Elementary Row Matrix for Inverse of Elementary Row Operation is Inverse
Theorem
Let $e$ be an elementary row operation.
Let $\mathbf E$ be the elementary row matrix corresponding to $e$.
Let $e'$ be the inverse of $e$.
Then the elementary row matrix corresponding to $e'$ is the inverse of $\mathbf E$.
Proof 1
We will demonstrate this for each of the $3$ types of elementary row operation.
Let $\sim$ denote row equivalence.
Let $\mathbf I$ denote the unit matrix.
$\text {ERO} 1$: Scalar Product of Row
Let $e_1$ be the elementary row operation:
- $e_1 := r_i \to \lambda r_i$
for an arbitrary row $r_i$ of $\mathbf I$.
Let $\mathbf E_1$ be the elementary row matrix created by applying $e_1$ to $\mathbf I$.
From Existence of Inverse Elementary Row Operation: Scalar Product of Row, the inverse $e_1'$ of $e_1$ is given by:
- $e_1' := r_i \to \dfrac 1 \lambda r_i$
Thus applying $e_1'$ to $\mathbf E_1$ transforms $\mathbf E_1$ back to $\mathbf I$.
From Elementary Row Operations as Matrix Multiplications, for every elementary row operation there exists a corresponding elementary row matrix.
Let $\mathbf E_1'$ be the elementary row matrix created by applying $e_1'$ to $\mathbf I$.
Then:
- $\mathbf E_1 \mathbf E_1' = \mathbf E_1' \mathbf E_1 = \mathbf I$
$\text {ERO} 2$: Add Scalar Product of Row to Another
Let $e_2$ be the elementary row operation:
- $e_2 := r_i \to r_i + \lambda r_j$
for arbitrary row $r_i$ and $r_j$ of $\mathbf I$ such that $i \ne j$.
Let $\mathbf E_2$ be the elementary row matrix created by applying $e_2$ to $\mathbf I$.
From Existence of Inverse Elementary Row Operation: Add Scalar Product of Row to Another, the inverse $e_2'$ of $e_2$ is given by:
- $e_2' := r_i \to r_i - \lambda r_j$
Thus applying $e_2'$ to $\mathbf E_2$ transforms $\mathbf E_2$ back to $\mathbf I$.
From Elementary Row Operations as Matrix Multiplications, for every elementary row operation there exists a corresponding elementary row matrix.
Let $\mathbf E_2'$ be the elementary row matrix created by applying $e_2'$ to $\mathbf I$.
Then:
- $\mathbf E_2 \mathbf E_2' = \mathbf E_2' \mathbf E_2 = \mathbf I$
$\text {ERO} 3$: Exchange Rows
Let $e_3$ be the elementary row operation:
- $e_3 := r_i \leftrightarrow r_j$
for arbitrary row $r_i$ and $r_j$ of $\mathbf I$ such that $i \ne j$.
Let $\mathbf E_3$ be the elementary row matrix created by applying $e_3$ to $\mathbf I$.
From Existence of Inverse Elementary Row Operation: Exchange Rows, the inverse $e_3'$ of $e_3$ is $e_3$ itself:
- $e_3' := r_i \leftrightarrow r_j = e_3$
Thus applying $e_3$ to $\mathbf E_3$ transforms $\mathbf E_3$ back to $\mathbf I$.
- $\mathbf E_3 \mathbf E_3 = \mathbf I$
Hence the result, from Proof by Cases.
$\blacksquare$
Proof 2
We will demonstrate this for each of the $3$ types of elementary row operation.
In the below:
- $e$ denotes a given elementary row operation
- $\mathbf E$ denotes the elementary row matrix corresponding to $e$
- $e'$ denotes the inverse of $e$
- $\mathbf E'$ denotes the elementary row matrix corresponding to $e'$.
Let $n$ denote the order of $\mathbf E$ and $\mathbf E'$.
The strategy is to demonstrate that:
- $\mathbf E \mathbf E' = \mathbf I$
where $\mathbf I$ denotes the unit matrix of order $n$.
Let $x_{i, j}$ and $y_{i, j}$ denote the elements of $\mathbf E$ and $\mathbf E'$ respectively at indices $\tuple {i, j}$.
