3 Matrices 3.7 Row operations 3.9 Row reduced echelon form

3.8 Elementary matrices

3.8.1 Definition of an elementary matrix

An elementary matrix is one you can get by doing a single row operation to an identity matrix.

Example 3.8.1.

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The elementary matrix $\begin{pmatrix}0&1\\ 1&0\end{pmatrix}$ results from doing the row operation $\textbf{r}_{1}\leftrightarrow\textbf{r}_{2}$ to $I_{2}$ .
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The elementary matrix $\begin{pmatrix}1&2&0\\ 0&1&0\\ 0&0&1\end{pmatrix}$ results from doing the row operation $\textbf{r}_{1}\mapsto\textbf{r}_{1}+2\textbf{r}_{2}$ to $I_{3}$ .
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The elementary matrix $\begin{pmatrix}-1&0\\ 0&1\end{pmatrix}$ results from doing the row operation $\textbf{r}_{1}\mapsto(-1)\textbf{r}_{1}$ to $I_{2}$ .

3.8.2 Doing a row operation is the same as multiplying by an elementary matrix

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 $r$ :

Theorem 3.8.1.

Let $r$ be a row operation and $A$ an $m\times n$ matrix. Then $r(A)=r(I_{m})A$ .

Proof.

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 $\mathbf{s}_{1},\ldots,\mathbf{s}_{m}$ and if $\mathbf{r}=\begin{pmatrix}a_{1}&\cdots&a_{m}\end{pmatrix}$ is a row vector then

\mathbf{r}A=a_{1}\mathbf{s}_{1}+\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.

1.

Let $r$ be $\textbf{r}_{j}\mapsto\textbf{r}_{j}+\lambda\textbf{r}_{i}$ . 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 $j^{\prime}\neq j$ , 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 $A$ .
2.

Let $r$ be $\textbf{r}_{j}\mapsto\lambda\textbf{r}_{j}$ . 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 $j^{\prime}\neq j$ , 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 $A$ .
3.

Let $r$ be $\textbf{r}_{i}\leftrightarrow\textbf{r}_{j}$ . 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

$\displaystyle\mathbf{r}_{i}A$ $\displaystyle=\mathbf{s}_{j}$

$\displaystyle\mathbf{r}_{j}A$ $\displaystyle=\mathbf{s}_{i}.$

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.

∎

Corollary 3.8.2.

Elementary matrices are invertible.

Proof.

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 $s(I_{n})$ . ∎

	$\displaystyle\mathbf{r}_{i}A$	$\displaystyle=\mathbf{s}_{j}$
	$\displaystyle\mathbf{r}_{j}A$	$\displaystyle=\mathbf{s}_{i}.$