Matrix is Row Equivalent to Reduced Echelon Matrix

Theorem
Every $m \times n$ matrix is row equivalent to an $$m \times n$$ reduced echelon matrix.

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

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) $$r_i \to \frac {r_i} {a_{i j}}$$
 * 2) $$r_1 \leftrightarrow r_i$$

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.