Section 2.2 Equations, Matrices, and Transformations Equations

2.2A

 

Definition: Consider the system of linear equations

 

$$\begin{array}{cccc}
a_{11}x_{1} a_{12}x_{2} \cdots a_{1n}x_{n} = b_{1}\\
a_{21}x_{1} a_{22}x_{2} \cdots a_{2n}x_{n} = b_{2}\\
\vdots \\
a_{m1}x_{1} a_{m2}x_{2} \cdots a_{mn}x_{n} = b_{m}\\
\end{array}$$

 

Such a system can be expressed in vector form or as a vector equation by using linear combinations of column vectors:

 

$$x_{1}\begin{bmatrix}a_{11}\\a_{21}\\\vdots \\a_{m1}\end{bmatrix} x_{2}\begin{bmatrix}a_{12}\\a_{22}\\\vdots \\a_{m2}\end{bmatrix} \cdots x_{n}\begin{bmatrix}a_{1n}\\a_{2n}\\\vdots \\a_{mn}\end{bmatrix} = \begin{bmatrix}b_{1}\\b_{2}\\\vdots \\b_{m}\end{bmatrix}$$

 

Example 1: Write a vector equation that is equivalent to the given linear
system of equations.

 

$$\begin{array}{ccc}
x_{1} x_{3} & = & 1\\
2x_{2}-x_{3} & = & 2\\
3x_{1} x_{2} 3x_{4} & = & 3
\end{array}$$

 

 

Exercise 1: Write a vector equation that is equivalent to the given linear system of equations.

 

$$\begin{array}{ccc}
3x_{2}-x_{3} & = & 4\\
x_{1}-x_{2} x_{3} & = & -3\\
2x_{1} x_{2}-2x_{3} & = & 2
\end{array}$$

 

Definition: If [latex]A[/latex] is an [latex]m \times n[/latex] matrix, with column vectors, [latex]\vec{v_{1}}, \cdots, \vec{v_{n}}[/latex] in [latex]\mathbb{R}^m[/latex], and [latex]\vec{x}[/latex] is a vector in [latex]\mathbb{R}^n[/latex], then the product of [latex]A[/latex] and [latex]\vec{x}[/latex], denoted by [latex]A\vec{x}[/latex] in [latex]\mathbb{R}^n[/latex], is the linear combination of the columns of [latex]A[/latex] using the corresponding entries in [latex]\vec{x}[/latex] as weights; that is,

 

$$A\vec{x} = \begin{bmatrix}\vec{v_{1}} & \cdots & \vec{v_{n}}\end{bmatrix}
\begin{bmatrix}x_{1}
\\\vdots\\
x_{n}
\end{bmatrix}
= x_{1}\vec{v_{1}} x_{2}\vec{v_{2}} \cdots x_{n}\vec{v_{n}}$$

 

Note:
[latex]A\vec{x}[/latex] is defined only if the number of columns of [latex]A[/latex] equals the number of entries in [latex]\vec{x}[/latex].

 

 

Example 2: [latex]A = \begin{bmatrix} 2 & 0 & -1\\ 0 & 1 & 0\\ 1 & 2 & -2\\ -2 & 1 & 3 \end{bmatrix}[/latex], [latex]\;\vec{x} = \begin{bmatrix} 2\\ 3\\ 4 \end{bmatrix}[/latex] find [latex]A\vec{x}[/latex].

 

Exercise 2: [latex]A = \begin{bmatrix} 2 & 0 & 1 & -1\\ 0 & 1 & 2 & 0\\ 1 & 2 & 3 & -2 \end{bmatrix}[/latex], [latex]\;\vec{x} = \begin{bmatrix} 2\\ 3\\ -1\\ 4 \end{bmatrix}[/latex] find [latex]A\vec{x}[/latex].

 

Example 3: Write [latex]-3\vec{u} 2\vec{v}[/latex] as a matrix times a vector where [latex]\vec{u} = \begin{bmatrix} 2\\-1\end{bmatrix}[/latex] and
[latex]\vec{v} = \begin{bmatrix} -3\\4\end{bmatrix}[/latex].

