The Timpson Lyceum · Linear Algebra (2026)

Linear Transformations, Matrices, and Linear Models – Lecture 4 (Linear Algebra)

Overview

This lecture finishes Chapter 1 with three sections (1.8, 1.9 and 1.10). It starts by treating multiplication by a matrix as a function: a rule $T$ that takes vectors in $\mathbb{R}^n$ as inputs and returns vectors in $\mathbb{R}^m$ . Such a linear transformation respects addition and scaling, always sends $\mathbf{0}$ to $\mathbf{0}$ , and can project, shear, rotate, or stretch space. The central result is Theorem 10: every linear transformation from $\mathbb{R}^n$ to $\mathbb{R}^m$ is multiplication by a unique matrix, and that matrix can be read off from where the transformation sends the columns of the identity matrix. The pivot structure of that matrix then decides whether $T$ is onto and whether it is one-to-one. The lecture closes with two applied linear models: a diet problem and an electrical circuit.

Key Concepts

Function vocabulary

A function gives each input exactly one output: if $a = b$ then $f(a) = f(b)$ . A circle fails the vertical line test, so it is not the graph of a function.
A function is one-to-one if equal outputs force equal inputs. $y = x^2$ is not one-to-one, since $(-1)^2 = 1^2$ . A non-horizontal line is.
A function is onto if every possible output is actually produced. $y = x^2$ never outputs $-4$ , so as a function from $\mathbb{R}$ to $\mathbb{R}$ it is not onto. The line $y = x + 2$ is onto.
The domain is the set of allowed inputs. The range (also called the image) is the set of outputs actually produced.

Transformations from $\mathbb{R}^n$ to $\mathbb{R}^m$

A transformation (also called a function or a mapping) $T : \mathbb{R}^n \to \mathbb{R}^m$ is a rule that assigns to each $\mathbf{x}$ 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. The range is the set of all images $T(\mathbf{x})$ . It lies inside the codomain but need not fill it. For example, $T(\mathbf{x}) = 0\mathbf{x}$ (the $2 \times 4$ zero matrix) sends all of $\mathbb{R}^4$ to the single point $\mathbf{0}$ in $\mathbb{R}^2$ .
A matrix transformation has the form $T(\mathbf{x}) = A\mathbf{x}$ , often written $\mathbf{x} \mapsto A\mathbf{x}$ . For $A\mathbf{x}$ to be defined with $\mathbf{x}$ in $\mathbb{R}^n$ and $A\mathbf{x}$ in $\mathbb{R}^m$ , the matrix $A$ must be $m \times n$ : one column per input entry and one row per output entry.
For example, with

A = \begin{bmatrix} 4 & -3 & 1 & 3 \\ 2 & 0 & 5 & 1 \end{bmatrix}, \qquad A\begin{bmatrix} 1 \\ 1 \\ 1 \\ 1 \end{bmatrix} = \begin{bmatrix} 5 \\ 8 \end{bmatrix}, \qquad A\begin{bmatrix} 1 \\ 4 \\ -1 \\ 3 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}

the map $\mathbf{x} \mapsto A\mathbf{x}$ goes from $\mathbb{R}^4$ to $\mathbb{R}^2$ .

Is $\mathbf{b}$ in the range of $\mathbf{x} \mapsto A\mathbf{x}$ ? That is the same question as whether $A\mathbf{x} = \mathbf{b}$ is consistent, so row reduce $[\,A \;\; \mathbf{b}\,]$ . A row $[\,0 \;\cdots\; 0 \;\; c\,]$ with $c \neq 0$ means no. If there is a solution and no free variable, then exactly one $\mathbf{x}$ maps to $\mathbf{b}$ .

