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# Comprehensive Guide on Elementary Matrices in Linear Algebra

schedule Aug 12, 2023
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Linear Algebra
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Definition.

# Definition of elementary matrices

Elementary matrices are obtained by performing a single elementary row operation on an identity matrix. Recall that an elementary row operation is one of the following:

• multiplying a row by a non-zero constant.

• swapping one row with another row.

• adding a multiple of a row to another to another row.

Because there are three types of elementary row operations, there are also three types of elementary matrices.

Example.

## Non-elementary matrices

Consider the following matrix:

$$\boldsymbol{A}_1= \begin{pmatrix} 4&0&0\\ 0&5&0\\ 0&0&1\\ \end{pmatrix},\;\;\;\;\; \boldsymbol{A}_2= \begin{pmatrix} 0&1&0\\ 0&0&1\\ 1&0&0\\ \end{pmatrix}$$

Show that these are not elementary matrices.

Solution. The $3\times3$ identity matrix is:

$$\boldsymbol{I}_3=\begin{pmatrix} 1&0&0\\ 0&1&0\\ 0&0&1\\ \end{pmatrix}$$

Matrix $\boldsymbol{A}_1$ can be obtained by performing two elementary row operations on the identity matrix:

1. multiply the first row of the identity matrix by $4$.

2. multiply the second row by $5$.

Since an elementary matrix is defined as a matrix that can be obtained from a single elementary operation, $\boldsymbol{A}_1$ is not an elementary matrix.

Similarly, $\boldsymbol{A}_2$ is not an elementary matrix because we must perform two swapping elementary operations on the identity matrix to obtain $\boldsymbol{A}_2$.

Example.

## Elementary matrices

Consider the following matrices:

$$\boldsymbol{E}_1= \begin{pmatrix} 1&0&0\\ 0&4&0\\ 0&0&1\\ \end{pmatrix},\;\;\;\;\; \boldsymbol{E}_2= \begin{pmatrix} 0&1&0\\ 1&0&0\\ 0&0&1\\ \end{pmatrix},\;\;\;\;\;\boldsymbol{E}_3= \begin{pmatrix} 1&0&4\\ 0&1&0\\ 0&0&1\\ \end{pmatrix}$$

Show that these are elementary matrices.

Solution. The $3\times3$ identity matrix is:

$$\boldsymbol{I}_3=\begin{pmatrix} 1&0&0\\ 0&1&0\\ 0&0&1\\ \end{pmatrix}$$

Note the following:

• matrix $\boldsymbol{E}_1$ can be obtained by the elementary row operation of multiplying the second row of the identity matrix by $4$.

• matrix $\boldsymbol{E}_2$ can be obtained by the elementary row operation of swapping the first row and the second row of the identity matrix.

• finally, $\boldsymbol{E}_3$ can be obtained by the elementary row operation of multiplying the third row by $4$ and then adding it to the first row.

Theorem.

## Effect of multiplying elementary matrix on a matrix

Let $\boldsymbol{E}$ be any elementary matrix and $\boldsymbol{A}$ be any matrix. Taking the matrix product $\boldsymbol{EA}$ is equivalent to performing a single elementary row operation on $\boldsymbol{A}$ where the type of the row operation corresponds to the type of the elementary matrix.

Sketch proof. Let's go through a sketch proof and assume that the shape of the elementary matrix $\boldsymbol{E}$ and matrix $\boldsymbol{A}$ are both $3\times3$. This sketch proof can easily be generalized to any square shape.

Consider the following matrices:

$$\boldsymbol{A}= \begin{pmatrix} a&b&c\\ d&e&f\\ g&h&i\\ \end{pmatrix},\;\;\;\;\; \boldsymbol{E}_1= \begin{pmatrix} 1&0&0\\ 0&4&0\\ 0&0&1\\ \end{pmatrix},\;\;\;\;\; \boldsymbol{E}_2= \begin{pmatrix} 0&1&0\\ 1&0&0\\ 0&0&1\\ \end{pmatrix},\;\;\;\;\;\boldsymbol{E}_3= \begin{pmatrix} 1&0&4\\ 0&1&0\\ 0&0&1\\ \end{pmatrix}$$

Here, $\boldsymbol{E}_1$, $\boldsymbol{E}_2$ and $\boldsymbol{E}_3$ are the elementary matrices from our previous examplelink. Recall that each of these matrices corresponds to a different type of elementary row operation. Our goal is to compute $\boldsymbol{E}_1\boldsymbol{A}$, $\boldsymbol{E}_2\boldsymbol{A}$ and $\boldsymbol{E}_3\boldsymbol{A}$ and show that these matrix products can be thought of as performing the corresponding elementary row operation on $\boldsymbol{A}$.