 

Exercise 3: Write [latex]4\vec{u} (-2)\vec{v} 3\vec{w}[/latex] as a matrix times a vector where [latex]\vec{u} = \begin{bmatrix} -1\\-3\end{bmatrix}[/latex],
[latex]\vec{v} = \begin{bmatrix} -2\\0\end{bmatrix}[/latex] and [latex]\vec{w} = \begin{bmatrix} 0\\2\end{bmatrix}[/latex].

 

Theorem: If [latex]A[/latex] is an [latex]m \times n[/latex] matrix, [latex]\vec{u}[/latex] and [latex]\vec{v}[/latex] are vectors in [latex]\mathbb{R}^n[/latex], and [latex]c[/latex] is a scalar, then

1. [latex]A(\vec{u} \vec{v}) = A\vec{u} A\vec{v}[/latex]

2. [latex]A(c\vec{u}) = cA\vec{u}.[/latex]

 

Definition: If [latex]A[/latex] is an [latex]m \times n[/latex] matrix, with column vectors, [latex]\vec{v_{1}}, \cdots,\vec{v_{n}}[/latex] in [latex]\mathbb{R}^m[/latex], and [latex]\vec{x}[/latex] is a vector in [latex]\mathbb{R}^n[/latex], and [latex]\vec{b}[/latex] is a vector in [latex]\mathbb{R}^m[/latex], then [latex]A\vec{x} = \vec{b}[/latex] is called a
matrix equation.

 

 

Theorem: If [latex]A[/latex] is an [latex]m \times n[/latex] matrix, with column vectors, [latex]\vec{v_{1}}, \cdots, \vec{v_{n}}[/latex] in [latex]\mathbb{R}^n[/latex], and [latex]\vec{x}[/latex] is a vector in [latex]\mathbb{R}^n[/latex], and [latex]\vec{b}[/latex] is a vector in [latex]\mathbb{R}^m[/latex], then the matrix equation [latex]A\vec{x} = \vec{b}[/latex] has the same solution set as the vector equation [latex]x_{1}\vec{v_{1}} \cdots x_{n}\vec{v_{n}} = \vec{b}[/latex], which, in turn, has the same solution set as the system of linear equations whose augmented matrix is [latex]\begin{bmatrix} \vec{v_{1}} & \cdots & \vec{v_{n}} & \vec{b} \end{bmatrix}[/latex].

 

Fact: The equation [latex]A\vec{x}=\vec{b}[/latex] has a solution if and only if [latex]\vec{b}[/latex] is a linear combination of the columns of [latex]A[/latex].

 

Example 4: Solve the matrix equation [latex]A\vec{x} = \vec{b}[/latex] where [latex]A = \begin{bmatrix} 2&1\\ 1& -2 \end{bmatrix}[/latex] and [latex]\vec{b} = \begin{bmatrix} 3\\ 4\end{bmatrix}[/latex].

Exercise 4: Solve the matrix equation [latex]A\vec{x} = \vec{b}[/latex] where [latex]A = \begin{bmatrix} 1&1\\ 3& -2 \end{bmatrix}[/latex] and [latex]\vec{b} = \begin{bmatrix} 1\\ -1\end{bmatrix}[/latex].

 

Theorem: Let [latex]A[/latex] be an [latex]m \times n[/latex] matrix. Then the following statements are logically equivalent. That is, for a particular [latex]A[/latex], either they are all true statements or they are all false.

1. For each [latex]\vec{b}[/latex] in [latex]\mathbb{R}^m[/latex], the equation [latex]A\vec{x} = \vec{b}[/latex] has a solution

 

2. Each [latex]\vec{b}[/latex] in [latex]\mathbb{R}^m[/latex] is a linear combination of the columns of [latex]A[/latex].

 

3. [latex]A[/latex] has a pivot position in every row.

 

Example 5: Show each [latex]\vec{b}[/latex] in [latex]\mathbb{R}^m[/latex] is a linear combination of the columns of [latex]A = \begin{bmatrix} 1&-1&2\\ 2&0&3\\ 0&1&-2 \end{bmatrix}[/latex]

 

 

Exercise 5: Show each [latex]\vec{b}[/latex] in [latex]\mathbb{R}^m[/latex] is a linear combination of the columns of [latex]A = \begin{bmatrix} 0&-1&2\\ 1&3&0\\ -2&2&-2 \end{bmatrix}[/latex]

 

Definition: For each [latex]n > 2[/latex], the identity matrix [latex]I_{n}[/latex] is the [latex]n \times n[/latex] matrix with 1’s on the main diagonal (upper left to lower right), and zeros elsewhere.

