MAT240H1/Unit 03
Unit 03 · Algebra I

Bases & Dimension

A basis is a minimal spanning set — just enough vectors to build everything, with none to spare. That its size is the same no matter which basis you choose is a genuine theorem, and it is what makes "dimension" a single well-defined number.

By the end you can
Why it matters & where it comes from
Why it exists

A basis gives every vector unique coordinates, turning an abstract space into something computable. Dimension is the single number that classifies the space.

In the real world

Choosing a basis is choosing coordinates — pixel bases for images, Fourier bases for audio, principal components for data. The right basis makes a problem easy.

Story & open questions

That every basis of a space has the same size is a theorem, not an assumption — the result that makes "dimension" well defined and underpins all of linear algebra.

Bases and unique representation

Definition Basis

A basis of $V$ is a linearly independent set that spans $V$. Standard examples: $e_1,\dots,e_n$ for $F^n$; the monomials $1,x,\dots,x^k$ for $P_k(F)$; the matrix units $E_{ij}$ for $M_{m\times n}(F)$.

Theorem Unique representation

$\beta=\{v_1,\dots,v_n\}$ is a basis of $V$ iff every $v\in V$ can be written as $v=a_1v_1+\cdots+a_nv_n$ in exactly one way. The $a_i$ are the coordinates of $v$ relative to $\beta$.

Proof
Basis ⇒ unique: spanning gives at least one representation. If $\sum a_iv_i=\sum b_iv_i$, then $\sum(a_i-b_i)v_i=0$, and independence forces $a_i=b_i$ for all $i$.
Unique ⇒ basis: existence of representations is spanning, and applying uniqueness to the vector $0$ (which has the all-zero representation) is exactly independence. ∎

The Replacement Theorem and dimension

Theorem Replacement (Steinitz exchange)

If $V$ is spanned by a set $G$ of $n$ vectors and $L$ is an independent set of $m$ vectors in $V$, then $m\le n$, and some $n-m$ vectors of $G$ can be adjoined to $L$ to produce a spanning set. In words: independent sets are never larger than spanning sets, and can be grown into them.

Corollary Invariance of dimension

Any two bases of a finite-dimensional $V$ have the same number of elements — each is independent and each spans, so each size bounds the other. That common size is $\dim V$.

Corollary Building bases

In a finite-dimensional space: every spanning set contains a basis (discard redundant vectors one at a time), and every independent set extends to a basis (adjoin replacement vectors). With $\dim V=n$: any $n$ independent vectors form a basis, any $n$ spanning vectors form a basis, and any $n+1$ vectors are dependent.

The one-number summary Dimension turns qualitative questions into counting. "Can these 4 vectors be independent in $\mathbb{R}^3$?" — no, $4>3$. "Do these 3 independent vectors span $\mathbb{R}^3$?" — yes, $3=\dim$. Learn to reach for the count before reaching for a computation.

Dimensions of the standard spaces

Examples
$\dim F^n=n$,   $\dim M_{m\times n}(F)=mn$,   $\dim P_k(F)=k+1$,   and $P(F)$ (all polynomials) is infinite-dimensional — no finite set can span it.
The field matters: $\mathbb{C}$ has dimension $1$ over $\mathbb{C}$ (basis $\{1\}$) but dimension $2$ over $\mathbb{R}$ (basis $\{1,i\}$).
Theorem Subspaces

If $W$ is a subspace of a finite-dimensional $V$, then $\dim W\le\dim V$, with equality iff $W=V$.

Proof idea
A basis of $W$ is an independent set in $V$, so its size is at most $\dim V$ (Replacement). If the sizes are equal, that basis is a maximal independent set in $V$, hence a basis of $V$, so $W=\operatorname{span}=V$. ∎
Common pitfalls

• A basis must be independent and spanning — neither alone suffices.
• Coordinates depend on the basis and on its order; write the basis down before writing coordinates.
• "Dimension" without naming the field is ambiguous — $\dim_{\mathbb{C}}\mathbb{C}=1$ but $\dim_{\mathbb{R}}\mathbb{C}=2$.
• To find a subspace's dimension, produce a basis; guessing the count without exhibiting independent spanners proves nothing.

