01.01.02 · foundations / linear-algebra

Dual space and double dual

shipped3 tiersLean: none

Anchor (Master): Halmos — Finite-Dimensional Vector Spaces §13–§17; Lang — Algebra Ch. III §6; Bourbaki — Algèbre Ch. II §2

Intuition Beginner

A dual space is the home of all the rulers you can hold up to a vector. A vector is an object you can add and stretch. A ruler is a rule that reads off a single number from any vector you point at, and reads it off in a way that respects adding and stretching: a longer vector gets a bigger reading, twice the vector gets twice the reading, and the sum of two vectors gets the sum of the two readings.

The collection of all such rulers for a given vector space $V$ is itself a vector space, called the dual space $V^{*}$ . You can add two rulers, you can scale a ruler, and the result is still a ruler. So the dual space is built from the same toolkit as the original, but its members are measuring devices rather than the things being measured.

Why does this matter? Every measurement of a physical vector is a ruler in disguise. Reading off one coordinate, integrating a function against a fixed weight, evaluating a polynomial at a fixed point — these are all rulers, and they all live in the dual space of whatever space the vectors come from.

Visual Beginner

A column of numbers on the left stands for a vector in $V$ . A row of numbers in the middle stands for a ruler in $V^{*}$ . Multiplying the row by the column collapses both into a single number — the reading. Two different rows give two different readings of the same column.

The picture also shows the second dual $V^{**}$ . Every vector $v$ in $V$ doubles as a ruler on rulers: hand it a ruler $f$ , and the vector reads back $f (v)$ , the number the ruler reads on $v$ . Roles swap: vectors become the things doing the measuring, and rulers become the things being measured.

Worked example Beginner

Work in the coordinate space of length-three columns. Take the vector

v = (2, 1, 3) .

A ruler on this space is a row of three numbers $(a, b, c)$ that reads off $a v_{1} + b v_{2} + c v_{3}$ from any column $v$ .

Take the ruler $e^{1} = (1, 0, 0)$ . Applying $e^{1}$ to $v$ gives $1 \cdot 2 + 0 \cdot 1 + 0 \cdot 3 = 2$ . So $e^{1}$ reads off the first coordinate of any column.

Take a second ruler $f = (0, 1, 2)$ . Applying $f$ to $v$ gives $0 \cdot 2 + 1 \cdot 1 + 2 \cdot 3 = 1 + 6 = 7$ .

Now add the two rulers: $e^{1} + f = (1, 1, 2)$ . Applying $(1, 1, 2)$ to $v$ gives $1 \cdot 2 + 1 \cdot 1 + 2 \cdot 3 = 2 + 1 + 6 = 9 = 2 + 7$ . Adding the rulers added the readings.

What this tells us: rulers are themselves vectors. The dual space is built from the same combine-and-scale operations as the original space, but its members are the measurement rules.

Check your understanding Beginner

Formal definition Intermediate+

Let $V$ be a vector space over a field $F$ . The dual space is the set $$ V^* = {f : V \to \mathbb{F} \mid f \text{ is } \mathbb{F}\text{-linear}} $$ of $F$ -linear maps from $V$ to $F$ . Pointwise addition and pointwise scalar multiplication make $V^{*}$ into an $F$ -vector space: $(f + g) (v) = f (v) + g (v)$ and $(c f) (v) = c \cdot f (v)$ for $f, g \in V^{*}$ , $c \in F$ , $v \in V$ . Elements of $V^{*}$ are called linear functionals, linear forms, or covectors; the value $f (v)$ is often written $⟨ f, v ⟩$ to emphasise the symmetric pairing $V^{*} \times V \to F$ .

When $V$ is finite-dimensional with ordered basis $e_{1}, \dots, e_{n}$ , the dual basis of $V^{*}$ consists of the functionals $e^{1}, \dots, e^{n}$ defined on the basis by the Kronecker rule $$ e^i(e_j) = \delta^i_j = \begin{cases} 1 & i = j, \ 0 & i \ne j, \end{cases} $$ and extended to all of $V$ by linearity: $e^{i} (\sum_{j} c^{j} e_{j}) = c^{i}$ . The functional $e^{i}$ reads off the $i$ -th coordinate of a vector relative to the chosen basis.

The double dual is $V^{**} = (V^{*})^{*}$ , the space of linear functionals on the dual space. There is a canonical map $$ \mathrm{ev} : V \to V^{**}, \qquad v \mapsto \mathrm{ev}_v, \qquad \mathrm{ev}_v(f) = f(v). $$ The notation does not name a basis: $ev_{v}$ is the rule that hands any functional $f$ back its value at $v$ . This map is $F$ -linear in $v$ (by linearity of $f$ ) and is natural in $V$ with respect to linear maps, in a sense made precise below.

For a subspace $W \subseteq V$ , the annihilator of $W$ is $$ W^\perp = {f \in V^* : f(w) = 0 \text{ for every } w \in W} \subseteq V^. $$ It is a subspace of $V^ $. S y mm e t r i c a l l y, f or a s u b s p a ce$ \Phi \subseteq V^* $t h e * * p r e - annihi l a t or * * i s$ \Phi^\perp = {v \in V : f(v) = 0 \text{ for every } f \in \Phi} \subseteq V $. U n d er t h ec an o ni c a l e v a l u a t i o nma p, t ak in g annihi l a t or tw i ceo na s u b s p a ce$ W \subseteq V $l an d s ba c k in s i d e$ V \cong V^{**} $, an d in f ini t e d im e n s i o n so n er eco v er s$ W$ exactly.

For a linear map $T : V \to W$ , the transpose or dual map is $$ T^* : W^* \to V^, \qquad T^(g) = g \circ T. $$ The arrow direction reverses: $T$ sends $V$ to $W$ , while $T^{*}$ sends $W^{*}$ to $V^{*}$ . Composition reverses, $(T \circ S)^{*} = S^{*} \circ T^{*}$ , and identities are preserved, $id_{V}^{*} = id_{V^{*}}$ . In matrix terms with respect to dual bases, the matrix of $T^{*}$ is the transpose of the matrix of $T$ — the operation that gives the construction its name.

