MAT240H1/Unit 01

Unit 01 · Algebra I

Fields & Vector Spaces

Linear algebra starts not with matrices but with axioms: a field of scalars and a vector space over it. Everything later — bases, dimension, eigenvalues — is a consequence of these rules, which is why the first proofs in the course are about the axioms themselves.

By the end you can

• State the field axioms and work in $\mathbb{Q}$, $\mathbb{R}$, $\mathbb{C}$, and $\mathbb{Z}_p$.
• State the eight vector space axioms.
• Prove the first consequences: $0v=0$, $(-1)v=-v$, cancellation.
• Verify whether a given set with given operations is a vector space.

Why it matters & where it comes from

Why it exists

Defining a vector space abstractly — by axioms, not as arrows or columns — means one theorem covers polynomials, functions, and data vectors at once. Abstraction buys reach.

In the real world

Vector spaces over $\mathbb{R}$ and $\mathbb{C}$ are the language of machine learning (data as vectors), quantum mechanics, and signal processing; finite fields power error-correcting codes and cryptography.

Story & open questions

Grassmann and Peano formalized vector spaces in the late 1800s; the abstract axiomatic view took hold in the 20th century and is now the default language of pure and applied mathematics alike.

Fields

Definition Field

A field $F$ is a set with two operations $+$ and $\cdot$ such that both are associative and commutative, distributivity holds, there are identities $0\ne1$, every element has an additive inverse, and every nonzero element has a multiplicative inverse. Examples: $\mathbb{Q}$, $\mathbb{R}$, $\mathbb{C}$, and the integers mod $p$, written $\mathbb{Z}_p$, for prime $p$.

Example Why $p$ must be prime

In $\mathbb{Z}_6$: $2\cdot3=6\equiv0$, yet $2\ne0$ and $3\ne0$. If $2$ had an inverse $u$, then $3=(u\cdot2)\cdot3=u\cdot0=0$ — false. So $\mathbb{Z}_6$ is not a field.

For prime $p$, every $a\not\equiv0$ has an inverse — e.g. in $\mathbb{Z}_5$, $3^{-1}=2$ because $3\cdot2=6\equiv1$. ∎

Why a field and not something weaker? Linear algebra constantly divides by scalars — to normalize a vector, to solve $ax=b$, to run row reduction. Division by every nonzero scalar is exactly what the field axioms guarantee and what $\mathbb{Z}$ or $\mathbb{Z}_6$ lack.

Vector spaces

Definition Vector space

A vector space over a field $F$ is a set $V$ with vector addition and scalar multiplication satisfying, for all $u,v,w\in V$ and $a,b\in F$: (VS1) $u+v=v+u$; (VS2) $(u+v)+w=u+(v+w)$; (VS3) a zero vector with $v+0=v$; (VS4) additive inverses; (VS5) $1v=v$; (VS6) $(ab)v=a(bv)$; (VS7) $a(u+v)=au+av$; (VS8) $(a+b)v=av+bv$.

Examples

$F^n$ (coordinate tuples), the $m\times n$ matrices $M_{m\times n}(F)$, the polynomials $P(F)$, and the functions from any set $S$ into $F$ — with pointwise operations — are all vector spaces over $F$. So is $\mathbb{C}$ as a vector space over $\mathbb{R}$: the field of scalars is part of the data.

First proofs from the axioms

Theorem Cancellation

If $u+w=v+w$ in a vector space, then $u=v$.

Proof

Add the inverse $-w$ (VS4) to both sides and use associativity (VS2): $u=(u+w)+(-w)=(v+w)+(-w)=v$. ∎ (Consequence: the zero vector and inverses are unique.)

Theorem Basic identities

For every $v\in V$ and $a\in F$: $0v=0$,   $a\,0=0$,   $(-1)v=-v$,   and $av=0$ implies $a=0$ or $v=0$.

Proof of two of them

$0v=(0+0)v=0v+0v$ by VS8; cancel $0v$ to get $0v=0$. ∎

$v+(-1)v=1v+(-1)v=(1+(-1))v=0v=0$, so $(-1)v$ is the additive inverse of $v$. ∎

If $av=0$ with $a\ne0$: $v=1v=(a^{-1}a)v=a^{-1}(av) =a^{-1}0=0$ — this is where "every nonzero scalar is invertible" earns its keep. ∎

How to verify (or refute) a vector space

Routine

First check the two closures: is $u+v$ in the set? is $av$ in the set? Most non-examples already fail here.

Locate the zero vector and check it belongs to the set.

Only then walk the eight axioms — for subsets of a known vector space with the inherited operations, the axioms are automatic and closure is the whole question (Unit 2).