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 · A plane

Find a basis for $W=\{(x,y,z)\in\mathbb{R}^3:x+y+z=0\}$ and state $\dim W$.

Answer
Solve for $x=-y-z$: every element is $y(-1,1,0)+z(-1,0,1)$, and those two vectors are independent. Basis $\{(-1,1,0),(-1,0,1)\}$, dimension $2$.
Practice 2 · Matrices

What is the dimension of $M_{2\times2}(\mathbb{R})$? Exhibit a basis.

Answer
$4$; the matrix units $E_{11},E_{12},E_{21},E_{22}$ (one entry $1$, rest $0$).
Practice 3 · Counting

Can $4$ vectors be independent in $\mathbb{R}^3$? Can $2$ vectors span it?

Answer
No and no: more than $\dim=3$ vectors are dependent, and fewer than $3$ cannot span.
Practice 4 · Unique representation

Prove: $\beta$ is a basis iff every vector has exactly one representation as a linear combination of $\beta$.

Solution
Spanning = at least one representation. Independence = uniqueness, since two representations subtract to a nontrivial combination equal to $0$. Conversely uniqueness at $v=0$ is precisely independence.
Practice 5 · Shrink to a basis

Find a basis for $\operatorname{span}\{(1,2,3),(2,4,6),(1,1,1)\}$.

Solution
$(2,4,6)=2(1,2,3)$ is redundant; $(1,2,3)$ and $(1,1,1)$ are not proportional, hence independent. Basis $\{(1,2,3),(1,1,1)\}$; the span is a plane (dimension $2$).
Practice 6 · Symmetric matrices

Find the dimension of the space of symmetric $2\times2$ real matrices.

Answer
$3$ — a symmetric matrix is determined by the entries $a_{11},a_{22},a_{12}$; a basis is $E_{11},\,E_{22},\,E_{12}+E_{21}$.
Practice 7 · Extend to a basis

Extend $\{(1,1,0)\}$ to a basis of $\mathbb{R}^3$.

Solution
Adjoin $(0,1,0)$ and $(0,0,1)$: then $a(1,1,0)+b(0,1,0)+c(0,0,1)=(a,\,a+b,\,c)=0$ forces $a=b=c=0$, and three independent vectors in a $3$-dimensional space form a basis.
Practice 8 · Full-size subspace

Prove: if $W$ is a subspace of finite-dimensional $V$ and $\dim W=\dim V$, then $W=V$.

Solution
A basis of $W$ is an independent set of size $\dim V$ in $V$, hence a basis of $V$. Then $V=\operatorname{span}\subseteq W$, so $W=V$.

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:

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 basis spine is: spanning gives existence of coordinates, independence gives uniqueness, and dimension counts how many coordinates are needed.

Study routine. When proving a list is a basis, decide whether a dimension shortcut is allowed; if not, prove spanning and independence directly.

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 · Basis check

Show $(1,0,0),(0,1,0),(0,0,1)$ is a basis of $F^3$.

Guide
Every vector has a unique coordinate expression in this list, so it spans and is independent.
Extension 2 · Dimension shortcut

In a 3-dimensional space, why do three independent vectors form a basis?

Guide
A linearly independent list of length equal to the dimension is automatically spanning by the extension theorem.
Extension 3 · Subspace dimension

If $W$ is a proper subspace of a finite-dimensional $V$, what can be said about dim $W$?

Guide
dim $W<\dim V$. If dimensions were equal, a basis of $W$ would extend to a basis of $V$ without adding vectors, forcing $W=V$.
Extension 4 · Coordinates

Find coordinates of $p(x)=2+x$ in basis $(1,x,x^2)$ for $P_2(F)$.

Guide
The coordinate vector is $(2,1,0)$ relative to that ordered basis.
← Unit 02Subspaces, span & independence Unit 04 →Linear transformations