I was going through Artin's Algebra and in Theorem 1.2.21 on Square Systems, Artin states that if A is an invertible Square matrix, then its row echelon form A' is an identity matrix. How does one arrive at that conclusion? Can anyone provide a proof of the statement?
Thanks in advance.
Suppose $A$ is an $n\times n$ matrix. Form the $n\times 2n$ augmented matrix $[A|I]$ and row reduce it to reduced row echelon form using elementary matrices: $$E_k\cdots E_1\cdot [A|I]=[A'|B]$$
Now, either $A'$ is $I$ or $A'$ is not $I$.
If $A'$ is not $I$, then there must be at least one zero row in $A'$, or in other words, there are strictly less than $n$ pivotal columns in $A'$. This means that $A'$ is not invertible, by rank-nullity. Since $A$ and $A'$ are row equivalent, then $A$ is not invertible.
If $A'=I$, then we have $$E_k\cdots E_1\cdot A= I\text{ and }E_k\cdots E_1\cdot I=B$$Since each elementary matrix is invertible, and the product of invertible matrices is invertible, then $B$ is invertible. Now, we get that $BA=I$, which implies that $A$ is invertible.
If $A$ is invertible, then we cannot have the first case, so we must have the second case.