Matrix is Row Equivalent to Reduced Echelon Matrix

Theorem
Let $\mathbf A = \sqbrk a_{m n}$ be a matrix of order $m \times n$ over a field $F$.

Then $A$ is row equivalent to a reduced echelon matrix of order $m \times n$.

Proof
Let the first column of $\mathbf A$ containing a non-zero element be column $j$.

Let such a non-zero element be in row $i$.

Take element $a_{i j} \ne 0$ and perform the elementary row operations:


 * $(1): \quad r_i \to \dfrac {r_i} {a_{i j}}$
 * $(2): \quad r_1 \leftrightarrow r_i$

This gives a matrix with $1$ in the $\tuple {1, j}$ position:


 * $\begin {bmatrix}

0 & \cdots &     0 &       1 & b_{1, j + 1} & \cdots & b_{1 n} \\ 0 & \cdots &     0 & b_{2 j} & b_{2, j + 1} & \cdots & b_{2 n} \\ \vdots & \ddots & \vdots & \vdots &       \vdots & \ddots &  \vdots \\ 0 & \cdots &     0 & b_{m j} & b_{m, j + 1} & \cdots & b_{m n} \\ \end {bmatrix}$

Now the elementary row operations $r_k \to r_k - b_{k j} r_1, k \in \set {2, 3, \ldots, m}$ gives the matrix:


 * $\begin{bmatrix}

0     & \cdots &      0 &      1 & c_{1, j + 1} & \cdots & c_{1 n} \\ 0     & \cdots &      0 &      0 & c_{2, j + 1} & \cdots & c_{2 n} \\ \vdots & \ddots & \vdots & \vdots &      \vdots & \ddots &  \vdots \\ 0     & \cdots &      0 &      0 & c_{m, j + 1} & \cdots & c_{m n} \\ \end{bmatrix}$

If some zero rows have appeared, do some further elementary row operations, that is row interchanges, to put them at the bottom.

We now repeat the process with the remaining however-many-there-are rows:


 * $\begin{bmatrix}

\cdots &     0 &      1 & d_{1, j + 1} & \cdots & d_{1, k - 1} & d_{1 k} & d_{1, k + 1} & \cdots & d_{1 n} \\ \cdots &     0 &      0 &            0 & \cdots &            0 &       1 & d_{2, k + 1} & \cdots & d_{2 n} \\ \cdots &     0 &      0 &            0 & \cdots &            0 & d_{3 k} & d_{3, k + 1} & \cdots & d_{3 n} \\ \ddots & \vdots & \vdots &      \vdots & \ddots &       \vdots &  \vdots &       \vdots & \ddots & \vdots  \\ \cdots &     0 &      0 &            0 & \cdots &            0 & d_{n k} & d_{m, k + 1} & \cdots & d_{m n} \\ \end{bmatrix}$

Then we can get the reduced echelon form by:


 * $r_i \to r_i - d_{i k} r_2, i \in \set {1, 3, 4, \ldots, m}$

as follows:
 * $\begin{bmatrix}

\cdots &     0 &      1 & {e_{1, j + 1 } } & \cdots & {e_{1, k - 1} } &      0 & {e_{1, k + 1} } & \cdots & {e_{1 n} } \\ \cdots &     0 &      0 &                0 & \cdots &               0 &      1 & {e_{2, k + 1} } & \cdots & {e_{2 n} } \\ \cdots &     0 &      0 &                0 & \cdots &               0 &      0 & {e_{3, k + 1} } & \cdots & {e_{3 n} } \\ \ddots & \vdots & \vdots &          \vdots & \ddots &          \vdots & \vdots &          \vdots & \ddots & \vdots     \\ \cdots &     0 &      0 &                0 & \cdots &               0 &      0 & {e_{m, k + 1} } & \cdots & {e_{m n} } \\ \end{bmatrix}$

Thus we progress, until the entire matrix is in reduced echelon form.