Elementary row operation and elementary matrix

An elementary row operation is a linear transformation $A'=TA$ , where the transformation matrix $T$ performs one of the following on $A$ :

If $A$ is the identity matrix $I$ , the transformed matrix is called an elementary matrix, which is denoted by $\varepsilon$ in place of $A'$ . In other words, an elementary matrix $\varepsilon$ is a square matrix that is related to an identity matrix by a single elementary row operation.

For example,

$\varepsilon_1=\begin{pmatrix}0&1\\1&0\end{pmatrix}\;\;\;\;\varepsilon_2=\begin{pmatrix}1&0\\0&7\end{pmatrix}\;\;\;\;\varepsilon_3=\begin{pmatrix}1&4\\0&1\end{pmatrix}$

are elementary matrices, where $\varepsilon_1,\varepsilon_2$ and $\varepsilon_3$ are obtained from $I=\begin{pmatrix}1&0\\0&1\end{pmatrix}$ by

Type 1. Swapping rows 1 and 2 of $I$ .
Type 2. Multiplying row 2 of $I$ by 7
Type 3. Adding 4 times row 2 of $I$ to row 1 of $I$

respectively.

Interestingly, $\varepsilon$ itself is a transformation matrix if $A=I$ because $\varepsilon=TA=TI=T$ . Therefore, when we multiply $\varepsilon$ by a matrix $M$ , we are performing an elementary row operation on $M$ . For example,

$\varepsilon_1M=\begin{pmatrix}0&1\\1&0\end{pmatrix}\begin{pmatrix}a&b&c\\s&t&u\end{pmatrix}=\begin{pmatrix}s&t&u\\a&b&c\end{pmatrix}$

$\varepsilon_2M=\begin{pmatrix}1&0\\0&7\end{pmatrix}\begin{pmatrix}a&b&c\\s&t&u\end{pmatrix}=\begin{pmatrix}a&b&c\\7s&7t&7u\end{pmatrix}$

$\varepsilon_3M=\begin{pmatrix}1&4\\0&1\end{pmatrix}\begin{pmatrix}a&b&c\\s&t&u\end{pmatrix}=\begin{pmatrix}a+4s&b+4t&c+4u\\s&t&u\end{pmatrix}$

An elementary matrix of dimension $(n\geq 2)\times(n\geq 2)$ has an inverse if $\varepsilon_i^{-1}\varepsilon_i=I=\varepsilon_i\varepsilon_i^{-1}$ , where the inverse $\varepsilon_i^{-1}$ is a matrix that reverses the transformation carried out by $\varepsilon_i$ . Every elementary matrix has an inverse because

Type 1. Two successive row swapping operations of a matrix $A$ returns $A$ , i.e. $\varepsilon_1\varepsilon_1A=A\implies \varepsilon_1\varepsilon_1=I$ . Comparing $\varepsilon_1\varepsilon_1=I$ with $\varepsilon_i^{-1}\varepsilon_i=I=\varepsilon_i\varepsilon_i^{-1}$ , we have $\varepsilon_1^{-1}=\varepsilon_1$ .

Type 2. It is always possible to satisfy $\varepsilon_i^{-1}\varepsilon_i=I=\varepsilon_i\varepsilon_i^{-1}$ when $\varepsilon_2$ and $\varepsilon_2^{-1}$ differ by one diagonal matrix element $(\varepsilon)_{ii}$ , with $(\varepsilon_2)_{ii}=k$ , $(\varepsilon_2^{-1})_{i'i'}=1/k$ and $i=i'$ .

Type 3. It is always possible to satisfy $\varepsilon_i^{-1}\varepsilon_i=I=\varepsilon_i\varepsilon_i^{-1}$ when $\varepsilon_3$ and $\varepsilon_3^{-1}$ differ by one matrix element $(\varepsilon)_{ij}$ , where $(\varepsilon_3)_{ij}=k$ and $(\varepsilon_3^{-1})_{i'j'}=-k$ with $i=i'$ , $j=j'$ , $i\neq j$ and $i'\neq j'$ .

Thus, all elementary matrices have corresponding inverses, which are themselves elementary matrices. For example, the inverses of $\varepsilon_1,\varepsilon_2$ and $\varepsilon_3$ are

$\varepsilon_1^{-1}=\begin{pmatrix}0&1\\1&0\end{pmatrix}\;\;\;\;\varepsilon_2^{-1}=\begin{pmatrix}1&0\\0&1/7\end{pmatrix}\;\;\;\;\varepsilon_3^{-1}=\begin{pmatrix}1&-4\\0&1\end{pmatrix}$

Finally, a non-singular matrix can always be expressed as a product of elementary matrices. The proof is as follows:

Let $A=\varepsilon_1\varepsilon_2\cdots\varepsilon_k$ . Since every elementary matrix is non-singular, we can multiply the inverses of the elementary matrices successively on the left of $A=\varepsilon_1\varepsilon_2\cdots\varepsilon_k$ to give:

$\varepsilon_1^{-1}A=\varepsilon_2\cdots\varepsilon_k$

$\varepsilon_2^{-1}\varepsilon_1^{-1}A=\varepsilon_3\cdots\varepsilon_k$

$\varepsilon_k^{-1}\cdots\varepsilon_2^{-1}\varepsilon_1^{-1}A=I\;\;\;\;\;\;\;\;3$

Similarly, we can multiply the inverses of the elementary matrices successively on the right of $A=\varepsilon_1\varepsilon_2\cdots\varepsilon_k$ to give:

$A\varepsilon_k^{-1}=\varepsilon_1\varepsilon_2\cdots\varepsilon_{k-1}$

$A\varepsilon_k^{-1}\varepsilon_{k-1}^{-1}=\varepsilon_1\varepsilon_2\cdots\varepsilon_{k-2}$

$A\varepsilon_k^{-1}\varepsilon_{k-1}^{-1}\cdots\varepsilon_1^{-1}=I\;\;\;\;\;\;\;\;4$

Combining eq3 and eq4, we have $A^{-1}A=I=AA^{-1}$ , where $A^{-1}=\varepsilon_k^{-1}\cdots\varepsilon_2^{-1}\varepsilon_1^{-1}$ , which completes the proof.

Elementary row operation and elementary matrix

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