LA-3.5

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Dual Spaces

The dual space $V^*$ consists of all linear functionals on $V$ —linear maps from $V$ to the base field. It's a "mirror" of $V$ that reveals deep structural insights.

3-4 hours Advanced 10 Objectives

Learning Objectives

Define linear functionals and the dual space V*
Understand the geometric meaning of linear functionals as hyperplanes
Construct the dual basis from any given basis
Prove dim(V*) = dim(V) for finite-dimensional V
Define and compute annihilators of subspaces
Understand the dimension formula for annihilators
Define the double dual V** and the evaluation map
Prove V ≅ V** naturally for finite-dimensional V
Distinguish natural vs. non-natural isomorphisms
Define and understand the dual (transpose) of a linear map

Prerequisites

Linear maps and L(V, W) (LA-3.1)
Isomorphisms and dimension (LA-3.4)
Basis and dimension (LA-2.4)
Kernel and image (LA-3.2)
Rank-Nullity Theorem (LA-3.3)

Historical Context

The concept of duality pervades mathematics. In geometry, points and lines are dual; in logic, AND and OR are dual. Linear algebra's dual space captures this idea perfectly.

The distinction between "natural" and "unnatural" isomorphisms was formalized by Samuel Eilenberg and Saunders Mac Lane when they invented category theory in the 1940s. The natural isomorphism $V \cong V^{**}$ was one of the motivating examples!

1. Linear Functionals

A linear functional is the simplest type of linear map: it takes vectors to scalars.

Definition 3.11: Linear Functional

A linear functional on a vector space $V$ over field $F$ is a linear map $\phi: V \to F$ .

Equivalently, $\phi$ satisfies:

$\phi(v + w) = \phi(v) + \phi(w)$ for all $v, w \in V$
$\phi(cv) = c\phi(v)$ for all $c \in F, v \in V$

Example 3.25: Examples of Linear Functionals

Coordinate extraction: $\phi(x_1, \ldots, x_n) = x_i$ (projects onto $i$ -th coordinate)
Dot product: $\phi_a(x) = a \cdot x$ for fixed $a \in \mathbb{R}^n$
Evaluation: $\text{ev}_c(p) = p(c)$ for polynomials $p$
Integration: $\phi(f) = \int_0^1 f(x)\,dx$ on $C[0,1]$
Trace: $\text{tr}(A) = \sum_i a_{ii}$ on $M_{n \times n}$

Remark 3.16: Geometric Meaning

A nonzero linear functional $\phi: V \to F$ defines a family of parallel hyperplanes:

H_c = \{v \in V : \phi(v) = c\}

The kernel $\ker \phi = H_0$ is the hyperplane through the origin. Since $\phi$ is nonzero and surjective onto $F$ , we have $\dim(\ker \phi) = n - 1$ (codimension 1).

2. The Dual Space

Definition 3.12: Dual Space

The dual space of $V$ is the vector space of all linear functionals on $V$ :

V^* = \mathcal{L}(V, F) = \{\phi: V \to F \mid \phi \text{ is linear}\}

Theorem 3.44: Dimension of Dual Space

If $V$ is a finite-dimensional vector space over $F$ , then:

\dim(V^*) = \dim(V)

Proof:

We have $V^* = \mathcal{L}(V, F)$ . The dimension formula for $\mathcal{L}(V, W)$ gives:

\dim(V^*) = \dim(V) \cdot \dim(F) = \dim(V) \cdot 1 = \dim(V)

∎

Corollary 3.7: V and V* are Isomorphic

For finite-dimensional $V$ : $V \cong V^*$ (as vector spaces).

Remark 3.17: Important Caveat

While $V \cong V^*$ , there is no canonical isomorphism! To construct an isomorphism, you must choose a basis. Different bases give different isomorphisms.

3. The Dual Basis

Given a basis of $V$ , we can construct a corresponding basis of $V^*$ called the dual basis.