Geometric examples

Projection. $A = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{bmatrix}$ sends $(x_1, x_2, x_3)$ to $(x_1, x_2, 0)$ . It collapses all of $\mathbb{R}^3$ onto the $x_1x_2$ -plane, like casting a shadow. For example, $(1, 3, 7) \mapsto (1, 3, 0)$ , and every point on the vertical line above $(1, 3, 0)$ has that same image.
Shear. $A = \begin{bmatrix} 1 & 3 \\ 0 & 1 \end{bmatrix}$ fixes $(2, 0)$ and sends $(0, 2)$ to $(6, 2)$ and $(2, 2)$ to $(8, 2)$ . The square with corners $(0,0)$ , $(2,0)$ , $(2,2)$ , $(0,2)$ is pushed sideways into a parallelogram.
Rotation. $A = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}$ rotates the plane $90^\circ$ counterclockwise. With $\mathbf{u} = (4, 1)$ and $\mathbf{v} = (2, 3)$ , it gives $T(\mathbf{u}) = (-1, 4)$ , $T(\mathbf{v}) = (-3, 2)$ , and $T(\mathbf{u} + \mathbf{v}) = T(6, 4) = (-4, 6) = T(\mathbf{u}) + T(\mathbf{v})$ .
Dilation. $T(\mathbf{x}) = 3\mathbf{x}$ scales every vector by 3.

Linear transformations

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

\begin{aligned} &\text{(i)} \quad T(\mathbf{u} + \mathbf{v}) = T(\mathbf{u}) + T(\mathbf{v}) \\ &\text{(ii)} \quad T(c\mathbf{u}) = cT(\mathbf{u}) \end{aligned}

Every matrix transformation is linear, because $A(\mathbf{u} + \mathbf{v}) = A\mathbf{u} + A\mathbf{v}$ and $A(c\mathbf{u}) = cA\mathbf{u}$ .
A linear transformation sends $\mathbf{0}$ to $\mathbf{0}$ . For any $\mathbf{u}$ in the domain,

T(\mathbf{0}) = T(0\mathbf{u}) = 0\,T(\mathbf{u}) = \mathbf{0}

One-step test. $T$ is linear if and only if, for all vectors $\mathbf{u}, \mathbf{v}$ and all scalars $c, d$ ,

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

Linearity gives this by applying (i) and then (ii). In the other direction, $c = d = 1$ recovers (i), and $d = 0$ recovers (ii). In practice this single identity is the one checked.

Superposition principle. Repeating the argument extends it to any number of terms:

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

Example: $T(\mathbf{x}) = 3\mathbf{x}$ is linear.

\begin{aligned} T(c\mathbf{u} + d\mathbf{v}) &= 3(c\mathbf{u} + d\mathbf{v}) \\ &= c(3\mathbf{u}) + d(3\mathbf{v}) \\ &= cT(\mathbf{u}) + dT(\mathbf{v}) \end{aligned}

The matrix of a linear transformation (Theorem 10)

$I_n$ is the $n \times n$ identity matrix, with 1s on the diagonal and 0s elsewhere. It satisfies $I_n\mathbf{x} = \mathbf{x}$ , just as $1 \cdot x = x$ . Only square identity matrices exist, since the output must have the same size as the input.
$\mathbf{e}_j$ is the $j$ th column of $I_n$ : a 1 in position $j$ and 0s elsewhere. The size $n$ is read from context.
Theorem 10. Let $T : \mathbb{R}^n \to \mathbb{R}^m$ be linear. Then there is a unique $m \times n$ matrix $A$ with $T(\mathbf{x}) = A\mathbf{x}$ for all $\mathbf{x}$ , namely the standard matrix

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

Existence. Write $\mathbf{x} = I_n\mathbf{x} = x_1\mathbf{e}_1 + \cdots + x_n\mathbf{e}_n$ and use linearity:

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}

Uniqueness. Suppose $T(\mathbf{x}) = B\mathbf{x}$ as well. Multiplying a matrix by $\mathbf{e}_j$ returns its $j$ th column, so $\mathbf{a}_j = A\mathbf{e}_j = T(\mathbf{e}_j) = B\mathbf{e}_j = \mathbf{b}_j$ for every $j$ . The two matrices have the same columns, so $B = A$ .
Every linear map between $\mathbb{R}^n$ and $\mathbb{R}^m$ is therefore a matrix transformation, and knowing where it sends $\mathbf{e}_1, \dots, \mathbf{e}_n$ determines it completely. For example, $T(\mathbf{x}) = 3\mathbf{x}$ on $\mathbb{R}^2$ has $T(\mathbf{e}_1) = (3, 0)$ and $T(\mathbf{e}_2) = (0, 3)$ , so its standard matrix is $\begin{bmatrix} 3 & 0 \\ 0 & 3 \end{bmatrix}$ .

