2.8 KiB
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
\psi_j(v_k)=\begin{cases}
1 \textup{ if }k=i\\
0 \textup{ if }k\neq i
\end{cases}
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,
v=\psi_1(v)v_1+...+\psi_n(v)v_n
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
T'(\mathscr{P}_3(\mathbb{F}))'\to (\mathscr{P}_2(\mathbb{F}))',T'(\psi)(f)=\psi(T(f))=\psi(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.
dim\ U^0=dim\ V-dim\ U=(dim\ V')-dim\ U
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)