 

Fact: For each [latex]n > 2[/latex], we have [latex]I_{n}\vec{x} = \vec{x}[/latex] for each n-vector [latex]\vec{x}[/latex] in [latex]\mathbb{R}^n[/latex].

 

Theorem: Let [latex]A[/latex] and [latex]B[/latex] be [latex]m \times n[/latex] matrices. If [latex]A\vec{x} = B\vec{x}[/latex] for all [latex]\vec{x}[/latex] in [latex]\mathbb{R}^n[/latex], then [latex]A = B[/latex].

 

 

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

a. The equation [latex]A\vec{x} = \vec{b}[/latex] is referred to as a vector equation.

 

b. A vector [latex]\vec{b}[/latex] is a linear combination of the columns of a matrix if and only if the equation [latex]A\vec{x} = \vec{b}[/latex] has at least one solution.

 

c. The equation [latex]A\vec{x} = \vec{b}[/latex] is consistent if the augmented matrix [latex]\begin{bmatrix} A & \vec{b}\end{bmatrix}[/latex] has a pivot position in every row.

 

d. The first entry in the product [latex]A\vec{x}[/latex] is a sum of products.

 

e. If [latex]A[/latex] is an [latex]m \times n[/latex] matrix and if the equation [latex]A\vec{x} = \vec{b}[/latex] is inconsistent for some [latex]\vec{b}[/latex] in [latex]\mathbb{R}^m[/latex] then [latex]A[/latex] cannot have a pivot position in every row.

 

f. The weight [latex]c_{1}, \cdots, c_{p}[/latex] in a linear combination [latex]c_{1}\vec{v_{1}} \cdots c_{p}\vec{v_{p}}[/latex] cannot all be zero.

 

Group Work 2: Find the solution of [latex]A\vec{x} = \vec{b}[/latex] where

 

[latex]A = \begin{bmatrix} 1 & 3 & 3 & 0\\ 0 & 0 & -1 & 1\\ 0 & 1 & 2 & 1 \end{bmatrix}[/latex] and [latex]\vec{b} = \begin{bmatrix} -2\\ 1\\ 1 \end{bmatrix}[/latex]

 

Group Work 3: In each case either show that the statement is true, or give
an example showing that it is false.

a. [latex]\begin{bmatrix}3\\2\end{bmatrix}[/latex] is a linear combination of [latex]\begin{bmatrix}1\\0\end{bmatrix}[/latex] and [latex]\begin{bmatrix}0\\1\end{bmatrix}[/latex]

 

b. [latex]A\vec{x}[/latex] has a zero entry, then [latex]A[/latex] has a row of zeros.

 

c. If [latex]A\vec{x} = 0[/latex] where [latex]\vec{x} \neq 0[/latex], then [latex]A = 0[/latex].

 

d. Every linear combination of vectors in [latex]\mathbb{R}^n[/latex] can be written in the form [latex]A\vec{x}[/latex].

 

e. If [latex]A = \begin{bmatrix}\vec{a_{1}}&\vec{a_{2}}&\vec{a_{3}}\end{bmatrix}[/latex] in terms of its columns, and if [latex]\vec{b} = 3\vec{a_{1}} -2\vec{a_{2}}[/latex], then the system [latex]A\vec{x} = \vec{b}[/latex] has a solution.

 

f. If [latex]A = \begin{bmatrix}\vec{a_{1}}&\vec{a_{2}}&\vec{a_{3}}\end{bmatrix}[/latex] in terms of its columns, and if [latex]A\vec{x} = \vec{b}[/latex] has a solution then
[latex]\vec{b}=s\vec{a_{1}} - t\vec{a_{2}}[/latex] for some [latex]s, t[/latex].

 

g. If [latex]A[/latex] is an [latex]m \times n[/latex] matrix and [latex]m < n[/latex], then [latex]A\vec{x} = \vec{b}[/latex] has a solution for every column [latex]\vec{b}[/latex].

