Prove that det($A$) is non-zero iff $A$ is row equivalent to the $n\times n$ identity matrix

Question

$A$ is an $n\times n$ matrix. Now if the row-reduced echelon form for this $A$ is $E$ then after all the row operations we have $\det(A)=M\det(E)$ where $M$ is a non-zero scalar from the field. If $E$ is the $n\times n$ identity matrix then $\det(A) \neq 0$. For the converse, $\det(A)$ is non-zero, so is $M$. Then obviously $\det(E)\neq0$. How to reach the conclusion that $E$ is the identity matrix,from here just need a little help with that.

Answer 1

You have reduced the proof to this: If $E$ is an $n\times n$ matrix in reduced row echelon form, and $\det(E) \neq0$, then $E$ is the identity matrix. To show this, I think you can work directly from the definition of RREF.

The leading coefficient in the first row must be in the first column. Otherwise $e_{11} = 0$ and $\det(E)$ would be zero. So $a_{11} = 1$ and there are only zeroes below it.

Suppose now that $e_{11} = e_{22} = \dots e_{kk} = 1$, and all entries above and below these are zero. That is, $E$ has a block decomposition $$ E = \begin{pmatrix} I_k & * \\ 0 & E' \end{pmatrix} $$ where $I_k$ is the $k\times k$ identity matrix, and $E'$ is a $(n-k)\times(n-k)$ matrix in RREF. By the same argument as in the previous paragraph, the leading coefficient of the first row of $E'$ must occur in the first position. Therefore $e_{k+1,k+1} = 1$, and all entries above and below it are zero.

Therefore by induction $e_{ii} = 1$ for all $i$ from $1$ to $n$, and all entries above and below these diagonal entries are zero. So $E$ is the identity matrix.