Math Core

Every matrix gives a linear transformation. This lesson shows the converse: every linear transformation from $\mathbb{R}^n$ to $\mathbb{R}^m$ is multiplication by some matrix, and you can write that matrix down just by watching what the transformation does to $n$ special vectors. That turns geometric descriptions like "rotate by $30^\circ$ " or "reflect across a line" into concrete matrices you can compute with.

The standard matrix

The columns of the $n \times n$ identity matrix are the standard basis vectors $\mathbf{e}_1, \dots, \mathbf{e}_n$ of $\mathbb{R}^n$ . In $\mathbb{R}^2$ , $\mathbf{e}_1 = (1, 0)$ and $\mathbf{e}_2 = (0, 1)$ . Every vector is a combination of them:

\mathbf{x} = \begin{bmatrix} x_1 \\ \vdots \\ x_n \end{bmatrix} = x_1\mathbf{e}_1 + \cdots + x_n\mathbf{e}_n.

If $T$ is linear, superposition gives

T(\mathbf{x}) = x_1T(\mathbf{e}_1) + \cdots + x_nT(\mathbf{e}_n) = \begin{bmatrix} T(\mathbf{e}_1) & \cdots & T(\mathbf{e}_n) \end{bmatrix}\mathbf{x}.

The standard matrix of a linear transformation

Let $T: \mathbb{R}^n \to \mathbb{R}^m$ be linear. Then there is exactly one $m \times n$ matrix $A$ with $T(\mathbf{x}) = A\mathbf{x}$ for every $\mathbf{x}$ in $\mathbb{R}^n$ , namely

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

$A$ is called the standard matrix of $T$ . Its $j$ th column is the image of the $j$ th standard basis vector.

A linear transformation is completely determined by what it does to $\mathbf{e}_1, \dots, \mathbf{e}_n$ . Find those $n$ images, stack them as columns, and you have the whole transformation.

Worked example: From a formula

Find the standard matrix of $T(x_1, x_2) = (x_1 - 2x_2, \; 3x_1, \; x_1 + x_2)$ .

$T$ maps $\mathbb{R}^2$ to $\mathbb{R}^3$ , so its matrix is $3 \times 2$ . Compute the images of $\mathbf{e}_1$ and $\mathbf{e}_2$ :

T(\mathbf{e}_1) = T(1, 0) = (1, 3, 1), \qquad T(\mathbf{e}_2) = T(0, 1) = (-2, 0, 1).

A = \begin{bmatrix} 1 & -2 \\ 3 & 0 \\ 1 & 1 \end{bmatrix}.

You can also read $A$ straight off the formula: row $i$ of $A$ holds the coefficients of $x_1, x_2$ in the $i$ th output.

Geometric transformations of the plane

For maps of $\mathbb{R}^2$ , find where $\mathbf{e}_1$ and $\mathbf{e}_2$ land, and you're done.

Rotation. Rotating counterclockwise by an angle $\varphi$ sends $\mathbf{e}_1$ to $(\cos\varphi, \sin\varphi)$ and sends $\mathbf{e}_2$ , which starts a quarter turn ahead, to $(-\sin\varphi, \cos\varphi)$ . So

\text{rotation by } \varphi: \quad \begin{bmatrix} \cos\varphi & -\sin\varphi \\ \sin\varphi & \cos\varphi \end{bmatrix}.

The same reasoning gives the whole table below. In each case the columns are the images of $\mathbf{e}_1$ and $\mathbf{e}_2$ .

