Lecture 17

Chapter III Linear maps

Assumption: $U,V,W$ are vector spaces (over $\mathbb{F}$)

Duality 3F

Definition 3.108

A linear functional on $V$ is a linear map from $V$ to $\mathbb{F}$.

Definition 3.110

The dual space of V denoted by $V'$ ($\check{V},V^*$) is given by $V'=\mathscr{L}(V,\mathbb{F})$.

The elements of $V'$ are also called linear functional.

Theorem 3.111

The $dim\ V'=dim\ V$.

Proof:

$dim\ \mathscr{L}(V,\mathbb{F})=dim\ V\cdot dim\ \mathbb{F}$

Definition 3.112

If $v_1,...,v_n$ is a basis for $V$, then the dual basis of $v_1,..,v_n$ is $\psi_1,...,\psi_n\in V'$ where

Example:

$V=\mathbb{R}^3$ $e_1,e_2,e_3$ the standard basis, the dual basis $\psi_1,\psi_2,\psi_3$ is given by $\psi_1 (x,y,z)=x,\psi_2 (x,y,z)=y,\psi_3 (x,y,z)=z$

Theorem 3.116

When $v_1,...,v_n$ a basis of $V$ the dual basis $\psi_1,...,\psi_n\in V'$ is a basis

Sketch of Proof:

$dim\ V'=dim\ V=n$, $\psi_1,...,\psi_n\in V'$ are linearly independent.

Theorem 3.114

Given $v_1,...,v_n$ a basis of $V$, and $\psi_1,...,\psi_n\in V'$ be dual basis of $V'$. then for $v\in V$,

Proof:

Let $V=a_1 v_1+...+a_n v_n$, consider $\psi_k(v)$, by definition $\psi_k(v)=\psi_k(a_1 v_1+...+a_n v_n)=a_1\psi_k( v_1)+...+a_n\psi_k( v_n)=a_k$

Definition 3.118

Suppose $T\in \mathscr{L}(V,W)$. The dual map $T'\in \mathcal{R}( W', V')$ defined by $T'(\psi)=\psi\circ T$. ($\psi\in W'=\mathcal{R}(W,\mathbb{F}), T'(\psi) \in V'=\mathscr{L}(V,\mathbb{F})$)

Example:

$T:\mathscr{P}_2(\mathbb{F})\to \mathscr{P}_3(\mathbb{F}),T(f)=xf$

Suppose $\psi(f)=f'(1)\to T(\psi)(f)=(xf)'(1)=f(1)+(xf')(1)=f(1)+f'(1)$

Theorem 3.120

Suppose $T\in \mathscr{L}(V,W)$

a) $(S+T)'=S'+T', \forall S\in \mathscr{L}(V,W)$
b) $(\lambda T)'=\lambda T', \forall \lambda\in \mathbb{F}$
c) $(ST)'=T'S', \forall S\in \mathscr{L}(V,W)$

Goal: find $range\ T'$ and $null\ T'$

Definition 3.121

Let $U\subseteq V$ be a subspace. The annihilator of $U$, denoted by $U^0$ is given by $U^0=\{ \psi\in V'\vert \psi(u)=0\forall u\in U\}$

Proposition 3.124

Given $U\subseteq V$ be a subspace. The annihilator of $U$, $U^0\subseteq V'$ is a subspace.

Sketch of proof:

look at $i:U\to V,i(u)=u$, compute $i':V'\to U'$ look at $null\ i'=U^0$

Theorem 3.128, 3.130

a) $null\ T'=(range\ T)^0$, $dim (null\ T')=dim\ null\ T+dim\ W-dim\ V$
b) $range\ T'=(null\ T)^0$, $dim (range\ T')=dim (range\ T)$