Phase 4 - Lesson 4.1

Vector Spaces, Span, Independence, Basis, Dimension

The abstract stage on which all of linear algebra plays out: closure, spanning sets, independence, and the invariant called dimension.

⏱ 50 min● Intermediate🔗 Prereqs: 3.x multivariable calculus

↖ Phase 4 hub

Builds on: Vectors in R^n from 3.1.
Leads to: Linear maps (4.2) act between the spaces defined here; every later structure is a special space.

Learning Objectives

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State the vector-space axioms and verify whether a given set is a subspace.
Compute the span of a set of vectors and decide linear independence.
Construct a basis and compute the dimension of a subspace.
Explain why every basis of a finite-dimensional space has the same size.

Key Vocabulary

Vector space: A set with addition and scalar multiplication satisfying the eight axioms (closure, associativity, identity, inverses, distributivity).

Subspace: A nonempty subset closed under addition and scalar multiplication; automatically contains \(0\).

Span: The set \(\operatorname{span}(v_1,\dots,v_k)=\{\sum c_i v_i\}\) of all linear combinations.

Linear independence: No nontrivial combination gives zero: \(\sum c_i v_i=0\Rightarrow c_i=0\) for all \(i\).

Basis: A linearly independent spanning set; every vector has unique coordinates in it.

Dimension: The number of vectors in any basis - an invariant of the space.

Intuition & Motivation

Intuition

A vector space is just a place where you can add things and scale them and stay inside. That abstraction pays off because polynomials, matrices, random variables, and stock-return vectors all obey the same eight rules - so one theory covers them all. Span is everything you can build from a set; independence is having no redundancy; a basis is the sweet spot - enough vectors to build everything, few enough that coordinates are unique. Dimension is the head-count of that minimal building set, and remarkably it does not depend on which basis you pick.

Spaces and subspaces

Definition - Subspace test

A nonempty subset \(U\subseteq V\) is a subspace iff it is closed under the two operations: \(u,w\in U\Rightarrow u+w\in U\) and \(c\in\F,\,u\in U\Rightarrow cu\in U\). Both together force \(0\in U\).

Examples: lines and planes through the origin in \(\R^3\); polynomials of degree \(\le n\); the set of portfolios that are fully hedged against a factor. A line not through the origin fails (no zero vector).

Span, independence, basis

A list \((v_1,\dots,v_k)\) is linearly independent when the only way to write \(0\) as a combination is with all coefficients zero. Otherwise some \(v_j\) is a combination of the others - redundant. A basis is an independent list that also spans; then every vector has unique coordinates.

Theorem - Dimension is well-defined

In a finite-dimensional space every basis has the same number of elements; call it \(\dim V\). Moreover any independent list has length \(\le\dim V\) and any spanning list has length \(\ge\dim V\).

Proof

The engine is the Steinitz exchange lemma: if \((u_i)\) is independent and \((w_j)\) spans, one can swap the \(u_i\) into the spanning list one at a time, proving length(independent) ≤ length(spanning). Applying this with two bases in both directions forces their lengths to be equal.

Worked Example - Is a set a basis of \(\R^3\)?

Test \(v_1=(1,0,1),\ v_2=(0,1,1),\ v_3=(1,1,0)\). Three vectors in a 3-D space: a basis iff independent.

Set \(c_1 v_1+c_2 v_2+c_3 v_3=0\): componentwise \(c_1+c_3=0,\ c_2+c_3=0,\ c_1+c_2=0\).

From the first two, \(c_1=c_2=-c_3\). Substitute into the third: \(-2c_3=0\Rightarrow c_3=0\), hence all zero.

Independent and of length \(3=\dim\R^3\), so it is a basis. Any vector now has unique coordinates in it.

Interactive: rank of a coordinate set

Common Mistakes to Avoid

Calling a set a basis because it spans, without checking independence (or vice-versa) - you need both.
Forgetting that a subspace must contain \(0\); affine sets (lines off the origin) are not subspaces.
Thinking more vectors always span more - beyond \(\dim V\) they must be dependent.
Confusing the number of components (ambient dimension) with the dimension of the subspace they span.

Quant Practitioner Tips

To test independence of \(k\) vectors, stack them and compute the rank - independent iff rank \(=k\).
In finite dimensions, \(\dim V\) vectors that are independent automatically span (and vice versa) - you only check one property.
Think of asset-return vectors: a set of factors is a basis of the return space only if it is independent and spans all realized returns.

Knowledge Check

Q1 Easy

Which set is a subspace of \(\R^2\)?

The line \(y=x+1\)

The line \(y=2x\)

The unit circle

The first quadrant

Q2 Medium

A list of 4 vectors in \(\R^3\) is:

Always independent

Always spanning

Necessarily dependent

Necessarily a basis

Q3 Medium

Why is the dimension of a finite-dimensional space well-defined?

Because R^n is the only space

Because every basis has the same number of elements

Because bases are unique

Because vectors have finitely many components

Practical Exercise

Let \(U=\{(x,y,z)\in\R^3: x+y+z=0\}\). (a) Show \(U\) is a subspace. (b) Find a basis and \(\dim U\).

▶ Show full solution

(a) If \(u,w\in U\) then their coordinates each sum to zero; so do \(u+w\) and \(cu\) (sums scale linearly). \(U\) is nonempty (\(0\in U\)) and closed, hence a subspace.

(b) Solve \(z=-x-y\) with \(x,y\) free: \((x,y,z)=x(1,0,-1)+y(0,1,-1)\). So \(\{(1,0,-1),(0,1,-1)\}\) spans \(U\), and it is independent (neither is a multiple of the other). Thus it is a basis and \(\dim U=2\) - a plane through the origin.

After the reveal, answer for yourself: One linear constraint on R^3 cut the dimension from 3 to 2; each independent constraint costs one dimension.

Lesson Summary

A vector space is any set closed under addition and scaling; a subspace must contain \(0\). Span builds everything from a set, independence forbids redundancy, and a basis is the minimal spanning set giving unique coordinates. Every basis has the same size, the dimension - the single most important invariant of a space.

Formula Sheet Additions

Independence via rank

\[v_1,\dots,v_k\ \text{independent}\iff \rank[v_1\,\cdots\,v_k]=k\]

Redundancy is exactly rank deficiency.

Retrieval Practice

Close the lesson and answer from memory before checking. This is deliberate, effortful recall - the single highest-yield study action.

▶ Show retrieval prompts & answers

Q: What two properties define a basis?
A: Linear independence and spanning; together they give every vector unique coordinates.

Q: Why must 4 vectors in R^3 be dependent?
A: Any linearly independent list has length at most the dimension (3), so a list of 4 cannot be independent.

Completion Checklist

I can explain the core ideas in my own words
I worked the derivations/examples by hand
I completed the interactive workbench(es)
I passed the knowledge check

Confidence / mastery rating

Personal notes

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