3F Duality


Topics

  • Dual Spaces and the Dual Map
  • Kernel and Range of the Dual
  • Matrix of the Dual



Dual Spaces and the Dual Map


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Definition: Linear Functional

A linear frunctional on VV is a linear map from VV to F\mathbb{F}. In other words, a linear functional is an element of L(V,F)\mathscr{L}(V, \mathbb{F}).

Definition: Dual Space, V'

The dual space of VV, denoted V′V', is the vector space of all linear functionals on VV. In other words, V′=L(V,F)V'=\mathscr{L}(V,\mathbb{F}).

Lemma: dim⁡(V′)=dim⁡(V)\dim(V')=\dim(V)

Let VV be a finite-dimensional vector space with basis β=(v1,…,vn)\beta = (v_1, \ldots, v_n). The dual basis of β\beta is the list (φ1,…,φn)(\varphi_1, \ldots, \varphi_n) of elements in V′V', where each φi\varphi_i sends viv_i to 11 and all other vjv_j's to 0.0.

Proof:

Let n=dim⁡(V)n = \dim(V). Then because V′=L(V,F)V' = \mathscr{L}(V,\mathbb{F}),

dim⁡(V′)=dim⁡(L(V,F))=dim⁡(V)⋅dim⁡(F)=n⋅1=n.\dim(V') = \dim \left( \mathscr{L}(V,\mathbb{F}) \right) = \dim(V) \cdot \dim(\mathbb{F}) = n \cdot 1 = n.

Definition: Dual Basis

Let VV be a finite-dimensional vector space with basis β=(v1,…,vn)\beta = (v_1, \ldots, v_n). The dual basis of β\beta is the list (φ1,…,φn)(\varphi_1, \ldots, \varphi_n) of elements in V′V', where each φi\varphi_i sends viv_i to 11 and all other vjv_j's to 0.0.

Lemma: Dual Basis and Linear Combination Scalars

Suppose β=(v1,…,vn)\beta = (v_1, \ldots, v_n) is a basis for VV and (φ1,…,φn)(\varphi_1, \ldots, \varphi_n) is its dual basis. Then for every v∈Vv\in V, v=φ1(v)v1+⋯+φn(v)vnv = \varphi_1(v) v_1 + \cdots + \varphi_n(v) v_n.

Proof:

Let v∈Vv \in V. Because β\beta is a basis for VV, there exist unique scalars c1,…,cnc_1, \ldots, c_n such that v=c1v1+⋯+cnvnv=c_1 v_1 + \cdots + c_n v_n. Now, apply any φi\varphi_i to this equation. Because φi\varphi_i is linear, we get

φ1(v)=φi(c1v1+⋯+cnvn)=c1φi(v1)+⋯ciφi(vi)+⋯+cnφi(vn) \varphi_1(v) = \varphi_i(c_1 v_1 + \cdots + c_n v_n) = c_1 \varphi_i(v_1) + \cdots c_i \varphi_i(v_i) + \cdots + c_n \varphi_i(v_n)

=0+⋯+ci⋅1+⋯+0=ci. = 0 + \cdots + c_i \cdot 1 + \cdots + 0 = c_i.

Lemma: Dual Basis is a Basis for the Dual Space

Suppose VV is a finite dimensional vector space and β=(v1,…,vn)\beta = (v_1, \ldots, v_n) is a basis for VV. Then (φ1,…,φn)(\varphi_1, \ldots, \varphi_n) is a basis for V′.V'.

Proof:

We will show linear independence, which is sufficient for showing it is a basis because of the list's length. Suppose c1φ1+⋯+cnφn=0c_1 \varphi_1 + \cdots + c_n \varphi_n = 0. Applying this map to any v∈Vv \in V must be 00, and also applying this map to any viv_i results in cic_i. So every cic_i must be 00.

Definition: Dual Map, T′T'

Let T∈L(V,W)T \in \mathscr{L}(V,W). The dual map of TT is the linear map T′∈L(W′,V′)T' \in \mathscr{L}(W', V') defined for each φ∈W′\varphi \in W' as

T′(φ)=φ∘T.T'(\varphi) = \varphi \circ T.

Lemma: Algebraic Properties of Dual Maps

Suppose T∈L(V,W)T \in \mathscr{L}(V, W). Then a) (S+T)′=S′+T′(S + T)' = S' + T' for every S∈L(V,W)S \in \mathscr{L}(V, W). b) (λT)′=λT′(\lambda T)' = \lambda T' for all λ∈F\lambda \in \mathbb{F}. c) (TS)′=S′T′(TS)' = S' T' for all S∈L(U,V)S \in \mathscr{L}(U,V).