Definition 3.13: Dual Basis

Let $B = \{e_1, \ldots, e_n\}$ be a basis of $V$ . The dual basis $B^* = \{e_1^*, \ldots, e_n^*\}$ of $V^*$ is defined by:

e_i^*(e_j) = \delta_{ij} = \begin{cases} 1 & \text{if } i = j \\ 0 & \text{if } i \neq j \end{cases}

Theorem 3.45: Dual Basis is a Basis

The dual basis $\{e_1^*, \ldots, e_n^*\}$ is indeed a basis for $V^*$ .

Proof:

Existence: For each $i$ , define $e_i^*$ by specifying its values on the basis: $e_i^*(e_j) = \delta_{ij}$ . By the extension theorem, this uniquely defines a linear functional.

Linear independence: Suppose $\sum_i c_i e_i^* = 0$ . Apply both sides to $e_j$ :

0 = \left(\sum_i c_i e_i^*\right)(e_j) = \sum_i c_i e_i^*(e_j) = \sum_i c_i \delta_{ij} = c_j

So all $c_j = 0$ , proving linear independence.

Spanning: Let $\phi \in V^*$ . For any $v = \sum_j a_j e_j$ :

\phi(v) = \sum_j a_j \phi(e_j) = \sum_j \phi(e_j) e_j^*(v) = \left(\sum_j \phi(e_j) e_j^*\right)(v)

So $\phi = \sum_j \phi(e_j) e_j^*$ , proving spanning.

∎

Remark 3.18: Coordinates via Dual Basis

The dual basis extracts coordinates! If $v = \sum_i a_i e_i$ , then:

e_j^*(v) = a_j

So $e_j^*$ picks out the $j$ -th coordinate of $v$ with respect to the basis $B$ .

Example 3.26: Dual Basis in R²

Problem: Find the dual basis to $\{(1, 1), (1, -1)\}$ in $\mathbb{R}^2$ .

Solution: Let $e_1 = (1, 1)$ and $e_2 = (1, -1)$ .

We need $e_1^*, e_2^*: \mathbb{R}^2 \to \mathbb{R}$ with:

$e_1^*(1, 1) = 1$ , $e_1^*(1, -1) = 0$
$e_2^*(1, 1) = 0$ , $e_2^*(1, -1) = 1$

Writing $e_1^*(x, y) = ax + by$ :

$a + b = 1$ and $a - b = 0$ , so $a = b = 1/2$ .

Answer: $e_1^*(x, y) = \frac{x + y}{2}$ , $e_2^*(x, y) = \frac{x - y}{2}$

4. The Double Dual V**

The double dual $V^{**} = (V^*)^*$ has a special relationship with $V$ .

Definition 3.14: Double Dual

The double dual of $V$ is:

V^{**} = (V^*)^* = \mathcal{L}(V^*, F)

Definition 3.15: Evaluation Map

For each $v \in V$ , define $\text{ev}_v: V^* \to F$ by:

\text{ev}_v(\phi) = \phi(v)

This is a linear functional on $V^*$ , so $\text{ev}_v \in V^{**}$ .

Theorem 3.46: Natural Isomorphism V ≅ V**

The map $\Phi: V \to V^{**}$ defined by $\Phi(v) = \text{ev}_v$ is:

Linear
Injective (for any V)
An isomorphism when V is finite-dimensional

This isomorphism is natural—it doesn't depend on any choices.

Proof:

Linearity: For $v, w \in V$ and $c \in F$ :

\Phi(v + w)(\phi) = \phi(v + w) = \phi(v) + \phi(w) = \Phi(v)(\phi) + \Phi(w)(\phi)

Similarly for scalar multiplication.

Injectivity: Suppose $\Phi(v) = 0$ , i.e., $\text{ev}_v = 0$ . This means $\phi(v) = 0$ for all $\phi \in V^*$ .

