Section 2.5 Elementary Matrices

Definition: An elementary matrix is one that is obtained by performing a single elementary row operation on an identity matrix.

Example 1: [latex]E_{1}=\begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ -2 & 0 & 1 \end{bmatrix}[/latex], [latex]E_{2}=\begin{bmatrix} 0 & 1 & 0\\ 1 & 0 & 0\\ 0 & 0 & 1 \end{bmatrix}[/latex], [latex]E_{3}=\begin{bmatrix} 1 & 0 & 0\\ 0 & 2 & 0\\ 0 & 0 & 1 \end{bmatrix}[/latex], and [latex]A=\begin{bmatrix} x_{11} & x_{12} & x_{13}\\ x_{21} & x_{22} & x_{23}\\ x_{31} & x_{32} & x_{33} \end{bmatrix}[/latex]. [latex]E_{1}[/latex], [latex]E_{2}[/latex] and [latex]E_{3}[/latex] are elementary matrices. Describe how to get [latex]E_{1}[/latex], [latex]E_{2}[/latex] and [latex]E_{3}[/latex] from identity matrix [latex]I_{3}[/latex] by elementary row operations. Compute [latex]E_{1}A[/latex], [latex]E_{2}A[/latex] and [latex]E_{3}A[/latex] and describe how these products can be obtained by elementary row operations.

Exercise 1: [latex]E_{1}=\begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & -3 & 1 \end{bmatrix}[/latex], [latex]E_{2}=\begin{bmatrix} 1 & 0 & 0\\ 0 & 0 & 1\\ 0 & 1 & 0 \end{bmatrix}[/latex], [latex]E_{3}=\begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 3 \end{bmatrix}[/latex], and [latex]A=\begin{bmatrix} x_{11} & x_{12} & x_{13}\\ x_{21} & x_{22} & x_{23}\\ x_{31} & x_{32} & x_{33} \end{bmatrix}[/latex]. [latex]E_{1}[/latex], [latex]E_{2}[/latex] and [latex]E_{3}[/latex] are elementary matrices. Describe how to get [latex]E_{1}[/latex], [latex]E_{2}[/latex] and [latex]E_{3}[/latex] from identity matrix [latex]I_{3}[/latex] by elementary row operations. Compute [latex]E_{1}A[/latex], [latex]E_{2}A[/latex] and [latex]E_{3}A[/latex] and describe how these products can be obtained by elementary row operations.

Fact: 1. If an elementary row operation is performed on an [latex]m \times n[/latex] matrix [latex]A[/latex], the resulting matrix can be written as [latex]EA[/latex], where the [latex]m \times m[/latex] matrix [latex]E[/latex] is created by performing the same row operation on [latex]I_{m}[/latex].

2. Each elementary matrix is invertible. The inverse of [latex]E[/latex] is the elementary matrix of the same type that transforms [latex]E[/latex] back into [latex]I[/latex].

Theorem: An [latex]n \times n[/latex] matrix is invertible if and only if [latex]A[/latex] is row equivalent to [latex]I_{n}[/latex] and in this case, any sequence of elementary row operations that reduces [latex]A[/latex] to [latex]I_{n}[/latex] also transforms [latex]I_{n}[/latex] into [latex]A^{-1}[/latex].

Proof:

Fact: If [latex]E_{p}E_{p-1} \cdots E_{1}A = I_{n}[/latex] then [latex]A^{-1} = (E_{p} \cdots E_{1})I_{n}[/latex] where [latex]E_{i}'s[/latex] are elementary matrices that transform [latex]A[/latex] into [latex]I_{n}[/latex].

Example 2: Find [latex]A^{-1}[/latex] where [latex]A=\begin{bmatrix} 1 & 0 & 3\\ 0 & 1 & 1\\ 2 & -1 & 2 \end{bmatrix}[/latex].

Exercise 2: Find [latex]A^{-1}[/latex] where [latex]A=\begin{bmatrix} 2 & 1 & 0\\ 1 & 1 & 2\\ 0 & -1 & 2 \end{bmatrix}[/latex].

Remark: Let [latex]A[/latex] be an invertible matrix and [latex]I = \begin{bmatrix} \vec{e_{1}} & \cdots & \vec{e_{n}} \end{bmatrix}[/latex] then there are [latex]\vec{u_{1}} \cdots \vec{u_{n}}[/latex] such that [latex]A\vec{u_{i}} = \vec{e_{i}}[/latex] for [latex]i = 1, \cdots, n[/latex]. The augmented matrix [latex]\begin{bmatrix} A\vec{e_{i}} \end{bmatrix}[/latex] will be equivalent to [latex]\begin{bmatrix} I_{n}\vec{u_{1}} \end{bmatrix}[/latex] by using the same elementary row operations, [latex]E_{p}, \cdots, E_{1}[/latex] for [latex]i = 1, \cdots, n[/latex]. Therefore we can write [latex]\begin{bmatrix} A | \vec{e_{1}} \cdots \vec{e_{n}} \end{bmatrix} = \begin{bmatrix} A | I_{n} \end{bmatrix}[/latex] is equivalent to [latex]\begin{bmatrix} I_{n} | \vec{u_{1}} \cdots \vec{u_{n}} \end{bmatrix}[/latex]. From the theorem and the example above, we know [latex]\begin{bmatrix} \vec{u_{1}} & \cdots & \vec{u_{n}} \end{bmatrix}[/latex] is the inverse matrix of [latex]A[/latex], i.e. [latex]A^{-1} = \begin{bmatrix} \vec{u_{1}} & \cdots & \vec{u_{n}} \end{bmatrix}[/latex], the solutions of [latex]A\vec{x} = \vec{e_i}[/latex] for [latex]i = 1, \cdots , n[/latex] form the columns of [latex]A^{-1}[/latex].