Let's start with $\boldsymbol{E}_1\boldsymbol{A}$. Recall that $\boldsymbol{E}_1$ can be obtained by multiplying the second row of the identity matrix by $4$. Let's see what $\boldsymbol{E}_1\boldsymbol{A}$ gives us:

\begin{align*} \boldsymbol{E}_1\boldsymbol{A}&= \begin{pmatrix} 1&0&0\\0&4&0\\0&0&1\\ \end{pmatrix}\begin{pmatrix} a&b&c\\d&e&f\\g&h&i\\\end{pmatrix}= \begin{pmatrix}a&b&c\\4d&4e&4f\\g&h&i\\ \end{pmatrix} \end{align*}

Multiplying $\boldsymbol{E}_1$ to $\boldsymbol{A}$ results in the second row of $\boldsymbol{A}$ getting multiplied by $4$ - just like the way $\boldsymbol{E}_1$ is obtained from the identity matrix! In other words, the product $\boldsymbol{E}_1\boldsymbol{A}$ represents an elementary row operation of multiplying a single row by a non-zero constant value.

Let's now move on to $\boldsymbol{E}_2\boldsymbol{A}$. Recall that $\boldsymbol{E}_2$ can be obtained by swapping the first row and second row of the identity matrix. Let's see what $\boldsymbol{E}_2\boldsymbol{A}$ gives us:

\begin{align*} \boldsymbol{E}_2\boldsymbol{A}&= \begin{pmatrix} 0&1&0\\1&0&0\\0&0&1\\ \end{pmatrix}\begin{pmatrix} a&b&c\\d&e&f\\g&h&i\\\end{pmatrix}= \begin{pmatrix}d&e&f\\a&b&c\\g&h&i\\ \end{pmatrix} \end{align*}

Notice how multiplying $\boldsymbol{E}_2$ to $\boldsymbol{A}$ also swaps the first row and the second row of $\boldsymbol{A}$! By multiplying $\boldsymbol{E}_2$ to $\boldsymbol{A}$, we're essentially performing an elementary row operation on $\boldsymbol{A}$ in which we swap the two rows.

Finally, let's compute $\boldsymbol{E}_3\boldsymbol{A}$. We know that $\boldsymbol{E}_3$ is obtained by multiplying the third row by $4$, and then adding it to the first row. Let's see what $\boldsymbol{E}_3\boldsymbol{A}$ gives us:

\begin{align*} \boldsymbol{E}_3\boldsymbol{A}&= \begin{pmatrix} 1&0&4\\0&1&0\\0&0&1\\ \end{pmatrix}\begin{pmatrix} a&b&c\\d&e&f\\g&h&i\\\end{pmatrix}= \begin{pmatrix}a+4g&b+4h&c+4i\\d&e&f\\g&h&i\\ \end{pmatrix} \end{align*}

Multiplying $\boldsymbol{E}_3$ to $\boldsymbol{A}$ results in multiplying the third row of $\boldsymbol{A}$ by $4$ and then adding it to the first row. Again, multiplying $\boldsymbol{E}_3$ to $\boldsymbol{A}$ can be treated as performing a single elementary row operation on $\boldsymbol{A}$ in which we add a multiple of one row to another.

We have just shown that for any of the three types of elementary matrix $\boldsymbol{E}$, multiplying $\boldsymbol{E}$ to some matrix $\boldsymbol{A}$ results in a single elementary row operation on $\boldsymbol{A}$ whose type corresponds to that of the elementary matrix.

Theorem.

## Elementary matrices are invertible

Any elementary matrix is invertible, and its inverse is also an elementary matrix. Specifically, the inverses of the three types of elementary matrices are:

Example of elementary matrix

Inverse of elementary matrix

Multiplying a row by a non-zero constant k

$$\boldsymbol{E}_1=\begin{pmatrix} 1&0&0\\ 0&k&0\\ 0&0&1\\ \end{pmatrix}$$
$$\boldsymbol{E}^{-1}_1 =\begin{pmatrix} 1&0&0\\ 0&1/k&0\\ 0&0&1\\ \end{pmatrix}$$

Swapping one row with another row

$$\boldsymbol{E}_2=\begin{pmatrix} 0&1&0\\ 1&0&0\\ 0&0&1\\ \end{pmatrix}$$
$$\boldsymbol{E}^{-1}_2 =\begin{pmatrix} 0&1&0\\ 1&0&0\\ 0&0&1\\ \end{pmatrix}$$

Adding a multiple of a row to another

$$\boldsymbol{E}_3= \begin{pmatrix} 1&0&4\\ 0&1&0\\ 0&0&1\\ \end{pmatrix}$$
$$\boldsymbol{E}_3^{-1} =\begin{pmatrix} 1&0&-4\\ 0&1&0\\ 0&0&1\\ \end{pmatrix}$$

Proof. Recall that elementary matrices are obtained by performing one of the following elementary row operations on an identity matrix:

• multiplying a row by a non-zero constant.