Counterexamples to common slips

The isomorphism $V ≅ V^{*}$ in finite dimensions is real but is not canonical: it requires a choice of basis (or of an inner product). Two different bases of $V$ give two different identifications $V \to V^{*}$ , and the difference is visible in coordinates. The isomorphism $V ≅ V^{**}$ via $v \mapsto ev_{v}$ requires no such choice.
For infinite-dimensional $V$ , the canonical map $V \to V^{**}$ is still injective but is no longer surjective. The algebraic double dual is strictly larger than $V$ already for $V$ a countable-dimensional vector space, and the topological double dual of a Banach space is larger still unless the space is reflexive.
The annihilator is taken inside the dual space, not inside the original. The orthogonal complement $W^{⊥} \subseteq V$ that one meets in inner-product geometry is a different object: it depends on a chosen inner product, while the dual-space annihilator is canonical.

Key theorem with proof Intermediate+

Theorem (dual-basis and dimension; Halmos §14, Shilov §4). Let $V$ be a finite-dimensional vector space over $F$ with ordered basis $e_{1}, \dots, e_{n}$ . The functionals $e^{1}, \dots, e^{n}$ defined by $e^{i} (e_{j}) = δ_{j}^{i}$ form a basis of $V^ $. I n p a r t i c u l a r$ \dim V^* = \dim V = n $, an d t h ee v a l u a t i o nma p$ \mathrm{ev} : V \to V^{**} $,$ v \mapsto \mathrm{ev}_v$, is an isomorphism.*

Proof. Three claims to establish in sequence: the $e^{i}$ are linearly independent in $V^{*}$ , the $e^{i}$ span $V^{*}$ , and the evaluation map $ev$ is an isomorphism.

Step 1: independence of the dual basis. Suppose scalars $c_{1}, \dots, c_{n} \in F$ satisfy $$ c_1 e^1 + c_2 e^2 + \cdots + c_n e^n = 0 $$ in $V^{*}$ . Apply both sides to the basis vector $e_{j}$ for an arbitrary $j$ . The left-hand side gives $$ \sum_{i=1}^n c_i , e^i(e_j) = \sum_{i=1}^n c_i , \delta^i_j = c_j, $$ and the right-hand side gives $0$ . Hence $c_{j} = 0$ for every $j$ , so the family ${e^{i}}$ is linearly independent.

Step 2: spanning. Let $f \in V^{*}$ be arbitrary. Define scalars $c_{i} = f (e_{i})$ for $i = 1, \dots, n$ , and consider the functional $$ g = \sum_{i=1}^n c_i , e^i \in V^. $$ For any basis vector $e_{j}$ , $$ g(e_j) = \sum_{i=1}^n c_i , e^i(e_j) = \sum_{i=1}^n c_i , \delta^i_j = c_j = f(e_j). $$ The two functionals $f$ and $g$ agree on a basis of $V$ , so they agree on all of $V$ by linearity. Therefore $f = g \in span {e^{1}, \dots, e^{n}}$ , establishing spanning. The $e^{i}$ form a basis of $V^ $an d$ \dim V^* = n = \dim V$.

Step 3: evaluation is an isomorphism. The map $ev : V \to V^{**}$ , $ev (v) (f) = f (v)$ , is $F$ -linear in $v$ : for $u, v \in V$ and $a, b \in F$ , $$ \mathrm{ev}(au + bv)(f) = f(au + bv) = a f(u) + b f(v) = a , \mathrm{ev}(u)(f) + b , \mathrm{ev}(v)(f), $$ so $ev (a u + b v) = a ev (u) + b ev (v)$ .

The map is injective: if $ev (v) = 0$ in $V^{**}$ , then $f (v) = 0$ for every $f \in V^{*}$ . Expanding $v = \sum_{j} c^{j} e_{j}$ and applying $e^{i}$ gives $e^{i} (v) = c^{i} = 0$ for every $i$ , so $v = 0$ .

By Step 1 and Step 2 applied to $V^{*}$ , the dual of $V^{*}$ has dimension $dim V^{*} = n$ , so $dim V^{**} = n = dim V$ . An injective linear map between finite-dimensional spaces of the same dimension is an isomorphism (rank-nullity, 01.01.05). The evaluation map is an isomorphism. $□$

Bridge. This builds toward every algebraic and analytic use of duality. The foundational reason the construction works is exactly the dual-basis pairing $e^{i} (e_{j}) = δ_{j}^{i}$ : it identifies $V$ and $V^{*}$ as two sides of one bilinear pairing $V \times V^{*} \to F$ , and that pairing is the structural input behind every subsequent duality theorem. The central insight is that the isomorphism $V ≅ V^{*}$ is dual to a choice of basis — change the basis and the identification changes — while the isomorphism $V ≅ V^{**}$ via $ev$ is canonical and natural. This pattern recurs in 01.01.15 (bilinear forms), where a non-degenerate bilinear form is exactly a basis-free isomorphism $V \to V^{*}$ , and it generalises to functional analysis, where the Riesz representation theorem identifies a Hilbert space with its own continuous dual under the inner product. Putting these together, the dual space identifies $V$ with the space of scalar measurements of $V$ , the double dual identifies $V$ with itself canonically, and the transpose map identifies a linear map with its action on rulers — three facets of one duality framework. The bridge is the recognition that algebraic duality, geometric pairing, and the index calculus of upper and lower indices in tensor algebra are different presentations of one biorthogonal pairing $V \times V^{*} \to F$ . This appears again in 03.01.01 (tensor product) and downstream in 01.01.15 (bilinear and quadratic forms), where the same pairing is the universal target of bilinear maps.

Exercises Intermediate+

Exercise 3 (medium, symbolic).

In $V = R^{3}$ with standard basis $e_{1}, e_{2}, e_{3}$ , take the new basis $v_{1} = e_{1} + e_{2}$ , $v_{2} = e_{2} + e_{3}$ , $v_{3} = e_{1} + e_{3}$ . Compute the dual basis $v^{1}, v^{2}, v^{3}$ of $V^{*}$ , expressed as row vectors in standard coordinates.

Hint

Write the matrix $M$ whose columns are the $v_{i}$ in standard coordinates, then invert. The rows of $M^{- 1}$ are the dual basis vectors.

Answer

The matrix of $v_{1}, v_{2}, v_{3}$ in standard coordinates is $$ M = \begin{pmatrix} 1 & 0 & 1 \ 1 & 1 & 0 \ 0 & 1 & 1 \end{pmatrix}, \qquad \det M = 2. $$ The inverse is $$ M^{-1} = \tfrac{1}{2} \begin{pmatrix} 1 & 1 & -1 \ -1 & 1 & 1 \ 1 & -1 & 1 \end{pmatrix}. $$ The dual basis functionals are the rows of $M^{- 1}$ : $$ v^1 = \tfrac{1}{2}(1, 1, -1), \quad v^2 = \tfrac{1}{2}(-1, 1, 1), \quad v^3 = \tfrac{1}{2}(1, -1, 1). $$ Verify: $v^{1} (v_{1}) = \frac{1}{2} (1 + 1 + 0) = 1$ , $v^{1} (v_{2}) = \frac{1}{2} (0 + 1 - 1) = 0$ , $v^{1} (v_{3}) = \frac{1}{2} (1 + 0 - 1) = 0$ . Rubric: full credit for the three rows with the verification.

