The Timpson Lyceum · Linear Algebra (2026)

Introduction to Linear Algebra– Lecture 0 (Linear Algebra)

Overview

This opening lecture is a fast tour of what linear algebra is about, meant to give a feel for the subject before the course starts over from the beginning. It revisits something already familiar — the equation of a line — and reframes it as a set of solutions, which turns the question "where do two lines cross?" into the question "what numbers satisfy both equations at once?" From there it defines what makes an equation linear, shows that linear equations are only lines in two dimensions, and introduces the subscript notation used throughout the course. The second half introduces vectors along with the two operations applied to them constantly — addition and scalar multiplication — and closes by rewriting the very same system of equations as a single vector equation. The two pictures answer identical questions, and moving fluidly between them is the central skill the course develops.

Key Concepts

An equation is a set of solutions

The solution set of an equation is every assignment of values to the variables that makes the equation true. For $y = 2x$ , the pairs $(0,0)$ , $(1,2)$ , $(-1,-2)$ all qualify, and there are infinitely many more.
Interpreting each solution as a point and plotting all of them produces a line. Strictly, the equation is not the line; the line is the geometric interpretation of the equation's solution set.

Systems of linear equations

To solve a system is to find values that satisfy every equation simultaneously. Geometrically, those are the points lying on every one of the graphs at once — the intersection.
For $y = 2x$ together with $y = -x + 3$ , the only pair on both lines is $(1,2)$ : one solution.
A system of two linear equations has exactly one of three outcomes:
- One solution — the lines cross at a single point, as above.
- No solution — the lines are parallel and never meet, e.g. $y = 2x$ and $y = 2x + 1$ .
- Infinitely many solutions — the equations describe the same line, e.g. $y = 2x$ and $2y = 4x$ , so every solution of one is a solution of the other.

What makes an equation linear

An equation is linear when every variable appears to the first power. In $x + y + 13z = 21$ each of $x$ , $y$ , $z$ carries an exponent of $1$ , so the equation is linear; replacing $x$ with $x^2$ destroys that.
"Linear" does not mean "line." The solution set of a single linear equation in $n$ variables is an $(n-1)$ -dimensional object sitting inside $n$ -dimensional space:

Variables	Solution set of one linear equation
1	a point (e.g. $x = 3$ )
2	a line
3	a plane
4	a 3-dimensional surface in 4-dimensional space
$n$	an $(n-1)$ -dimensional surface in $n$ -dimensional space

Lines are just what the two-variable case happens to look like; the familiar picture is a special case, not the general rule.

Standard notation for linear equations

The alphabet runs out long before the variables do — a problem may involve 87 unknowns. Instead of $x$ , $y$ , $z$ , the course indexes variables: $x_1$ , $x_2$ , $x_3$ , and so on.
A typical linear equation is written $2x_1 + 7x_2 + 13x_3 = 9$
Equations are also kept in the form with all variables on the left and the constant on the right — $2x - y = 0$ rather than $y = 2x$ . Slope-intercept form is convenient for graphing, but this form is far more natural for the manipulations of linear algebra.

Vectors

Working definition: a vector is an ordered collection of numbers. (A deeper and more general definition arrives in Chapter 4; this one is enough to get started.)
Vectors are written as columns, with order mattering — the top entry is first, the second entry next. Textbooks mark a vector with a bold letter; handwritten work uses an arrow, as in $\vec{u}$ , to signal that the object is a vector and not a number.

\vec{u} = \begin{bmatrix} 2 \\ 1 \end{bmatrix}, \qquad \vec{v} = \begin{bmatrix} -1 \\ 1 \end{bmatrix}, \qquad \begin{bmatrix} 7 \\ \pi \\ e \\ 19 \end{bmatrix}

A vector's length is its number of entries, and that number is its dimension — the last vector above is a four-dimensional vector.

The two basic operations

Vector addition works entry by entry. For $\vec{u} = \begin{bmatrix} a_1 \\ a_2 \end{bmatrix}$ and $\vec{v} = \begin{bmatrix} b_1 \\ b_2 \end{bmatrix}$ :

\begin{aligned} \vec{u} + \vec{v} &= \begin{bmatrix} a_1 + b_1 \\ a_2 + b_2 \end{bmatrix} \\ \begin{bmatrix} 2 \\ 1 \end{bmatrix} + \begin{bmatrix} -1 \\ 1 \end{bmatrix} &= \begin{bmatrix} 1 \\ 2 \end{bmatrix} \end{aligned}

Scalar multiplication by a real number $c$ scales every entry:

\begin{aligned} c\vec{u} &= \begin{bmatrix} c\,a_1 \\ c\,a_2 \end{bmatrix} \\ -3\begin{bmatrix} 2 \\ 1 \end{bmatrix} &= \begin{bmatrix} -6 \\ -3 \end{bmatrix} \end{aligned}

The $+$ joining two vectors is not the $+$ of the real numbers — it is a new operation, defined above, that happens to be built out of real-number addition applied to the entries. Texts write both with the same symbol and leave the reader to tell them apart from context; the same caution applies to multiplication, where $c\vec{u}$ and $c_1c_2$ mean different things.
There is no multiplication of a vector by a vector in this general setting. The cross product is defined only in three dimensions, and the dot product, though it does appear later, is set aside for now. Addition and scalar multiplication are the operations that matter.

