# 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.

$\blacksquare$

## Examples

### Arbitrary Matrix $1$

Let $\mathbf A = \begin {bmatrix} 0 & 1 & 1 \\ 0 & 1 & 0 \\ 1 & 1 & 0 \\ \end {bmatrix}$

This can be converted into the echelon form:

$\mathbf E = \begin {bmatrix} 1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end {bmatrix}$

### Arbitrary Matrix $2$

Let $\mathbf A = \begin {bmatrix} 1 & 1 & 1 & 1 \\ 2 & 3 & 4 & 5 \\ 3 & 4 & 5 & 6 \\ \end {bmatrix}$

This can be converted into the echelon form:

$\mathbf E = \begin {bmatrix} 1 & 1 & 1 & 1 \\ 0 & 1 & 2 & 3 \\ 0 & 0 & 0 & 0 \\ \end {bmatrix}$

### Arbitrary Matrix $3$

Let $\mathbf A = \begin {bmatrix} 1 & 2 & 3 & 5 \\ 1 & 2 & 3 & 4 \\ 0 & 0 & 1 & 1 \\ \end {bmatrix}$

This can be converted into the echelon form:

$\mathbf E = \begin {bmatrix} 1 & 2 & 3 & 5 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 1 \\ \end {bmatrix}$

### Arbitrary Matrix $4$

Let $\mathbf A = \begin {bmatrix} 1 & 1 & 2 & 3 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 3 & 3 \\ \end {bmatrix}$

This can be converted into the echelon form:

$\mathbf E = \begin {bmatrix} 1 & 1 & 2 & 3 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 0 \\ \end {bmatrix}$

### Arbitrary Matrix $5$

Let $\mathbf A = \begin {bmatrix} -1 & 0 & 1 & 2 & 3 \\ 0 & 1 & 2 & 3 & 4 \\ -1 & -2 & -3 & -4 & -5 \\ \end {bmatrix}$

This can be converted into the echelon form:

$\mathbf E = \begin {bmatrix} 1 & 0 & -1 & -2 & -3 \\ 0 & 1 & 2 & 3 & 4 \\ 0 & 0 & 0 & 0 & 0 \\ \end {bmatrix}$

## 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.