If $v \neq 0$ , extend $\{v\}$ to a basis and define $\phi$ with $\phi(v) = 1$ . This contradicts $\phi(v) = 0$ . So $v = 0$ .

Surjectivity (finite dim): $\dim(V^{**}) = \dim(V^*) = \dim(V)$ . Since $\Phi$ is injective and dimensions match, $\Phi$ is surjective.

∎

Remark 3.19: Why 'Natural'?

The isomorphism $\Phi: V \to V^{**}$ is defined by the same formula for all vector spaces. No basis choice is needed. In contrast, $V \cong V^*$ requires choosing a basis.

This is formalized in category theory: $\Phi$ is a natural transformation.

5. Annihilators

Annihilators provide a correspondence between subspaces of $V$ and subspaces of $V^*$ .

Definition 3.16: Annihilator

Let $W \subseteq V$ be a subspace. The annihilator of $W$ is:

W^0 = \{\phi \in V^* : \phi(w) = 0 \text{ for all } w \in W\}

Theorem 3.47: Annihilator is a Subspace

$W^0$ is a subspace of $V^*$ .

Theorem 3.48: Dimension of Annihilator

For finite-dimensional $V$ with subspace $W \subseteq V$ :

\dim(W^0) = \dim(V) - \dim(W)

Proof:

Extend a basis $\{w_1, \ldots, w_k\}$ of $W$ to a basis $\{w_1, \ldots, w_k, v_{k+1}, \ldots, v_n\}$ of $V$ .

Let $\{w_1^*, \ldots, w_k^*, v_{k+1}^*, \ldots, v_n^*\}$ be the dual basis.

Then $\phi \in W^0$ iff $\phi(w_i) = 0$ for all $i$ .

Writing $\phi = \sum_j c_j w_j^* + \sum_l d_l v_l^*$ :

\phi(w_i) = c_i = 0 \text{ for all } i

So $W^0 = \text{span}\{v_{k+1}^*, \ldots, v_n^*\}$ , which has dimension $n - k$ .

∎

Example 3.27: Computing an Annihilator

Problem: Let $W = \text{span}\{(1, 0, 1)\} \subseteq \mathbb{R}^3$ . Find $W^0$ .

Solution:

$\phi(x, y, z) = ax + by + cz \in W^0$ iff $\phi(1, 0, 1) = 0$ .

So $a + c = 0$ , giving $c = -a$ .

Answer: $W^0 = \{(x, y, z) \mapsto ax + by - az : a, b \in \mathbb{R}\}$

$\dim(W^0) = 3 - 1 = 2$ ✓

Theorem 3.49: Double Annihilator

Under the natural identification $V \cong V^{**}$ :

(W^0)^0 = W

6. The Dual Map (Transpose)

Every linear map $T: V \to W$ induces a map between the dual spaces in the opposite direction.

Definition 3.17: Dual Map

Let $T: V \to W$ be linear. The dual map (or transpose) is:

T^*: W^* \to V^*, \quad T^*(\phi) = \phi \circ T

That is, $T^*(\phi)(v) = \phi(T(v))$ for all $v \in V$ .

Theorem 3.50: Properties of Dual Maps

For linear maps $S, T$ and scalar $c$ :

$(S + T)^* = S^* + T^*$
$(cT)^* = cT^*$
$(ST)^* = T^* S^*$ (note the reversal!)
$(I_V)^* = I_{V^*}$
If $T$ is invertible, then $(T^*)^{-1} = (T^{-1})^*$

Theorem 3.51: Kernel and Image of Dual Map

For $T: V \to W$ :

$\ker(T^*) = (\text{im } T)^0$
$\text{im}(T^*) = (\ker T)^0$

Corollary 3.8: Rank of Dual Map

$\text{rank}(T^*) = \text{rank}(T)$

Remark 3.20: Matrix Interpretation

If $T$ is represented by matrix $A$ (with respect to bases $B_V$ and $B_W$ ), then $T^*$ is represented by $A^T$ (with respect to dual bases $B_W^*$ and $B_V^*$ ).

