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

Define a basis and prove the unique-representation property.
State the Replacement Theorem and derive invariance of dimension.
Shrink spanning sets to bases and extend independent sets to bases.
Compute dimensions of the standard spaces and of subspaces.

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:

Bases. Use bases as independent spanning lists and as coordinate systems.
Dimension. Prove all bases have the same size and use dimension to compare subspaces.
Exchange ideas. Use replacement/exchange lemmas to move between spanning lists and independent lists.
Finite-dimensional structure. Know maximal independent equals minimal spanning in finite dimensions.

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