3.3 KiB
Lecture 28
Chapter VI Inner Product Spaces
Orthonormal basis 6B
Example:
Find a polynomial q\in \mathscr{P}_2(\mathbb{R}) such that
\int^1_{-1}p(t)cos(\pi t)dt=\int^1_{-1}p(t)q(t)dt
for p\in \mathscr{P}_2(\mathbb{R})
note that \varphi(p)=\int^1_{-1}p(t)cos(\pi t)cos(\pi t)dt is a linear functional. Thus by Riesz Representation Theorem, \exists unique q such that \varphi (p)=\langle p,q \rangle=\int^1_{-1}pq
q=\overline{\varphi(e_0)}e_0+\overline{\varphi(e_z)}e_z
where e_0,e_1,e_z is an orthonormal basis.
and q=\frac{15}{2\pi^2}(1-3x^2)
Orthogonal Projection and Minimization
Definition 6.46
If U is a subset of V, then the orthogonal complement of U denoted U^\perp
U^\perp=\{v\in V\vert \langle u,v\rangle =0,\forall u\in U\}
The set of vectors orthogonal to every vector in U.
Theorem 6.48
Let U be a subset of V.
(a) U^\perp is a subspace of V.
(b) \{0\}^\perp=V
(c) V^\perp =\{0\}
(d) U\cap U^\perp\subseteq\{0\}
(e) If G,H subsets of V with G\subseteq H, then H^\perp\subseteq G^\perp
Example:
Two perpendicular line in 2D plane.
Let e_1,...,e_m be an orthonormal list, let u=Span(e_1,...,e_m) How do I find U^\perp?
Extend to an orthonormal basis e_1,...,e_m,f_1...,f_n. U^\perp=Span(f_1,...,f_n)
Theorem 6.40
Suppose U is finite dimensional subspace of V, then V=U\oplus U^\perp
Proof:
Note U\cap U^\perp=\{0\}, so it suffices to show U+U^\perp=V. Fix an orthonormal basis e_1,...,e_m of U. Let v\in V, let u=\langle v,e_1\rangle e_1+...+\langle v,e_m\rangle e_m let w=v-u, then v=u+W we need to check that w\in U^\perp
\langle w,e_k \rangle=\langle v,e_k\rangle-\langle u,e_k\rangle=\langle v,e_k\rangle-\langle v,e_k\rangle=0
So w\in U^\perp
Corollary 6.51
dim\ U^\perp=dim\ V-dim\ U
Theorem 6.52
Let U be a finite dimensional of a vector space V. Then (U^\perp)^\perp=U
Proof:
First let u\in U we want to show u\in (U^\perp)^\perp, then \langle u,w \rangle=0 for all w\in U^\perp but then u\in (U^\perp)^\perp
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Corollary 6.54
U^\vert=\{0\}\iff U=V
Proof:
(U^\perp)^\perp=\{0\}^\perp\implies U=V
Definition 6.55
Given U a finite dimensional subspace of V. The orthogonal projection of V onto $U$ is the operator P_u\in \mathscr{L}(V) defined by: For each v write v=u+w where u\in U and w\in U^\perp then P_u v=u
Formula:
Let e_1,...,e_n an orthonormal basis of U.
P_u v=\langle v,e_1\rangle e_1+...+\langle v,e_m\rangle e_m
Theorem 6.57
(a) P_u is linear.
(b) P_u u=U,\forall u\in U
(c) P_u w=0,\forall w\in U^\perp
(d) range\ P_u=U
(e) null\ P_u=U^\vert
(f) v-P_u v\in U^\perp
(g) P_u^2=P_u
(h) ||P_u v||\leq ||v||
Proof:
(a) Let v,v'\in V and suppose v=u+w,v'=u'+w', then v+u'=(u+u')+(w+w') this implies that P_u(v+v')=u+u'=P_u v+ P_u v'
...
Theorem 6.58 Riesz Representation Theorem
Let V be a finite dimensional vector space for v\in V define \varphi_v\in V' by \varphi_v(u)=\langle u,v \rangle. Then the map v\to \varphi_v is a bijection.
Proof:
Surjectivity Ideal is let w\in (null\ \varphi)^\perp
v=\frac{\varphi(w)}{||w||^2}w,\varphi(v)=||v||^2
make sense \varphi_v=\varphi