This is why $T^*$ is sometimes called the transpose of $T$ .

7. Worked Examples

Dual of Differentiation

Problem: Let $D: P_2 \to P_1$ be differentiation. Describe $D^*: P_1^* \to P_2^*$ .

Solution:

For $\phi \in P_1^*$ and $p \in P_2$ :

D^*(\phi)(p) = \phi(D(p)) = \phi(p')

If $\phi(q) = \int_0^1 q(x)\,dx$ (integration), then:

D^*(\phi)(a + bx + cx^2) = \phi(b + 2cx) = \int_0^1 (b + 2cx)\,dx = b + c

Annihilator of Row Space

Problem: Let $A$ be an $m \times n$ matrix. How is $(\text{row}(A))^0$ related to $\ker(A)$ ?

Solution:

Identify $(\mathbb{R}^n)^* \cong \mathbb{R}^n$ via the dot product.

Then $\phi_a(x) = a \cdot x \in (\text{row}(A))^0$ iff $a \cdot r = 0$ for all rows $r$ .

This means $A^T a = 0$ , so $a \in \ker(A^T)$ .

Result: $(\text{row}(A))^0 \cong \ker(A^T) = (\text{col}(A))^\perp$

Dual Basis Computation

Problem: Find the dual basis to $\{1, 1+x, 1+x+x^2\}$ in $P_2$ .

Solution:

Need $e_1^*, e_2^*, e_3^*$ with $e_i^*(e_j) = \delta_{ij}$ .

Write $e_i^*(p) = a_i p(0) + b_i p'(0) + c_i p''(0)/2$ (Taylor coefficients).

Solving the system:

$e_1^*(p) = p(0) - p'(0) + p''(0)/2$
$e_2^*(p) = p'(0) - p''(0)$
$e_3^*(p) = p''(0)/2$

8. Common Mistakes

Mistake 1: V = V*

$V \cong V^*$ does NOT mean they're equal! $V$ contains vectors, $V^*$ contains linear functionals. They're different types of objects.

Mistake 2: Dual Map Direction

If $T: V \to W$ , then $T^*: W^* \to V^*$ goes the opposite direction! This reversal is essential and easy to forget.

Mistake 3: Natural vs. Non-Natural

$V \cong V^{**}$ is natural (canonical), but $V \cong V^*$ is NOT. Don't assume there's a "preferred" isomorphism between $V$ and $V^*$ .

Mistake 4: Infinite Dimensions

In infinite dimensions, $V^*$ can be much larger than $V$ ! The natural map $V \to V^{**}$ is still injective but may not be surjective.

9. Key Takeaways

Dual Space

$V^* = \mathcal{L}(V, F)$ , $\dim(V^*) = \dim(V)$

Dual Basis

$e_i^*(e_j) = \delta_{ij}$

Double Dual

$V \cong V^{**}$ naturally via $v \mapsto \text{ev}_v$

Annihilator

$\dim(W^0) = \dim(V) - \dim(W)$

10. Additional Practice Problems

Problem 1

Let $V = P_3$ and $\phi(p) = p(0) + p(1)$ . Find $\ker(\phi)$ and describe its dual basis.

Problem 2

Prove that if $W_1 \subseteq W_2$ , then $W_2^0 \subseteq W_1^0$ .

Problem 3

Show that $(W_1 + W_2)^0 = W_1^0 \cap W_2^0$ and $(W_1 \cap W_2)^0 = W_1^0 + W_2^0$ .

Problem 4

If $T: V \to W$ is surjective, prove $T^*: W^* \to V^*$ is injective.

Problem 5

Let $\{\phi_1, \phi_2, \phi_3\}$ be a basis of $(\mathbb{R}^3)^*$ . Find the basis of $\mathbb{R}^3$ dual to this.

