1.8 Introduction to Linear Transformations

Theory: Introduction to Linear Transformations

Transformations (Lay §1.8)

A transformation (function, mapping) $T$ from $\mathbb{R}^n$ to $\mathbb{R}^m$ , written $T: \mathbb{R}^n \to \mathbb{R}^m$ , assigns to each $\mathbf{x}$ in $\mathbb{R}^n$ a vector $T(\mathbf{x})$ in $\mathbb{R}^m$ .

\operatorname{Range}(T) = \{\, \underbrace{T(\mathbf{x})}_{\text{image}} \in \underbrace{\mathbb{R}^m}_{\text{codomain}} \;:\; \mathbf{x} \in \underbrace{\mathbb{R}^n}_{\text{domain}} \,\}

Domain $\mathbb{R}^n$ : where the inputs live.
Codomain $\mathbb{R}^m$ : where the outputs are allowed to live.
Image of $\mathbf{x}$ : the output $T(\mathbf{x})$ .
Range: the set of all images — a subset of the codomain (possibly smaller than it).

Matrix transformations

If $A$ is $m \times n$ , then $T(\mathbf{x}) = A\mathbf{x}$ defines $T: \mathbb{R}^n \to \mathbb{R}^m$ (number of columns = input size, number of rows = output size). The range of $T$ is the set of all linear combinations of the columns of $A$ , i.e. the span of the columns.

How to translate the typical questions about $T(\mathbf{x}) = A\mathbf{x}$ :

Question	Becomes
Find the image of $\mathbf{u}$	compute $A\mathbf{u}$
Find $\mathbf{x}$ whose image is $\mathbf{b}$	solve $A\mathbf{x} = \mathbf{b}$ (row reduce $[A \ \mathbf{b}]$ )
Is there more than one such $\mathbf{x}$ ?	does $A\mathbf{x} = \mathbf{b}$ have free variables? (unique?)
Is $\mathbf{c}$ in the range of $T$ ?	is $A\mathbf{x} = \mathbf{c}$ consistent?

Linear Transformations

Definition (Lay §1.8): A transformation $T$ is linear if

$T(\mathbf{u} + \mathbf{v}) = T(\mathbf{u}) + T(\mathbf{v})$ for all $\mathbf{u}, \mathbf{v}$ in the domain of $T$ ;

$T(c\mathbf{u}) = cT(\mathbf{u})$ for all scalars $c$ and all $\mathbf{u}$ in the domain of $T$ .

Every matrix transformation is linear (because $A(\mathbf{u} + \mathbf{v}) = A\mathbf{u} + A\mathbf{v}$ and $A(c\mathbf{u}) = cA\mathbf{u}$ ). Consequences of the definition:

T(\mathbf{0}) = \mathbf{0} \qquad\qquad T(c\mathbf{u} + d\mathbf{v}) = cT(\mathbf{u}) + dT(\mathbf{v})

and more generally $T(c_1\mathbf{v}_1 + \cdots + c_p\mathbf{v}_p) = c_1T(\mathbf{v}_1) + \cdots + c_pT(\mathbf{v}_p)$ (the superposition principle). In words: a linear transformation preserves linear combinations. So if you know $T$ on a few vectors, you know it on every combination of them (see Exercise 3 below).

Quick “not linear” test: if $T(\mathbf{0}) \ne \mathbf{0}$ , then $T$ is not linear. For example $T(x_1, x_2) = (x_1 + 1, x_2)$ sends $\mathbf{0}$ to $(1, 0)$ , so it is not linear. (The converse fails: $T(\mathbf{0}) = \mathbf{0}$ alone does not prove linearity.)

