Mathematics.

abstract linear algebra

Dual Vector Spaces

Linear Algebra40 minDifficulty6 out of 10

You should know: vector space, linear transformation

Overview

The dual space V* of a vector space V is the set of all linear functionals on V — linear maps from V to its scalar field. Every finite-dimensional vector space is isomorphic to its dual, but the isomorphism is not canonical (it depends on a choice of basis). The double dual V** is canonically isomorphic to V, which captures a deep symmetry in linear algebra.

Intuition

Think of dual vectors as 'measurement devices' for vectors. Each linear functional f in V* takes a vector v in V and returns a number — it measures v in some way. The standard basis vectors measure components: the i-th dual basis vector e^i extracts the i-th coordinate. The dual is the space of all possible linear measurements you can make on vectors.

Formal Definition

Definition

Let V be a vector space over field F. The dual space V* is defined as:

V∗={f:V→F∣f is linear}V^* = \{ f : V \to F \mid f \text{ is linear} \}

V* consists of all linear maps from V to the scalar field F.

Dual space
If {e1,…,en} is a basis of V, then ei(ej)=δji\text{If } \{e_1,\ldots,e_n\} \text{ is a basis of } V, \text{ then } e^i(e_j) = \delta^i_j

The dual basis {e^1,...,e^n} satisfies e^i(e_j) = δ_{ij} (Kronecker delta).

Dual basis
dim⁡(V∗)=dim⁡(V)\dim(V^*) = \dim(V)

For finite-dimensional V, V* has the same dimension as V.

Dimension

Notation

NotationMeaning
V∗V^*Dual space of V: all linear functionals f: V → F
f(v)f(v)Evaluation of linear functional f on vector v; also written ⟨f, v⟩
eie^ii-th dual basis vector: e^i(e_j) = δ_{ij}

Properties

V** ≅ V canonically

There is a canonical isomorphism V→V∗∗ given by v↦(f↦f(v)).\text{There is a canonical isomorphism } V \to V^{**} \text{ given by } v \mapsto (f \mapsto f(v)).

Dual of a linear map

If T:V→W is linear, its transpose (dual map) T∗:W∗→V∗ is defined by (T∗g)(v)=g(Tv).\text{If } T: V \to W \text{ is linear, its transpose (dual map) } T^*: W^* \to V^* \text{ is defined by } (T^* g)(v) = g(Tv).

Annihilator

For a subspace U⊆V, its annihilator U0={f∈V∗∣f(u)=0 for all u∈U} satisfies dim⁡(U0)=dim⁡(V)−dim⁡(U).\text{For a subspace } U \subseteq V, \text{ its annihilator } U^0 = \{f \in V^* \mid f(u)=0 \text{ for all } u \in U\} \text{ satisfies } \dim(U^0) = \dim(V) - \dim(U).

Worked Examples

  1. 1

    By definition, e^i(e_j) = δ_{ij}. So e^1 must satisfy e^1(e_1)=1 and e^1(e_2)=0.

    e1(x1,x2)=1⋅x1+0⋅x2=x1e^1(x_1, x_2) = 1 \cdot x_1 + 0 \cdot x_2 = x_1
  2. 2

    Similarly e^2 extracts the second coordinate.

    e2(x1,x2)=x2e^2(x_1, x_2) = x_2
  3. 3

    Evaluate e^1 on (3,5).

    e1(3,5)=3e^1(3,5) = 3

✓ Answer

e^1(x_1,x_2) = x_1, e^2(x_1,x_2) = x_2; e^1(3,5) = 3.

Practice Problems

Mediumfree response

Let V = R^3. Write an explicit formula for the dual basis vector e^2 (corresponding to the second standard basis vector) and compute e^2(1,4,9).

Mediumfree response

Let V = R^2 and U = span{(1,1)}. Find the annihilator U^0 ⊆ V* and confirm that dim(U^0) = dim(V) - dim(U).

Common Mistakes

Common Mistake

Thinking V and V* are the same space (not just isomorphic).

V and V* are isomorphic for finite-dimensional V, but the isomorphism requires a choice of basis — it is not canonical. Only V** is canonically isomorphic to V.

Quiz

The dual space V* consists of:
For finite-dimensional V, which isomorphism is canonical (basis-independent)?
If T: V → W is linear, the transpose T*: W* → V* is defined by:

Summary

  • The dual space V* is the vector space of all linear functionals f: V → F.
  • A basis {e_1,...,e_n} of V determines a dual basis {e^1,...,e^n} by e^i(e_j) = δ_{ij}.
  • dim(V*) = dim(V) for finite-dimensional V; V and V* are isomorphic but not canonically.
  • V** is canonically isomorphic to V via v ↦ (f ↦ f(v)).
  • The transpose (dual map) T*: W* → V* of T: V → W satisfies (T*g)(v) = g(Tv).

References

  1. WebsiteWikipedia — Dual space