Matrices are transformations

Day: 5 of 40
Length: 2:21
Textbook: Lay §1.8, 1.9
Syllabus: Groundwork before Lesson 10

A matrix moves every point of the plane, and its columns record where the two basis arrows land.

Yesterday you multiplied a matrix by one vector at a time. Today asks what a matrix does to every vector at once. Read that way, the product is a rule: feed in any $\yellow{\mathbf x}$ and the matrix hands back $A\yellow{\mathbf x}$ , and doing that to every point at once moves the whole plane. The rest of the course leans on this picture of a matrix as a motion, starting tomorrow, when Day 6 runs two motions in a row.

Each section below opens with a problem. Try it before you open the walkthrough, even if you only get partway.

Transformations

Let $A = \begin{bmatrix} \green{1} & \red{1} \\ \green{0} & \red{1} \end{bmatrix}$ . Compute $A\yellow{\mathbf x}$ for $\yellow{\mathbf x} = (1, 2)$ . Then compute $A(0, 2)$ and $A(3, 0)$ . What does $A$ do to a point, in words?

Walk through the solution

By the row rule, $A(x_1, x_2) = (x_1 + x_2,\ x_2)$ . So

A\begin{bmatrix} 1 \\ 2 \end{bmatrix} = \begin{bmatrix} 3 \\ 2 \end{bmatrix}, \qquad A\begin{bmatrix} 0 \\ 2 \end{bmatrix} = \begin{bmatrix} 2 \\ 2 \end{bmatrix}, \qquad A\begin{bmatrix} 3 \\ 0 \end{bmatrix} = \begin{bmatrix} 3 \\ 0 \end{bmatrix}.

Look at what changed. The second entry, the height, never changes. The first entry grows by the height: $(1, 2)$ moved $2$ to the right, $(0, 2)$ also moved $2$ to the right, and $(3, 0)$ sits at height $0$ and did not move at all. So $A$ slides each point sideways by its height.

A shear. The blue grid lines lean to the right while the horizontal lines stay put. The green arrow e1 stays at (1, 0), the red arrow e2 lands on (1, 1), and the yellow arrow Ax ends at (3, 2). The matrix A has green first column (1, 0) and red second column (1, 1).

Doing that to every point moves the whole grid. This matrix is a shear : the base line stays still and the grid leans over. The grid lines stay straight, parallel and evenly spaced, and the origin stays put.

Transformation

A transformation $T$ from $\mathbb{R}^n$ to $\mathbb{R}^m$ is a rule that assigns to each vector $\mathbf x$ in $\mathbb{R}^n$ a vector $T(\mathbf x)$ in $\mathbb{R}^m$ . We write $T : \mathbb{R}^n \to \mathbb{R}^m$ . The set $\mathbb{R}^n$ is the domain, $\mathbb{R}^m$ is the codomain, and $T(\mathbf x)$ is the image of $\mathbf x$ . The set of all images is the range of $T$ .

When $A$ is an $m \times n$ matrix, $T(\mathbf x) = A\mathbf x$ is a matrix transformation. Its domain is $\mathbb{R}^n$ because $\mathbf x$ needs one entry per column of $A$ , and its codomain is $\mathbb{R}^m$ because each column has $m$ entries.

Let $A$ be a $2 \times 3$ matrix. The map $T(\mathbf x) = A\mathbf x$ goes

The standard matrix

Often you know a move in words, such as a quarter turn, and want its matrix. This section finds a recipe for that matrix that needs no algebra.

Turn the plane a quarter turn counterclockwise about the origin. Where do $\green{\mathbf e_1} = (1, 0)$ and $\red{\mathbf e_2} = (0, 1)$ land? Use that to find where $(2, 1)$ lands.

Walk through the solution

The arrow $\green{\mathbf e_1}$ points right, and a quarter turn counterclockwise makes it point up, so it lands on $(0, 1)$ . The arrow $\red{\mathbf e_2}$ points up and turns to point left, so it lands on $(-1, 0)$ .