Exercise 4 (medium, short-answer).

Let $V = R_{2} [x]$ , the space of real polynomials of degree at most $2$ , and consider the three evaluation functionals $f_{0} (p) = p (- 1)$ , $f_{1} (p) = p (0)$ , $f_{2} (p) = p (1)$ . Show that $f_{0}, f_{1}, f_{2}$ form a basis of $V^{*}$ and compute the dual basis $ℓ_{0}, ℓ_{1}, ℓ_{2} \in V$ .

Hint

The dual basis vectors $ℓ_{k} \in V$ are the Lagrange interpolation polynomials at the nodes $- 1, 0, 1$ , characterised by $ℓ_{k} (a_{j}) = δ_{j}^{k}$ .

Answer

$V$ has dimension $3$ (basis $1, x, x^{2}$ ), so $dim V^{*} = 3$ . To show $f_{0}, f_{1}, f_{2}$ is a basis it suffices to exhibit a dual basis in $V$ . The Lagrange polynomial $ℓ_{k}$ at nodes $a_{0} = - 1, a_{1} = 0, a_{2} = 1$ is $$ \ell_k(x) = \prod_{j \ne k} \frac{x - a_j}{a_k - a_j}. $$ Explicitly, $$ \ell_0(x) = \frac{(x - 0)(x - 1)}{(-1 - 0)(-1 - 1)} = \tfrac{1}{2} x(x - 1) = \tfrac{1}{2}(x^2 - x), $$ $$ \ell_1(x) = \frac{(x + 1)(x - 1)}{(0 + 1)(0 - 1)} = -(x^2 - 1) = 1 - x^2, $$ $$ \ell_2(x) = \frac{(x + 1)(x - 0)}{(1 + 1)(1 - 0)} = \tfrac{1}{2} x(x + 1) = \tfrac{1}{2}(x^2 + x). $$ By construction $f_{j} (ℓ_{k}) = ℓ_{k} (a_{j}) = δ_{j}^{k}$ , so ${f_{0}, f_{1}, f_{2}}$ and ${ℓ_{0}, ℓ_{1}, ℓ_{2}}$ are a dual pair. Since ${ℓ_{k}}$ is a basis of $V$ (any quadratic equals $p (- 1) ℓ_{0} + p (0) ℓ_{1} + p (1) ℓ_{2}$ by checking at three points), ${f_{k}}$ is a basis of $V^{*}$ . Rubric: full credit for the three Lagrange polynomials plus the biorthogonality verification.

Exercise 5 (medium, short-answer).

Let $T : R^{3} \to R^{2}$ be the linear map with matrix $A = (1021 0 - 1)$ in the standard bases. Compute the matrix of the transpose $T^{*} : (R^{2})^{*} \to (R^{3})^{*}$ in the dual bases, and verify the matrix-transpose relation.

Hint

Apply $T^{*}$ to each dual basis vector of the target and read off the coordinates in the dual basis of the source.

Answer

Let $e^{1}, e^{2}$ be the dual basis of $(R^{2})^{*}$ . Then $T^{*} (e^{1}) (v) = e^{1} (T v)$ is the first row of $A$ applied to $v$ , namely $(1, 2, 0) \cdot v$ , so in the dual basis of $(R^{3})^{*}$ its coordinates are $(1, 2, 0)$ . Similarly $T^{*} (e^{2})$ has coordinates $(0, 1, - 1)$ . The matrix of $T^{*}$ with columns the coordinate vectors of $T^{*} (e^{1}), T^{*} (e^{2})$ is $$ B = \begin{pmatrix} 1 & 0 \ 2 & 1 \ 0 & -1 \end{pmatrix} = A^\top. $$ The matrix of $T^{*}$ in dual bases is the transpose of the matrix of $T$ . Rubric: full credit for the matrix plus the identification with $A^{⊤}$ .

Exercise 6 (medium, short-answer).

Show that for any subspace $W$ of a finite-dimensional vector space $V$ , the double annihilator $W^{⊥⊥} = {v \in V : f (v) = 0 for every f \in W^{⊥}}$ equals $W$ .

Hint

Show $W \subseteq W^{⊥⊥}$ directly, then count dimensions using $dim W^{⊥} = dim V - dim W$ applied twice.

Answer

Inclusion $W \subseteq W^{⊥⊥}$ is immediate: for $w \in W$ and $f \in W^{⊥}$ , $f (w) = 0$ by definition of $W^{⊥}$ , so $w$ annihilates every $f \in W^{⊥}$ , meaning $w \in W^{⊥⊥}$ .

Dimension count: $dim W^{⊥} = dim V - dim W$ from the dual-pairing rank theorem (proved by extending a basis of $W$ to a basis of $V$ and taking the corresponding dual basis: exactly the $dim V - dim W$ dual basis vectors not paired with $W$ form a basis of $W^{⊥}$ ). Applied to $W^{⊥} \subseteq V^{*}$ , the annihilator $W^{⊥⊥} \subseteq V^{**} = V$ has dimension $dim V^{*} - dim W^{⊥} = n - (n - dim W) = dim W$ . A subspace of $V$ contained in $W^{⊥⊥}$ with the same dimension as $W$ equals $W$ . Rubric: full credit for the inclusion plus the dimension argument.

Exercise 7 (hard, short-answer).

Let $V$ be a finite-dimensional vector space and let $T : V \to V$ be a linear operator. Prove that $ker T^{*} = (im T)^{⊥}$ and $im T^{*} = (ker T)^{⊥}$ , and deduce $rank T = rank T^{*}$ .

Hint

For the first identity, unwind the definition $T^{*} (g) = g \circ T$ . For the rank identity, apply the dimension formula $dim W + dim W^{⊥} = dim V$ to the kernel and image identities.

Answer

A functional $g \in V^{*}$ lies in $ker T^{*}$ iff $T^{*} (g) = g \circ T = 0$ iff $g (T v) = 0$ for every $v \in V$ iff $g$ vanishes on $im T$ iff $g \in (im T)^{⊥}$ . So $ker T^{*} = (im T)^{⊥}$ .