Common pitfalls

• Closure under both operations must be checked first — it is not an axiom you get for free.
• The zero vector must actually lie in the set; a set missing $0$ is disqualified immediately.
• Scalars come from the field: over $\mathbb{Z}_p$ arithmetic wraps around, and facts like "$2\ne0$" can fail (in $\mathbb{Z}_2$, $2=0$).
• $0v=0$ and $(-1)v=-v$ are theorems, not axioms — a proof may not assume them before proving them.

Practice

Original problems written for this guide in the style of Friedberg, Insel & Spence exercises. Write a full solution before opening the answer.

Practice 1 · Subset check

Is $\{(x,y):x+y=0\}\subseteq\mathbb{R}^2$, with the usual operations, a vector space?

Answer

Yes. It contains $(0,0)$, and sums and scalar multiples of solutions of $x+y=0$ still satisfy it.

Practice 2 · Subset check

Is $\{(x,y):x+y=1\}$ a vector space?

Answer

No — it does not contain $(0,0)$, and it is not closed: the sum of two solutions has $x+y=2$.

Practice 3 · Modular arithmetic

Compute $3\cdot4$ and $3+4$ in $\mathbb{Z}_5$.

Answer

$12\equiv2$ and $7\equiv2\pmod 5$ — both equal $2$.

Practice 4 · From the axioms

Prove from the axioms that $0v=0$ for every vector $v$.

Solution

$0v=(0+0)v=0v+0v$ by VS8; adding $-(0v)$ to both sides (cancellation) leaves $0=0v$.

Practice 5 · From the axioms

Prove $(-1)v=-v$, i.e. that $(-1)v$ is the additive inverse of $v$.

Solution

$v+(-1)v=(1+(-1))v=0v=0$ using VS5, VS8, and Practice 4; uniqueness of inverses (cancellation) finishes it.

Practice 6 · Inverses mod p

Find $4^{-1}$ in $\mathbb{Z}_7$.

Answer

$4\cdot2=8\equiv1 \pmod 7$, so $4^{-1}=2$.

Practice 7 · Not a field

Show that $\mathbb{Z}_6$ is not a field.

Solution

$2\cdot3\equiv0$ with $2,3\ne0$. A field has no such zero divisors: if $2^{-1}$ existed, then $3=2^{-1}(2\cdot3)=0$, a contradiction.

Practice 8 · The field matters

Explain why $\mathbb{R}$ is a vector space over $\mathbb{Q}$, and why $\mathbb{Q}$ is not a vector space over $\mathbb{R}$ with the usual operations.

Answer

Real numbers can be added and multiplied by rational scalars, and all eight axioms hold. But $\mathbb{Q}$ is not closed under multiplication by real scalars: $\sqrt2\cdot1\notin\mathbb{Q}$.

Expanded topic checklist

Course coverage. This unit is part of the complete MAT240H1 topic map. It should cover the following items before you move on:

• Fields. Work over arbitrary fields, including $\mathbb{R}$, $\mathbb{C}$, and finite fields such as $\mathbb{Z}_p$.
• Vector spaces. Verify all vector-space axioms and recognize standard/nonstandard examples.
• Scalars matter. Track how the chosen field changes linear combinations and dimensions.
• Examples. Use coordinate spaces, polynomial spaces, function spaces, and matrix spaces as recurring models.

Textbook/source alignment

Use the current UofT MAT240H1 calendar topics as the coverage spine, and pair lecture notes with a Friedberg-Insel-Spence-style linear algebra text for extra reading. The exact assigned textbook should still be checked against the instructor syllabus.

Core theorem map

The algebra spine is: choose the field, define addition and scalar multiplication, verify closure and axioms, then reason about vectors abstractly.

Study routine. Before solving any linear algebra problem, write the field in the margin; many MAT240 traps come from changing the field without noticing.

Deep practice extensions

Use these after the main practice set. They are designed to force definition-level reasoning, theorem selection, and mixed application rather than routine calculation only.

Extension 1 · Field check

Why is $\mathbb{Z}_p$ a field when $p$ is prime?

Guide

Every nonzero residue mod $p$ has a multiplicative inverse because gcd$(a,p)=1$, so Bezout gives $ax+py=1$ and hence $ax\equiv1\pmod p$.

Extension 2 · Vector space example

Show the set of polynomials of degree at most $3$ over $F$ is a vector space.

Guide

Closure under addition and scalar multiplication holds, and the usual polynomial operations inherit the vector-space axioms from coefficients in $F$.

Extension 3 · Non-example

Why are polynomials of exactly degree $3$ not a vector space?

Guide

The zero polynomial is not degree $3$, and sums can drop degree, e.g. $x^3+(-x^3)=0$.

Extension 4 · Field dependence

Compare $\mathbb{C}$ as a vector space over $\mathbb{C}$ and over $\mathbb{R}$.

Guide

Over $\mathbb{C}$ it has dimension $1$ with basis $\{1\}$; over $\mathbb{R}$ it has dimension $2$ with basis $\{1,i\}$.

← IndexAll chapters Unit 02 →Subspaces, span & independence