Math Core

So far you have read $A\mathbf{x} = \mathbf{b}$ as a question: which $\mathbf{x}$ produce $\mathbf{b}$ ? There is a second, more dynamic reading. The matrix $A$ is a machine that takes in a vector $\mathbf{x}$ and outputs a new vector $A\mathbf{x}$ . Thinking of matrices as functions on vectors is what lets linear algebra describe rotations, projections, computer graphics and much more.

Transformations

Definition

Transformation

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

$\mathbb{R}^n$ is the domain and $\mathbb{R}^m$ is the codomain.
$T(\mathbf{x})$ is the image of $\mathbf{x}$ .
The set of all images $T(\mathbf{x})$ is the range of $T$ .

The range is a subset of the codomain, and it may be much smaller. The codomain just says where outputs live; the range says which outputs actually occur.

Matrix transformations

Every $m \times n$ matrix $A$ defines a transformation $T(\mathbf{x}) = A\mathbf{x}$ , often written $\mathbf{x} \mapsto A\mathbf{x}$ . Since $\mathbf{x}$ must have $n$ entries for $A\mathbf{x}$ to be defined and the result has $m$ entries, the domain is $\mathbb{R}^n$ and the codomain is $\mathbb{R}^m$ . Notice the order: an $m \times n$ matrix maps $\mathbb{R}^n$ to $\mathbb{R}^m$ .

Because $A\mathbf{x}$ is a linear combination of the columns of $A$ , the range of $\mathbf{x} \mapsto A\mathbf{x}$ is exactly the span of the columns of $A$ . Asking "is $\mathbf{b}$ in the range of $T$ ?" is the same as asking "is $A\mathbf{x} = \mathbf{b}$ consistent?"

Worked example: Images and preimages

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

(a) Find $T(\mathbf{u})$ for $\mathbf{u} = (2, -1)$ .

T(\mathbf{u}) = 2\begin{bmatrix} 1 \\ 3 \\ -1 \end{bmatrix} - \begin{bmatrix} -3 \\ 5 \\ 7 \end{bmatrix} = \begin{bmatrix} 5 \\ 1 \\ -9 \end{bmatrix}.

(b) Find an $\mathbf{x}$ whose image is $\mathbf{b} = (3, 2, -5)$ , and decide whether it is unique.

Row reduce the augmented matrix:

\begin{bmatrix} 1 & -3 & 3 \\ 3 & 5 & 2 \\ -1 & 7 & -5 \end{bmatrix} \sim \begin{bmatrix} 1 & -3 & 3 \\ 0 & 14 & -7 \\ 0 & 4 & -2 \end{bmatrix} \sim \begin{bmatrix} 1 & -3 & 3 \\ 0 & 1 & -\tfrac{1}{2} \\ 0 & 0 & 0 \end{bmatrix}.

So $x_2 = -\tfrac{1}{2}$ and $x_1 = 3 + 3x_2 = \tfrac{3}{2}$ . There are no free variables, so $\mathbf{x} = \left(\tfrac{3}{2}, -\tfrac{1}{2}\right)$ is the only vector mapped to $\mathbf{b}$ .

Seeing a transformation

For a map $\mathbb{R}^2 \to \mathbb{R}^2$ , a good way to see what it does is to watch the unit square with corners $(0,0)$ , $(1,0)$ , $(1,1)$ and $(0,1)$ . Take $A = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}$ . Then

A\begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = \begin{bmatrix} x_1 + x_2 \\ x_2 \end{bmatrix}.

Each point slides to the right by an amount equal to its height. The corners go to $(0,0)$ , $(1,0)$ , $(2,1)$ and $(1,1)$ , and the square becomes a parallelogram. This is a shear.

The unit square and its image under the shear, a slanted parallelogram with corners (0,0), (1,0), (2,1), (1,1).Open in grapher →

Other simple matrices give other familiar motions. $\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}$ projects every point straight down onto the $x_1$ -axis. $\begin{bmatrix} 3 & 0 \\ 0 & 3 \end{bmatrix}$ stretches everything away from the origin by a factor of $3$ . $\begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}$ reflects across the $x_1$ -axis.

Linearity

Matrix transformations have two properties that come straight from the algebra of $A\mathbf{x}$ : $A(\mathbf{u} + \mathbf{v}) = A\mathbf{u} + A\mathbf{v}$ and $A(c\mathbf{u}) = cA\mathbf{u}$ . Those two properties turn out to be the essential ones.

Definition

Linear transformation

A transformation $T$ is linear if, for all vectors $\mathbf{u}, \mathbf{v}$ in its domain and all scalars $c$ ,

$T(\mathbf{u} + \mathbf{v}) = T(\mathbf{u}) + T(\mathbf{v})$ , and
$T(c\mathbf{u}) = cT(\mathbf{u})$ .

