Matrix is Row Equivalent to 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 an echelon matrix of order $m \times n$.

Proof
Using the operation Row Operation to Clear First Column of Matrix, $\mathbf A$ is converted to $\mathbf B$, which will be in the form:


 * $\begin{bmatrix}

0     & \cdots &      0 &      1 & b_{1, j + 1} & \cdots & b_{1 n} \\ 0     & \cdots &      0 &      0 & b_{2, j + 1} & \cdots & b_{2 n} \\ \vdots & \ddots & \vdots & \vdots &      \vdots & \ddots &  \vdots \\ 0     & \cdots &      0 &      0 & b_{m, j + 1} & \cdots & b_{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 then address our attention to the submatrix:


 * $\begin{bmatrix}

b_{2, j + 1} & b_{2, j + 2} & \cdots & b_{2 n} \\ b_{3, j + 1} & b_{3, j + 2} & \cdots & b_{3 n} \\ \vdots &      \vdots & \ddots &  \vdots \\ b_{m, j + 1} & b_{m, j + 2} & \cdots & b_{m n} \\ \end{bmatrix}$

and perform the same operation on that.

This results in the submatrix being transformed into the form:


 * $\begin{bmatrix}

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

Again, we process the submatrix:


 * $\begin{bmatrix}

c_{3, j + 2} & \cdots & c_{3 n} \\ \vdots & \ddots & \vdots \\ c_{m, j + 2} & \cdots & c_{m n} \\ \end{bmatrix}$

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

Also presented as
Some sources use the non-unity variant of the echelon matrix, and so do not seek to ensure that the leading coefficients are equal to $1$.

However, it is often left as an optional step to perform the necessary elementary row operation to multiply each row by the reciprocal of its leading coefficients so as to make it so.