NoteNextra-origin/content/Math429/Math429_L25.md

Lecture 25

Chapter VI Inner Product Spaces

Inner Products and Norms 6A

Dot Product (Euclidean Inner Product)

v\cdot w=v_1w_1+...+v_n w_n

-\cdot -:\mathbb{R}^n\times \mathbb{R}^n\to \mathbb{R}

Some properties

• v\cdot v\geq 0
• v\cdot v=0\iff v=0
• (u+v)\cdot w=u\cdot w+v\cdot w
• (c\cdot v)\cdot w=c\cdot(v\cdot w)

Definition 6.2

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: \langle u+v,w\rangle=\langle u,w\rangle+\langle v,w\rangle

Homogeneity: \langle \lambda u, v\rangle=\lambda\langle u,v\rangle

Conjugate symmetry: \langle u,v\rangle=\overline{\langle v,u\rangle}

Note: the dot product on \mathbb{R}^n satisfies these properties

Example:

V=C^0([-1,-])

L_2 - inner product.

\langle f,g\rangle=\int^1_{-1} f\cdot g

\langle f,f\rangle=\int ^1_{-1}f^2\geq 0

\langle f+g,h\rangle=\langle f,h\rangle+\langle g,h\rangle

\langle \lambda f,g\rangle=\lambda\langle f,g\rangle

\langle f,g\rangle=\int^1_{-1} f\cdot g=\int^1_{-1} g\cdot f=\langle g,f\rangle

The result is in real vector space so no conjugate...

Theorem 6.6

For \langle,\rangle an inner product

(a) Fix V, then the map given by u\mapsto \langle u,v\rangle is a linear map (Warning: if \mathbb{F}=\mathbb{C}, then u\mapsto\langle u,v\rangle is not linear).

(b,c) \langle 0,v\rangle=\langle v,0\rangle=0

(d) \langle u,v+w\rangle=\langle u,v\rangle+\langle u,w\rangle (second terms are additive.)

(e) \langle u,\lambda v\rangle=\bar{\lambda}\langle u,v\rangle

Definition 6.4

An inner product space is a pair of vector space and inner product on it. (v,\langle,\rangle). In practice, we will say "V is an inner product space" and treat V as the vector space.

For the remainder of the chapter. V,W are inner product vector spaces...

Definition 6.7

For v\in V the norm of $V$ is given by ||v||:=\sqrt{\langle v,v\rangle}

Theorem 6.9

Suppose v\in V.

(a) ||v||=0\iff v=0
(b) ||\lambda v||=|\lambda|\ ||v||

Proof:

||\lambda v||^2=\langle \lambda v,\lambda v\rangle =\lambda\langle v,\lambda v\rangle=\lambda\bar{\lambda}\langle v,v\rangle

So |\lambda|^2 \langle v,v\rangle=|\lambda|^2||v||^2, ||\lambda v||=|\lambda|\ ||v||

Definition 6.10

v,u\in V are orthogonal if \langle v,u\rangle=0.

Theorem 6.12 (Pythagorean Theorem)

If u,v\in V are orthogonal, then ||u+v||^2=||u||^2+||v||

Proof:

\begin{aligned}
    ||u+v||^2&=\langle u+v,u+v\rangle\\
    &=\langle u,u+v\rangle+\langle v,u+v\rangle\\
    &=\langle u,u\rangle+\langle u,v\rangle+\langle v,u\rangle+\langle v,v\rangle\\
    &=||u||^2+||v||^2
\end{aligned}

Theorem 6.13

Suppose u,v\in V, v\neq 0, set c=\frac{<u,v>}{||v||^2}, then let w=u-v\cdot v, then v and w are orthogonal.

Theorem 6.14 (Cauchy-Schwarz)

Let u,v\in V, then |<u,v>|\leq ||u||\ ||v|| where equality occurs only u,v are parallel...

Proof:

Take the square norm of u=\frac{<u,v>}{||u||^2}v+w.

Theorem 6.17 Triangle Inequality

If u,v\in V, then ||u+v||\leq ||u||+||v||

Proof:

\begin{aligned}
    ||u+v||^2&=<u+v,u+v>\\
    &=<u,u>+<u,v>+<v,u>+<v,v>\\
    &=||u||^2+||v||^2+2Re(<u,v>)\\
    &\leq ||u||^2+||v||^2+2|<u,v>|\\
    &\leq  ||u||^2+||v||^2+2||u||\ ||v||\\
    &\leq (||u||+||v ||)^2
\end{aligned}