Lecture 25

Chapter VI Inner Product Spaces

Inner Products and Norms 6A

Dot Product (Euclidean Inner Product)

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:

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: