An elementary matrix is one you can get by doing a single row operation to an identity matrix.
The elementary matrix $\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$ results from doing the row operation ${\text{\mathbf{r}}}_{1}\leftrightarrow {\text{\mathbf{r}}}_{2}$ to ${I}_{2}$.
The elementary matrix $\left(\begin{array}{ccc}1& 2& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right)$ results from doing the row operation ${\text{\mathbf{r}}}_{1}\mapsto {\text{\mathbf{r}}}_{1}+2{\text{\mathbf{r}}}_{2}$ to ${I}_{3}$.
The elementary matrix $\left(\begin{array}{cc}-1& 0\\ 0& 1\end{array}\right)$ results from doing the row operation ${\text{\mathbf{r}}}_{1}\mapsto (-1){\text{\mathbf{r}}}_{1}$ to ${I}_{2}$.
Doing a row operation $r$ to a matrix has the same effect as multiplying that matrix on the left by the elementary matrix corresponding to $r$:
Let $r$ be a row operation and $A$ an $m\mathrm{\times}n$ matrix. Then $r\mathit{}\mathrm{(}A\mathrm{)}\mathrm{=}r\mathit{}\mathrm{(}{I}_{m}\mathrm{)}\mathit{}A$.
We will use the fact that matrix multiplication happens rowwise. Specifically, we use Proposition 3.2.5 which says that if the rows of $A$ are ${\mathbf{s}}_{1},\mathrm{\dots},{\mathbf{s}}_{m}$ and if $\mathbf{r}=\left(\begin{array}{ccc}{a}_{1}& \mathrm{\cdots}& {a}_{m}\end{array}\right)$ is a row vector then
$$\mathbf{r}A={a}_{1}{\mathbf{s}}_{1}+\mathrm{\cdots}+{a}_{m}{\mathbf{s}}_{m}$$ |
and Theorem 3.2.6, which tells us that the rows of $r({I}_{m})A$ are ${\mathbf{r}}_{1}A$, …, ${\mathbf{r}}_{m}A$ where ${\mathbf{r}}_{j}$ is the $j$th row of $r({I}_{m})$. We deal with each row operation separately.
Let $r$ be ${\text{\mathbf{r}}}_{j}\mapsto {\text{\mathbf{r}}}_{j}+\lambda {\text{\mathbf{r}}}_{i}$. Row $j$ of $r({I}_{m})$ has a 1 in position $j$, a $\lambda $ in position $i$, and zero everywhere else, so by the Proposition mentioned above
$${\mathbf{r}}_{j}A={\mathbf{s}}_{j}+\lambda {\mathbf{s}}_{i}.$$ |
For ${j}^{\prime}\ne j$, row ${j}^{\prime}$ of $r({I}_{m})$ has a 1 at position ${j}^{\prime}$ and zeroes elsewhere, so
$${\mathbf{r}}_{{j}^{\prime}}A={\mathbf{s}}_{{j}^{\prime}}.$$ |
The theorem mentioned above tells us that these are the rows of $r({I}_{m})A$, but they are exactly the result of doing $r$ to $A$.
Let $r$ be ${\text{\mathbf{r}}}_{j}\mapsto \lambda {\text{\mathbf{r}}}_{j}$. Row $j$ of $r({I}_{m})$ has a $\lambda $ in position $j$ and zero everywhere else, so
$${\mathbf{r}}_{j}A=\lambda {\mathbf{s}}_{j}.$$ |
For ${j}^{\prime}\ne j$, row ${j}^{\prime}$ of $r({I}_{m})$ has a 1 at position ${j}^{\prime}$ and zeroes elsewhere, so
$${\mathbf{r}}_{{j}^{\prime}}A={\mathbf{s}}_{{j}^{\prime}}.$$ |
As before, these are the rows of $r({I}_{m})A$ and they show that this is the same as the result of doing $r$ to $A$.
Let $r$ be ${\text{\mathbf{r}}}_{i}\leftrightarrow {\text{\mathbf{r}}}_{j}$. Row $i$ of $r({I}_{m})$ has a 1 in position $j$ and zeroes elsewhere, and row $j$ of $r({I}_{m})$ has a 1 in position $i$ and zeroes elsewhere, so rows $i$ and $j$ of $r({I}_{m})A$ are given by
${\mathbf{r}}_{i}A$ | $={\mathbf{s}}_{j}$ | ||
${\mathbf{r}}_{j}A$ | $={\mathbf{s}}_{i}.$ |
As in the previous two cases, all other rows of $r({I}_{m})A$ are the same as the corresponding row of $A$. The result follows.
∎
Elementary matrices are invertible.
Let $r$ be a row operation, $s$ be the inverse row operation to $r$, and let ${I}_{n}$ an identity matrix. By Theorem 3.8.1, $r({I}_{n})s({I}_{n})=r(s({I}_{n}))$. Because $s$ is inverse to $r$, this is ${I}_{n}$. Similarly, $s({I}_{n})r({I}_{n})=s(r({I}_{n}))={I}_{n}$. It follows that $r({I}_{n})$ is invertible with inverse $s({I}_{n})$. ∎