Proof:

The proofs of (a) and (b) are left to the reader. To prove (c), we use the definition:

(TS)′(φ)=φ∘T∘S=S′(φ∘T)=S′(T′(φ))=(S′∘T′)(φ). (TS)'(\varphi) = \varphi \circ T \circ S = S'(\varphi \circ T) = S'(T'(\varphi))=(S' \circ T')(\varphi).


Kernel and Range of a Dual


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We start with a slight deviation from the book.

Definition: Span (for infinite sets)

Let U⊆VU \subseteq V. Then

span(U)={c1u1+⋯+cnun ∣ u1,…,un∈U and c1,…,cn∈F}.span(U) = \left\{ c_1 u_1 + \cdots + c_n u_n \ | \ u_1, \ldots, u_n \in U \text{ and } c_1, \ldots, c_n \in \mathbb{F} \right\}.

Lemma: Smallest Subspace

For any subset U⊆VU \subseteq V, span(U)span(U) is the smallest subspace containing UU.

Proof:

(Sketch) a) To show that span(U)span(U) contains the zero vector, that it has closure under vector addition and also closure under scalar multiplication, we note that all three statements can be expressed as linear combinations of elements in UU. That latter two involve adding and scaling linear combinations. b) For the smallest subspace claim, clearly span(U)span(U) conains UU because 1⋅u∈U1 \cdot u \in U for every u∈Uu \in U, so we just need to show that every subspace containing U must contain everything in span(U). This follows from subspaces containing all linear combinations of their elements from closure under vector addition and scalar multiplication.

Definition: Annihilator, U0U^0

Let UU be a subset of VV. The annihilator of UU is defined as the set of linear functionals that "annihilate" everything in UU, specifically:

U0={φ∈V′ ∣ φ(u)=0 ∀u∈U}. U^0 = \left\{ \varphi \in V' \ | \ \varphi(u) = 0 \ \forall u \in U \right\} .

Theorem: Annihilators are the same for generating sets of subspaces.

Let S⊆VS \subseteq V and let U=span(S)U=span(S). Then U0=S0U^0=S^0.

Proof:

The inclusion U0⊆S0U^0 \subseteq S^0 follows from the fact that any φ∈U0\varphi \in U^0 sends everything in UU to 00, so it automatically sends everything in S⊆US \subseteq U to 00 as well because everything in SS is also in UU. The other inclusion S0⊆U0S^0 \subseteq U^0 follows from the linearity of the functionals. Suppose φ∈S0\varphi \in S^0 and let u∈Uu \in U. Then u=c1v1+⋯+unvn for some v1,…,vn∈S.u = c_1 v_1 + \cdots + u_n v_n \text{ for some } v_1, \ldots, v_n \in S. And therefore

φ(u)=φ(c1v1+⋯+cnvn)=c1φ(v1)+⋯+cnφn(vn)\varphi(u) = \varphi(c_1 v_1 + \cdots + c_n v_n ) = c_1 \varphi(v_1) + \cdots + c_n \varphi_n(v_n)

=c1⋅0+⋯+cn⋅0=0. = c_1 \cdot 0 + \cdots + c_n \cdot 0 = 0.

Lemma: The annihilator is a subspace

Let U⊆VU \subseteq V. Then U0U^0 is a subspace.

Proof:

We use the usual subspace test. Zero vector: The zero function sends everything to 00, so it is in the annihilator of every subset of VV. Vector Addition: Suppose φ1,φ2∈U0\varphi_1, \varphi_2 \in U^0 and let u∈Uu \in U. Then

(φ1+φ2)(u)=φ1(u)+φ2(u)=0+0=0.\left( \varphi_1 + \varphi_2 \right)(u) = \varphi_1(u) + \varphi_2(u) = 0 + 0 = 0.

Scalar Multiplication: Suppose λ∈F\lambda \in \mathbb{F} and φ∈U0\varphi \in U^0. Let u∈Uu \in U. Then

(λφ)(u)=λ(φ(u))=λ(0)=0.(\lambda \varphi)(u) = \lambda \left( \varphi(u) \right) = \lambda(0) = 0.

We now deviate more from the book, starting with a claim about a basis for the annihilator.