Example 3: Find the third column of [latex]A^{-1}[/latex] without computing the other columns, where [latex]A=\begin{bmatrix} 3 & 4 & -3\\ 0 & 1 & 2\\ -1 & 0 & 4 \end{bmatrix}[/latex].

Exercise 3: Find the third column of [latex]A^{-1}[/latex] without computing the other columns, where [latex]A=\begin{bmatrix} 1 & 4 & 3\\ -2 & 0 & 2\\ 0 & -1 & 5 \end{bmatrix}[/latex].

Theorem: Suppose [latex]A[/latex] is [latex]m \times n[/latex] and [latex]B[/latex] is obtained by [latex]A[/latex] by elementary row operations.

1. [latex]B = UA[/latex] where [latex]U[/latex] is an [latex]m \times m[/latex] invertible matrix

2. [latex]U[/latex] can be computed by [latex]\begin{bmatrix}A|I_{m}\end{bmatrix} \rightarrow \begin{bmatrix}B|U\end{bmatrix}[/latex] using the operations carrying [latex]A[/latex] to [latex]B[/latex].

3. where [latex]E_{1}, \cdots, E_{k}[/latex] are the elementary matrices corresponding (in order) to the elementary row operations carrying [latex]A[/latex] to [latex]B[/latex].

Example 4: If [latex]A=\begin{bmatrix} 1 & 3 & 2\\ 2 & 1 & -1 \end{bmatrix}[/latex], express the reduced row-echelon form [latex]R[/latex] of [latex]A[/latex] as [latex]R = UA[/latex] where [latex]U[/latex] is invertible.

Exercise 4: If [latex]A=\begin{bmatrix} 0 & 3 & 1\\ 1 & 1 & 1 \end{bmatrix}[/latex], express the reduced row-echelon form [latex]R[/latex] of [latex]A[/latex] as [latex]R = UA[/latex] where [latex]U[/latex] is invertible.

Theorem: A square matrix is invertible if and only if it is a product of elementary matrices.

Example 5: Express [latex]A=\begin{bmatrix} 1 & 3\\ 2 & 1 \end{bmatrix}[/latex] as product of elementary matrices.

Exercise 5: Express [latex]A=\begin{bmatrix} -1 & 1\\ 1 & 2 \end{bmatrix}[/latex] as product of elementary matrices.

Group Work 1: Mark each statement True or False. Justify each answer.

a. If [latex]A[/latex] is an invertible [latex]n \times n[/latex] matrix then the equation [latex]A\vec{x} = \vec{b}[/latex] is consistent for each [latex]\vec{b}[/latex] in [latex]\mathbb{R}^n[/latex].

b. Each elementary matrix is invertible.

c. If [latex]A[/latex] is invertible, then elementary row operations that reduce [latex]A[/latex] to the identity [latex]I_{n}[/latex] also reduce [latex]A^{-1}[/latex] to [latex]I_{n}[/latex].

d. If [latex]A[/latex] is invertible then the inverse of [latex]A^{-1}[/latex] is [latex]A[/latex].

e. [latex]A[/latex] is an [latex]n \times n[/latex] matrix and [latex]A\vec{x} = \vec{e_{i}}[/latex] is consistent for [latex]i = 1, \cdots, n[/latex] and [latex]\vec{e_{i}}[/latex] is the i-th column of [latex]I_{n}[/latex]. Then [latex]A[/latex] is invertible.

Group Work 2: Let [latex]E[/latex] be an elementary matrix. Show that [latex]E^T[/latex] is also an elementary matrix.

Group Work 3: Let [latex]A[/latex] and [latex]B[/latex] be [latex]m \times n[/latex] and [latex]n \times m[/latex] matrices, respectively. If [latex]m > n[/latex], show that [latex]AB[/latex] is not invertible. Hint: Use [latex]B\vec{x} = 0[/latex] has a non-trivial solution.

Group Work 4: Mark each statement True or False. Justify each answer.

a. If [latex]A[/latex] can be row reduced to identity matrix then [latex]A[/latex] is invertible.

b. [latex]0[/latex] an elementary matrix.

c. [latex]I[/latex] an elementary matrix.

d. [latex]\begin{bmatrix}P|Q\end{bmatrix}[/latex] is obtained by row operations from [latex]\begin{bmatrix}A|I\end{bmatrix}[/latex] then [latex]P=QA[/latex].

e. If [latex]A\vec{x} = 0[/latex] has trivial solution only then [latex]A[/latex] is a product of elementary matrices.