• swapping one row with another row.

• adding a multiple of a row to another row.

Once again, we will use the elementary matrices below that represent each type of operation:

$$\boldsymbol{E}_1= \begin{pmatrix} 1&0&0\\ 0&4&0\\ 0&0&1\\ \end{pmatrix},\;\;\;\;\; \boldsymbol{E}_2= \begin{pmatrix} 0&1&0\\ 1&0&0\\ 0&0&1\\ \end{pmatrix},\;\;\;\;\;\boldsymbol{E}_3= \begin{pmatrix} 1&0&4\\ 0&1&0\\ 0&0&1\\ \end{pmatrix}$$

Let's now show that each type of elementary matrix has an inverse. For the first type, the elementary matrix will be of the following form:

$$\boldsymbol{E}_1=\begin{pmatrix} 1&0&0\\ 0&4&0\\ 0&0&1\\ \end{pmatrix}$$

Remember that the inverse of $\boldsymbol{E}_1$, denoted by $\boldsymbol{E}^{-1}_1$, is a matrix such that $\boldsymbol{E}^{-1}_1\boldsymbol{E}_1= \boldsymbol{E}_1\boldsymbol{E}^{-1}_1 =\boldsymbol{I}_n$. Our goal is to find $\boldsymbol{E}^{-1}$ below:

$$\boldsymbol{E}^{-1}_1\boldsymbol{E}_1=\boldsymbol{I}_n \;\;\;\;\;\Longleftrightarrow\;\;\;\;\; \boldsymbol{E}^{-1}_1 \begin{pmatrix} 1&0&0\\0&4&0\\0&0&1\\ \end{pmatrix}= \begin{pmatrix} 1&0&0\\0&1&0\\0&0&1\\ \end{pmatrix}$$

By definition, an elementary matrix is obtained by performing a single elementary row operation on the identity matrix. This means that we can perform a single elementary row operation that reverses this to give us back the identity matrix. Therefore, the inverse matrix $\boldsymbol{E}^{-1}$ must also be an elementary matrix! In this specific case, the inverse $\boldsymbol{E}^{-1}$ is:

$$\boldsymbol{E}^{-1}_1= \begin{pmatrix} 1&0&0\\0&1/4&0\\0&0&1\\ \end{pmatrix}$$

Multiplying this elementary matrix $\boldsymbol{E}^{-1}_1$ to $\boldsymbol{E}_1$ will result in multiplying the second row of $\boldsymbol{E}_1$ by $1/4$, thereby giving us back the identity matrix!

Next, let's find the inverse of the elementary matrix $\boldsymbol{E}_2$ below:

$$\boldsymbol{E}^{-1}_2\boldsymbol{E}_2=\boldsymbol{I}_n \;\;\;\;\;\Longleftrightarrow\;\;\;\;\; \boldsymbol{E}^{-1}_2 \begin{pmatrix} 0&1&0\\1&0&0\\0&0&1\\ \end{pmatrix}= \begin{pmatrix} 1&0&0\\0&1&0\\0&0&1\\ \end{pmatrix}$$

We know that $\boldsymbol{E}_2$ swaps the first row and the second row. Therefore, its inverse matrix should be another elementary matrix that swaps back the first row and the second row:

$$\boldsymbol{E}^{-1}_2=\begin{pmatrix} 0&1&0\\ 1&0&0\\ 0&0&1\\ \end{pmatrix}$$

Finally, let's look at the inverse of $\boldsymbol{E}_3$ below:

$$\boldsymbol{E}^{-1}_3\boldsymbol{E}_3=\boldsymbol{I}_n \;\;\;\;\;\Longleftrightarrow\;\;\;\;\; \boldsymbol{E}^{-1}_3 \begin{pmatrix}1&0&4\\0&1&0\\0&0&1\\\end{pmatrix} = \begin{pmatrix} 1&0&0\\0&1&0\\0&0&1\\ \end{pmatrix}$$

The inverse matrix $\boldsymbol{E}_3^{-1}$ is an elementary matrix that corresponds to the elementary row operation of multiplying the third row by $-4$ and then adding it to the first row:

$$\boldsymbol{E}^{-1}_3=\begin{pmatrix} 1&0&-4\\ 0&1&0\\ 0&0&1\\ \end{pmatrix}$$

We have managed to show that:

• every elementary matrix has an inverse and is thus invertible.