Claim 1: Basis for the Annihilator

Let VV be a finite dimensional vector space, UU a subspace of VV, (u1,…,uk)(u_1, \ldots, u_k) a basis for UU, and (u1,…,uk,v1,…,vℓ)(u_1, \ldots, u_k, v_1, \ldots, v_\ell) an basis extension for VV. Let (μ1,…,μk,φ1,…,φℓ)(\mu_1, \ldots, \mu_k, \varphi_1, \ldots, \varphi_\ell) be the corresponding dual basis vectors, a basis for V′V'. Then list of dual extension vectors, (φ1,…,φℓ)(\varphi_1, \ldots, \varphi_\ell), is a basis for U0U^0.

Proof:

This should remind you of the proof of the rank-nullity theorem. (φ1,…,φℓ)(\varphi_1, \ldots, \varphi_\ell) is linearly independent because it is part of a basis. Each φi\varphi_i annihilates UU because it sends all of the basis vectors u1,…,uku_1, \ldots, u_k to 00. Finally, (φ1,…,φℓ)(\varphi_1, \ldots, \varphi_\ell) is a spanning set for the following reason: For any φ∈U0\varphi \in U^0, φ∈V′\varphi \in V' means that there exist scalars c1,…,ck,d1,…,dℓ∈Fc_1, \ldots, c_k, d_1, \ldots, d_\ell \in \mathbb{F} such that φ=∑i=0kciμi+∑j=1ℓdjφj\varphi = \sum_{i=0}^k c_i \mu_i + \sum_{j=1}^\ell d_j \varphi_j. But each cic_i must be 00 because φ\varphi sends every uiu_i to 00. So φ=∑j=1ℓdjφj\varphi = \sum_{j=1}^\ell d_j \varphi_j.


Picture of the Setup

Theorem: Dimension of an Annihilator

Suppose VV is finite-dimensional and UU is a subspace of VV. Then

dim⁡(U0)=dim⁡(V)−dim⁡(U). \dim(U^0) = \dim(V) - \dim(U).

Proof:

In Claim 1, k=dim⁡(U)k = \dim(U), ℓ=dim⁡(U0)\ell = \dim(U^0), and k+ℓ=dim⁡(V)k + \ell = \dim(V).

Corollary: Condition for the annihilator to equal {0}\{ 0 \} or the whole space

Suppose VV is a finite dimensional vector space and UU is a subspace of VV. Then a) U0={0}  ⟺  U=VU^0 = \{ 0 \} \iff U = V b) U0=V′  ⟺  U={0}U^0 = V' \iff U = \{ 0 \}

Proof:

Both conditions in part (a) happen if and only if k=dim⁡(V)k = \dim(V) and ℓ=0\ell = 0 in Claim 1. Both conditions in part (b) happen if and only if k=0k=0 and ℓ=dim⁡(V)\ell = \dim(V).

We now have two more claims related to the book's results, followed by a setup that will take care of all the remaining proofs.

Claim 2: ker⁡(T′)=T(V)0\ker(T') = T(V)^0

Let T∈L(V,W)T \in \mathscr{L}(V, W). Then ker⁡(T′)=T(V)0\ker(T') = T(V)^0.

Proof:

On the one hand, ker⁡(T′)\ker(T') is the set of dual vectors ψ∈W′\psi \in W' such that (ψ∘T)(v)=0(\psi \circ T)(v) = 0 for every v∈Vv \in V. On the other hand, T(V)0T(V)^0 is the set of dual vectors ψ∈W′\psi \in W' such that ψ(T(v))=0\psi(T(v)) = 0 for every v∈Vv \in V.

Claim 3: T′(W′)⊆ker⁡(T)0T'(W') \subseteq \ker(T)^0

Proof:

Let ψ∈W′\psi \in W' to represent an arbitrary element T′(ψ)∈T′(W′)T'(\psi) \in T'(W'). Let u∈ker⁡(T)u \in \ker(T). Then (T′(ψ))(u)=(ψ∘T)(u)=ψ(0)=0.(T'(\psi))(u) = (\psi \circ T)(u) = \psi(0) = 0. Thus, T′(ψ)T'(\psi) annihilates every u∈Uu \in U.