Taking dimensions, $dim ker T^{*} = dim V - dim im T = dim V - rank T$ . Rank-nullity for $T^{*} : V^{*} \to V^{*}$ gives $dim V^{*} = dim ker T^{*} + dim im T^{*}$ , hence $dim im T^{*} = dim V - (dim V - rank T) = rank T$ . So $rank T^{*} = rank T$ .

For the second identity: $im T^{*} \subseteq (ker T)^{⊥}$ because any $g \circ T$ vanishes on $ker T$ ; the two subspaces of $V^{*}$ both have dimension $rank T = dim V - dim ker T = dim (ker T)^{⊥}$ , so they coincide. Rubric: full credit for the two identities plus the rank identification.

Exercise 8 (hard, short-answer).

Let $V$ be an infinite-dimensional vector space over $F$ with a Hamel basis ${e_{α}}_{α \in I}$ . Show that the canonical evaluation map $ev : V \to V^{**}$ is injective but not surjective.

Hint

For injectivity, repeat the finite-dimensional argument coordinate by coordinate. For non-surjectivity, exhibit an element of $V^{**}$ not represented by any vector in $V$ by using the dual basis to produce a functional supported on infinitely many basis elements.

Answer

Injectivity. If $ev (v) = 0$ , expand $v = \sum_{α \in F} c^{α} e_{α}$ as a finite sum. Apply the dual functional $e^{β}$ defined by $e^{β} (e_{α}) = δ_{α}^{β}$ for each $β \in F$ : $e^{β} (v) = c^{β} = 0$ . So all coordinates vanish and $v = 0$ .

Non-surjectivity. The dual space $V^{*}$ has dimension at least $∣ I ∣$ — exactly $∣ F ∣^{∣ I ∣}$ in fact, since a functional is determined by its values on the basis, which can be arbitrary. Define $Φ \in V^{**}$ by $Φ (f) = \sum_{α} f (e_{α})$ for $f$ supported on a fixed finite subset, and zero otherwise; more concretely, fix a countable subset ${e_{α_{n}}}_{n \in N}$ of the basis, let $e^{α_{n}}$ be the dual functionals, and define $Φ$ on $V^{*}$ by $Φ (f) = \sum_{n} f (e_{α_{n}})$ whenever the sum is finite, extended via Hamel-basis arguments to all of $V^{*}$ . Such a $Φ$ exists in $V^{**}$ but is not of the form $ev_{v}$ for any $v \in V$ : every $ev_{v}$ vanishes on dual functionals not in the finite support of $v$ in the dual basis, while $Φ$ has infinite support. Rubric: full credit for the injectivity argument plus a construction of a non-evaluation element of $V^{**}$ .

Lean formalization Intermediate+

Mathlib provides the dual-space infrastructure as Module.Dual K V := V →ₗ[K] K and the canonical double-dual map via Module.Dual.eval and Module.evalEquiv. The intended formalisation reads:

import Mathlib.LinearAlgebra.Dual
import Mathlib.LinearAlgebra.FiniteDimensional

variable {K V : Type*} [Field K] [AddCommGroup V] [Module K V]

/-- Dual basis of a finite basis. -/
example (b : Basis (Fin n) K V) : Basis (Fin n) K (Module.Dual K V) :=
  b.dualBasis

/-- Dimension of dual equals dimension of original for finite-dim V. -/
example [FiniteDimensional K V] :
    FiniteDimensional.finrank K (Module.Dual K V) =
      FiniteDimensional.finrank K V :=
  Module.finrank_dual_eq_finrank K V

/-- Canonical map V → V** is an isomorphism in finite dimensions. -/
noncomputable example [FiniteDimensional K V] :
    V ≃ₗ[K] Module.Dual K (Module.Dual K V) :=
  Module.evalEquiv K V

/-- Transpose / dual map; functoriality. -/
example {W : Type*} [AddCommGroup W] [Module K W] (T : V →ₗ[K] W) :
    Module.Dual K W →ₗ[K] Module.Dual K V :=
  T.dualMap

The proof gap declared in Mathlib gap analysis lies one level downstream. Mathlib has the abstract dual-basis API but does not ship the explicit Lagrange-interpolation duality witness for $V = K_{n} [x]$ paired with point-evaluation functionals at $n + 1$ distinct nodes. The double-annihilator identity $W^{\circ\circ} = W$ for a submodule of a finite-dimensional space is recoverable from Submodule.dualAnnihilator plus Module.evalEquiv plus rank-counting, but is not packaged as a single simp lemma. Both are short upstream-contribution targets.

Advanced results Master

Proposition (functoriality of $(-)^{*}$ ). The assignment $V \mapsto V^ $,$ T \mapsto T^ $i s a co n t r a v a r ian t a dd i t i v e f u n c t or f r o m t h ec a t e g or y o f$ \mathbb{F} $- v ec t or s p a ces t o i t se l f . C o m p os i t i o ni sr e v er se d,$ (T \circ S)^ = S^* \circ T^ $, i d e n t i t i es a r e p r eser v e d,$ \mathrm{id}_V^ = \mathrm{id}_{V^} $, an d t h e f u n c t or se n d s d i r ec t s u m s t o d i r ec t s u m s,$ (V \oplus W)^ \cong V^* \oplus W^$, with the obvious projections / injections swapped.

The composite $(-)^{**}$ is therefore a covariant functor. The natural transformation $ev : id \Rightarrow (-)^{**}$ has components $ev_{V} : V \to V^{**}$ , $ev_{V} (v) (f) = f (v)$ , and is a natural isomorphism on the full subcategory of finite-dimensional vector spaces. Naturality means: for every linear map $T : V \to W$ , the square $$ \begin{array}{ccc} V & \xrightarrow{\mathrm{ev}_V} & V^{} \ T \downarrow & & \downarrow T^{} \ W & \xrightarrow{\mathrm{ev}_W} & W^{} \end{array} $$ commutes. The verification is one line: $T^{}(\mathrm{ev}_V(v))(g) = \mathrm{ev}_V(v)(T^(g)) = (T^(g))(v) = g(T(v)) = \mathrm{ev}_W(T(v))(g) $. N a t u r a l i t y i s t h e p r ec i ser e a so n t h e i so m or p hi s m$ V \cong V^{**} $i s d escr ib e d a sc an o ni c a l : i t co mm u t es w i t h e v er y l in e a r ma p in s i g h t, w hi l e t h e ba s i s - d e p e n d e n t i so m or p hi s m$ V \cong V^*$ does not.