• the inverse of an elementary matrix is also an elementary matrix.

This completes the proof.

Theorem.

# Transpose of an elementary matrix is also an elementary matrix

The transpose of an elementary matrix is also an elementary matrix.

Proof. Let's consider the three types of elementary matrices:

• matrix $\boldsymbol{E}_1$ corresponds to multiplying a single row by $k$.

• matrix $\boldsymbol{E}_2$ corresponds to interchanging two rows.

• matrix $\boldsymbol{E}_3$ corresponds to adding a multiple of one row to another row.

We shall consider $3\times3$ elementary matrices but this proof can be easily extended to a general $n\times{n}$ shape. Here are some examples of each type of elementary matrix:

$$\boldsymbol{E}_1= \begin{pmatrix} 4&0&0\\ 0&1&0\\ 0&0&1\\ \end{pmatrix},\;\;\;\;\; \boldsymbol{E}_2= \begin{pmatrix} 0&1&0\\ 1&0&0\\ 0&0&1\\ \end{pmatrix},\;\;\;\;\;\boldsymbol{E}_3= \begin{pmatrix} 1&0&4\\ 0&1&0\\ 0&0&1\\ \end{pmatrix}$$

Elementary matrices of type $\boldsymbol{E}_1$ is a diagonal matrix, which means that their transpose is itself.

Next, elementary matrices of type $\boldsymbol{E}_2$ are symmetric because swapping two rows, say the $i$-th and $j$-th row, is equivalent to swapping the $i$-th and $j$-th column. You can confirm this using $\boldsymbol{E}_2$ above. The transpose of a symmetric matrix is itself.

Finally, elementary matrices of type $\boldsymbol{E}_3$ involves adding a multiple $k$ of the $i$-th row to some other $j$-th row. The transpose of $\boldsymbol{E}_3$ is:

$$\boldsymbol{E}_3^T= \begin{pmatrix} 1&0&0\\ 0&1&0\\ 4&0&1\\ \end{pmatrix}$$

This is yet another elementary matrix corresponding to adding a multiple $k$ of the $j$-th row to the $i$-th row.

This completes the proof.

Theorem.

# Equivalent statements of matrix invertibility

If $\boldsymbol{A}$ is an $n\times{n}$ matrix, then the following statements are equivalent:

1. $\boldsymbol{A}$ is invertible.

2. $\boldsymbol{Ax}=\boldsymbol{0}$ has only the trivial solution of $\boldsymbol{x}=\boldsymbol{0}$.

3. the reduced row echelon form of $\boldsymbol{A}$ is the identity matrix $\boldsymbol{I}_n$. Equivalently, the reduced row echelon form of $\boldsymbol{A}$ does not contain a row with all zeros.

4. $\boldsymbol{A}$ can be expressed as a product of elementary matrices.

Note the following:

• if one of these statements is true, then all statements will be true.

• if one of these statements is false, then all statements will be false.

Proof. Our proof will follow the following flow:

$$1\implies2\implies3\implies4\implies1$$

$\color{blue}1\implies2$ - Let's begin by assuming that statement $1$ is true, that is, matrix $\boldsymbol{A}$ is invertible. In other words, the inverse $\boldsymbol{A}^{-1}$ exists. To prove statement $2$, suppose $\boldsymbol{x}_0$ is any solution to $\boldsymbol{Ax}=0$. Multiplying both sides by the inverse matrix $\boldsymbol{A}^{-1}$ yields:

\begin{align*} \boldsymbol{A}^{-1}\boldsymbol{A}\boldsymbol{x}_0&=\boldsymbol{A}^{-1}\boldsymbol{0}\\ \boldsymbol{I}_n\boldsymbol{x}_0&=\boldsymbol{0}\\ \boldsymbol{x}_0&=\boldsymbol{0}\\ \end{align*}

We have shown that any solution $\boldsymbol{x}_0$ must be the zero vector, which is the trivial solution.