Now write $(2, 1)$ as a walk made of those arrows: $(2, 1) = 2\green{\mathbf e_1} + 1\red{\mathbf e_2}$ . The turn carries the whole walk along rigidly, so the turned point is two turned $\green{\mathbf e_1}$ steps plus one turned $\red{\mathbf e_2}$ step:

2\begin{bmatrix} 0 \\ 1 \end{bmatrix} + 1\begin{bmatrix} -1 \\ 0 \end{bmatrix} = \begin{bmatrix} -1 \\ 2 \end{bmatrix}.

By Day 4, that combination is a matrix times $(2, 1)$ , and the matrix has the two landing spots as its columns:

\begin{bmatrix} \green{0} & \red{-1} \\ \green{1} & \red{0} \end{bmatrix}\begin{bmatrix} 2 \\ 1 \end{bmatrix} = \begin{bmatrix} -1 \\ 2 \end{bmatrix}.

A quarter turn counterclockwise. The green arrow e1 now points up to (0, 1) and the red arrow e2 points left to (−1, 0). Their coordinates form the columns of A = [[0, −1], [1, 0]].

\mathbf x

Theorem 10

Let $T : \mathbb{R}^n \to \mathbb{R}^m$ be a linear transformation. Then there is exactly one 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) & \cdots & T(\mathbf e_n) \end{bmatrix},

where $\mathbf e_j$ is column $j$ of the $n \times n$ identity matrix. This $A$ is the standard matrix of $T$ .

The word "linear" is defined precisely in the section after next. For now it means the map carries walks along the way the quarter turn did.

Drag the green and red tips to build a quarter turn clockwise. Picture where

\mathbf e_1

and

\mathbf e_2

should end up first.

\begin{bmatrix} \textcolor{#7BC96F}{1} & \textcolor{#F2665E}{0} \\ \textcolor{#7BC96F}{0} & \textcolor{#F2665E}{1} \end{bmatrix}

Rotation

The quarter turn was one angle. The same recipe gives a turn through any angle $\varphi$ .

Rotate the plane counterclockwise through an angle $\varphi$ . Where do $\green{\mathbf e_1}$ and $\red{\mathbf e_2}$ land? Write the matrix, and check it against Problem 2 with $\varphi = 90^\circ$ .

Walk through the solution

The arrow $\green{\mathbf e_1}$ has length $1$ and ends up at angle $\varphi$ from the positive $x_1$ -axis. A point on the unit circle at angle $\varphi$ is $(\cos\varphi, \sin\varphi)$ , so that is where $\green{\mathbf e_1}$ lands.

The arrow $\red{\mathbf e_2}$ starts a quarter turn ahead of $\green{\mathbf e_1}$ , so it ends at angle $\varphi + 90^\circ$ . Turning $(\cos\varphi, \sin\varphi)$ by a quarter turn gives $(-\sin\varphi, \cos\varphi)$ , by the same move as Problem 2. The matrix is

\begin{bmatrix} \green{\cos\varphi} & \red{-\sin\varphi} \\ \green{\sin\varphi} & \red{\cos\varphi} \end{bmatrix}.

At $\varphi = 90^\circ$ , $\cos\varphi = 0$ and $\sin\varphi = 1$ , so the matrix is $\begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}$ , the quarter turn from Problem 2.

Find the angle that makes the F land on its dashed outline. Read the first column of the target matrix as a point on the dashed unit circle.

Slide the angle

\varphi

, in degrees, until the F covers the dashed outline. The outline is the F moved by

\begin{bmatrix} \textcolor{#7BC96F}{-0.71} & \textcolor{#F2665E}{-0.71} \\ \textcolor{#7BC96F}{0.71} & \textcolor{#F2665E}{-0.71} \end{bmatrix}

\varphi = 0

\begin{bmatrix} \textcolor{#7BC96F}{\cos\varphi} & \textcolor{#F2665E}{-\sin\varphi} \\ \textcolor{#7BC96F}{\sin\varphi} & \textcolor{#F2665E}{\cos\varphi} \end{bmatrix}

= \begin{bmatrix} \textcolor{#7BC96F}{1} & \textcolor{#F2665E}{0} \\ \textcolor{#7BC96F}{0} & \textcolor{#F2665E}{1} \end{bmatrix}

Linearity

Problem 2 relied on the turn carrying a walk of $\green{\mathbf e_1}$ and $\red{\mathbf e_2}$ steps along with it. Naming that property tells you exactly which moves a matrix can make.