Proposition (rank of the transpose). For a linear map $T : V \to W$ between finite-dimensional vector spaces, $\mathrm{rank}, T^ = \mathrm{rank}, T $,$ \ker T^* = (\mathrm{im}, T)^\perp $, an d$ \mathrm{im}, T^* = (\ker T)^\perp$.*

The matrix-theoretic content of this proposition is the row-rank equals column-rank theorem: for any matrix $A$ , $rank (A) = rank (A^{⊤})$ . The duality proof exhibits the matrix theorem as a corollary of a categorical fact about the dual functor.

Proposition (annihilator duality). For finite-dimensional $V$ and a subspace $W \subseteq V$ , the dual map of the inclusion $ι : W ↪ V$ has kernel $W^{⊥}$ , and the induced map $V^/W^\perp \to W^ $i s ani so m or p hi s m . S y mm e t r i c a l l y, t h e q u o t i e n t$ V \to V/W $ha s d u a l ma p$ (V/W)^ \to V^ $w h ose ima g e i se x a c tl y$ W^\perp $. S o b o t h$ W^ $an d$ (V/W)^ $a r eco m p u t ab l e f r o m$ V^ $an d$ W^\perp$ via the short exact sequence* $$ 0 \to W^\perp \to V^* \to W^* \to 0. $$

Theorem (algebraic-versus-topological dual; Banach 1932). For an infinite-dimensional Banach space $X$ with continuous dual $X^{'} = {f : X \to F ∣ f continuous linear}$ , the canonical evaluation map $ev : X \to X^{''}$ is an isometric embedding. The space $X$ is called reflexive when $ev$ is surjective. Reflexive examples: every Hilbert space (by Riesz representation), every $L^{p} (μ)$ for $1 < p < \infty$ , every finite-dimensional Banach space. Non-reflexive examples: $L^{1} (μ)$ for infinite $μ$ has continuous dual $L^{\infty} (μ)$ whose continuous dual is strictly larger than $L^{1}$ ; $C ([0, 1])$ and $ℓ^{1}$ are also non-reflexive.

The algebraic double dual of an infinite-dimensional vector space is always strictly larger than the original, by a cardinality argument on Hamel bases. The topological double dual of a Banach space can be either larger than or canonically isomorphic to the original, with reflexivity the dividing line.

Theorem (Riesz representation; Hilbert-space case). Let $H$ be a Hilbert space over $F = R$ or $C$ with inner product $⟨ -, - ⟩$ . For every continuous linear functional $f \in H^{'}$ there is a unique $u \in H$ with $f (v) = ⟨ v, u ⟩$ for every $v \in H$ , and $∥ f ∥_{H^{'}} = ∥ u ∥_{H}$ . The map $H \to H^{'}$ , $u \mapsto ⟨ -, u ⟩$ , is an isometric anti-linear isomorphism (linear in the real case).

Riesz representation realises the inner product as a canonical identification $H ≅ H^{'}$ in the topological category — the analogue, for Hilbert spaces, of the basis-free identification $V ≅ V^{**}$ in the algebraic category. The inner-product structure plays the role that the choice of basis plays in the algebraic finite-dimensional case, but here the identification with $H^{'}$ is canonical, not basis-dependent, because the inner product itself is the canonical extra datum that distinguishes a Hilbert space from a general Banach space.

Theorem (Bourbaki dual pairing). Let $V, W$ be $F$ -vector spaces and let $⟨ -, - ⟩ : V \times W \to F$ be a bilinear pairing. The pairing is non-degenerate if the induced maps $V \to W^ $,$ v \mapsto \langle v, - \rangle $, an d$ W \to V^ $,$ w \mapsto \langle -, w \rangle $, a r e b o t hinj ec t i v e . I n t h e f ini t e - d im e n s i o na l c a se, n o n - d e g e n er a cy i se q u i v a l e n tt o b o t hin d u ce d ma p s b e in g i so m or p hi s m s, an d t h e n$ \dim V = \dim W$.

The non-degenerate bilinear pairing is the canonical input to all of bilinear-form theory: symmetric non-degenerate pairings give quadratic forms (01.01.15), antisymmetric non-degenerate pairings give symplectic forms, and the Bourbaki framing makes the universal-property structure transparent.

Theorem (tensor algebra incarnation). Let $V$ be a finite-dimensional vector space over $F$ . The space of bilinear forms $V \times V \to F$ is canonically identified with $V^ \otimes V^ $, t h e t e n sor s q u a r eo f t h e d u a l . M or e g e n er a l l y, t h es p a ceo f$ k $- m u l t i l in e a r f or m s$ V^k \to \mathbb{F} $i s i d e n t i f i e d w i t h t h e$ k $- f o l d t e n sor p o w er$ (V^)^{\otimes k} $, an d t h es p a ceo f mi x e d t e n sor so f t y p e$ (p, q) $—$ p $- m u l t i l in e a r in co v ec t or s an d$ q $- m u l t i l in e a r in v ec t or s — i s$ V^{\otimes p} \otimes (V^)^{\otimes q} $. T h e d u a l i t y$ V \cong V^{**} $a l l o w s f r ee p a ss a g e b e tw ee n u pp er - in d e x (co n t r a v a r ian t, in$ V $) an d l o w er - in d e x (co v a r ian t, in$ V^ $) t e n sor co m p o n e n t s v iain d e x l o w er in g an d r ai s in g w i t ha f i x e d bi l in e a r f or m o n$ V$.*

This proposition is the algebraic substrate of the abstract index calculus in differential geometry and general relativity: upper indices live in $V$ , lower indices live in $V^{*}$ , contraction is the pairing $V \otimes V^{*} \to F$ , and a metric tensor on $V$ provides the basis-free isomorphism $V \to V^{*}$ that lowers and raises indices in a coordinate-free way.

Synthesis. Duality is the foundational reason linear algebra has the symmetric structure it does. The central insight is that every vector space $V$ carries a canonical companion $V^{*}$ of linear functionals, that the companion has the same dimension in the finite-dimensional case, and that the second iteration $V^{**}$ returns to $V$ via a canonical map — basis-free and natural in the functorial sense. This builds toward the entire theory of bilinear forms (01.01.15), tensor products (03.01.01), differential forms on manifolds, and the topological duality theory of Banach and locally convex spaces. The bridge is that algebraic duality and analytic duality are the same construction at different levels of structural enrichment: the algebraic dual is the universal vector space of linear functionals, while the topological dual restricts to continuous functionals, and the Riesz representation theorem identifies the topological dual of a Hilbert space with itself canonically, generalising the basis-free finite-dimensional identification $V ≅ V^{**}$ . Putting these together, every classical pairing in linear algebra — the matrix transpose, the annihilator, the bilinear form, the inner product, the integration pairing on function spaces — is dual to itself in a precise sense: applying the dual construction twice returns the original, and the canonical map back is the universal datum that organises all of duality.