Solved Examples (Textbook Questions)

Exercise 1

(Adapted from Lay §1.8, Example 1)

Let $A = \begin{bmatrix} 1 & -3 \\ 3 & 5 \\ -1 & 7 \end{bmatrix}$ , $\mathbf{u} = \begin{bmatrix} 2 \\ -1 \end{bmatrix}$ , $\mathbf{b} = \begin{bmatrix} 3 \\ 2 \\ -5 \end{bmatrix}$ , $\mathbf{c} = \begin{bmatrix} 3 \\ 2 \\ 5 \end{bmatrix}$ , and define $T: \mathbb{R}^2 \to \mathbb{R}^3$ by $T(\mathbf{x}) = A\mathbf{x}$ .

(a) Find $T(\mathbf{u})$ . (b) Find an $\mathbf{x}$ whose image under $T$ is $\mathbf{b}$ . (c) Is there more than one such $\mathbf{x}$ ? (d) Is $\mathbf{c}$ in the range of $T$ ?

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Solution

Translate first: (b) $T(\mathbf{x}) = \mathbf{b}$ means solve $A\mathbf{x} = \mathbf{b}$ ; (c) asks is that solution unique?; (d) asks is $A\mathbf{x} = \mathbf{c}$ consistent?

(a)

T(\mathbf{u}) = A\mathbf{u} = \begin{bmatrix} 1 & -3 \\ 3 & 5 \\ -1 & 7 \end{bmatrix}\begin{bmatrix} 2 \\ -1 \end{bmatrix} = \begin{bmatrix} 2 + 3 \\ 6 - 5 \\ -2 - 7 \end{bmatrix} = \begin{bmatrix} 5 \\ 1 \\ -9 \end{bmatrix}

(b) Row reduce $[A \ \mathbf{b}]$ : $R_2 \leftarrow R_2 - 3R_1$ , $R_3 \leftarrow R_3 + R_1$ ; then $R_2 \leftarrow \tfrac{1}{14}R_2$ , $R_3 \leftarrow R_3 - 4R_2$ ; then $R_1 \leftarrow R_1 + 3R_2$ :

\left[\begin{array}{cc|c} 1 & -3 & 3 \\ 3 & 5 & 2 \\ -1 & 7 & -5 \end{array}\right] \sim \left[\begin{array}{cc|c} 1 & -3 & 3 \\ 0 & 14 & -7 \\ 0 & 4 & -2 \end{array}\right] \sim \left[\begin{array}{cc|c} 1 & -3 & 3 \\ 0 & 1 & -0.5 \\ 0 & 0 & 0 \end{array}\right] \sim \left[\begin{array}{cc|c} 1 & 0 & 1.5 \\ 0 & 1 & -0.5 \\ 0 & 0 & 0 \end{array}\right]

So $\mathbf{x} = \begin{bmatrix} 1.5 \\ -0.5 \end{bmatrix}$ .

(c) Both columns have pivots → no free variables → the solution is unique. Exactly one $\mathbf{x}$ maps to $\mathbf{b}$ .

(d) Row reduce $[A \ \mathbf{c}]$ (same first step, then swap $R_2 \leftrightarrow R_3$ and scale the new $R_2$ by $\tfrac14$ , then $R_3 \leftarrow R_3 - 14R_2$ ):

\left[\begin{array}{cc|c} 1 & -3 & 3 \\ 3 & 5 & 2 \\ -1 & 7 & 5 \end{array}\right] \sim \left[\begin{array}{cc|c} 1 & -3 & 3 \\ 0 & 14 & -7 \\ 0 & 4 & 8 \end{array}\right] \sim \left[\begin{array}{cc|c} 1 & -3 & 3 \\ 0 & 1 & 2 \\ 0 & 14 & -7 \end{array}\right] \sim \left[\begin{array}{cc|c} 1 & -3 & 3 \\ 0 & 1 & 2 \\ 0 & 0 & -35 \end{array}\right]

Row 3 says $0 = -35$ → inconsistent → $\mathbf{c}$ is not in the range of $T$ . (So $T$ is not onto $\mathbb{R}^3$ — as expected for a $3 \times 2$ matrix.)