Building geometric matrices from $T(\mathbf{e}_j)$

Picture where each $\mathbf{e}_j$ lands and write those images down as the columns:

Transformation	$T(\mathbf{e}_1)$	$T(\mathbf{e}_2)$	Standard matrix
Rotate $90^\circ$ counterclockwise	$(0, 1)$	$(-1, 0)$	$\begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}$
Rotate $90^\circ$ clockwise	$(0, -1)$	$(1, 0)$	$\begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}$
Rotate $180^\circ$	$(-1, 0)$	$(0, -1)$	$\begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix}$
Reflect across the $x_1$ -axis	$(1, 0)$	$(0, -1)$	$\begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}$
Reflect across the $x_2$ -axis	$(-1, 0)$	$(0, 1)$	$\begin{bmatrix} -1 & 0 \\ 0 & 1 \end{bmatrix}$

Reflection through the origin sends $\mathbf{x}$ to $-\mathbf{x}$ . It has the same standard matrix as rotation by $180^\circ$ , so the two are the same transformation.
Rotation by $\theta$ . Counterclockwise rotation sends $\mathbf{e}_1$ to $(\cos\theta, \sin\theta)$ . It sends $\mathbf{e}_2$ to the point at angle $\theta + 90^\circ$ , which is $(\cos(\theta + 90^\circ), \sin(\theta + 90^\circ)) = (-\sin\theta, \cos\theta)$ :

A = \begin{bmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{bmatrix}

Setting $\theta = 90^\circ$ or $\theta = 180^\circ$ recovers the rows of the table above.

The same method works in $\mathbb{R}^3$ . Rotating $90^\circ$ about the $x_3$ -axis, so that $\mathbf{e}_1 \mapsto \mathbf{e}_2$ , $\mathbf{e}_2 \mapsto -\mathbf{e}_1$ and $\mathbf{e}_3$ stays put, gives

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

Onto and one-to-one

$T : \mathbb{R}^n \to \mathbb{R}^m$ is onto $\mathbb{R}^m$ if every $\mathbf{b}$ in $\mathbb{R}^m$ is the image of at least one $\mathbf{x}$ in $\mathbb{R}^n$ . Several inputs may share an image.
$T$ is one-to-one if every $\mathbf{b}$ in $\mathbb{R}^m$ is the image of at most one $\mathbf{x}$ . Some $\mathbf{b}$ may have no preimage at all.
A map that is both one-to-one and onto is a bijection: each $\mathbf{b}$ is the image of exactly one $\mathbf{x}$ . A rotation of $\mathbb{R}^2$ is a bijection. The projection of $\mathbb{R}^3$ onto a plane is neither one-to-one nor onto $\mathbb{R}^3$ .
Theorem 11. A linear $T$ is one-to-one if and only if $T(\mathbf{x}) = \mathbf{0}$ has only the trivial solution. If the homogeneous equation has only $\mathbf{x} = \mathbf{0}$ , then each consistent $T(\mathbf{x}) = \mathbf{b}$ has its solution set shifted from that single point, so it has exactly one solution.
Theorem 12. Let $T(\mathbf{x}) = A\mathbf{x}$ with $A$ an $m \times n$ matrix. (a) $T$ is onto $\mathbb{R}^m$ if and only if the columns of $A$ span $\mathbb{R}^m$ . (b) $T$ is one-to-one if and only if the columns of $A$ are linearly independent.
In terms of pivots:

Property of $T$	Columns of $A$	Pivot pattern of $A$
Onto $\mathbb{R}^m$	Span $\mathbb{R}^m$	A pivot in every row
One-to-one	Linearly independent	A pivot in every column

For example, $A = \begin{bmatrix} 1 & -4 & 8 & 1 \\ 0 & 2 & -1 & 3 \\ 0 & 0 & 0 & 5 \end{bmatrix}$ has a pivot in every row, so $\mathbf{x} \mapsto A\mathbf{x}$ maps $\mathbb{R}^4$ onto $\mathbb{R}^3$ . Column 3 has no pivot, so there is a free variable, and the map is not one-to-one.
Coordinate notation. Lay often writes a transformation as $T(x_1, x_2)$ in place of $T\!\left(\begin{bmatrix} x_1 \\ x_2 \end{bmatrix}\right)$ . To show such a $T$ is linear, pull each variable out into its own column:

T(x_1, x_2) = \begin{bmatrix} 3x_1 + x_2 \\ 5x_1 + 7x_2 \\ x_1 + 3x_2 \end{bmatrix} = \begin{bmatrix} 3 & 1 \\ 5 & 7 \\ 1 & 3 \end{bmatrix}\begin{bmatrix} x_1 \\ x_2 \end{bmatrix}

This is a matrix transformation from $\mathbb{R}^2$ to $\mathbb{R}^3$ , so it is linear.

Linear models (Section 1.10)

With applied problems, the difficult part is translating the data into a matrix. After that, row reduction finishes the job.
Diet. Per 100 g, nonfat milk supplies $(36, 52, 0)$ grams of (protein, carbohydrate, fat), soy flour supplies $(51, 34, 7)$ , and whey supplies $(13, 74, 1.1)$ . Let $x_1, x_2, x_3$ be the numbers of 100 g units eaten. The target of 33 g protein, 45 g carbohydrate and 3 g fat becomes

x_1\begin{bmatrix} 36 \\ 52 \\ 0 \end{bmatrix} + x_2\begin{bmatrix} 51 \\ 34 \\ 7 \end{bmatrix} + x_3\begin{bmatrix} 13 \\ 74 \\ 1.1 \end{bmatrix} = \begin{bmatrix} 33 \\ 45 \\ 3 \end{bmatrix}

Solving gives $x_1 \approx 0.277$ , $x_2 \approx 0.392$ , $x_3 \approx 0.233$ , or roughly 27.7 g of nonfat milk, 39.2 g of soy flour and 23.3 g of whey.

Electrical networks. Two physical laws turn a circuit into a linear system. Ohm's law gives the voltage drop across a resistor as $V = RI$ . Kirchhoff's voltage law says that around any closed loop, the sum of the $RI$ drops equals the sum of the battery voltages, so the net change in potential around the loop is zero. Assign a loop current to each loop and write one equation per loop. A resistor on a branch shared with a neighboring loop carries both loop currents, with opposite signs when they run in opposite directions.

Worked Example

Finding a standard matrix from $T(\mathbf{e}_1)$ and $T(\mathbf{e}_2)$ . A linear transformation $T : \mathbb{R}^2 \to \mathbb{R}^3$ satisfies $T(\mathbf{e}_1) = (5, -7, 2)$ and $T(\mathbf{e}_2) = (-3, 8, 0)$ . Find a formula for $T(\mathbf{x})$ .

\begin{aligned} T(\mathbf{x}) &= x_1T(\mathbf{e}_1) + x_2T(\mathbf{e}_2) \\ &= x_1\begin{bmatrix} 5 \\ -7 \\ 2 \end{bmatrix} + x_2\begin{bmatrix} -3 \\ 8 \\ 0 \end{bmatrix} = \begin{bmatrix} 5 & -3 \\ -7 & 8 \\ 2 & 0 \end{bmatrix}\begin{bmatrix} x_1 \\ x_2 \end{bmatrix} \end{aligned}

Knowing only these two images fixes the image of every vector in $\mathbb{R}^2$ .

Range and preimages. Let

A = \begin{bmatrix} 1 & -3 \\ 3 & 5 \\ -1 & 7 \end{bmatrix},\quad \mathbf{u} = \begin{bmatrix} 2 \\ -1 \end{bmatrix},\quad \mathbf{b} = \begin{bmatrix} 3 \\ 2 \\ -5 \end{bmatrix},\quad \mathbf{c} = \begin{bmatrix} 3 \\ 2 \\ 5 \end{bmatrix}

and $T(\mathbf{x}) = A\mathbf{x}$ , a map from $\mathbb{R}^2$ to $\mathbb{R}^3$ .