Setup for the rest of this section

  • Let T∈L(V,W)T \in \mathscr{L}(V,W), where VV and WW are finite dimensional
  • Let (u1,…,uℓ)(u_1, \ldots, u_\ell) be a basis for ker⁡(T)\ker(T).
  • Let (u1,…,uk,v1,…,vℓ)(u_1, \ldots, u_k, v_1, \ldots, v_\ell) be an extended basis from UU to VV. It follows from the proof of the rank-nullity theorem that (w1,…,wℓ)=(T(v1),…,T(vℓ))(w_1, \ldots, w_\ell) = \left( T(v_1), \ldots, T(v_\ell) \right) is a basis for T(V)T(V).
  • Let (w1,…,wℓ,y1,…,ym)(w_1, \ldots, w_\ell, y_1, \ldots, y_m) be an extended basis forT(V)T(V) to WW.
  • Let (ω1,…,ωℓ,ψ1,…,ψm)(\omega_1, \ldots, \omega_\ell, \psi_1, \ldots, \psi_m) be the dual basis for W′W'.
  • Observe that (ψ1,…,ψm)(\psi_1, \ldots, \psi_m) is a basis for T(V)0T(V)^0 by Claim 1 and for ker⁡(T′)\ker(T') by Claim 2.
  • Observe that the dual vectors (φ1,…,φℓ)=(T′(ω1),…,T′(ωℓ))(\varphi_1, \ldots, \varphi_\ell) = \left( T'(\omega_1), \ldots, T'(\omega_\ell) \right) in V′V' are a basis for T′(W′)T'(W') by the proof the rank-nullity theorem.

Picture of the Setup


Lemma: The kernel of T′T'

Suppose VV and WW are finite dimensional vector spaces and T∈L(V,W)T \in \mathscr{L}(V,W). Then a) ker⁡(T′)=T(V)0\ker(T') = T(V)^0. b) dim⁡(ker⁡(T′))=dim⁡(ker⁡(T))+dim⁡(W)−dim⁡(V)\dim(\ker(T'))=\dim(\ker(T)) + \dim(W) - \dim(V).

Proof:

a) This is Claim 2. b) LHS = mm, RHS = (k)+(ℓ+m)−(k+ℓ)=m(\cancel{k}) + (\cancel{\ell} + m) - (\cancel{k} + \cancel{\ell}) = m

Theorem: TT is surjective   ⟺  \iff T′T' is injective.

Suppose VV and WW are finite-dimensional and T∈L(V,W)T \in \mathscr{L}(V,W). Then TT is surjective   ⟺  \iff T′T' is injective.

Proof:

Both conditions happen if and only if m=0m = 0.

Lemma: The range of T′T'

Suppose VV and WW are finite-dimensional and T∈L(V,W)T \in \mathscr{L}(V,W). Then a) dim⁡(T′(W′))=dim⁡(T(V))\dim(T'(W')) = \dim(T(V)). b) T′(W′)=ker⁡(T)0T'(W') = \ker(T)^0

Proof:

a) Both are ℓ\ell. b) T′(W)⊆ker⁡(T)0T'(W) \subseteq \ker(T)^0 by Claim 3, and we have the other containment from having the same dimension of ℓ\ell.

Theorem: TT is injective   ⟺  \iff T′T' is surjective

Suppose VV and WW are finite-dimensional and T∈L(V,W).T \in \mathscr{L}(V,W). Then TT is injective   ⟺  \iff T′T' is surjective.

Proof:

Both conditions happen if and only if k=0k = 0.

Matrix of a Dual


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Theorem: Matrix of T′T' is the transpose of the matrix of TT

Suppose VV and WW are finite dimensional vector spaces and T∈L(V,W)T \in \mathscr{L}(V,W). Then

M(T′)=(M(T))T. \mathcal{M}(T') = \left( \mathcal{M}(T) \right)^T.

Proof:

(Summary) There are isomorphisms between the vector spaces VV and V′V' and between the vector spaces WW and W′W' that extend linearly from the bijection between a given basis and its corresponding dual basis. We start by observing that every column vector with respect to this basis in either VV or WW corresponds to its transpose as a functional. Now, since T′T' is pre-composition with TT, a dual vector in W′W' is the transpose of a vector in WW and T′T' of that vector is that transposed vector preceeded by M(T)T\mathcal{M}(T)^T. To verify this, put any dual basis vector in W′W' into the matrix and examine the corresponding column.

Theorem: Column Rank == Row Rank

Suppose A∈FA \in \mathbb{F}. Then the column rank of AA equals the row rank of AA.

Proof:

Let V=FnV = \mathbb{F}^n, W=FpW = \mathbb{F}^p, and let TT be the linear operator obtained by applying the matrix AA. Then the column rank of AA is the dimsion of T(V)T(V). Recall the setup for the proofs for the previous section titled "Setup for the Rest of this Section". In that setup, ℓ\ell is shown to be both the dimension of both T(V)T(V) and T′(W′)T'(W'). So the row rank of AA, which is the column rank of ATA^T, must be the same as the rank of AA because AA has the same rank as that of AT=M(T′).A^T = \mathcal{M}(T').