Exercise 2

(Adapted from Lay §1.8, Exercises 3–6)

With $T$ defined by $T(\mathbf{x}) = A\mathbf{x}$ , find a vector $\mathbf{x}$ whose image under $T$ is $\mathbf{b}$ , and determine whether $\mathbf{x}$ is unique.

3. $A = \begin{bmatrix} 1 & 0 & -2 \\ -2 & 1 & 6 \\ 3 & -2 & -5 \end{bmatrix}, \ \mathbf{b} = \begin{bmatrix} -1 \\ 7 \\ -3 \end{bmatrix}$ $\qquad$ 4. $A = \begin{bmatrix} 1 & -3 & 2 \\ 0 & 1 & -4 \\ 3 & -5 & -9 \end{bmatrix}, \ \mathbf{b} = \begin{bmatrix} 6 \\ -7 \\ -9 \end{bmatrix}$

5. $A = \begin{bmatrix} 1 & -5 & -7 \\ -3 & 7 & 5 \end{bmatrix}, \ \mathbf{b} = \begin{bmatrix} -2 \\ -2 \end{bmatrix}$ $\qquad$ 6. $A = \begin{bmatrix} 1 & -2 & 1 \\ 3 & -4 & 5 \\ 0 & 1 & 1 \\ -3 & 5 & -4 \end{bmatrix}, \ \mathbf{b} = \begin{bmatrix} 1 \\ 9 \\ 3 \\ -6 \end{bmatrix}$

Show solution ↓

Solution

In every part: row reduce $[A \ \mathbf{b}]$ to reduced echelon form; unique iff no free variables.

3. $R_2 \leftarrow R_2 + 2R_1$ , $R_3 \leftarrow R_3 - 3R_1$ ; then $R_3 \leftarrow R_3 + 2R_2$ ; then $R_3 \leftarrow \tfrac15 R_3$ and back-substitute ( $R_2 \leftarrow R_2 - 2R_3$ , $R_1 \leftarrow R_1 + 2R_3$ ):

\left[\begin{array}{ccc|c} 1 & 0 & -2 & -1 \\ -2 & 1 & 6 & 7 \\ 3 & -2 & -5 & -3 \end{array}\right] \sim \left[\begin{array}{ccc|c} 1 & 0 & -2 & -1 \\ 0 & 1 & 2 & 5 \\ 0 & -2 & 1 & 0 \end{array}\right] \sim \left[\begin{array}{ccc|c} 1 & 0 & -2 & -1 \\ 0 & 1 & 2 & 5 \\ 0 & 0 & 5 & 10 \end{array}\right] \sim \left[\begin{array}{ccc|c} 1 & 0 & 0 & 3 \\ 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & 2 \end{array}\right]

$\mathbf{x} = (3, 1, 2)$ , unique (pivot in every column).

4. $R_3 \leftarrow R_3 - 3R_1$ ; then $R_3 \leftarrow R_3 - 4R_2$ ; then back-substitute ( $R_2 \leftarrow R_2 + 4R_3$ , $R_1 \leftarrow R_1 - 2R_3$ , then $R_1 \leftarrow R_1 + 3R_2$ ):

\left[\begin{array}{ccc|c} 1 & -3 & 2 & 6 \\ 0 & 1 & -4 & -7 \\ 3 & -5 & -9 & -9 \end{array}\right] \sim \left[\begin{array}{ccc|c} 1 & -3 & 2 & 6 \\ 0 & 1 & -4 & -7 \\ 0 & 4 & -15 & -27 \end{array}\right] \sim \left[\begin{array}{ccc|c} 1 & -3 & 2 & 6 \\ 0 & 1 & -4 & -7 \\ 0 & 0 & 1 & 1 \end{array}\right] \sim \left[\begin{array}{ccc|c} 1 & 0 & 0 & -5 \\ 0 & 1 & 0 & -3 \\ 0 & 0 & 1 & 1 \end{array}\right]

$\mathbf{x} = (-5, -3, 1)$ , unique.