Problem 6

Show that the trace map $\text{tr}: M_n \to F$ is the unique (up to scalar) linear functional that vanishes on all commutators $[A, B] = AB - BA$ .

11. More Worked Examples

Integration as Linear Functional

Problem: On $P_2$ , define $\phi_1(p) = \int_0^1 p$ , $\phi_2(p) = \int_0^2 p$ , $\phi_3(p) = \int_1^2 p$ . Are these linearly independent?

Solution:

Notice $\phi_3 = \phi_2 - \phi_1$ (by additivity of integrals over intervals).

So $\phi_1 - \phi_2 + \phi_3 = 0$ .

Answer: No, they are linearly dependent.

Dual of Projection

Problem: Let $P: V \to V$ be a projection ( $P^2 = P$ ). What is $P^*$ ?

Solution:

$(P^*)^2 = (P^2)^* = P^*$

So $P^*$ is also a projection!

$\ker(P^*) = (\text{im } P)^0$ and $\text{im}(P^*) = (\ker P)^0$ .

Hyperplane Determined by Functional

Problem: Two functionals $\phi, \psi \in V^*$ have the same kernel. What can we conclude?

Solution:

If $\ker \phi = \ker \psi$ and both are nonzero, then both kernels are hyperplanes of the same dimension.

Since $\dim(\ker \phi) = n - 1$ , the quotient $V / \ker \phi$ is 1-dimensional.

Both $\phi$ and $\psi$ factor through this quotient, so:

Answer: $\psi = c\phi$ for some nonzero $c \in F$ .

Annihilator of Kernel

Problem: For nonzero $\phi \in V^*$ , find $(\ker \phi)^0$ .

Solution:

$\dim(\ker \phi)^0 = \dim V - \dim(\ker \phi) = n - (n-1) = 1$

Any $\psi \in (\ker \phi)^0$ vanishes on $\ker \phi$ , so $\ker \phi \subseteq \ker \psi$ .

This forces $\psi = c\phi$ for some $c$ .

Answer: $(\ker \phi)^0 = \text{span}\{\phi\}$

12. Connections to Other Topics

Inner Products

An inner product $\langle \cdot, \cdot \rangle$ on $V$ gives a canonical isomorphism:

V \to V^*, \quad v \mapsto \langle v, \cdot \rangle

This is called the Riesz representation theorem in infinite dimensions.

Quantum Mechanics: Bra-Ket Notation

In Dirac notation:

$|\psi\rangle$ is a ket (vector in $V$ )
$\langle \phi|$ is a bra (linear functional in $V^*$ )
$\langle \phi | \psi \rangle$ is the pairing (inner product)

Differential Geometry: Cotangent Space

The cotangent space $T_p^* M$ at a point $p$ of a manifold $M$ is the dual of the tangent space.

Tangent vectors = velocities of curves
Cotangent vectors = differentials of functions
The differential $df$ of a function $f$ is a cotangent vector

Optimization: Duality Theory

Linear programming has a dual problem involving:

Primal variables ↔ Dual constraints
Primal constraints ↔ Dual variables
Strong duality: optimal values are equal

This extends to convex optimization (Lagrange duality).

13. Theoretical Insights

Theorem 3.52: Annihilator and Quotient

For any subspace $W \subseteq V$ :

W^0 \cong (V/W)^*

Proof:

Define $\Psi: W^0 \to (V/W)^*$ by $\Psi(\phi)([v]) = \phi(v)$ .

This is well-defined: if $[v] = [v']$ , then $v - v' \in W$ , so $\phi(v) = \phi(v')$ .

Injectivity: If $\Psi(\phi) = 0$ , then $\phi(v) = 0$ for all $v$ , so $\phi = 0$ .

Surjectivity: Any $\psi \in (V/W)^*$ lifts to $\phi = \psi \circ \pi \in W^0$ where $\pi: V \to V/W$ .

