1.7-1.9 Linear Independence & Transformations

Theory: Linear Independence & Linear Transformations

Linear Independence (Lay §1.7)

An indexed set of vectors $\{\mathbf{v}_1, \dots, \mathbf{v}_p\}$ in $\mathbb{R}^n$ is said to be linearly independent if the vector equation:

c_1 \mathbf{v}_1 + c_2 \mathbf{v}_2 + \cdots + c_p \mathbf{v}_p = \mathbf{0}

has only the trivial solution $c_1 = c_2 = \cdots = c_p = 0$ .

The set is linearly dependent if there exist weights $c_1, \dots, c_p$ , not all zero, such that the equation holds.

[!IMPORTANT] Matrix Criterion: The columns of a matrix $A$ are linearly independent if and only if the equation $A\mathbf{x} = \mathbf{0}$ has only the trivial solution — that is, if and only if $A$ has a pivot position in every column (no free variables).

Quick Inspection Rules

Set of two vectors $\{\mathbf{u}, \mathbf{v}\}$ : Linearly dependent if and only if one vector is a scalar multiple of the other.
Set containing the zero vector: Any set $\{\mathbf{v}_1, \dots, \mathbf{v}_p\}$ containing $\mathbf{0}$ is always linearly dependent (Theorem 9).
More vectors than dimensions ( $p > n$ in $\mathbb{R}^n$ ): Any set of $p$ vectors in $\mathbb{R}^n$ with $p > n$ is always linearly dependent (Theorem 8).

Linear Transformations (Lay §1.8 & §1.9)

A transformation $T: \mathbb{R}^n \to \mathbb{R}^m$ is linear if:

$T(\mathbf{u} + \mathbf{v}) = T(\mathbf{u}) + T(\mathbf{v})$ for all $\mathbf{u}, \mathbf{v} \in \mathbb{R}^n$ ,
$T(c\mathbf{u}) = c T(\mathbf{u})$ for all scalars $c$ and all $\mathbf{u} \in \mathbb{R}^n$ .

Every linear transformation $T: \mathbb{R}^n \to \mathbb{R}^m$ is a matrix transformation $T(\mathbf{x}) = A\mathbf{x}$ , where the standard matrix $A$ is given by:

A = \begin{bmatrix} T(\mathbf{e}_1) & T(\mathbf{e}_2) & \cdots & T(\mathbf{e}_n) \end{bmatrix}

where $\mathbf{e}_j$ is the $j$ -th column of the identity matrix $I_n$ .

One-to-One and Onto Mappings (Theorem 11 & 12)

Let $T: \mathbb{R}^n \to \mathbb{R}^m$ be a linear transformation with standard matrix $A$ :

One-to-One (Injective): $T$ is one-to-one $\iff T(\mathbf{x}) = \mathbf{0}$ has only the trivial solution $\iff$ the columns of $A$ are linearly independent $\iff A$ has a pivot in every column.
Onto $\mathbb{R}^m$ (Surjective): $T$ maps $\mathbb{R}^n$ onto $\mathbb{R}^m \iff$ the columns of $A$ span $\mathbb{R}^m$ $\iff A$ has a pivot in every row.

Solved Examples (Textbook Questions)

Exercise 1

(Adapted from Lay §1.7, Exercise 15 & 19)

Determine all values of $h$ for which the following vectors are linearly dependent:

\mathbf{v}_1 = \begin{bmatrix} 1 \\ -3 \\ 2 \end{bmatrix}, \quad \mathbf{v}_2 = \begin{bmatrix} -3 \\ 9 \\ -6 \end{bmatrix}, \quad \mathbf{v}_3 = \begin{bmatrix} 5 \\ -7 \\ h \end{bmatrix}

Explain by inspection (without row reduction) why the set $S = \left\{ \begin{bmatrix} 1 \\ 4 \end{bmatrix}, \begin{bmatrix} -2 \\ 3 \end{bmatrix}, \begin{bmatrix} 5 \\ 6 \end{bmatrix} \right\}$ in $\mathbb{R}^2$ must be linearly dependent.

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Solution

Part 1: Notice immediately that:

\mathbf{v}_2 = \begin{bmatrix} -3 \\ 9 \\ -6 \end{bmatrix} = -3 \begin{bmatrix} 1 \\ -3 \\ 2 \end{bmatrix} = -3 \mathbf{v}_1

Since $\mathbf{v}_2$ is a scalar multiple of $\mathbf{v}_1$ , we have:

3\mathbf{v}_1 + 1\mathbf{v}_2 + 0\mathbf{v}_3 = 3\mathbf{v}_1 + (-3\mathbf{v}_1) + \mathbf{0} = \mathbf{0}

This provides a non-trivial linear combination with weights $c_1 = 3, c_2 = 1, c_3 = 0$ . Therefore, the vectors $\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3$ are linearly dependent for all real values of $h$ ( $h \in \mathbb{R}$ ).

Part 2: The set $S$ contains $p = 3$ vectors in $\mathbb{R}^2$ ( $n = 2$ ). By Theorem 8 (Lay §1.7), if a set contains more vectors than there are entries in each vector ( $p > n$ ), then the set is linearly dependent. Since $3 > 2$ , $S$ is linearly dependent by inspection.

Exercise 2

(Adapted from Lay §1.9, Exercise 3 & 7)

Find the standard matrix $A$ of the linear transformation $T: \mathbb{R}^2 \to \mathbb{R}^2$ that first rotates points through $\pi/2$ radians counterclockwise, and then reflects points through the horizontal $x_1$ -axis.

