Matrix is Row Equivalent to Echelon Matrix

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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.


Sources