Let $z_{i j}$ denote the element of $\mathbf E \mathbf E'$ at indices $\tuple {i, j}$.
$\text {ERO} 1$: Scalar Product of Row
Let $e$ be the elementary row operation:
- $e := r_k \to \lambda r_k$
where $\lambda \ne 0$.
From Elementary Matrix corresponding to Elementary Row Operation, $\mathbf E$ is of the form:
- $x_{a b} = \begin {cases} \delta_{a b} & : a \ne k \\ \lambda \cdot \delta_{a b} & : a = k \end {cases}$
where:
- $\delta_{a b}$ is the Kronecker delta:
- $\delta_{a b} = \begin {cases} 1 & : \text {if $a = b$} \\ 0 & : \text {if $a \ne b$} \end {cases}$
From Existence of Inverse Elementary Row Operation: Scalar Product of Row, $e'$ is the elementary row operation:
- $e' := r_k \to \dfrac 1 \lambda r_k$
From Elementary Matrix corresponding to Elementary Row Operation, $\mathbf E'$ is of the form:
- $y_{a b} = \begin {cases} \delta_{a b} & : a \ne k \\ \dfrac 1 \lambda \cdot \delta_{a b} & : a = k \end {cases}$
By definition of matrix product:
- $\ds \forall a, b \in \set {1, 2, \ldots, n}: z_{a b} = \sum_{p \mathop = 1}^n x_{a p} y_{p b}$
Thus $z_{a b} \ne 0$ if and only if $a = p$ and $b = p$.
When $a \ne k$:
- $x_{a a} = y_{a a} = 1$
and so:
- $z_{a a} = 1 \times 1 = 1$
When $a = k$:
- $x_{a a} = \lambda$, $y_{l b} = \dfrac 1 \lambda$
and so:
- $z_{a a} = \lambda \times \dfrac 1 \lambda = 1$
and for all $z_{a b}$ where $a \ne b$:
- $z_{a b} = 0$
That is:
- $z_{a b} = \delta_{a b}$
and by definition:
- $\mathbf E \mathbf E' = \mathbf I$
$\Box$
$\text {ERO} 2$: Add Scalar Product of Row to Another
Let $e$ be the elementary row operation:
- $e := r_i \to r_i + \lambda r_j$
From Elementary Matrix corresponding to Elementary Row Operation, $\mathbf E$ is of the form:
- $x_{a b} = \delta_{a b} + \lambda \cdot \delta_{a i} \cdot \delta_{j b}$
From Existence of Inverse Elementary Row Operation: Add Scalar Product of Row to Another, $e'$ is the elementary row operation:
- $e' := r_i \to r_i - \lambda r_j$
From Elementary Matrix corresponding to Elementary Row Operation, $\mathbf E'$ is of the form:
- $y_{a b} = \delta_{a b} - \lambda \cdot \delta_{a i} \cdot \delta_{j b}$
| \(\ds \forall a, b \in \set {1, 2, \ldots, n}: \, \) | \(\ds z_{a b}\) | \(=\) | \(\ds \sum_{p \mathop = 1}^n x_{a p} y_{p b}\) | |||||||||||
| \(\ds \) | \(=\) | \(\ds \sum_{p \mathop = 1}^n \paren {\delta_{a p} + \lambda \cdot \delta_{a i} \cdot \delta_{j p} } \cdot \paren {\delta_{p b} - \lambda \cdot \delta_{p i} \cdot \delta_{j b} }\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \sum_{p \mathop = 1}^n \paren {\delta_{a p} \cdot \delta_{p b} + \lambda \cdot \delta_{a i} \cdot \delta_{j p} \cdot \delta_{p b} - \lambda \cdot \delta_{p i} \cdot \delta_{j b} \cdot \delta_{a p} - \lambda \cdot \delta_{a i} \cdot \delta_{j p} \cdot \lambda \cdot \delta_{p i} \cdot \delta_{j b} }\) |
We have that:
| \(\ds \sum_{p \mathop = 1}^n \delta_{a p} \cdot \delta_{p b}\) | \(=\) | \(\ds \delta_{a b}\) | ||||||||||||
| \(\ds \sum_{p \mathop = 1}^n \lambda \cdot \delta_{p i} \cdot \delta_{j b} \cdot \delta_{a p}\) | \(=\) | \(\ds \lambda \cdot \delta_{a i} \cdot \delta_{j b}\) | ||||||||||||
| \(\ds -\sum_{p \mathop = 1}^n \lambda \cdot \delta_{a i} \cdot \delta_{j p} \cdot \delta_{p b}\) | \(=\) | \(\ds -\lambda \cdot \delta_{j b} \cdot \delta_{a i}\) | ||||||||||||
| \(\ds \sum_{p \mathop = 1}^n \lambda \cdot \delta_{a i} \cdot \delta_{j p} \cdot \lambda \cdot \delta_{p i} \cdot \delta_{j b}\) | \(=\) | \(\ds \lambda^2 \sum_{p \mathop = 1}^n \delta_{a i} \cdot \delta_{j p} \cdot \delta_{p i} \cdot \delta_{j b}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \lambda^2 \delta_{a i} \cdot \delta_{j i} \cdot \delta_{j b}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds 0\) | as $i \ne j$ |
But:
- $\lambda \cdot \delta_{j b} \cdot \delta_{a i} - \lambda \cdot \delta_{j b} \cdot \delta_{a i} = 0$
So everything vanishes except $\delta_{a b}$, and so:
- $z_{a b} = \delta_{a b}$
and by definition, again:
- $\mathbf E \mathbf E' = \mathbf I$
$\Box$
$\text {ERO} 3$: Exchange Rows
Let $e$ be the elementary row operation:
- $e := r_i \leftrightarrow r_j$
From Elementary Matrix corresponding to Elementary Row Operation, $\mathbf E$ is of the form:
- $x_{a b} = \begin {cases} \delta_{a b} & : \text {if $a \ne i$ and $a \ne j$} \\ \delta_{j b} & : \text {if $a = i$} \\ \delta_{i b} & : \text {if $a = j$} \end {cases}$
From Existence of Inverse Elementary Row Operation: Exchange Rows, $e'$ is the elementary row operation:
- $e' := r_i \leftrightarrow r_j$
From Elementary Matrix corresponding to Elementary Row Operation, $\mathbf E'$ is of the form:
- $y_{a b} = \begin {cases} \delta_{a b} & : \text {if $a \ne i$ and $a \ne j$} \\ \delta_{j b} & : \text {if $a = i$} \\ \delta_{i b} & : \text {if $a = j$} \end {cases}$
By definition of matrix product:
- $\ds \forall a, b \in \set {1, 2, \ldots, n}: z_{a b} = \sum_{p \mathop = 1}^n x_{a p} y_{p b}$
When $a \ne i$ and $b \ne j$ we have:
| \(\ds z_{a b}\) | \(=\) | \(\ds \sum_{p \mathop = 1}^n \delta_{a p} \delta_{p b}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \delta_{a b}\) |
When $a = i$ and $b \ne i$ we have:
| \(\ds z_{a b}\) | \(=\) | \(\ds \sum_{p \mathop = 1}^n \delta_{j p} \delta_{p b}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds 0\) |
When $a = j$ and $b \ne j$ we have:
| \(\ds z_{a b}\) | \(=\) | \(\ds \sum_{p \mathop = 1}^n \delta_{i p} \delta_{p b}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds 0\) |
When $a = b = i$ we have:
| \(\ds z_{i i}\) | \(=\) | \(\ds \sum_{p \mathop = 1}^n \delta_{j p} \delta_{p j}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \delta_{j j}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds 1\) |
When $a = b = j$ we have:
| \(\ds z_{j j}\) | \(=\) | \(\ds \sum_{p \mathop = 1}^n \delta_{i p} \delta_{p i}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \delta_{i i}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds 1\) |
Hence by definition, again:
- $\mathbf E \mathbf E' = \mathbf I$
$\Box$
Thus in all cases:
- $\mathbf E \mathbf E' = \mathbf I$
Hence the result.
$\blacksquare$