5. $R_2 \leftarrow R_2 + 3R_1$ gives $(0, -8, -16 \mid -8)$ ; $R_2 \leftarrow -\tfrac18 R_2$ ; $R_1 \leftarrow R_1 + 5R_2$ :

\left[\begin{array}{ccc|c} 1 & -5 & -7 & -2 \\ -3 & 7 & 5 & -2 \end{array}\right] \sim \left[\begin{array}{ccc|c} 1 & -5 & -7 & -2 \\ 0 & 1 & 2 & 1 \end{array}\right] \sim \left[\begin{array}{ccc|c} 1 & 0 & 3 & 3 \\ 0 & 1 & 2 & 1 \end{array}\right]

Write the equations: $x_1 + 3x_3 = 3$ , $x_2 + 2x_3 = 1$ , $x_3$ free. So

\mathbf{x} = \begin{bmatrix} 3 - 3x_3 \\ 1 - 2x_3 \\ x_3 \end{bmatrix} = \begin{bmatrix} 3 \\ 1 \\ 0 \end{bmatrix} + x_3\begin{bmatrix} -3 \\ -2 \\ 1 \end{bmatrix}

One choice ( $x_3 = 0$ ): $\mathbf{x} = (3, 1, 0)$ . Not unique — $x_3$ is free (as must happen for a $2 \times 3$ matrix).

A solution is not $(3, 1)$ ! $T$ eats vectors in $\mathbb{R}^3$ , so $\mathbf{x}$ needs 3 entries. Always write out the equations and solve for the basic variables.

6. $R_2 \leftarrow R_2 - 3R_1$ , $R_4 \leftarrow R_4 + 3R_1$ :

\left[\begin{array}{ccc|c} 1 & -2 & 1 & 1 \\ 3 & -4 & 5 & 9 \\ 0 & 1 & 1 & 3 \\ -3 & 5 & -4 & -6 \end{array}\right] \sim \left[\begin{array}{ccc|c} 1 & -2 & 1 & 1 \\ 0 & 2 & 2 & 6 \\ 0 & 1 & 1 & 3 \\ 0 & -1 & -1 & -3 \end{array}\right] \sim \left[\begin{array}{ccc|c} 1 & 0 & 3 & 7 \\ 0 & 1 & 1 & 3 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right]

(Second step: $R_2 \leftarrow \tfrac12 R_2$ , then $R_3 \leftarrow R_3 - R_2$ , $R_4 \leftarrow R_4 + R_2$ , $R_1 \leftarrow R_1 + 2R_2$ .) So $x_1 = 7 - 3x_3$ , $x_2 = 3 - x_3$ , $x_3$ free:

\mathbf{x} = \begin{bmatrix} 7 \\ 3 \\ 0 \end{bmatrix} + x_3\begin{bmatrix} -3 \\ -1 \\ 1 \end{bmatrix}

One choice: $\mathbf{x} = (7, 3, 0)$ . Not unique.

Exercise 3

(Adapted from Lay §1.8, Exercises 17, 19, 20)

17. $T: \mathbb{R}^2 \to \mathbb{R}^2$ is linear and maps $\mathbf{u} = \begin{bmatrix} 5 \\ 2 \end{bmatrix}$ into $\begin{bmatrix} 2 \\ 1 \end{bmatrix}$ and $\mathbf{v} = \begin{bmatrix} 1 \\ 3 \end{bmatrix}$ into $\begin{bmatrix} -1 \\ 3 \end{bmatrix}$ . Find the images of $3\mathbf{u}$ , $2\mathbf{v}$ , and $3\mathbf{u} + 2\mathbf{v}$ .