Image of $\mathbf{u}$ . $A\mathbf{u}$ is the combination of the columns weighted by the entries of $\mathbf{u}$ :

\begin{aligned} T(\mathbf{u}) &= u_1\mathbf{a}_1 + u_2\mathbf{a}_2 \\ &= 2\begin{bmatrix} 1 \\ 3 \\ -1 \end{bmatrix} - 1\begin{bmatrix} -3 \\ 5 \\ 7 \end{bmatrix} = \begin{bmatrix} 5 \\ 1 \\ -9 \end{bmatrix} \end{aligned}

An $\mathbf{x}$ with $T(\mathbf{x}) = \mathbf{b}$ . Row reduce $[\,A \;\; \mathbf{b}\,]$ :

\begin{aligned} \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 & -\tfrac{1}{2} \\ 0 & 0 & 0 \end{array}\right] \sim \left[\begin{array}{cc|c} 1 & 0 & \tfrac{3}{2} \\ 0 & 1 & -\tfrac{1}{2} \\ 0 & 0 & 0 \end{array}\right] \end{aligned}

so $\mathbf{x} = \left(\tfrac{3}{2}, -\tfrac{1}{2}\right)$ . Check the first entry: $\tfrac{3}{2}(1) - \tfrac{1}{2}(-3) = 3$ . Both columns are pivot columns, so there is no free variable, and this is the only $\mathbf{x}$ whose image is $\mathbf{b}$ .

Is $\mathbf{c}$ in the range? Only the last entry has changed. The same row operations give

\begin{aligned} \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 & -\tfrac{1}{2} \\ 0 & 0 & \tfrac{5}{2} \end{array}\right] \end{aligned}

The last row says $0 = \tfrac{5}{2}$ , so the system is inconsistent and $\mathbf{c}$ is not in the range of $T$ . A $3 \times 2$ matrix has at most two pivots, so it cannot have a pivot in every row, and $T$ cannot be onto $\mathbb{R}^3$ .

A three-loop circuit. Applying Kirchhoff's voltage law to each loop gives one equation per loop current:

\begin{aligned} 11I_1 - 3I_2 \phantom{{}- I_3} &= 30 \\ -3I_1 + 6I_2 - I_3 &= 5 \\ -I_2 + 3I_3 &= -25 \end{aligned} \quad\Longrightarrow\quad \left[\begin{array}{ccc|c} 11 & -3 & 0 & 30 \\ -3 & 6 & -1 & 5 \\ 0 & -1 & 3 & -25 \end{array}\right]

Row reduction gives $I_1 = 3$ A, $I_2 = 1$ A and $I_3 = -8$ A. The negative sign means the third current runs opposite to the direction assumed, at 8 A. As a check on the first loop, the shared branch carries $I_1 - I_2 = 2$ A through its $3\,\Omega$ resistor. The drops are $4(3) + 4(3) + 3(2) = 30$ volts, which match the 30 V battery.

Summary

A transformation $T : \mathbb{R}^n \to \mathbb{R}^m$ assigns each input in the domain $\mathbb{R}^n$ an output in the codomain $\mathbb{R}^m$ . The range is the set of outputs actually produced, and $\mathbf{b}$ is in the range of $\mathbf{x} \mapsto A\mathbf{x}$ exactly when $A\mathbf{x} = \mathbf{b}$ is consistent.
$T$ is linear when $T(c\mathbf{u} + d\mathbf{v}) = cT(\mathbf{u}) + dT(\mathbf{v})$ . Every linear $T$ sends $\mathbf{0}$ to $\mathbf{0}$ and obeys the superposition principle.
Theorem 10: every linear $T : \mathbb{R}^n \to \mathbb{R}^m$ equals $\mathbf{x} \mapsto A\mathbf{x}$ for exactly one matrix, $A = [\,T(\mathbf{e}_1) \;\cdots\; T(\mathbf{e}_n)\,]$ .
Rotations, reflections, shears, projections and dilations can each be built by tracking where the columns of the identity matrix land. Rotation by $\theta$ has columns $(\cos\theta, \sin\theta)$ and $(-\sin\theta, \cos\theta)$ .
$T(\mathbf{x}) = A\mathbf{x}$ is onto exactly when $A$ has a pivot in every row (its columns span $\mathbb{R}^m$ ). It is one-to-one exactly when $A$ has a pivot in every column (its columns are linearly independent).
Diet planning and circuit analysis reduce to systems $A\mathbf{x} = \mathbf{b}$ . The modeling step is turning the data, or Ohm's and Kirchhoff's laws, into the right matrix.