Existence of Inverse Elementary Row Operation

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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

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