19. Let $\mathbf{e}_1 = \begin{bmatrix} 1 \\ 0 \end{bmatrix}$ , $\mathbf{e}_2 = \begin{bmatrix} 0 \\ 1 \end{bmatrix}$ , $\mathbf{y}_1 = \begin{bmatrix} 2 \\ 5 \end{bmatrix}$ , $\mathbf{y}_2 = \begin{bmatrix} -1 \\ 6 \end{bmatrix}$ , and let $T: \mathbb{R}^2 \to \mathbb{R}^2$ be linear with $T(\mathbf{e}_1) = \mathbf{y}_1$ , $T(\mathbf{e}_2) = \mathbf{y}_2$ . Find the images of $\begin{bmatrix} 5 \\ -3 \end{bmatrix}$ and $\begin{bmatrix} x_1 \\ x_2 \end{bmatrix}$ .

20. Let $\mathbf{x} = \begin{bmatrix} x_1 \\ x_2 \end{bmatrix}$ , $\mathbf{v}_1 = \begin{bmatrix} -2 \\ 5 \end{bmatrix}$ , $\mathbf{v}_2 = \begin{bmatrix} 7 \\ -3 \end{bmatrix}$ , and let $T: \mathbb{R}^2 \to \mathbb{R}^2$ map $\mathbf{x}$ into $x_1\mathbf{v}_1 + x_2\mathbf{v}_2$ . Find a matrix $A$ such that $T(\mathbf{x}) = A\mathbf{x}$ for each $\mathbf{x}$ .

Show solution ↓

Solution

All three use the same idea: linearity lets you pull out scalars and split sums.

17.

T(3\mathbf{u}) = 3T(\mathbf{u}) = 3\begin{bmatrix} 2 \\ 1 \end{bmatrix} = \begin{bmatrix} 6 \\ 3 \end{bmatrix}, \qquad T(2\mathbf{v}) = 2T(\mathbf{v}) = 2\begin{bmatrix} -1 \\ 3 \end{bmatrix} = \begin{bmatrix} -2 \\ 6 \end{bmatrix}

T(3\mathbf{u} + 2\mathbf{v}) = 3T(\mathbf{u}) + 2T(\mathbf{v}) = \begin{bmatrix} 6 \\ 3 \end{bmatrix} + \begin{bmatrix} -2 \\ 6 \end{bmatrix} = \begin{bmatrix} 4 \\ 9 \end{bmatrix}

19. Write the input as a combination of $\mathbf{e}_1, \mathbf{e}_2$ : $\begin{bmatrix} 5 \\ -3 \end{bmatrix} = 5\mathbf{e}_1 - 3\mathbf{e}_2$ . Then

T\left(\begin{bmatrix} 5 \\ -3 \end{bmatrix}\right) = 5\mathbf{y}_1 - 3\mathbf{y}_2 = \begin{bmatrix} 10 \\ 25 \end{bmatrix} - \begin{bmatrix} -3 \\ 18 \end{bmatrix} = \begin{bmatrix} 13 \\ 7 \end{bmatrix}

In general $\mathbf{x} = x_1\mathbf{e}_1 + x_2\mathbf{e}_2$ , so

T(\mathbf{x}) = x_1\mathbf{y}_1 + x_2\mathbf{y}_2 = x_1\begin{bmatrix} 2 \\ 5 \end{bmatrix} + x_2\begin{bmatrix} -1 \\ 6 \end{bmatrix} = \begin{bmatrix} 2x_1 - x_2 \\ 5x_1 + 6x_2 \end{bmatrix}

(This is a preview of Theorem 10: the standard matrix is $[\mathbf{y}_1 \ \mathbf{y}_2] = \begin{bmatrix} 2 & -1 \\ 5 & 6 \end{bmatrix}$ .)

20. By the definition of a matrix–vector product, $x_1\mathbf{v}_1 + x_2\mathbf{v}_2 = [\mathbf{v}_1 \ \mathbf{v}_2]\mathbf{x}$ . So

T(\mathbf{x}) = \begin{bmatrix} -2 & 7 \\ 5 & -3 \end{bmatrix}\begin{bmatrix} x_1 \\ x_2 \end{bmatrix}, \qquad A = \begin{bmatrix} -2 & 7 \\ 5 & -3 \end{bmatrix}