The duality also identifies several constructions that look distinct at first inspection. The transpose of a linear map and the matrix transpose of its matrix are not two different operations but one: the latter is the matrix presentation of the former in dual bases. The annihilator of a subspace and the orthogonal complement under an inner product are not two different operations but one: the latter is the image of the former under the Riesz isomorphism $V ≅ V^{*}$ . The cotangent space at a point of a smooth manifold and the dual of the tangent space are not two different objects but one: differential forms are sections of the cotangent bundle, and the entire de Rham complex is a tensor-power construction over the dual functor. This is exactly the unifying observation that appears again in 01.01.15 (bilinear forms), where a non-degenerate bilinear form is by construction an isomorphism $V \to V^{*}$ , and in 03.01.01 (tensor product), where the tensor algebra is built from $V$ and $V^{*}$ together as the natural ambient for all multilinear constructions on the underlying vector space.

Full proof set Master

Proposition (functoriality of $(-)^{*}$ ), proof. For linear maps $T : V \to W$ and $S : U \to V$ , the composite $T \circ S : U \to W$ satisfies $$ (T \circ S)^(g) = g \circ (T \circ S) = (g \circ T) \circ S = T^(g) \circ S = S^(T^(g)) $$ for any $g \in W^{*}$ . So $(T \circ S)^{*} = S^{*} \circ T^{*}$ . For $id_{V} : V \to V$ , $id_{V}^{*} (f) = f \circ id_{V} = f$ , so $id_{V}^{*} = id_{V^{*}}$ . Additivity: $(T_{1} + T_{2})^{*} (g) = g \circ (T_{1} + T_{2}) = g \circ T_{1} + g \circ T_{2} = T_{1}^{*} (g) + T_{2}^{*} (g)$ . Direct sum: the pair of projections $π_{V} : V \oplus W \to V$ and $π_{W} : V \oplus W \to W$ has duals $π_{V}^{*} : V^{*} \to (V \oplus W)^{*}$ and $π_{W}^{*} : W^{*} \to (V \oplus W)^{*}$ whose images are complementary subspaces of $(V \oplus W)^{*}$ with intersection $0$ , exhibiting the splitting $(V \oplus W)^{*} = V^{*} \oplus W^{*}$ . The component identifications are: $V^{*} \subseteq (V \oplus W)^{*}$ corresponds to functionals supported on the $V$ summand. $□$

Proposition (rank of the transpose), proof. A functional $g \in W^{*}$ lies in $ker T^{*}$ iff $T^{*} (g) = g \circ T = 0$ iff $g$ vanishes on $im T$ iff $g \in (im T)^{⊥}$ . The dimension identity $dim (im T)^{⊥} = dim W - dim im T = dim W - rank T$ combined with rank-nullity for $T^{*} : W^{*} \to V^{*}$ — $dim W^{*} = dim ker T^{*} + dim im T^{*}$ , i.e. $dim W = (dim W - rank T) + dim im T^{*}$ — gives $dim im T^{*} = rank T$ , so $rank T^{*} = rank T$ . For the image identity, $im T^{*} \subseteq (ker T)^{⊥}$ is immediate from $T^{*} (g) (v) = g (T v)$ , which vanishes whenever $v \in ker T$ ; equality follows by dimension count, $dim im T^{*} = rank T = dim V - dim ker T = dim (ker T)^{⊥}$ . $□$

Proposition (annihilator duality), proof. Let $ι : W ↪ V$ be the inclusion. The dual map $ι^{*} : V^{*} \to W^{*}$ restricts a functional on $V$ to a functional on $W$ . A functional $f \in V^{*}$ lies in $ker ι^{*}$ iff its restriction to $W$ is zero iff $f \in W^{⊥}$ . So $ker ι^{*} = W^{⊥}$ . The first isomorphism theorem applied to $ι^{*} : V^{*} \to W^{*}$ gives an injection $V^{*} / W^{⊥} ↪ W^{*}$ . The image is all of $W^{*}$ : any functional $g \in W^{*}$ extends to a functional on $V$ by extending a basis of $W$ to a basis of $V$ and assigning arbitrary values on the new basis elements; the extension restricts back to $g$ . So $V^{*} / W^{⊥} ≅ W^{*}$ .

For the quotient case, the projection $π : V \to V / W$ has dual $π^{*} : (V / W)^{*} \to V^{*}$ taking a functional $\overset{ˉ}{f}$ on $V / W$ to $\overset{ˉ}{f} \circ π$ , a functional on $V$ that vanishes on $W$ (since $π$ kills $W$ ). So $im π^{*} \subseteq W^{⊥}$ . Conversely, any $h \in W^{⊥}$ factors through $V / W$ as $h = \overset{ˉ}{h} \circ π$ with $\overset{ˉ}{h}$ well-defined on cosets by the vanishing on $W$ . So $im π^{*} = W^{⊥}$ , and $π^{*}$ gives an isomorphism $(V / W)^{*} ≅ W^{⊥}$ . Combining the two identifications, the short exact sequence $0 \to W^{⊥} \to V^{*} \to W^{*} \to 0$ is the dual of the short exact sequence $0 \to W \to V \to V / W \to 0$ . $□$

Theorem (algebraic-versus-topological dual; Banach 1932), proof sketch. The Hahn-Banach extension theorem ^{[Hahn 1927]} says: every bounded linear functional on a subspace of a normed space $X$ extends to a bounded linear functional on $X$ with the same norm. Applied to the one-dimensional subspace spanned by a fixed nonzero $x \in X$ , with the functional $t x \mapsto t ∥ x ∥$ of norm $1$ , Hahn-Banach produces a functional $f \in X^{'}$ with $∥ f ∥ = 1$ and $f (x) = ∥ x ∥$ . The evaluation map $ev : X \to X^{''}$ , $ev_{x} (g) = g (x)$ , then has $∣ ev_{x} (g) ∣ = ∣ g (x) ∣ \leq ∥ g ∥∥ x ∥$ , so $∥ ev_{x} ∥ \leq ∥ x ∥$ , while the Hahn-Banach functional witnesses $∣ ev_{x} (f) ∣ = ∥ x ∥$ , giving $∥ ev_{x} ∥ = ∥ x ∥$ . The map is an isometric embedding. Surjectivity is the definition of reflexivity and is not automatic: $L^{1}$ has continuous dual $L^{\infty}$ whose continuous dual is strictly larger than $L^{1}$ (Banach 1932 Ch. IV ^{[Banach 1932]}). $□$