Then, calculate the image $T(\mathbf{u})$ of $\mathbf{u} = \begin{bmatrix} 3 \\ -4 \end{bmatrix}$ .

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Solution

The standard matrix is $A = [T(\mathbf{e}_1) \; T(\mathbf{e}_2)]$ , where $\mathbf{e}_1 = \begin{bmatrix} 1 \\ 0 \end{bmatrix}$ and $\mathbf{e}_2 = \begin{bmatrix} 0 \\ 1 \end{bmatrix}$ .

Step 1: Track $\mathbf{e}_1 = \begin{bmatrix} 1 \\ 0 \end{bmatrix}$ .

Rotate $\mathbf{e}_1$ counterclockwise by $\pi/2$ ( $90^\circ$ ): $\begin{bmatrix} 1 \\ 0 \end{bmatrix} \xrightarrow{\text{rotate } 90^\circ} \begin{bmatrix} 0 \\ 1 \end{bmatrix}$
Reflect the result across the horizontal axis ( $x_1$ -axis, which sends $(x, y) \mapsto (x, -y)$ ): $\begin{bmatrix} 0 \\ 1 \end{bmatrix} \xrightarrow{\text{reflect}} \begin{bmatrix} 0 \\ -1 \end{bmatrix}$

Therefore, $T(\mathbf{e}_1) = \begin{bmatrix} 0 \\ -1 \end{bmatrix}$ .

Step 2: Track $\mathbf{e}_2 = \begin{bmatrix} 0 \\ 1 \end{bmatrix}$ .

Rotate $\mathbf{e}_2$ counterclockwise by $\pi/2$ : $\begin{bmatrix} 0 \\ 1 \end{bmatrix} \xrightarrow{\text{rotate } 90^\circ} \begin{bmatrix} -1 \\ 0 \end{bmatrix}$
Reflect across the horizontal axis: $\begin{bmatrix} -1 \\ 0 \end{bmatrix} \xrightarrow{\text{reflect}} \begin{bmatrix} -1 \\ 0 \end{bmatrix}$

Therefore, $T(\mathbf{e}_2) = \begin{bmatrix} -1 \\ 0 \end{bmatrix}$ .

Step 3: Construct the standard matrix $A$ .

A = \begin{bmatrix} T(\mathbf{e}_1) & T(\mathbf{e}_2) \end{bmatrix} = \begin{bmatrix} 0 & -1 \\ -1 & 0 \end{bmatrix}

(Note: This is the reflection across the line $x_2 = -x_1$ .)

Step 4: Compute $T(\mathbf{u})$ .

T\left(\begin{bmatrix} 3 \\ -4 \end{bmatrix}\right) = \begin{bmatrix} 0 & -1 \\ -1 & 0 \end{bmatrix} \begin{bmatrix} 3 \\ -4 \end{bmatrix} = \begin{bmatrix} 0(3) + (-1)(-4) \\ -1(3) + 0(-4) \end{bmatrix} = \begin{bmatrix} 4 \\ -3 \end{bmatrix}

Exercise 3

(Adapted from Lay §1.9, Exercise 25)

Let $T: \mathbb{R}^3 \to \mathbb{R}^3$ be the linear transformation defined by:

T(x_1, x_2, x_3) = \begin{bmatrix} x_1 - 2x_2 + 4x_3 \\ 2x_1 - 3x_2 + 5x_3 \\ -3x_1 + 5x_2 - 7x_3 \end{bmatrix}

Find the standard matrix $A$ of $T$ .
Determine whether $T$ is one-to-one.
Determine whether $T$ maps $\mathbb{R}^3$ onto $\mathbb{R}^3$ .

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Solution

Step 1: Write down the standard matrix $A$ . From the coefficients of $x_1, x_2, x_3$ :

A = \begin{bmatrix} 1 & -2 & 4 \\ 2 & -3 & 5 \\ -3 & 5 & -7 \end{bmatrix}

Step 2: Row reduce $A$ to echelon form.

$R_2 \leftarrow R_2 - 2R_1$ : $[2, -3, 5] - 2[1, -2, 4] = [0, 1, -3]$
$R_3 \leftarrow R_3 + 3R_1$ : $[-3, 5, -7] + 3[1, -2, 4] = [0, -1, 5]$

Matrix:

\begin{bmatrix} 1 & -2 & 4 \\ 0 & 1 & -3 \\ 0 & -1 & 5 \end{bmatrix}

$R_3 \leftarrow R_3 + R_2$ : $[0, -1, 5] + [0, 1, -3] = [0, 0, 2]$

The echelon form is:

\begin{bmatrix} 1 & -2 & 4 \\ 0 & 1 & -3 \\ 0 & 0 & 2 \end{bmatrix}

Step 3: Analyze pivots.

Column 1 has a pivot ( $1$ ).
Column 2 has a pivot ( $1$ ).
Column 3 has a pivot ( $2$ ).

Every column has a pivot, and every row has a pivot:

Is $T$ one-to-one? Yes. Since $A$ has a pivot in every column, the columns of $A$ are linearly independent, and $A\mathbf{x} = \mathbf{0}$ has only the trivial solution. By Theorem 12, $T$ is one-to-one.
Does $T$ map $\mathbb{R}^3$ onto $\mathbb{R}^3$ ? Yes. Since $A$ has a pivot in every row (3 pivots in 3 rows), the columns of $A$ span $\mathbb{R}^3$ (Theorem 4). By Theorem 12, $T$ maps $\mathbb{R}^3$ onto $\mathbb{R}^3$ .