4.2 Dual vector spaces

To any vector space $V$ we can associate a new vector space, its dual:

Definition · Dual vector space

For $V$ a vector space over a field $k$ , the dual vector space $V^{*}$ is the vector space of all linear maps $V \to k$ , i.e.,

$V^{*} = {l : V \to k such that l (α v + β w) = α l (v) + β l (w)}$

for $α, β \in k, v, w \in V$ .

Given a linear transformation $L$ acting on $V,$ we can define:

Definition · Transpose transformation

The transpose of $L$ is the linear transformation

$L^{t} : V^{*} \to V^{*}$

given by

$(L^{t} l) (v) = l (Lv) (4.1)$

for $l \in V^{*}, v \in V .$

For any choice of basis ${e_{j}}$ of $V,$ there is a dual basis ${e_{j}^{*}}$ of $V^{*}$ that satisfies

e_{j}^{*} (e_{k}) = δ_{j k}

Coordinates on $V$ with respect to a basis are linear functions, and thus elements of $V^{*}$ . The coordinate function $v_{j}$ can be identified with the dual basis vector $e_{j}^{*}$ since

e_{j}^{*} (v) = e_{j}^{*} (v_{1} e_{1} + v_{2} e_{2} + \dots + v_{n} e_{n}) = v_{j}

It can easily be shown that the elements of the matrix for $L$ in the basis $e_{j}$ are given by

L_{j k} = e_{j}^{*} (L e_{k})

and that the matrix for the transpose map (with respect to the dual basis) is the matrix transpose

(L^{T})_{j k} = L_{k j}

Matrix notation can be used to write elements

l = l_{1} e_{1}^{*} + l_{2} e_{2}^{*} + \dots + l_{n} e_{n}^{*} \in V^{*}

of $V^{*}$ as row vectors

(l_{1} l_{2} \dots l_{n})

of coordinates on $V^{*}$ . Evaluation of $l$ on a vector $v$ is then given by matrix multiplication

l (v) = (l_{1} l_{2} \dots l_{n}) v_{1} v_{2} ⋮ v_{n} = l_{1} v_{1} + l_{2} v_{2} + \dots + l_{n} v_{n}

One can equally well interpret this formula as the expansion of $l \in V^{*}$ in the dual basis of coordinate functions $v_{j} = e_{j}^{*}$ .

For any representation $(π, V)$ of a group $G$ on $V$ , we can define a corresponding representation on $V^{*}$ :

Definition · Dual or contragredient representation

The dual or contragredient representation on $V^{*}$ is given by taking as linear operators

$(π^{- 1})^{t} (g) : V^{*} \to V^{*} (4.2)$

These satisfy the homomorphism property since

(π^{- 1} (g_{1}))^{t} (π^{- 1} (g_{2}))^{t} = (π^{- 1} (g_{2}) π^{- 1} (g_{1}))^{t} = ((π (g_{1}) π (g_{2}))^{- 1})^{t}

One way to characterize this representation is as the action on $V^{*}$ such that pairings between elements of $V^{*}$ and $V$ are invariant, since

l (v) \to ((π^{- 1} (g))^{t} l) (π (g) v) = l (π (g)^{- 1} π (g) v) = l (v)

Choosing a basis of $V$ , a representation operator $π (g)$ becomes a matrix $P$ , acting on $V$ by

v_{1} v_{2} ⋮ v_{n} \to P v_{1} v_{2} ⋮ v_{n}

The action on the dual space $V^{*}$ will then be given by (interpreting the $v_{j}$ as the dual basis elements for $V^{*}$ )

l = (l_{1} l_{2} \dots l_{n}) v_{1} v_{2} ⋮ v_{n} \to (l_{1} l_{2} \dots l_{n}) P^{- 1} v_{1} v_{2} ⋮ v_{n} .

This can be read as saying that $π (g)$ acts by the matrix $(P^{- 1})^{T}$ on $l \in V^{*}$ or as $P^{- 1}$ on the $v_{j}$ , interpreted as basis elements of $V^{*}$

来源与版本

正文：英文 · 原著转录

核对状态：AI 辅助转录核对，未作人工审阅

原书 PDF · 印刷页 36、37

来源版本：2025-10-20

来源 PDF SHA-256：5a1941b2443b54d5db3d055f1e5ba390429b7a728475258017aaac87ee85a837

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4.2 Dual vector spaces