The Norm and Dot Product

Norm §

The norm (aka length or magnitude) of $\overrightarrow{x}$ = $\left[\begin{array}{c}{x}_1\\ x_2\\ ...\\ x_k\end{array}\right]$ |$\vec{x}$| = $\sqrt{x_1^2+\cdots +x_k^2}$

$\vec{x}$ = $\left[\begin{array}{c}1\\ 2\end{array}\right]$ $\in$ ${\mathbb{R}}^2$ then | $\vec{x}$ | = $\sqrt{1^2+2^2}$ = $\sqrt{5}$

Properties of the Norm §

Let $\vec{x},\vec{y}\in {\mathbb{R}}^n$ and c $\in \mathbb{R}$. Then

1. \| $\vec{x}$ | $\ge$ 0 with equality if and only if $\vec{x}=\vec{0}$
2. \| c $\vec{x}$ | = | c | | $\vec{x}$ |
3. \| $\vec{x}+\vec{y}$ | $\le$ | $\vec{x}$ | + | $\vec{y}$ | (The Triangle Inequality)

2 nonzero vectors in ${\mathbb{R}}^n$ are parallel if they are scalar multiples of one other

A vector $\vec{x}$ is a unit vector if | $\vec{x}$ | = 1

$\hat{x}=\frac{1}{\Vert \vec{x}\Vert }\vec{x}$

is called the normalization of $\vec{x}$

Dot Product §

Let $\vec{x}$ = $\left[\begin{array}{c}{x}_1\\ ...\\ x_n\end{array}\right]$ and $\vec{y}$ = $\left[\begin{array}{c}{y}_1\\ ...\\ y_n\end{array}\right]$ be vectors in ${\mathbb{R}}^n$. The dot product of $\vec{x}$ and $\vec{y}$ is the real number

$\vec{x}\cdot \vec{y}$ = $x_1{y}_1+\cdots +x_n{y}_n$

Properties of Dot Product §

Let $\vec{w},\vec{x},\vec{y}\in {\mathbb{R}}^n$ and c $\in \mathbb{R}$

1. $\vec{x}\cdot \vec{y}\in \mathbb{R}$
2. $\vec{x}\cdot \vec{y}$ = $\vec{y}\cdot \vec{x}$
3. $\vec{x}\cdot \vec{0}$ = 0
4. $\vec{x}\cdot \vec{x}$ = | $\vec{x}$ |${}^2$
5. $(c\vec{x})\cdot \vec{y}=c(\vec{x}\cdot \vec{y})=\vec{x}\cdot (c\vec{y})$
6. $\vec{w}\cdot (\vec{x}\pm \vec{y})=\vec{w}\cdot \vec{x}\pm \vec{w}\cdot \vec{y}$

Cauchy–Schwarz Inequalty §

This is how the theorem is spelled in the textbook

For any two vectors $\vec{x},\vec{y}\in {\mathbb{R}}^n$ we have

| $\vec{x}\cdot \vec{y}\Vert \le \Vert \vec{x}\Vert \Vert \vec{y}\Vert$

Two vectors are said to be orthogonal if $\vec{x}\cdot \vec{y}=\vec{0}$

$\theta$ = arccos ($\frac{\vec{x}\cdot \vec{y}}{\Vert \vec{x}\Vert \Vert \vec{y}\Vert })$

• $\vec{x}\cdot \vec{y}>0⟺0\le \theta <\frac{\pi }{2}⟺\vec{x}$ and $\vec{y}$ determine an acute angle
• $\vec{x}\cdot \vec{y}=\theta ⟺0=\frac{\pi }{2}⟺\vec{x}$ and $\vec{y}$ are perpendicular
• $\vec{x}\cdot \vec{y}<0⟺\frac{\pi }{2}<\theta \le \pi ⟺\vec{x}$ and $\vec{y}$ determine an obtuse angle