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

Content

Definition: Linear Functional
Definition: Dual Space, $V'$
Lemma: $\dim(V')=\dim(V)$
Definition: Dual Basis
Lemma: Dual Basis and Linear Combination Scalars
Lemma: Dual Basis is a Basis for the Dual Space
Definition: Dual Map, $T'$
Lemma: Algebraic Properties of Dual Maps

Definition: Linear Functional

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

Definition: Dual Space, V'

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

Lemma: $\dim(V')=\dim(V)$

Let $V$ be a finite-dimensional vector space with basis $\beta = (v_1, \ldots, v_n)$ . The dual basis of $\beta$ is the list $(\varphi_1, \ldots, \varphi_n)$ of elements in $V'$ , where each $\varphi_i$ sends $v_i$ to $1$ and all other $v_j$ 's to $0.$

Proof:

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

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

Definition: Dual Basis

Lemma: Dual Basis and Linear Combination Scalars

Suppose $\beta = (v_1, \ldots, v_n)$ is a basis for $V$ and $(\varphi_1, \ldots, \varphi_n)$ is its dual basis. Then for every $v\in V$ , $v = \varphi_1(v) v_1 + \cdots + \varphi_n(v) v_n$ .

Proof:

Let $v \in V$ . Because $\beta$ is a basis for $V$ , there exist unique scalars $c_1, \ldots, c_n$ such that $v=c_1 v_1 + \cdots + c_n v_n$ . Now, apply any $\varphi_i$ to this equation. Because $\varphi_i$ is linear, we get

\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 + \cdots + c_i \cdot 1 + \cdots + 0 = c_i.

Lemma: Dual Basis is a Basis for the Dual Space

Suppose $V$ is a finite dimensional vector space and $\beta = (v_1, \ldots, v_n)$ is a basis for $V$ . Then $(\varphi_1, \ldots, \varphi_n)$ is a basis for $V'.$

Proof:

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

Definition: Dual Map, $T'$

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

T'(\varphi) = \varphi \circ T.

Lemma: Algebraic Properties of Dual Maps

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

Proof:

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

(TS)'(\varphi) = \varphi \circ T \circ S = S'(\varphi \circ T) = S'(T'(\varphi))=(S' \circ T')(\varphi).

Kernel and Range of a Dual

Content

Definition: Span (of an infinite space)
Lemma: For any subset $U \subseteq V$ , $\text{span}(U)$ is the smallest subspace containing $U$ .
Definition: Annihilator, $U^0$
Theorem: Annihilators are the same for generating sets a subspaces
Lemma: The annihilator is a subspace
Lemma: Basis for an Annihilator
Theorem: Dimension of an Annihilator
Corrollary: Condition for the annihilator to equal $\left\{ \vec{0} \right\}$ or the whole space
Lemma: The Kernel of $T'$
Theorem: $T$ is surjective $\iff T'$ is injective.
Lemma: The range of $T'$
Theorem: $T$ is injective $\iff$ $T'$ is surjective

We start with a slight deviation from the book.

Definition: Span (for infinite sets)

Let $U \subseteq V$ . Then

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 $U \subseteq V$ , $span(U)$ is the smallest subspace containing $U$ .

Proof:

(Sketch) a) To show that $span(U)$ 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 $U$ . That latter two involve adding and scaling linear combinations. b) For the smallest subspace claim, clearly $span(U)$ conains $U$ because $1 \cdot u \in U$ for every $u \in U$ , 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, $U^0$

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

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 $S \subseteq V$ and let $U=span(S)$ . Then $U^0=S^0$ .

Proof:

The inclusion $U^0 \subseteq S^0$ follows from the fact that any $\varphi \in U^0$ sends everything in $U$ to $0$ , so it automatically sends everything in $S \subseteq U$ to $0$ as well because everything in $S$ is also in $U$ . The other inclusion $S^0 \subseteq U^0$ follows from the linearity of the functionals. Suppose $\varphi \in S^0$ and let $u \in U$ . Then $u = c_1 v_1 + \cdots + u_n v_n \text{ for some } v_1, \ldots, v_n \in S.$ And therefore

\varphi(u) = \varphi(c_1 v_1 + \cdots + c_n v_n ) = c_1 \varphi(v_1) + \cdots + c_n \varphi_n(v_n)

= c_1 \cdot 0 + \cdots + c_n \cdot 0 = 0.

Lemma: The annihilator is a subspace

Let $U \subseteq V$ . Then $U^0$ is a subspace.

Proof:

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

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

Scalar Multiplication: Suppose $\lambda \in \mathbb{F}$ and $\varphi \in U^0$ . Let $u \in U$ . Then

(\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 of the Annihilator

Let $V$ be a finite dimensional vector space, $U$ a subspace of $V$ , $(u_1, \ldots, u_k)$ a basis for $U$ , and $(u_1, \ldots, u_k, v_1, \ldots, v_\ell)$ an basis extension for $V$ . Let $(\mu_1, \ldots, \mu_k, \varphi_1, \ldots, \varphi_\ell)$ be the corresponding dual basis vectors, a basis for $V'$ . Then list of dual extension vectors, $(\varphi_1, \ldots, \varphi_\ell)$ , is a basis for $U^0$ .