A map $T$ carries walks along, and it sends $\green{\mathbf e_1}$ to $(1, 2)$ and $\red{\mathbf e_2}$ to $(-3, 1)$ . Find $T(4, -1)$ and the standard matrix of $T$ .

Walk through the solution

Write the input as a walk: $(4, -1) = 4\green{\mathbf e_1} - 1\red{\mathbf e_2}$ . The map moves each step, and the walk follows:

\begin{aligned} T\begin{bmatrix} 4 \\ -1 \end{bmatrix} &= 4\,\green{T(\mathbf e_1)} - 1\,\red{T(\mathbf e_2)} \\ &= \begin{bmatrix} 4 \\ 8 \end{bmatrix} - \begin{bmatrix} -3 \\ 1 \end{bmatrix} = \begin{bmatrix} 7 \\ 7 \end{bmatrix}. \end{aligned}

The standard matrix has the two images as columns, $\begin{bmatrix} \green{1} & \red{-3} \\ \green{2} & \red{1} \end{bmatrix}$ . Check with the row rule: $1 \cdot 4 + (-3)(-1) = 7$ and $2 \cdot 4 + 1 \cdot (-1) = 7$ .

The grid moved by the matrix [[2, −1], [1, 1]]. Two green steps T(e1) = (2, 1) and one red step T(e2) = (−1, 1) are placed tip to tail, and the yellow arrow T(x) runs from the origin to their end at (3, 3).

\yellow{\mathbf x} = (2, 1)

Linear transformation

A transformation $T$ is linear when, for all vectors $\mathbf u, \mathbf v$ in its domain and every scalar $c$ ,

(i) $T(\mathbf u + \mathbf v) = T(\mathbf u) + T(\mathbf v)$ , and

(ii) $T(c\,\mathbf u) = c\,T(\mathbf u)$ .

Every matrix transformation is linear, because $A(\mathbf u + \mathbf v) = A\mathbf u + A\mathbf v$ and $A(c\mathbf u) = cA\mathbf u$ from Day 4.

What linearity guarantees

Using (i) and then (ii) gives $T(c\,\mathbf u + d\,\mathbf v) = c\,T(\mathbf u) + d\,T(\mathbf v)$ . So the images of any two vectors determine the images of all their combinations, as in Problem 4.

Taking $c = 0$ in (ii) gives $T(\mathbf 0) = 0\,T(\mathbf u) = \mathbf 0$ . A linear map never moves the origin. A slide such as $T(\mathbf x) = \mathbf x + (1, 1)$ keeps grid lines straight, but it sends $\mathbf 0$ to $(1, 1)$ , so it is not linear and no matrix can do it. On Day 20 these same two rules define a linear map between any vector spaces, so differentiation counts too.

The vectors $\mathbf u$ and $\mathbf v$ below are not $\green{\mathbf e_1}$ and $\red{\mathbf e_2}$ , but the same reasoning works for them.

You only know where

T

sends

\mathbf u

and

\mathbf v

. Drag the teal point to where

T(2\,\mathbf u + 1\,\mathbf v)

must land.

T(\mathbf u) = \textcolor{#FFD447}{\begin{bmatrix} 2 \\ 1 \end{bmatrix}}

T(\mathbf v) = \textcolor{#5BC0EB}{\begin{bmatrix} -1 \\ 2 \end{bmatrix}}

\text{guess} = \textcolor{#3DD6B5}{\begin{bmatrix} 0 \\ 0 \end{bmatrix}}

Which of these maps is linear?

The range

Going backwards asks which $\mathbf x$ has a given image $\mathbf b$ . That is Day 4's equation $A\mathbf x = \mathbf b$ , seen as a picture.