$\color{blue}2\implies3$ - We assume that the only solution to $\boldsymbol{Ax}=\boldsymbol{0}$ is the zero vector $\boldsymbol{x}=\boldsymbol{0}$. The augmented matrix of $\boldsymbol{Ax}=\boldsymbol{0}$ is:

$$\begin{pmatrix} a_{1,1}&a_{1,2}&\cdots&a_{1,n}&\color{purple}0\\ a_{2,1}&a_{2,2}&\cdots&a_{2,n}&\color{purple}0\\ \vdots&\vdots&\smash\ddots&\vdots&\color{purple}\vdots\\ a_{n,1}&a_{n,2}&\cdots&a_{n,n}&\color{purple}0\\ \end{pmatrix}$$

Performing elementary row operations on this augmented matrix does not affect columns with all zeros. After Gaussian elimination, we are guaranteed to end up with an identity matrix:

$$\begin{equation}\label{eq:YsTnZmrYsImILLMMSka} \begin{pmatrix} 1&0&\cdots&0&\color{purple}0\\ 0&1&\cdots&0&\color{purple}0\\ \vdots&\vdots&\smash\ddots&\vdots&\color{purple}0\\ 0&0&\cdots&1&\color{purple}0\\ \end{pmatrix} \end{equation}$$

To understand why, consider the following augmented matrix:

$$\begin{pmatrix} 1&0&0&\color{purple}0\\ 0&1&0&\color{purple}0\\ 0&0&0&\color{purple}0\\ \end{pmatrix}$$

Here, the first two rows imply that $x_1=0$ and $x_2=0$. However, the last row implies that $0x_3=0$, which means $x_3$ can take on infinitely many values - not just $0$. This contradicts our assumption that the only solution to $\boldsymbol{A}\boldsymbol{x}=\boldsymbol{0}$ is the zero vector $\boldsymbol{x}=\boldsymbol{0}$. This means that the reduced row echelon form cannot have a row with all zeros. In other words, the reduced row echelon form must look like \eqref{eq:YsTnZmrYsImILLMMSka}.

$\color{blue}3\implies4$ - We assume that the reduced row echelon form of $\boldsymbol{A}$ is the identity matrix:

$$\begin{pmatrix} 1&0&\cdots&0&\color{purple}0\\ 0&1&\cdots&0&\color{purple}0\\ \vdots&\vdots&\smash\ddots&\vdots&\color{purple}0\\ 0&0&\cdots&1&\color{purple}0\\ \end{pmatrix}$$

This means that we managed to perform elementary row operations on $\boldsymbol{A}$ to obtain the identity matrix. From theoremlink, we know that performing elementary row operations on $\boldsymbol{A}$ is equivalent to multiplying corresponding elementary matrices $\boldsymbol{E}_1$, $\boldsymbol{E}_2$, $\cdots$, $\boldsymbol{E}_m$ to $\boldsymbol{A}$, that is:

$$\begin{equation}\label{eq:Gtg9cIjbX0SgWBZFJC4} \boldsymbol{E}_1\boldsymbol{E}_2\cdots\boldsymbol{E_m} \boldsymbol{A}=\boldsymbol{I}_n \end{equation}$$

We know from theoremlink that all elementary matrices are invertible, which means that each elementary matrix has an inverse. Therefore, \eqref{eq:Gtg9cIjbX0SgWBZFJC4} can be written as:

\begin{align*} \boldsymbol{E}_1\boldsymbol{E}_2\cdots\boldsymbol{E_m} \boldsymbol{A}&=\boldsymbol{I}_n\\ \boldsymbol{E}_2\cdots\boldsymbol{E_m} \boldsymbol{A}&=\boldsymbol{E}_1^{-1}\boldsymbol{I}_n\\ \boldsymbol{A}&=\boldsymbol{E}^{-1}_m\cdots\boldsymbol{E}^{-1}_2\boldsymbol{E}^{-1}_1\boldsymbol{I}_n\\ \boldsymbol{A}&=\boldsymbol{E}^{-1}_m\cdots\boldsymbol{E}^{-1}_2\boldsymbol{E}^{-1}_1\\ \end{align*}

We have also shown in theoremlink that these inverse matrices are also elementary matrices. Therefore, we have managed to write $\boldsymbol{A}$ as a product of elementary matrices!

$\color{blue}4\implies1$ - We assume that $\boldsymbol{A}$ can be written as a product of elementary matrices, say:

$$\boldsymbol{A} =\boldsymbol{E}^{-1}_m\cdots\boldsymbol{E}^{-1}_2\boldsymbol{E}^{-1}_1$$

All elementary matrices are invertible, and by theorem, we know that the product of invertible matrices is also invertible.

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