Every matrix transformation is linear. Two consequences follow immediately from the definition:

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

For the first, $T(\mathbf{0}) = T(0 \cdot \mathbf{0}) = 0 \cdot T(\mathbf{0}) = \mathbf{0}$ . Applying the second property repeatedly gives the superposition principle: $T$ carries any linear combination of inputs to the same linear combination of outputs.

T(c_1\mathbf{v}_1 + \cdots + c_p\mathbf{v}_p) = c_1T(\mathbf{v}_1) + \cdots + c_pT(\mathbf{v}_p).

Linear maps respect linear combinations

If you know what a linear transformation does to a few vectors, you know what it does to every linear combination of them. You don't need a formula for $T$ at all.

Worked example: Using only linearity

$T: \mathbb{R}^2 \to \mathbb{R}^2$ is linear, $T(\mathbf{u}) = (2, 1)$ and $T(\mathbf{v}) = (-1, 3)$ . Find $T(3\mathbf{u} - 2\mathbf{v})$ .

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} 8 \\ -3 \end{bmatrix}.

Worked example: Testing for linearity

Which of these maps $\mathbb{R}^2 \to \mathbb{R}^2$ are linear?

S(x_1, x_2) = (x_1 + 1, \; x_2), \qquad R(x_1, x_2) = (x_1x_2, \; x_2), \qquad U(x_1, x_2) = (4x_1 - x_2, \; 2x_2).

$S(0, 0) = (1, 0) \ne \mathbf{0}$ , so $S$ is not linear. (Translations are not linear.)
$R(0,0) = (0,0)$ , so that quick test doesn't help. Try scaling: $R(2, 2) = (4, 2)$ , but $2R(1,1) = 2(1, 1) = (2, 2)$ . Not linear.
$U(\mathbf{x}) = \begin{bmatrix} 4 & -1 \\ 0 & 2 \end{bmatrix}\mathbf{x}$ is a matrix transformation, so it is linear.

Common mistake

$T(\mathbf{0}) = \mathbf{0}$ is necessary for linearity but not sufficient. The map $R$ above passes that test and still fails. To show a map is linear, either verify both defining properties for arbitrary vectors or write it as $\mathbf{x} \mapsto A\mathbf{x}$ . To show it is not linear, one specific counterexample is enough.

Tip

A formula for $T(\mathbf{x})$ is linear exactly when each output entry is a sum of constant multiples of the input entries: no constant terms, no products of variables, no powers, absolute values or functions like $\sin$ .

Practice

Practice 1

Let $A = \begin{bmatrix} 2 & 0 & -1 \\ 1 & 3 & 4 \end{bmatrix}$ and $T(\mathbf{x}) = A\mathbf{x}$ . Find $T(\mathbf{u})$ for $\mathbf{u} = (1, -2, 3)$ .

Enter a point like (2, -3)

Practice 2

$A$ is a $4 \times 7$ matrix and $T(\mathbf{x}) = A\mathbf{x}$ . Which describes $T$ ?

Practice 3

Which transformation $\mathbb{R}^2 \to \mathbb{R}^2$ is linear?

Practice 4

$T$ is linear with $T(\mathbf{u}) = (1, 4)$ and $T(\mathbf{v}) = (-2, 1)$ . Find $T(2\mathbf{u} + 5\mathbf{v})$ .

Enter a point like (2, -3)

Practice 5

Let $A = \begin{bmatrix} 1 & -2 \\ 2 & -3 \end{bmatrix}$ and $T(\mathbf{x}) = A\mathbf{x}$ . Find the vector $\mathbf{x}$ with $T(\mathbf{x}) = (4, 5)$ .

Enter a point like (2, -3)

Practice 6

Let $A = \begin{bmatrix} 1 & 3 \\ 0 & 1 \\ 2 & 6 \end{bmatrix}$ and $T(\mathbf{x}) = A\mathbf{x}$ . Is $\mathbf{b} = (1, 1, 3)$ in the range of $T$ ?

Practice 7

Find the image of the point $(2, 3)$ under the shear $\mathbf{x} \mapsto \begin{bmatrix} 1 & -2 \\ 0 & 1 \end{bmatrix}\mathbf{x}$ .

Enter a point like (2, -3)

Practice 8

$T: \mathbb{R}^2 \to \mathbb{R}^2$ is linear with $T(1, 1) = (3, 1)$ and $T(1, -1) = (1, 5)$ . Find $T(4, 0)$ .

Enter a point like (2, -3)