Transformation	Image of $\mathbf{e}_1$	Image of $\mathbf{e}_2$	Standard matrix
Reflection across the $x_1$ -axis	$(1, 0)$	$(0, -1)$	$\begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}$
Reflection across the $x_2$ -axis	$(-1, 0)$	$(0, 1)$	$\begin{bmatrix} -1 & 0 \\ 0 & 1 \end{bmatrix}$
Reflection across $x_2 = x_1$	$(0, 1)$	$(1, 0)$	$\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$
Reflection across $x_2 = -x_1$	$(0, -1)$	$(-1, 0)$	$\begin{bmatrix} 0 & -1 \\ -1 & 0 \end{bmatrix}$
Reflection through the origin	$(-1, 0)$	$(0, -1)$	$\begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix}$
Horizontal shear by $k$	$(1, 0)$	$(k, 1)$	$\begin{bmatrix} 1 & k \\ 0 & 1 \end{bmatrix}$
Vertical shear by $k$	$(1, k)$	$(0, 1)$	$\begin{bmatrix} 1 & 0 \\ k & 1 \end{bmatrix}$
Horizontal stretch by $k$	$(k, 0)$	$(0, 1)$	$\begin{bmatrix} k & 0 \\ 0 & 1 \end{bmatrix}$
Projection onto the $x_1$ -axis	$(1, 0)$	$(0, 0)$	$\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}$

Don't memorize the table. Rebuild any row in seconds by asking where $\mathbf{e}_1$ and $\mathbf{e}_2$ go.

Worked example: A quarter turn

Find the standard matrix of the counterclockwise rotation by $90^\circ$ , and describe the image of the unit square.

With $\varphi = 90^\circ$ , $\cos\varphi = 0$ and $\sin\varphi = 1$ , so

A = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}, \qquad A\begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = \begin{bmatrix} -x_2 \\ x_1 \end{bmatrix}.

The corners $(0,0)$ , $(1,0)$ , $(1,1)$ , $(0,1)$ go to $(0,0)$ , $(0,1)$ , $(-1,1)$ , $(-1,0)$ : the square swings into the second quadrant.

The unit square in the first quadrant and its image after a 90° counterclockwise rotation, in the second quadrant.Open in grapher →

Composition is multiplication

If you apply $T$ with standard matrix $A$ and then $S$ with standard matrix $B$ , the combined map sends $\mathbf{x}$ to $B(A\mathbf{x}) = (BA)\mathbf{x}$ . So the standard matrix of "first $T$ , then $S$ " is $BA$ : the matrix of the first step goes on the right. This is exactly why matrix multiplication was defined as it was.

Worked example: Reflect, then shear

$T: \mathbb{R}^2 \to \mathbb{R}^2$ first reflects across the line $x_2 = x_1$ and then applies the horizontal shear $\begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix}$ . Find the standard matrix of $T$ .

The reflection has matrix $R = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$ and is applied first, so

A = \begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix}\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} = \begin{bmatrix} 2 & 1 \\ 1 & 0 \end{bmatrix}.

Check with $\mathbf{e}_1$ : the reflection sends it to $\mathbf{e}_2 = (0,1)$ , and the shear sends $(0, 1)$ to $(2, 1)$ , which is the first column of $A$ .

The unit square and its image under T, the parallelogram with corners (0,0), (2,1), (3,1), (1,0).Open in grapher →

Common mistake

Order matters. "First $A$ , then $B$ " is the product $BA$ , not $AB$ . In the example above, doing the shear first and the reflection second gives $\begin{bmatrix} 0 & 1 \\ 1 & 2 \end{bmatrix}$ , a different transformation.

One-to-one and onto

Definition

Onto and one-to-one

A mapping $T: \mathbb{R}^n \to \mathbb{R}^m$ is onto $\mathbb{R}^m$ if every $\mathbf{b}$ in $\mathbb{R}^m$ is the image of at least one $\mathbf{x}$ in $\mathbb{R}^n$ . It is one-to-one if every $\mathbf{b}$ in $\mathbb{R}^m$ is the image of at most one $\mathbf{x}$ .

