$, where each $a^{T}_{j}\in\textbf{F}^{n}$ represents the $j$-th column of $A$. Consequently, given $\textbf{x} = (x_{1},x_{2},\ldots,x_{n}) \in\textbf{F}^{n}$, one has that \begin{align*} A\textbf{x} = x_{1}a^{T}_{1} + x_{2}a^{T}_{2} + \ldots + x_{n}a^{T}_{n} = 0 \end{align*} Since $A$ is invertible, it results that $\textbf{x} = A^{-1}0 = 0$, that is to say, the set $\{a^{T}_{1},a^{T}_{2},\ldots,a^{T}_{n}\}$ is LI, from whence we conclude that $\text{rank}(A) = n$, as desired.
The converse implication $(\Rightarrow)$ can be proven using the suggestion of @Paul, since a linear operator (defined on a finite dimensional vector space) is invertible iff it is nonsingular iff it is surjective.