Existence of Inverse Elementary Row Operation
Theorem
Let $\map \MM {m, n}$ be a metric space of order $m \times n$ over a field $K$.
Let $\mathbf A \in \map \MM {m, n}$ be a matrix.
Let $\map e {\mathbf A}$ be an elementary row operation which transforms $\mathbf A$ to a new matrix $\mathbf A' \in \map \MM {m, n}$.
Let $\map {e'} {\mathbf A'}$ be the inverse of $e$.
Then $e'$ is an elementary row operation which always exists and is unique.
Proof
Let us take each type of elementary row operation in turn.
For each $\map e {\mathbf A}$, we will construct $\map {e'} {\mathbf A'}$ which will transform $\mathbf A'$ into a new matrix $\mathbf A'' \in \map \MM {m, n}$, which will then be demonstrated to equal $\mathbf A$.
In the below, let:
- $r_k$ denote row $k$ of $\mathbf A$
- $r'_k$ denote row $k$ of $\mathbf A'$
- $r''_k$ denote row $k$ of $\mathbf A''$
for arbitrary $k$ such that $1 \le k \le m$.
By definition of elementary row operation:
- only the row or rows directly operated on by $e$ is or are different between $\mathbf A$ and $\mathbf A'$
and similarly:
- only the row or rows directly operated on by $e'$ is or are different between $\mathbf A'$ and $\mathbf A''$.
Hence it is understood that in the following, only those rows directly affected will be under consideration when showing that $\mathbf A = \mathbf A''$.
$\text {ERO} 1$: Scalar Product of Row
Let $\map e {\mathbf A}$ be the elementary row operation:
- $e := r_k \to \lambda r_k$
where $\lambda \ne 0$.
Then $r'_k$ is such that:
- $\forall a'_{k i} \in r'_k: a'_{k i} = \lambda a_{k i}$
Now let $\map {e'} {\mathbf A'}$ be the elementary row operation which transforms $\mathbf A'$ to $\mathbf A''$:
- $e' := r_k \to \dfrac 1 \lambda r_k$
Because it is stipulated in the definition of an elementary row operation that $\lambda \ne 0$, it follows by definition of a field that $\dfrac 1 \lambda$ exists.
Hence $e'$ is defined.
So applying $e'$ to $\mathbf A'$ we get:
\(\ds \forall a''_{k i} \in r''_k: \, \) | \(\ds a''_{k i}\) | \(=\) | \(\ds \dfrac 1 \lambda a'_{k i}\) | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac 1 \lambda \paren {\lambda a_{k i} }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds a_{k i}\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \forall a''_{k i} \in r''_k: \, \) | \(\ds a''_{k i}\) | \(=\) | \(\ds a_{k i}\) | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds r''_k\) | \(=\) | \(\ds r_k\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \mathbf A''\) | \(=\) | \(\ds \mathbf A\) |
It is noted that for $e'$ to be an elementary row operation, the only possibility is for it to be as defined.
$\Box$
$\text {ERO} 2$: Add Scalar Product of Row to Another
Let $\map e {\mathbf A}$ be the elementary row operation:
- $e := r_k \to r_k + \lambda r_l$
Then $r'_k$ is such that:
- $\forall a'_{k i} \in r'_k: a'_{k i} = a_{k i} + \lambda a_{l i}$
Now let $\map {e'} {\mathbf A'}$ be the elementary row operation which transforms $\mathbf A'$ to $\mathbf A''$:
- $e' := r'_k \to r'_k - \lambda r'_l$
Applying $e'$ to $\mathbf A'$ we get:
\(\ds \forall a''_{k i} \in r''_k: \, \) | \(\ds a''_{k i}\) | \(=\) | \(\ds a'_{k i} - \lambda a'_{l i}\) | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {a_{k i} + \lambda a_{l i} } - \lambda a'_{l i}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \paren {a_{k i} + \lambda a_{l i} } - \lambda a_{l i}\) | as $\lambda a'_{l i} = \lambda a_{l i}$: row $l$ was not changed by $e$ | |||||||||||
\(\ds \) | \(=\) | \(\ds a_{k i}\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \forall a''_{k i} \in r''_k: \, \) | \(\ds a''_{k i}\) | \(=\) | \(\ds a_{k i}\) | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds r''_{k i}\) | \(=\) | \(\ds r_{k i}\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \mathbf A''\) | \(=\) | \(\ds \mathbf A\) |
It is noted that for $e'$ to be an elementary row operation, the only possibility is for it to be as defined.
$\Box$
$\text {ERO} 3$: Exchange Rows
Let $\map e {\mathbf A}$ be the elementary row operation:
- $e := r_k \leftrightarrow r_l$
Thus we have:
\(\ds r'_k\) | \(=\) | \(\ds r_l\) | ||||||||||||
\(\, \ds \text {and} \, \) | \(\ds r'_l\) | \(=\) | \(\ds r_k\) |
Now let $\map {e'} {\mathbf A'}$ be the elementary row operation which transforms $\mathbf A'$ to $\mathbf A''$:
- $e' := r'_k \leftrightarrow r'_l$
Applying $e'$ to $\mathbf A'$ we get:
\(\ds r''_k\) | \(=\) | \(\ds r'_l\) | ||||||||||||
\(\, \ds \text {and} \, \) | \(\ds r''_l\) | \(=\) | \(\ds r'_k\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds r''_k\) | \(=\) | \(\ds r_k\) | |||||||||||
\(\, \ds \text {and} \, \) | \(\ds r''_l\) | \(=\) | \(\ds r_l\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \mathbf A''\) | \(=\) | \(\ds \mathbf A\) |
It is noted that for $e'$ to be an elementary row operation, the only possibility is for it to be as defined.
$\Box$
Thus in all cases, for each elementary row operation which transforms $\mathbf A$ to $\mathbf A'$, we have constructed the only possible elementary row operation which transforms $\mathbf A'$ to $\mathbf A$.
Hence the result.
$\blacksquare$
Also see
Sources
- 1982: A.O. Morris: Linear Algebra: An Introduction (2nd ed.) ... (previous) ... (next): Chapter $1$: Linear Equations and Matrices: $1.3$ Applications to Linear Equations: Theorem $1.7$