Quant Academy Math · Probability · Quant Finance
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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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 \(\R^3\)?
1
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.
2
Set \(c_1 v_1+c_2 v_2+c_3 v_3=0\): componentwise \(c_1+c_3=0,\ c_2+c_3=0,\ c_1+c_2=0\).
3
From the first two, \(c_1=c_2=-c_3\). Substitute into the third: \(-2c_3=0\Rightarrow c_3=0\), hence all zero.
4
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.

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Source References

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