 

h. If [latex]A\vec{x} = \vec{b}[/latex] has a solution for some column [latex]\vec{b}[/latex], then it has a solution for every column [latex]\vec{b}[/latex].

2.2B

 

Definition: 1. A transformation (or function or mapping) [latex]T[/latex] from [latex]\mathbb{R}^n[/latex] to [latex]\mathbb{R}^m[/latex] is a rule that assigns to each vector [latex]\vec{x}[/latex] in [latex]\mathbb{R}^n[/latex] to a vector [latex]T(\vec{x})[/latex] in [latex]\mathbb{R}^m[/latex].

 

2. The set [latex]\mathbb{R}^n[/latex] is called the domain of [latex]T[/latex] and [latex]\mathbb{R}^m[/latex] is called the codomain of [latex]T[/latex].

 

3. [latex]T(\vec{x})[/latex] is called the image of [latex]\vec{x}[/latex]. The set {[latex]T(\vec{x})[/latex] for any [latex]\vec{x}[/latex] in [latex]\mathbb{R}^n[/latex]} is called the range of [latex]T[/latex].

 

4. Given a [latex]m \times n[/latex] matrix [latex]A[/latex], we can define a transformation from
[latex]\mathbb{R}^n[/latex] to [latex]\mathbb{R}^M[/latex] by defining [latex]T(\vec{x}) = A\vec{x}[/latex]. This kind of transformation is called a matrix transformation. [latex]T[/latex] is the matrix transformation induced by A.

 

 

Example 1: Let [latex]A = \begin{bmatrix} 1 & -2\\ -2 & 4\\ 3 & -6 \end{bmatrix}[/latex], [latex]\vec{b} = \begin{bmatrix} 3\\ 4\end{bmatrix}[/latex], [latex]\vec{c} = \begin{bmatrix} 3\\ -6\\ 9\end{bmatrix}[/latex], and [latex]\vec{d} = \begin{bmatrix} 2\\ -6\\ 9\end{bmatrix}[/latex].

Let [latex]T[/latex] be the transformation defined by [latex]T(\vec{x}) = A\vec{x}[/latex] from [latex]\mathbb{R}^2[/latex] to [latex]\mathbb{R}^3[/latex].

(a) Find [latex]T(\vec{b})[/latex].

(b) Find an [latex]\vec{x}[/latex] such that the image under [latex]T[/latex] is [latex]\vec{c}[/latex].

(c) Is there more than one [latex]\vec{x}[/latex] such that whose image under [latex]T[/latex] is [latex]\vec{c}[/latex]?

(d) Is there an [latex]\vec{x}[/latex] such that its image under [latex]T[/latex] is [latex]\vec{d}[/latex]?

 

 

Exercise 1: Let [latex]A = \begin{bmatrix} -1 & 3\\ 2 & -5\\ 2 & -4 \end{bmatrix}[/latex], [latex]\vec{b} = \begin{bmatrix} 5\\ 6\end{bmatrix}[/latex], [latex]\vec{c} = \begin{bmatrix} -2\\ 5\\ 6\end{bmatrix}[/latex], and [latex]\vec{d} = \begin{bmatrix} 2\\ -6\\ 5\end{bmatrix}[/latex].

Let [latex]T[/latex] be the transformation defined by [latex]T(\vec{x}) = A\vec{x}[/latex] from [latex]\mathbb{R}^2[/latex] to [latex]\mathbb{R}^3[/latex].

(a) Find [latex]T(\vec{b})[/latex].

(b) Find an [latex]\vec{x}[/latex] such that the image under [latex]T[/latex] is [latex]\vec{c}[/latex].

(c) Is there more than one [latex]\vec{x}[/latex] such that whose image under [latex]T[/latex] is [latex]\vec{c}[/latex]?

(d) Is there an [latex]\vec{x}[/latex] such that its image under [latex]T[/latex] is [latex]\vec{d}[/latex]?

 

Definition: Suppose [latex]S:\mathbb{R}^n\rightarrow \mathbb{R}^m[/latex] and [latex]T:\mathbb{R}^n\rightarrow \mathbb{R}^m[/latex] are transformations. Then [latex]S = T[/latex] if and only if [latex]S(\vec{x}) = T(\vec{x})[/latex] for every [latex]\vec{x} \in \mathbb{R}^n[/latex].