Let $T(\mathbf x) = A\mathbf x$ with $A = \begin{bmatrix} 2 & -1 \\ 1 & 1 \end{bmatrix}$ . Is $\mathbf b = (1, 5)$ in the range of $T$ ? If so, find an $\mathbf x$ with $T(\mathbf x) = \mathbf b$ .

Walk through the solution

$\mathbf b$ is in the range exactly when $A\mathbf x = \mathbf b$ has a solution, so row reduce $\begin{bmatrix} A & \mathbf b \end{bmatrix}$ .

1
Swap the rows so the pivot is a $1$ , which avoids fractions:
$\begin{bmatrix} 2 & -1 & 1 \\ 1 & 1 & 5 \end{bmatrix} \sim \begin{bmatrix} 1 & 1 & 5 \\ 2 & -1 & 1 \end{bmatrix}$
2
3

The moved grid is made of copies of $\green{T(\mathbf e_1)}$ and $\red{T(\mathbf e_2)}$ , so you can also count how many of each step reach $\mathbf b$ .

Drag

\mathbf x

until its image

T(\mathbf x) = A\mathbf x

lands on the ringed point

\mathbf b = (1, 5)

. Count steps along the blue grid.

\begin{bmatrix} \textcolor{#7BC96F}{2} & \textcolor{#F2665E}{-1} \\ \textcolor{#7BC96F}{1} & \textcolor{#F2665E}{1} \end{bmatrix}\textcolor{#FFD447}{\begin{bmatrix} 1 \\ 0 \end{bmatrix}} = \textcolor{#3DD6B5}{\begin{bmatrix} 2 \\ 1 \end{bmatrix}}

= 1\textcolor{#7BC96F}{\begin{bmatrix} 2 \\ 1 \end{bmatrix}} + 0\textcolor{#F2665E}{\begin{bmatrix} -1 \\ 1 \end{bmatrix}}

A gallery of moves

A horizontal shear keeps $\green{\mathbf e_1}$ fixed and sends $\red{\mathbf e_2}$ to $(2, 1)$ . Write its matrix, and find where the corner $(1, 1)$ of the unit square lands.

Walk through the solution

The columns are the landing spots, so the matrix is $\begin{bmatrix} \green{1} & \red{2} \\ \green{0} & \red{1} \end{bmatrix}$ . The corner is $\green{\mathbf e_1} + \red{\mathbf e_2}$ , so it lands on $(1, 0) + (2, 1) = (3, 1)$ . The top edge of the square slid $2$ to the right while the bottom edge stayed put.

Four small grids, each with the unit square outlined in gray and its image filled. A stretch doubles the square's width. A reflection flips it to the left and fills it pink. A vertical shear tilts it upward. A projection flattens the whole grid onto the horizontal axis. Each matrix is printed below its grid.

Each matrix above is $\begin{bmatrix} T(\mathbf e_1) & T(\mathbf e_2) \end{bmatrix}$ , so you can rebuild it from the picture. From left to right they are $\begin{bmatrix} 2 & 0 \\ 0 & 1 \end{bmatrix}$ , $\begin{bmatrix} -1 & 0 \\ 0 & 1 \end{bmatrix}$ , $\begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix}$ and $\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}$ . A pink square means the map flipped the plane over, and on Day 25 the sign of the determinant detects exactly that flip.

The standard catalog

\begin{array}{ll} \textsf{Reflection across the } x_1\textsf{-axis} & \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix} \\[10pt] \textsf{Reflection across the line } x_2 = x_1 & \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} \\[10pt] \textsf{Horizontal shear by } k & \begin{bmatrix} 1 & k \\ 0 & 1 \end{bmatrix} \\[10pt] \textsf{Vertical stretch by } k & \begin{bmatrix} 1 & 0 \\ 0 & k \end{bmatrix} \\[10pt] \textsf{Projection onto the } x_1\textsf{-axis} & \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} \end{array}

You never need to memorize this table. Picture where $\green{\mathbf e_1}$ and $\red{\mathbf e_2}$ go and write them down as columns.