Theorem (Riesz representation), proof sketch. Let $f \in H^{'}$ be a continuous linear functional on the Hilbert space $H$ . If $f = 0$ , take $u = 0$ . Otherwise $ker f$ is a closed proper subspace of $H$ ; by Hilbert-space orthogonality, $H = ker f \oplus (ker f)^{⊥}$ and $(ker f)^{⊥}$ is one-dimensional (since $f$ restricted to it is a nonzero linear functional from a one-dimensional space to $F$ ). Pick a unit vector $w \in (ker f)^{⊥}$ and let $α = f (w)$ . Set $u = \overset{α}{ˉ} w$ . For any $v \in H$ decompose $v = v_{0} + β w$ with $v_{0} \in ker f$ , $β \in F$ . Then $f (v) = β α$ and $⟨ v, u ⟩ = ⟨ v_{0} + β w, \overset{α}{ˉ} w ⟩ = β α$ (using $⟨ v_{0}, w ⟩ = 0$ and $∥ w ∥ = 1$ ). The two agree, and $u$ is the desired representative. Uniqueness: if $⟨ v, u_{1} ⟩ = ⟨ v, u_{2} ⟩$ for every $v$ , then $⟨ v, u_{1} - u_{2} ⟩ = 0$ for every $v$ , in particular for $v = u_{1} - u_{2}$ , giving $∥ u_{1} - u_{2} ∥^{2} = 0$ and $u_{1} = u_{2}$ . The norm equality $∥ f ∥_{H^{'}} = ∥ u ∥_{H}$ comes from Cauchy-Schwarz $∣ f (v) ∣ = ∣ ⟨ v, u ⟩ ∣ \leq ∥ v ∥∥ u ∥$ saturated at $v = u /∥ u ∥$ . $□$

Proposition (tensor algebra incarnation), proof sketch. A bilinear form $B : V \times V \to F$ is by the universal property of $V \otimes V$ (proved in 03.01.01) the same as a linear map $\tilde{B} : V \otimes V \to F$ , equivalently an element of $(V \otimes V)^{*} ≅ V^{*} \otimes V^{*}$ in finite dimensions. The natural isomorphism $(V \otimes W)^{*} ≅ V^{*} \otimes W^{*}$ in finite dimensions is verified by exhibiting bases: a basis of $(V \otimes W)^{*}$ is dual to the basis ${e_{i} \otimes f_{j}}$ of $V \otimes W$ , and the dual functional $(e_{i} \otimes f_{j})^{*}$ pairs to $δ_{j}^{i}$ exactly as $e^{i} \otimes f^{j}$ would. The dimensions agree, $dim V \cdot dim W$ , so the bilinear-functor argument gives an isomorphism. Iterating produces the identification of $k$ -multilinear forms with $(V^{*})^{\otimes k}$ and of mixed tensors with $V^{\otimes p} \otimes (V^{*})^{\otimes q}$ . $□$

Connections Master

Vector space 01.01.03. The dual space is built from the same field $F$ and the same combine-and-scale operations as $V$ , but its elements are scalar-valued linear maps rather than vectors. Without the vector-space axioms there is no notion of linear functional, and without the field $F$ as codomain there is no target for the functionals to land in. The dual construction is the simplest nontrivial functor out of the category of $F$ -vector spaces.
Subspace, basis, dimension 01.01.04. The dual-basis construction $e^{i} (e_{j}) = δ_{j}^{i}$ depends on having a basis to start with, and the dimension identity $dim V^{*} = dim V$ rests on the equivalence of finite-dimensional bases. The annihilator dimension formula $dim W^{⊥} = dim V - dim W$ is a direct consequence of extending a basis of $W$ to a basis of $V$ and reading off the dual-basis support — making the dimension theory of subspaces the foundation for the dimension theory of annihilators.
Linear transformation, rank-nullity 01.01.05. The transpose / dual map $T^{*}$ is a linear map between dual spaces, and rank-nullity for $T^{*}$ gives the rank equality $rank T^{*} = rank T$ that is the duality content of the row-rank equals column-rank theorem for matrices. The kernel-image identities $ker T^{*} = (im T)^{⊥}$ and $im T^{*} = (ker T)^{⊥}$ are the dual-space packaging of the four fundamental subspaces of a matrix.
Bilinear and quadratic forms 01.01.15. A bilinear form $B : V \times V \to F$ is the same datum as a linear map $V \to V^{*}$ , and $B$ is non-degenerate iff this map is an isomorphism. So bilinear-form theory is built directly on top of the dual space: symmetric forms give quadratic forms, antisymmetric forms give symplectic forms, and Hermitian forms give inner products. The classification of bilinear forms is a classification of basis-free isomorphisms $V \to V^{*}$ .
Tensor product 03.01.01. The tensor square $V^{*} \otimes V^{*}$ is canonically identified with the space of bilinear forms on $V$ in finite dimensions, and more generally $(V^{*})^{\otimes k}$ classifies $k$ -multilinear forms. The mixed tensor space $V^{\otimes p} \otimes (V^{*})^{\otimes q}$ is the abstract home of tensors of type $(p, q)$ in differential geometry, with index raising and lowering performed by a fixed bilinear form on $V$ . The dual space is half the data of every tensor construction.

Historical & philosophical context Master

The notion of a linear functional emerged in two strands. The algebraic strand began with the recognition that the row vectors and column vectors of a matrix were two presentations of related-but-distinct objects: rows acted on columns to produce scalars, and column-on-row was a different operation. Hermann Grassmann's Die lineale Ausdehnungslehre (1844) ^[Bourbaki] introduced the algebra of $n$ -dimensional extension that implicitly treated linear forms as objects distinct from vectors, though the formal vector-space concept stabilised only with Giuseppe Peano's Calcolo geometrico secondo l'Ausdehnungslehre di H. Grassmann (1888). Élie Cartan's tensor calculus, developed across his 1894–1922 papers on Lie groups and differential geometry, used the index notation with upper and lower indices to encode the distinction between vectors and covectors at the level of computation.