 

Definition: 1. Given a matrix [latex]A = \begin{bmatrix} 1 & m\\ 0 & 1\end{bmatrix}[/latex], where [latex]m[/latex] is a constant then the transformation [latex]T[/latex] defined by [latex]T(\vec{x}) = A\vec{x}[/latex] is called a shear transformation. This kind of transformation send the vector [latex]\begin{bmatrix} x_{1}\\x_{2}\end{bmatrix}[/latex] to [latex]\begin{bmatrix} x_{1} mx_{2}\\x_{2}\end{bmatrix}[/latex], i.e sending the point [latex](x_{1}, x_{2})[/latex] to the point [latex](x_{1} mx_{2}, x_{2})[/latex]. [latex](x_{1}, x_{2})[/latex] is moving horizontally to the right if [latex]m > 0[/latex] and [latex]x_{2} > 0[/latex] or [latex]m 0[/latex] and [latex]x_{2} 0[/latex], otherwise is moving horizontally to the left.

 

 

2. Let [latex]r[/latex] be a scalar then the transformation [latex]T:\mathbb{R}^n\rightarrow \mathbb{R}^n[/latex] defined by [latex]T(\vec{x}) = r\vec{x}[/latex] is called contraction when [latex]0 \leq r 1[/latex] and dilation when [latex]r > 1[/latex].

 

Example 2: Find the image of [latex]\vec{u} = \begin{bmatrix} 2\\ -3\end{bmatrix}[/latex] and [latex]\vec{v} = \begin{bmatrix} -2\\ 3\end{bmatrix}[/latex] under the matrix transformation defined by [latex]A = \begin{bmatrix} 1 & 4\\0 & 1\end{bmatrix}[/latex]. Then graph the corresponding points on the plane.

 

 

Exercise 2: Find the image of [latex]\vec{u} = \begin{bmatrix} -1\\ 2\end{bmatrix}[/latex] and [latex]\vec{v} = \begin{bmatrix} 2\\ -1\end{bmatrix}[/latex] under the matrix transformation [latex]T[/latex], defined by [latex]A = \begin{bmatrix} 1 & -3\\0 & 1\end{bmatrix}[/latex]. Describe geometrically what [latex]T[/latex] does to each vector.

 

Example 3: Find the image of [latex]\vec{u} = \begin{bmatrix} 1\\ -2\end{bmatrix}[/latex] and [latex]\vec{v} = \begin{bmatrix} -3\\ 1\end{bmatrix}[/latex] under the matrix transformation [latex]T[/latex], defined by [latex]A = \begin{bmatrix} 1 & 0\\0 & -1\end{bmatrix}[/latex]. Then graph the corresponding points on the plane. Describe geometrically what [latex]T[/latex] does to each vector.

 

Exercise 3: Find the image of [latex]\vec{u} = \begin{bmatrix} -1\\ 2\end{bmatrix}[/latex] and [latex]\vec{v} = \begin{bmatrix} 3\\ -5\end{bmatrix}[/latex] under the matrix transformation [latex]T[/latex], defined by [latex]A = \begin{bmatrix} 0 & 1\\1 & 0\end{bmatrix}[/latex]. Then graph the corresponding points on the plane. Describe geometrically what [latex]T[/latex] does to each vector.

 

Definition: The transformation [latex]R_{\theta} : \mathbb{R}^2\rightarrow \mathbb{R}^2[/latex] denotes counterclockwise rotation about the origin through an angle of [latex]\theta[/latex].

 

Example 4: The transformation [latex]R_{\pi/2} : \mathbb{R}^2\rightarrow \mathbb{R}^2[/latex] denotes a counterclockwise rotation about the origin through an angle of [latex]\pi/2[/latex] radians. Find the matrix of [latex]R_{\pi/2}[/latex].

 

 

Exercise 4: The transformation [latex]R_{\pi} : \mathbb{R}^2\rightarrow \mathbb{R}^2[/latex] denotes a counterclockwise rotation about the origin through an angle of [latex]\pi[/latex] radians. Find the matrix of [latex]R_{\pi}[/latex].