Onto is an existence question (is $A\mathbf{x} = \mathbf{b}$ always consistent?), and one-to-one is a uniqueness question (can $A\mathbf{x} = \mathbf{b}$ ever have two solutions?). Translating with what you already know about $A\mathbf{x} = \mathbf{b}$ :

$T$ is onto $\mathbb{R}^m$ $\iff$ the columns of $A$ span $\mathbb{R}^m$ $\iff$ $A$ has a pivot in every row.
$T$ 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.

The middle criterion for one-to-one uses linearity: if $T(\mathbf{u}) = T(\mathbf{v})$ with $\mathbf{u} \ne \mathbf{v}$ , then $T(\mathbf{u} - \mathbf{v}) = \mathbf{0}$ with $\mathbf{u} - \mathbf{v} \ne \mathbf{0}$ .

Counting pivots gives quick size limits. A map $\mathbb{R}^n \to \mathbb{R}^m$ with $n > m$ can't be one-to-one (too many columns for each to get a pivot), and one with $n < m$ can't be onto. For $n = m$ , the Invertible Matrix Theorem says one-to-one and onto happen together, exactly when $A$ is invertible.

Worked example: Checking both properties

Let $T(x_1, x_2, x_3) = (x_1 - x_2 + 2x_3, \; 2x_1 + x_2 - x_3)$ . Is $T$ one-to-one? Is it onto $\mathbb{R}^2$ ?

The standard matrix and its echelon form are

A = \begin{bmatrix} 1 & -1 & 2 \\ 2 & 1 & -1 \end{bmatrix} \sim \begin{bmatrix} 1 & -1 & 2 \\ 0 & 3 & -5 \end{bmatrix}.

There is a pivot in both rows, so $T$ is onto $\mathbb{R}^2$ . Column 3 has no pivot, so $x_3$ is free, $T(\mathbf{x}) = \mathbf{0}$ has nontrivial solutions, and $T$ is not one-to-one.

Tip

To check a standard matrix you built, test it on one vector that is not $\mathbf{e}_1$ or $\mathbf{e}_2$ . For example, a rotation by $90^\circ$ should send $(1, 1)$ to $(-1, 1)$ , and indeed $\begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}\begin{bmatrix} 1 \\ 1 \end{bmatrix} = \begin{bmatrix} -1 \\ 1 \end{bmatrix}$ .

Practice

Practice 1

Let $T(x_1, x_2) = (2x_1 + x_2, \; x_1 - 3x_2, \; -x_2)$ . What is the second column of the standard matrix of $T$ ? Enter it as $(a, b, c)$ .

Enter a point like (2, -3)

Practice 2

$T: \mathbb{R}^2 \to \mathbb{R}^2$ is linear with $T(\mathbf{e}_1) = (1, 2)$ and $T(\mathbf{e}_2) = (-1, 4)$ . Find $T(3, -2)$ .

Enter a point like (2, -3)

Practice 3

Find the image of $(2, 0)$ under the counterclockwise rotation of $\mathbb{R}^2$ by $60^\circ$ . Enter exact values; you may type sqrt(3).

Enter a point like (2, -3)

Practice 4

Find the image of $(3, 1)$ under reflection across the line $x_2 = -x_1$ .

Enter a point like (2, -3)

Practice 5

$T$ first rotates $\mathbb{R}^2$ counterclockwise by $90^\circ$ and then reflects across the $x_1$ -axis. Which single transformation is $T$ ?

Practice 6

Let $T(\mathbf{x}) = A\mathbf{x}$ with $A = \begin{bmatrix} 1 & 2 \\ 0 & 1 \\ 3 & -1 \end{bmatrix}$ . Which statement is true?

Practice 7

$T: \mathbb{R}^5 \to \mathbb{R}^3$ is linear, and its standard matrix has $3$ pivot positions. Which statement is true?

Practice 8

A horizontal shear $\begin{bmatrix} 1 & k \\ 0 & 1 \end{bmatrix}$ sends the point $(1, 2)$ to $(7, 2)$ . Find $k$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.