Drag the green and red tips so the F lands exactly on the dashed outline.

A = \begin{bmatrix} \textcolor{#7BC96F}{1} & \textcolor{#F2665E}{0} \\ \textcolor{#7BC96F}{0} & \textcolor{#F2665E}{1} \end{bmatrix}

Tomorrow you run two of these moves in a row, and the matrix of the combined move turns out to be a product of matrices.

Practice

Warm-up

A $3 \times 5$ matrix $A$ defines $T(\mathbf x) = A\mathbf x$ as a map $T : \mathbb{R}^n \to \mathbb{R}^m$ . Find $n$ and $m$ . (Modeled on Lay §1.8 #21.)

Show the solution

The product $A\mathbf x$ needs one entry of $\mathbf x$ for each of the $5$ columns, so $n = 5$ . Each column has $3$ entries, so $A\mathbf x$ lives in $\mathbb{R}^3$ and $m = 3$ .

Let $A = \begin{bmatrix} 2 & -1 \\ 0 & 3 \\ 1 & 1 \end{bmatrix}$ and $T(\mathbf x) = A\mathbf x$ . Find the image of $\mathbf u = (3, 2)$ . (Modeled on Lay §1.8 #1.)

Show the solution

By the row rule, $T(\mathbf u) = (2 \cdot 3 - 1 \cdot 2,\ 0 \cdot 3 + 3 \cdot 2,\ 1 \cdot 3 + 1 \cdot 2) = (4, 6, 5)$ . The input lives in $\mathbb{R}^2$ and the image lives in $\mathbb{R}^3$ .

Which matrix reflects the plane across the $x_1$ -axis?

Core

$T$ first performs a vertical shear that sends $\mathbf e_1$ to $\mathbf e_1 + 2\mathbf e_2$ and keeps $\mathbf e_2$ fixed, then reflects through the line $x_2 = x_1$ . Find $T(1, 1)$ . (Modeled on Lay §1.9 #9.)

Show a hint

Follow $\mathbf e_1$ and $\mathbf e_2$ through both steps, in order. The reflection swaps the two coordinates.

Show the solution

Track each basis vector through both steps.

\begin{aligned} \green{\mathbf e_1} &\to (1, 2) \to (2, 1), \\ \red{\mathbf e_2} &\to (0, 1) \to (1, 0), \end{aligned}

so $A = \begin{bmatrix} \green{2} & \red{1} \\ \green{1} & \red{0} \end{bmatrix}$ and $T(1, 1) = (2, 1) + (1, 0) = (3, 1)$ .

A linear map $T$ sends $\mathbf u = (1, 1)$ to $(4, 2)$ and $\mathbf v = (1, -1)$ to $(2, -4)$ . Find $T(\mathbf e_1)$ . (Modeled on Lay §1.8 #17.)

Show a hint

Write $\mathbf e_1$ as a combination of $\mathbf u$ and $\mathbf v$ . Adding them gives $(2, 0)$ .

Show the solution

Since $\mathbf e_1 = \tfrac12(\mathbf u + \mathbf v)$ , linearity gives

T(\mathbf e_1) = \tfrac12\bigl(T(\mathbf u) + T(\mathbf v)\bigr) = \tfrac12\begin{bmatrix} 6 \\ -2 \end{bmatrix} = \begin{bmatrix} 3 \\ -1 \end{bmatrix}.

In the same way $\mathbf e_2 = \tfrac12(\mathbf u - \mathbf v)$ gives $T(\mathbf e_2) = (1, 3)$ , so the standard matrix is $\begin{bmatrix} 3 & 1 \\ -1 & 3 \end{bmatrix}$ .

Let $T(\mathbf x) = A\mathbf x$ with $A = \begin{bmatrix} 1 & 2 \\ 0 & 1 \\ 2 & 3 \end{bmatrix}$ . Find $\mathbf x$ with $T(\mathbf x) = (4, 1, 7)$ . (Modeled on Lay §1.8 #9.)

