NoteNextra-origin/content/Math429/Math429_L26.md

Lecture 26

Chapter VI Inner Product Spaces

Inner Products and Norms 6A

---

Review

Dot products

Inner product

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

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

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

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

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

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

Norm

||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 e_1,...,e_m\in V is orthonormal if $<e_j,e_k>=1\begin{cases} 1 \textup{ if } j=k\ 0 \textup{ if }j\neq k \end{cases}$.

Example:

• Standard basis in \mathbb{F}^n is orthonormal.
• (\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 \mathbb{F}^3 is orthonormal.
• For <p,q>=\int^1_{-1}pq on \mathscr{P}_2(\mathbb{R}). The standard basis (1,x,x^2) is not orthonormal.

Theorem 6.24

Suppose e_1,...,e_m is an orthonormal list, then ||a_1 e_1+...+a_m e_m||^2=|a_1|^2+...+|a_m|^2

Proof:

Using induction of m.

m=1, clear (||e_1||^2=1) m>1, ||a_1 e_1+...a_{m-1}e_{m-1}||^2=|a_1|^2+...+|a_{m-1}|^2 and <a_1 e_1+...+a_{m-1} e_{m-1},a_m e_m>=0 by Pythagorean Theorem. ||(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:

||a_1 e_1+...+a_m e_m||^2=0, then |a_1|^2+...+|a_m|^2=0, then a_1=...=a_m=0

Theorem 6.28

Every orthonormal list of length dim\ 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 e_1,...,e_m is an orthonormal list v\in V

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

Proof:

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

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

||w+u||^2=||w||^2+||u||^2\\
||v||^2\geq ||u||^2

Theorem 6.30

Suppose e_1,...,e_n is an orthonormal basis, and u,v\in V, then

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

Proof:

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

\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 v_1,...,v_m be a linearly independent list.

Define f_k\in V by f_1=v_1,f_k=v_k-\sum_{j=1}^{k-1}\frac{<v_k,f_j>}{||f_j||^2}f_j

Define e_k=\frac{f_k}{||f_k||}, then e_1,...,e_m is orthonormal Span(v_1,...,v_m)=Span(f_1,...,f_m)