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Math429
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Math429 L26

Lecture 26

Chapter VI Inner Product Spaces

Inner Products and Norms 6A


Review

Dot products

Inner product

An inner product ,:V×VF\langle,\rangle:V\times V\to \mathbb{F}

Positivity: v,v0\langle v,v\rangle\geq 0

Definiteness: v,v=0    v=0\langle v,v\rangle=0\iff v=0

Additivity: <u+v,w>=<u,w>+<v,w><u+v,w>=<u,w>+<v,w>

Homogeneity: <λu,v>=λ<u,v><\lambda u, v>=\lambda<u,v>

Conjugate symmetry: <u,v>=<v,u><u,v>=\overline{<v,u>}

Norm

v=<v,v>||v||=\sqrt{<v,v>}


New materials

Orthonormal basis 6B

Definition 6.22

A list of vectors is orthonormal if each vector has norm = 1, and is orthogonal to every other vectors in the list.

if a list e1,...,emVe_1,...,e_m\in V is orthonormal if <ej,ek>=1{1 if j=k0 if jk<e_j,e_k>=1\begin{cases} 1 \textup{ if } j=k\\ 0 \textup{ if }j\neq k \end{cases}.

Example:

  • Standard basis in Fn\mathbb{F}^n is orthonormal.
  • (13,13,13),(12,12,0),(16,16,26)(\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}),(\frac{-1}{\sqrt{2}},\frac{1}{\sqrt{2}},0),(\frac{1}{\sqrt{6}},\frac{1}{\sqrt{6}},\frac{-2}{\sqrt{6}}) in F3\mathbb{F}^3 is orthonormal.
  • For <p,q>=11pq<p,q>=\int^1_{-1}pq on P2(R)\mathscr{P}_2(\mathbb{R}). The standard basis (1,x,x2)(1,x,x^2) is not orthonormal.

Theorem 6.24

Suppose e1,...,eme_1,...,e_m is an orthonormal list, then a1e1+...+amem2=a12+...+am2||a_1 e_1+...+a_m e_m||^2=|a_1|^2+...+|a_m|^2

Proof:

Using induction of mm.

m=1m=1, clear (e12=1||e_1||^2=1) m>1m>1, a1e1+...am1em12=a12+...+am12||a_1 e_1+...a_{m-1}e_{m-1}||^2=|a_1|^2+...+|a_{m-1}|^2 and <a1e1+...+am1em1,amem>=0<a_1 e_1+...+a_{m-1} e_{m-1},a_m e_m>=0 by Pythagorean Theorem. (a1e1+...am1em1)+amem2=a1e1+...am1em12+amem2=a12+...+am12+am2||(a_1 e_1+...a_{m-1}e_{m-1})+a_m e_m||^2=||a_1 e_1+...a_{m-1}e_{m-1}||^2+||a_m e_m||^2=|a_1|^2+...+|a_{m-1}|^2+|a_m|^2

Theorem 6.25

Every orthonormal list is linearly independent.

Proof:

a1e1+...+amem2=0||a_1 e_1+...+a_m e_m||^2=0, then a12+...+am2=0|a_1|^2+...+|a_m|^2=0, then a1=...=am=0a_1=...=a_m=0

Theorem 6.28

Every orthonormal list of length dim Vdim\ V is a basis.

Definition 6.27

An orthonormal basis is a basis that is an orthonormal list.

Theorem 6.26 Bessel's Inequality

Suppose e1,...,eme_1,...,e_m is an orthonormal list vVv\in V

<v,e1>2+...+<v,em>2v2|<v,e_1>|^2+...+|<v,e_m>|^2\leq ||v||^2

Proof:

Let vVv\in V, then let n=<v,e1>e1+...+<v,em>emn=<v,e_1>e_1+...+<v,e_m>e_m,

let w=vuw=v-u, Note that <u,ek>=<v,ek><u,e_k>=<v,e_k>, thus <w,ek>=0,<w,u>=0<w,e_k>=0, <w,u>=0, apply Pythagorean Theorem.

w+u2=w2+u2v2u2||w+u||^2=||w||^2+||u||^2\\ ||v||^2\geq ||u||^2

Theorem 6.30

Suppose e1,...,ene_1,...,e_n is an orthonormal basis, and u,vVu,v\in V, then

(a) v=<v,e1>e1+...+<v,en>env=<v,e_1>e_1+...+<v,e_n>e_n
(b) v2=<v,e1>2+...+<v,en>2||v||^2=|<v,e_1>|^2+...+|<v,e_n>|^2
(c) <u,v>=<u,e1><v,e1>+...+<u,en><v,en><u,v>=<u,e_1>\overline{<v,e_1>}+...+<u,e_n>\overline{<v,e_n>}

Proof:

(a) let a1,...,anFa_1,...,a_n\in \mathbb{F} such that v=a1e1+...+anenv=a_1 e_1+...+a_n e_n.

<v,ek>=<a,e1,ek>+...+<akek,ek>+...+<anen,en>=<akek,ek>=ak\begin{aligned} <v,e_k>&=<a,e_1,e_k>+...+<a_k e_k,e_k>+...+<a_n e_n,e_n>\\ &=<a_k e_k,e_k>\\ &= a_k \end{aligned}

Note 6.30 (c) means up to change of basis, every inner product on a finite dimensional vector space "looks like" an euclidean inner products...

Theorem 6.32 Gram-Schmidt

Let v1,...,vmv_1,...,v_m be a linearly independent list.

Define fkVf_k\in V by f1=v1,fk=vkj=1k1<vk,fj>fj2fjf_1=v_1,f_k=v_k-\sum_{j=1}^{k-1}\frac{<v_k,f_j>}{||f_j||^2}f_j

Define ek=fkfke_k=\frac{f_k}{||f_k||}, then e1,...,eme_1,...,e_m is orthonormal Span(v1,...,vm)=Span(f1,...,fm)Span(v_1,...,v_m)=Span(f_1,...,f_m)