Show a hint

Row 2 of the system says $x_2 = 1$ directly. Use it in row 1, and then check row 3.

Show the solution

The rows are $x_1 + 2x_2 = 4$ , $\ x_2 = 1$ and $2x_1 + 3x_2 = 7$ . Row 2 gives $x_2 = 1$ , and then row 1 gives $x_1 = 2$ . Row 3 checks: $4 + 3 = 7$ . So $\mathbf x = (2, 1)$ . Had the third entry of $\mathbf b$ been anything other than $7$ , $\mathbf b$ would not be in the range.

Find the standard matrix of a quarter turn clockwise. Enter it row by row. (Modeled on Lay §1.9 #3.)

Show a hint

Clockwise sends $\mathbf e_1$ pointing down. Alternatively, use the rotation matrix with $\varphi = -90^\circ$ .

Show the solution

A clockwise quarter turn sends $\mathbf e_1 = (1, 0)$ to $(0, -1)$ and $\mathbf e_2 = (0, 1)$ to $(1, 0)$ . With those as columns,

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

The rotation formula agrees: $\cos(-90^\circ) = 0$ and $\sin(-90^\circ) = -1$ .

Let $T(x_1, x_2) = (x_1 + 4x_2,\ 3x_1 - x_2)$ . Find $T(\mathbf e_1)$ and $T(\mathbf e_2)$ , write the standard matrix, and use it to find $T(2, 1)$ . (Modeled on Lay §1.9 #19.)

Show a hint

Put $(1, 0)$ into the formula to get the first column, and $(0, 1)$ to get the second.

Show the solution

$T(1, 0) = (1, 3)$ and $T(0, 1) = (4, -1)$ , so

A = \begin{bmatrix} 1 & 4 \\ 3 & -1 \end{bmatrix}, \qquad A\begin{bmatrix} 2 \\ 1 \end{bmatrix} = \begin{bmatrix} 2 + 4 \\ 6 - 1 \end{bmatrix} = \begin{bmatrix} 6 \\ 5 \end{bmatrix}.

The formula gives the same thing directly: $(2 + 4,\ 6 - 1) = (6, 5)$ .

Challenge

$T$ first rotates the plane a quarter turn counterclockwise and then reflects it across the $x_1$ -axis. Find the standard matrix of $T$ and enter it row by row. Which single reflection is it?

Show a hint

Track $\mathbf e_1$ and $\mathbf e_2$ through the turn first, then flip the sign of each second entry.

Show the solution

The turn sends $\mathbf e_1 \to (0, 1)$ and $\mathbf e_2 \to (-1, 0)$ . The reflection flips the sign of the second entry, so $\mathbf e_1 \to (0, -1)$ and $\mathbf e_2 \to (-1, 0)$ :

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

This matrix swaps the two coordinates and negates both, so it reflects the plane across the line $x_2 = -x_1$ . Two moves in a row made a single move, which is tomorrow's topic.

A linear map $T$ has $T(\mathbf u) = (1, 2)$ . What is $T(-3\mathbf u)$ ?

Show that $T(\mathbf x) = \mathbf x + (1, 1)$ is not linear. Then explain why every linear map must send $\mathbf 0$ to $\mathbf 0$ .

Show a hint

Compute $T(\mathbf 0)$ . For the second part, write $\mathbf 0 = 0\,\mathbf u$ for any $\mathbf u$ and use rule (ii).

Show the solution

$T(\mathbf 0) = (1, 1)$ , but a linear map sends $\mathbf 0$ to $\mathbf 0$ , so $T$ is not linear. Rule (i) fails directly too: $T(\mathbf 0 + \mathbf 0) = (1, 1)$ , while $T(\mathbf 0) + T(\mathbf 0) = (2, 2)$ .

For any linear $T$ and any vector $\mathbf u$ , $\mathbf 0 = 0\,\mathbf u$ , so rule (ii) gives $T(\mathbf 0) = T(0\,\mathbf u) = 0\,T(\mathbf u) = \mathbf 0$ .