 

Example 5: Let [latex]T(\vec{x}) = A\vec{x}[/latex] where [latex]\begin{bmatrix} 1 & 0 & 5\\ 2 & 3 & 0\end{bmatrix}[/latex]. Find a vector [latex]\vec{x}[/latex] such that [latex]T(\vec{x}) = \vec{b} = \begin{bmatrix} 3\\ -1\end{bmatrix}[/latex].

 

 

Exercise 5: Let [latex]T(\vec{x}) = A\vec{x}[/latex] where [latex]\begin{bmatrix} 0 & 1 & -3\\ -2 & 2 & 0\end{bmatrix}[/latex]. Find a vector [latex]\vec{x}[/latex] such that [latex]T(\vec{x}) = \vec{b} = \begin{bmatrix} -2\\ 1\end{bmatrix}[/latex].

 

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

a. A linear transformation is a special type of function.

 

b. If [latex]A[/latex] is a [latex]3 \times 5[/latex] matrix and [latex]T[/latex] is a transformation defined by [latex]T(\vec{x}) = A\vec{x}[/latex] then the domain of [latex]T[/latex] is [latex]\mathbb{R}^5[/latex].

 

c. If [latex]A[/latex] is an [latex]m \times n[/latex] matrix, then the range of the transformation [latex]\vec{x} \rightarrow A\vec{x}[/latex] is [latex]\mathbb{R}^m[/latex].

 

d. The range of the transformation [latex]\vec{x} \rightarrow A\vec{x}[/latex] is the set of all linear combinations of the columns of [latex]A[/latex].

 

e. If [latex]\vec{x_{1}}[/latex] and [latex]\vec{x_{2}}[/latex] are solutions to [latex]A\vec{x} = \vec{b}[/latex], then [latex]\vec{x_{1}}[/latex] – [latex]\vec{x_{2}}[/latex] is a solution to [latex]A\vec{x} = \vec{0}[/latex].

 

f. Let [latex]A = \begin{bmatrix}\vec{a_{1}}\vec{a_{2}}\vec{a_{3}}\end{bmatrix}[/latex] in terms of its columns. If [latex]\vec{a_{3}} = s\vec{a_{1}} t\vec{a_{2}}[/latex], then [latex]A\vec{x} = 0[/latex], where [latex]\vec{x} = \begin{bmatrix}s\\t\\-1\end{bmatrix}[/latex].

 

Group Work 2: Mark each statement True or False. Justify each answer.

Let [latex]T: \mathbb{R}^2\rightarrow \mathbb{R}^2[/latex] be a transformation. In each case show that [latex]T[/latex] is induced by a matrix and find the matrix.

a. [latex]T[/latex] is a reflection in the [latex]y[/latex] axis.

 

b. [latex]T[/latex] is a reflection in the line [latex]y = x[/latex].

 

c. [latex]T[/latex] is a reflection in the line [latex]y = -x[/latex].

 

d. [latex]T[/latex] is a clockwise rotation through [latex]\pi/2[/latex].

 

Group Work 3: In each case either show that the statement is true, or give
an example showing that it is false.

a. If a vector [latex]\vec{b}[/latex] is a linear combination of the columns of [latex]A[/latex], then the system [latex]A\vec{x} = \vec{b}[/latex] is consistent.

 

b. Let [latex]\vec{x_{1}}[/latex] and [latex]\vec{x_{2}}[/latex] be solutions to the homogeneous system [latex]A\vec{x} = \vec{0}[/latex] then [latex]\vec{x_{1}} \vec{x_{2}}[/latex] is a solution to [latex]A\vec{x} = \vec{0}[/latex].

 

c. Let [latex]A[/latex] and [latex]B[/latex] be matrices of the same size. If [latex]\vec{x}[/latex] is a solution to both the system [latex]A\vec{x} = \vec{0}[/latex] and the system [latex]B\vec{x} = \vec{0}[/latex], then [latex]\vec{x}[/latex] is a solution to the system [latex](A B)\vec{x} = \vec{0}[/latex].

 

d. Let [latex]A[/latex] be [latex]m \times n[/latex] and let [latex]\vec{x}[/latex] be in [latex]\mathbb{R}^n[/latex]. If [latex]A\vec{x}[/latex] has a zero entry then [latex]A[/latex] has a row of zeros.