The geometric picture

The vector $\begin{bmatrix} 2 \\ 1 \end{bmatrix}$ is drawn as an arrow running two units in the $x$ direction and one unit in the $y$ direction.
Arrows are conventionally drawn with their tail at the origin because that is the simplest place to put them, but position carries no meaning — the same vector may be drawn anywhere. Only direction and length matter.
A vector is not a ray; it has a definite, finite length, computed with the Pythagorean theorem: $\left\lVert \vec{u} \right\rVert = \sqrt{2^2 + 1^2} = \sqrt{5}$ and $\left\lVert \vec{v} \right\rVert = \sqrt{(-1)^2 + 1^2} = \sqrt{2}$ .
Addition is tip-to-tail: draw $\vec{u}$ , then draw $\vec{v}$ starting where $\vec{u}$ ends; the arrow from the origin to that final point is $\vec{u} + \vec{v}$ . Order does not matter, so $\vec{u} + \vec{v} = \vec{v} + \vec{u}$ .
Subtraction reverses an arrow: $\vec{u} - \vec{v}$ is $\vec{u}$ followed by $\vec{v}$ pointed the opposite way. Unlike addition, it is not commutative.

Linear combinations — the bridge between the two pictures

Given vectors $\vec{u}$ and $\vec{v}$ , a natural question is how many of each are needed to reach some target $\vec{w}$ — that is, which scalars $x$ and $y$ satisfy $x\vec{u} + y\vec{v} = \vec{w}$ . Such an expression is a linear combination of $\vec{u}$ and $\vec{v}$ .
This is not a new question. Writing a system with all variables on the left and reading it down the columns produces exactly that vector equation: the coefficients of $x$ become one vector, the coefficients of $y$ become another, and the constants become the target.
So "where do these two lines intersect?" and "how much of $\vec{u}$ and how much of $\vec{v}$ builds $\vec{w}$ ?" are the same question in two different languages. The same three outcomes appear here too — a unique combination, no combination at all, or infinitely many.

Worked Example

Solve the system below, then read the identical problem as a vector equation.

First put both equations in standard form — all variables left, constant right — and eliminate $y$ by adding the equations:

\begin{aligned} 2x - y &= 0 \\ x + y &= 3 \\ (2x - y) + (x + y) &= 0 + 3 \\ 3x &= 3 \\ x &= 1 \\ y &= 2x = 2 \end{aligned}

The solution is $(1,2)$ , the single point where the lines $y = 2x$ and $y = -x + 3$ cross.

Now read the same system down its columns. The $x$ -coefficients $2$ and $1$ form one vector, the $y$ -coefficients $-1$ and $1$ form another, and the constants $0$ and $3$ form the target:

\begin{aligned} x\begin{bmatrix} 2 \\ 1 \end{bmatrix} + y\begin{bmatrix} -1 \\ 1 \end{bmatrix} &= \begin{bmatrix} 0 \\ 3 \end{bmatrix} \\ 1\begin{bmatrix} 2 \\ 1 \end{bmatrix} + 2\begin{bmatrix} -1 \\ 1 \end{bmatrix} &= \begin{bmatrix} 2 + (-2) \\ 1 + 2 \end{bmatrix} \\ &= \begin{bmatrix} 0 \\ 3 \end{bmatrix} \end{aligned}

The same pair $x = 1$ , $y = 2$ solves both versions. Geometrically, stepping once along $\vec{u}$ and then twice along $\vec{v}$ lands exactly on the tip of $\vec{w}$ — the tip-to-tail picture of the intersection point found algebraically.

Summary

An equation is best understood as its set of solutions; graphing that set is one geometric interpretation of the equation, and solving a system means finding the values that satisfy every equation at once.
A system of linear equations has exactly one solution, no solution, or infinitely many — never any other count.
An equation is linear when every variable appears to the first power; its solution set is an $(n-1)$ -dimensional object in $n$ -dimensional space, so it is a line only in the two-variable case.
Variables are indexed as $x_1, x_2, x_3, \dots$ so that problems in arbitrarily many dimensions can be written down, and equations are kept with all variables on the left.
A vector is an ordered collection of numbers, supporting entry-wise addition and scalar multiplication; geometrically it is an arrow with direction and length whose position is irrelevant, added tip-to-tail.
Reading a system down its columns turns it into a single vector equation $x\vec{u} + y\vec{v} = \vec{w}$ — the same problem asked about linear combinations instead of intersections.