∎

Theorem 3.53: Duality Between Subspaces

The maps $W \mapsto W^0$ and $U \mapsto U^0$ (in the other direction) are mutually inverse bijections between:

Subspaces of $V$
Subspaces of $V^*$

This is an order-reversing correspondence (inclusion ↔ reverse inclusion).

Remark 3.21: Functor Properties

Taking duals is a contravariant functor:

It reverses the direction of maps
It reverses composition: $(S \circ T)^* = T^* \circ S^*$
It preserves identity: $(I_V)^* = I_{V^*}$

14. Study Tips

Tip 1: Think of Duals as Coordinates

The dual basis functionals $e_i^*$ extract coordinates. This makes $V^*$ very concrete.

Tip 2: Annihilators = Constraints

Think of $W^0$ as all linear equations that $W$ satisfies. Each $\phi \in W^0$ is a constraint.

Tip 3: Dimension Formulas

Use $\dim(W^0) = n - \dim(W)$ and $\text{rank}(T) = \text{rank}(T^*)$ freely.

Tip 4: Direction Reversal

Always remember: $T: V \to W$ gives $T^*: W^* \to V^*$ (opposite direction!).

15. Quick Reference Summary

Concept	Definition/Result
Dual space	$V^* = \mathcal{L}(V, F)$
Dimension	$\dim(V^*) = \dim(V)$
Dual basis	$e_i^*(e_j) = \delta_{ij}$
Double dual	$V \cong V^{**}$ naturally
Annihilator dim	$\dim(W^0) = n - \dim(W)$
Dual map	$T^: W^ \to V^$ , $T^(\phi) = \phi \circ T$
Rank preserved	$\text{rank}(T^*) = \text{rank}(T)$

16. Challenge Problems

Challenge 1: Reflexivity

Problem: A Banach space $V$ is called reflexive if the natural map $V \to V^{**}$ is surjective. Give an example of a non-reflexive Banach space.

Hint: Consider $\ell^1$ or $c_0$ .

Challenge 2: Annihilator Properties

Problem: Prove that $(W_1 \cap W_2)^0 = W_1^0 + W_2^0$ by computing dimensions on both sides and showing inclusion.

Challenge 3: Dual of Direct Sum

Problem: Prove that $(V \oplus W)^* \cong V^* \oplus W^*$ naturally.

Hint: Define the isomorphism by restriction.

17. Geometric Interpretation

Points and Hyperplanes

In projective geometry, there's a perfect duality:

Points in $\mathbb{P}^n$ ↔ Hyperplanes in $(\mathbb{P}^n)^*$
Lines ↔ Codimension-2 subspaces
"Two points determine a line" ↔ "Two hyperplanes intersect in codim-2"

Covectors as Gradients

A linear functional $\phi: \mathbb{R}^n \to \mathbb{R}$ can be written as:

\phi(x) = \nabla f \cdot x

where $f(x) = \phi(x)$ . So $\phi$ is the "gradient" of a linear function.

The level sets $\phi^{-1}(c)$ are the hyperplanes perpendicular to this gradient.

Pairing Diagram

The pairing between $V$ and $V^*$ is the bilinear map:

\langle \cdot, \cdot \rangle: V^* \times V \to F, \quad \langle \phi, v \rangle = \phi(v)

This is non-degenerate: if $\langle \phi, v \rangle = 0$ for all $\phi$ , then $v = 0$ .

18. Historical Note

The concept of duality has ancient roots. In projective geometry, Gergonne and Poncelet (early 1800s) discovered the duality between points and lines.

Stefan Banach (1920s-30s) developed the theory of dual spaces in the infinite-dimensional setting, leading to the powerful Hahn-Banach theorem.

The categorical perspective—distinguishing natural from unnatural isomorphisms—was formalized by Eilenberg and Mac Lane in their 1945 paper "General Theory of Natural Equivalences," which founded category theory. The isomorphism $V \cong V^{**}$ was one of their key examples!

What's Next?