The functional-analytic strand grew out of integral equation theory. Salvatore Pincherle's Mémoire sur le calcul fonctionnel distributif (Math. Annalen 49, 1897, 325–382) ^{[Pincherle 1897]} developed an early calculus of linear operators on function spaces, treating linear functionals as fundamental objects. Maurice Fréchet's 1906 thesis introduced function spaces with norm-style structure, and David Hilbert's 1904–1910 lectures on integral equations developed the spectral theory that demanded a coherent duality framework. Hans Hahn's Über lineare Gleichungssysteme in linearen Räumen (J. reine angew. Math. 157, 1927, 214–229) ^{[Hahn 1927]} proved what is now the Hahn-Banach extension theorem, the foundational result that makes the topological dual of a Banach space a substantive object. Stefan Banach's Théorie des opérations linéaires (Monografie Matematyczne 1, 1932) ^{[Banach 1932]} gave the systematic treatment of normed-space duality, including reflexivity, the weak and weak-star topologies, and the topological double dual.

The modern algebraic framing was assembled by the Bourbaki collective. Algèbre Ch. II (1947) presented the dual of a vector space as the universal example of a contravariant additive functor, and Algèbre Multilinéaire (Ch. III) absorbed the tensor algebra into a single functor-of-many-arguments framework. Frigyes Riesz's 1907 paper on $L^{2}$ functionals and his 1909 paper on $C ([a, b])$ functionals (Bull. Soc. Math. France) gave the representation theorems that identify these spaces with their own continuous duals; Riesz-Fischer theory and the abstract Hilbert-space Riesz representation followed. Saunders Mac Lane's categorical framing in Categories for the Working Mathematician (1971) made naturality of the double-dual embedding $V \to V^{**}$ a textbook example of a natural transformation. Halmos's Finite-Dimensional Vector Spaces (Princeton 1942, 2nd ed. 1958) became the standard pedagogical source for the algebraic finite-dimensional theory; Shilov's Linear Algebra §4–§5 (English ed. Dover 1977) is the immediate source for the present unit's framing of linear forms, dual bases, and biorthogonal systems.

Bibliography Master

@book{Shilov1977LinearAlgebra,
  author    = {Shilov, Georgii E.},
  title     = {Linear Algebra},
  publisher = {Dover Publications},
  year      = {1977},
  note      = {Translated from the 1971 Russian edition by Richard A. Silverman}
}

@book{Halmos1958FDVS,
  author    = {Halmos, Paul R.},
  title     = {Finite-Dimensional Vector Spaces},
  publisher = {Van Nostrand},
  edition   = {2},
  year      = {1958}
}

@book{LangAlgebra,
  author    = {Lang, Serge},
  title     = {Algebra},
  publisher = {Springer-Verlag},
  series    = {Graduate Texts in Mathematics},
  volume    = {211},
  edition   = {Revised 3rd},
  year      = {2002}
}

@book{BourbakiAlgebreII,
  author    = {Bourbaki, Nicolas},
  title     = {{\'E}l{\'e}ments de math{\'e}matique: Alg{\`e}bre, Chapitres 1 {\`a} 3},
  publisher = {Hermann},
  year      = {1970}
}

@article{Pincherle1897,
  author  = {Pincherle, Salvatore},
  title   = {M{\'e}moire sur le calcul fonctionnel distributif},
  journal = {Math. Annalen},
  volume  = {49},
  year    = {1897},
  pages   = {325--382}
}

@article{Hahn1927,
  author  = {Hahn, Hans},
  title   = {{\"U}ber lineare Gleichungssysteme in linearen R{\"a}umen},
  journal = {J. reine angew. Math.},
  volume  = {157},
  year    = {1927},
  pages   = {214--229}
}

@book{Banach1932,
  author    = {Banach, Stefan},
  title     = {Th{\'e}orie des op{\'e}rations lin{\'e}aires},
  publisher = {Monografie Matematyczne},
  volume    = {1},
  year      = {1932}
}

@article{Riesz1907,
  author  = {Riesz, Frigyes},
  title   = {Sur une esp{\`e}ce de g{\'e}om{\'e}trie analytique des syst{\`e}mes de fonctions sommables},
  journal = {C. R. Acad. Sci. Paris},
  volume  = {144},
  year    = {1907},
  pages   = {1409--1411}
}

@book{Grassmann1844,
  author    = {Grassmann, Hermann},
  title     = {Die lineale Ausdehnungslehre, ein neuer Zweig der Mathematik},
  publisher = {Otto Wigand, Leipzig},
  year      = {1844}
}

@book{Axler2015LADR,
  author    = {Axler, Sheldon},
  title     = {Linear Algebra Done Right},
  publisher = {Springer},
  edition   = {3},
  year      = {2015}
}

@book{HoffmanKunze1971,
  author    = {Hoffman, Kenneth and Kunze, Ray},
  title     = {Linear Algebra},
  publisher = {Prentice-Hall},
  edition   = {2},
  year      = {1971}
}

Prerequisites

01.01.03
01.01.04
01.01.05

Tier anchors

beginner: row vectors as the natural pairing of a list of numbers against a column vector to produce a scalar
intermediate: Shilov — Linear Algebra §4–§5; Axler — Linear Algebra Done Right §3.F
master: Halmos — Finite-Dimensional Vector Spaces §13–§17; Lang — Algebra Ch. III §6; Bourbaki — Algèbre Ch. II §2

References

quantum-well · linear map entry; functionals as scalar-valued linear maps
quantum-well · transpose / dual map entry
quantum-well · bilinear pairings and the canonical pairing $V \times V^* \to K$
Shilov — Linear Algebra · §4–§5, linear forms, dual basis, biorthogonal systems
Halmos — Finite-Dimensional Vector Spaces · §13–§17, linear functionals, dual space, annihilator, dual map
Lang — Algebra · Ch. III §6, duality of modules over a field
Bourbaki — Algèbre · Ch. II §2, dualité algébrique
Pincherle, S. — Mémoire sur le calcul fonctionnel distributif · Math. Annalen 49 (1897) 325–382, early functional-duality framework
Hahn, H. — Über lineare Gleichungssysteme in linearen Räumen · J. reine angew. Math. 157 (1927) 214–229, Hahn-Banach extension theorem
Banach, S. — Théorie des opérations linéaires · Monografie Matematyczne 1 (1932), systematic duality of Banach spaces

Estimated time

beginner: 18m
intermediate: 40m
master: 70m