Matrix is Row Equivalent to Reduced Echelon Matrix

Theorem
Let $\mathbf A = \left[{a}\right]_{m n}$ be an $m \times n$ matrix over a field $F$.

Then $A$ is row equivalent to an $m \times n$ reduced echelon matrix.

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 $\left({1, j}\right)$ position:


 * $\begin{bmatrix}

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

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

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

If some zero rows have appeared, do some 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_{1k} & d_{1,k+1} & \cdots & d_{1n} \\ \cdots & 0 & 0 & 0 & \cdots & 0 & 1 & d_{2,k+1} & \cdots & d_{2n} \\ \cdots & 0 & 0 & 0 & \cdots & 0 & d_{3k} & d_{3,k+1} & \cdots & d_{3n} \\ \ddots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & \vdots \\ \cdots & 0 & 0 & 0 & \cdots & 0 & d_{nk} & d_{m,k+1} & \cdots & d_{mn} \end{bmatrix}$

Then we can get the reduced echelon form by:


 * $r_i \to r_i - d_{ik} r_2, i \in \left\{{1, 3, 4, \ldots, m}\right\}$

as follows:
 * $\begin{bmatrix}

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

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