I'm studying proof of "Invertible Matrix Theorem". It says like below
Let $A$ be a square $n\times n$ matrix,
(g) The equation $Ax=b$ has at least one solution for each $b$ in $\Bbb{R}^n$.
(a) $A$ is an invertible matrix(g) implies (a).
I read a solution, but I couldn't understand.
Since $Ax=b$ is consistent, then there is solution for every $Ax=e_i$, where $e_i$ are $n\times 1$ matrices with all zeros except position $(i,1)$ where is $1$.
This means that $A$ can be row reduced to identity matrix - inverse exist.
But why $Ax=e_i$ means $A$ can be row reduced to identity matrix? I tried to understand how can derive it, but I totally don't know. How can I approach?
If $Ax=e_i$ has a solution for each $i$, then $A$ is surjective and has full rank (since the $e_i$ form a basis for the range). Therefore $A$ is invertible and can be row reduced to the identity.