I'm wondering why $\text{rank}(A)=n$ means $A$ is invertible.
Since invertible means one-to-one and onto, we have to prove that.
$\text{rank}(A)=n$ means $\dim(N(A))=0$ which means one-to-one.
Now, we have to show onto.
$\text{rank}(A)=\dim(R(A))=n$ and it means onto.
So we complete the proof.
Is that right?
[ADDITION]
I want to check this:
$\text{rank}(A)=\text{rank}(L_A)=\dim(R(L_A))=n$ and it means onto?
I mean, range and codomain have same dimension means they are same?
I know even though they have same dimension but still their spaces could be different, like row space and column space.
But the above statement "range and codomain have the same dimension means they are same" is true?
I solve it. It is true since $R(L_A)$ is a subspace of $F^n$.
Refer to theorem $1.11$ in textbook "linear algebra" by friedberg.
Best Answer
If $rank(A)=n$ then $A$ is an equivalent matrix to $I_{n\times n}$. This means that there are some elementary matrices $E_1,E_2,...,E_s$ such that $E_s...E_2E_1A=I$ and so $$A=E_1^{-1}...E_s^{-1}I$$ But $E_i^{-1}$ are also invertible as $I$ so $A$ is invertible as well.