Row Operation to Clear First Column of Matrix

Theorem
Let $\mathbf A = \sqbrk a_{m n}$ be an $m \times n$ matrix over a field $K$.

Then there exists a row operation to convert $\mathbf A$ into another $m \times n$ matrix $\mathbf B = \sqbrk b_{m n}$ with the following properties:


 * $(1): \quad$ Except possibly for element $b_{1 1}$, all the elements of column $1$ are $0$


 * $(2): \quad$ If $b_{1 1} \ne 0$, then $b_{1 1} = 1$.

This process is referred to as clearing the first column.

Proof
The following algorithm generates a sequence of elementary row operations which convert $\mathbf A$ to $\mathbf B$.

Let $\mathbf A' = \sqbrk {a'}_{m n}$ denote the state of $\mathbf A$ after having processed the latest step.

After each step, an implicit step can be included that requires that the form of $\mathbf A'$ is inspected to see if it is in the form $\mathbf B$, and if so, terminating the algorithm, but this is not essential.


 * $(1): \quad$ Are all elements in the first column of $\mathbf A$ equal to $0$?
 * If so, there is nothing to do, and the required row operation is the unit matrix $\mathbf I_m$.
 * Otherwise, move on to step $(2)$.


 * $(2): \quad$ Is element $a_{1 1}$ equal to $0$?
 * If so:
 * $\text (a): \quad$ find the smallest $k$ such that row $k$ of $\mathbf A$ such that $a_{k 1} \ne 0$
 * $\text (b): \quad$ use the elementary row operation $r_1 \leftrightarrow r_k$ which will result $a'_{1 1} = a_{k 1}$ and $a'_{k 1} = 0$.
 * Move on to step $(3)$.


 * $(3): \quad$ Is element $a'_{1 1}$ equal to $1$?
 * If so, use the elementary row operation $r_1 \to \lambda r_1$ where $\lambda = \dfrac 1 {a'_{1 1} }$, which will result $a'_{1 1} = 1$.
 * Move on to step $4$


 * $(4): \quad$ For each row $j$ from $2$ to $m$, do the following:
 * Is $a_{j 1} \ne 0$?
 * If so, use the elementary row operation $r_j \leftrightarrow r_j + \mu r_1$, where $\mu = -\dfrac {a'_{j 1} } {a'{1 1} }$, which will result in $a'_{j 1} = 0$.

This will result in an $m \times n$ matrix in the required form.

Exercising the above algorithm will have generated a sequence of elementary row operations $e_1, e_2, \ldots, e_t$.

For each $e_k$ we create the elementary matrix $\mathbf E_k$.

We then assemble the matrix product:
 * $\mathbf R := \mathbf E_t \mathbf E_{t - 1} \mathbf E_{t - 2} \dotsm \mathbf E_2 \mathbf E_1$

From Row Operation is Equivalent to Product of Elementary Matrices, $\mathbf R$ is the resulting $m \times m$ matrix corresponding to the row operation which is used to convert $\mathbf A$ to $\mathbf B$.

Also presented as
Some sources are not concerned about making $b_{1 1}$ equal to $1$.

Hence they do not use step $3$ of the algorithm.