Proof:

This should remind you of the proof of the rank-nullity theorem. $(\varphi_1, \ldots, \varphi_\ell)$ is linearly independent because it is part of a basis. Each $\varphi_i$ annihilates $U$ because it sends all of the basis vectors $u_1, \ldots, u_k$ to $0$ . Finally, $(\varphi_1, \ldots, \varphi_\ell)$ is a spanning set for the following reason: For any $\varphi \in U^0$ , $\varphi \in V'$ means that there exist scalars $c_1, \ldots, c_k, d_1, \ldots, d_\ell \in \mathbb{F}$ such that $\varphi = \sum_{i=0}^k c_i \mu_i + \sum_{j=1}^\ell d_j \varphi_j$ . But each $c_i$ must be $0$ because $\varphi$ sends every $u_i$ to $0$ . So $\varphi = \sum_{j=1}^\ell d_j \varphi_j$ .

Theorem: Dimension of an Annihilator

Suppose $V$ is finite-dimensional and $U$ is a subspace of $V$ . Then

\dim(U^0) = \dim(V) - \dim(U).

Proof:

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

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

Suppose $V$ is a finite dimensional vector space and $U$ is a subspace of $V$ . Then a) $U^0 = \{ 0 \} \iff U = V$ b) $U^0 = V' \iff U = \{ 0 \}$

Proof:

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

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: $\ker(T') = T(V)^0$

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

Proof:

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

Claim 3: $T'(W') \subseteq \ker(T)^0$

Proof:

Let $\psi \in W'$ to represent an arbitrary element $T'(\psi) \in T'(W')$ . Let $u \in \ker(T)$ . Then $(T'(\psi))(u) = (\psi \circ T)(u) = \psi(0) = 0.$ Thus, $T'(\psi)$ annihilates every $u \in U$ .

Setup for the rest of this section

Let $T \in \mathscr{L}(V,W)$ , where $V$ and $W$ are finite dimensional
Let $(u_1, \ldots, u_\ell)$ be a basis for $\ker(T)$ .
Let $(u_1, \ldots, u_k, v_1, \ldots, v_\ell)$ be an extended basis from $U$ to $V$ . It follows from the proof of the rank-nullity theorem that $(w_1, \ldots, w_\ell) = \left( T(v_1), \ldots, T(v_\ell) \right)$ is a basis for $T(V)$ .
Let $(w_1, \ldots, w_\ell, y_1, \ldots, y_m)$ be an extended basis for $T(V)$ to $W$ .
Let $(\omega_1, \ldots, \omega_\ell, \psi_1, \ldots, \psi_m)$ be the dual basis for $W'$ .
Observe that $(\psi_1, \ldots, \psi_m)$ is a basis for $T(V)^0$ by Claim 1 and for $\ker(T')$ by Claim 2.
Observe that the dual vectors $(\varphi_1, \ldots, \varphi_\ell) = \left( T'(\omega_1), \ldots, T'(\omega_\ell) \right)$ in $V'$ are a basis for $T'(W')$ by the proof the rank-nullity theorem.

Lemma: The kernel of $T'$

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

Proof:

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

Theorem: $T$ is surjective $\iff$ $T'$ is injective.

Suppose $V$ and $W$ are finite-dimensional and $T \in \mathscr{L}(V,W)$ . Then $T$ is surjective $\iff$ $T'$ is injective.

Proof:

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

Lemma: The range of $T'$

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

Proof:

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

Theorem: $T$ is injective $\iff$ $T'$ is surjective

Suppose $V$ and $W$ are finite-dimensional and $T \in \mathscr{L}(V,W).$ Then $T$ is injective $\iff$ $T'$ is surjective.

Proof:

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

Matrix of a Dual

Content

Theorem: Matrix of $T'$ is the transpose of the matrix of $T$
Theorem: Column Rank = Row Rank

Theorem: Matrix of $T'$ is the transpose of the matrix of $T$

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

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

Proof:

(Summary) There are isomorphisms between the vector spaces $V$ and $V'$ and between the vector spaces $W$ and $W'$ 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 $V$ or $W$ corresponds to its transpose as a functional. Now, since $T'$ is pre-composition with $T$ , a dual vector in $W'$ is the transpose of a vector in $W$ and $T'$ of that vector is that transposed vector preceeded by $\mathcal{M}(T)^T$ . To verify this, put any dual basis vector in $W'$ into the matrix and examine the corresponding column.

Theorem: Column Rank = Row Rank

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

Proof:

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

3F Duality

Topics

Dual Spaces and the Dual Map

Content

Definition: Linear Functional

Definition: Dual Space, V'

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

Definition: Dual Basis

Lemma: Dual Basis and Linear Combination Scalars

Lemma: Dual Basis is a Basis for the Dual Space

Definition: Dual Map, T′T'T′

Lemma: Algebraic Properties of Dual Maps

Kernel and Range of a Dual

Content

Definition: Span (for infinite sets)

Lemma: Smallest Subspace

Definition: Annihilator, U0U^0U0

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

Lemma: The annihilator is a subspace

Claim 1: Basis of the Annihilator

Theorem: Dimension of an Annihilator

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

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

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

Setup for the rest of this section

Lemma: The kernel of T′T'T′

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

Lemma: The range of T′T'T′

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

Matrix of a Dual

Content

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

Theorem: Column Rank = Row Rank

Lemma: $\dim(V')=\dim(V)$

Definition: Dual Map, $T'$

Definition: Annihilator, $U^0$

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

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

Claim 3: $T'(W') \subseteq \ker(T)^0$

Lemma: The kernel of $T'$

Theorem: $T$ is surjective $\iff$ $T'$ is injective.

Lemma: The range of $T'$

Theorem: $T$ is injective $\iff$ $T'$ is surjective

Theorem: Matrix of $T'$ is the transpose of the matrix of $T$