With dual spaces mastered, you're ready for:

Part IV: Matrices: Represent linear maps as matrices
Matrix Transpose: The matrix representation of the dual map
Bilinear Forms: Connections between V and V* via inner products
Spectral Theory: Eigenvectors and eigenvalues

Dual spaces may seem abstract, but they're essential in advanced mathematics—from differential geometry to quantum mechanics to optimization.

Dual Spaces Practice

Questions

Correct

Accuracy

What is

\dim(V^*)

\dim(V) = n

Easy

Not attempted

\{e_1, e_2\}

is a basis of

V

, what is

e_1^*(e_2)

Easy

Not attempted

The dual of a linear map

T: V \to W

T^*: ? \to ?

Medium

Not attempted

Is the map

V \to V^{**}

given by

v \mapsto \text{ev}_v

injective?

Medium

Not attempted

What is

\dim((\mathbb{R}^3)^{**})

Easy

Not attempted

The annihilator of a subspace

W \subseteq V

is:

Medium

Not attempted

\dim(V) = n

and

\dim(W) = k

, what is

\dim(W^0)

Hard

Not attempted

V \cong V^*

Hard

Not attempted

\phi \in V^*

and

\phi \neq 0

, what is

\dim(\ker \phi)

Medium

Not attempted

The kernel of a nonzero linear functional is called:

Easy

Not attempted

For the dual map

T^*

, we have

\ker(T^*) = ?

Hard

Not attempted

\{e_1^*, e_2^*, e_3^*\}

is the dual basis to

\{e_1, e_2, e_3\}

, then

(e_1^* + e_2^*)(e_1 + e_2 + e_3) = ?

Medium

Not attempted

Frequently Asked Questions

What's a linear functional?

A linear map from V to the base field F. It assigns a scalar to each vector, linearly. Examples: evaluation at a point, integration, trace of a matrix.

Why is V ≅ V** 'natural' but V ≅ V* is not?

The isomorphism V → V** via v ↦ ev_v doesn't depend on any choices—it's defined the same way for all vector spaces. But V → V* requires choosing a basis (or an inner product). 'Natural' means 'functorial' in category theory.

What's the dual basis used for?

It gives coordinates! If {e₁,...,eₙ} is a basis and {e₁*,...,eₙ*} is dual, then v = Σeᵢ*(v)eᵢ. The functionals eᵢ* extract the i-th coordinate.

How do annihilators relate to solving equations?

W⁰ consists of all linear constraints satisfied by W. If W is the solution space of Ax = 0, then W⁰ is spanned by the rows of A (in a sense).

What happens to duals in infinite dimensions?

V* can be much larger than V! The algebraic dual may not be useful; instead, we use continuous duals in functional analysis, where V* consists only of continuous linear functionals.

What is the geometric meaning of a linear functional?

A nonzero linear functional φ defines a family of parallel hyperplanes: the level sets {v : φ(v) = c} for various constants c. The kernel φ⁻¹(0) is the hyperplane through the origin.

How does the dual map T* relate to the transpose matrix?

When we represent T by a matrix A with respect to chosen bases, the dual map T* is represented by the transpose Aᵀ with respect to the dual bases. This is why T* is sometimes called the 'transpose' of T.

What is the double annihilator (W⁰)⁰?

For finite-dimensional V, the double annihilator (W⁰)⁰ in V** corresponds to W under the natural isomorphism V ≅ V**. So (W⁰)⁰ 'is' W in a precise sense.

Why do we care about dual spaces?

Dual spaces are essential in: (1) defining coordinates, (2) linear programming and optimization (duality theory), (3) differential geometry (cotangent spaces), (4) quantum mechanics (bra-ket notation), (5) functional analysis.

Is every subspace of V* an annihilator?

Yes! For finite-dimensional V, every subspace U ⊆ V* equals W⁰ for some W ⊆ V. Specifically, U = (U⁰)⁰ where U⁰ is taken in V.