38 - Norm

Norm / Magnitude / Length

Let $V$ be an inner product space
For $v\in V$ the norm of $v$ is

$$||v||:=\sqrt{⟨v,v⟩}$$

And then the distance between two vectors $v,w\in V$ is

$$d(v,w)=||v-w||$$

Properties of the distance function

For $u,v,w\in V$

1. $d(v,w)\ge 0$ and $d(v,w)=0⟺v=w$
2. $d(u,w)=d(w,u)$
3. $d(u,w)\le d(u,v)+d(v,w)$

Cauchy-Schwarz Inequality

For $v,w$ in an inner product space $V$ then

$$|⟨v,w⟩|\le ||v||||w||$$

Equality holds if $v$ and $w$ are linearly independent

Proof

If $w=0$ then the result is obvious so assume $w\ne 0$
For $t\in \mathbb{R}$ then

$$\begin{array}{rl}0& \le ||v+tw|{|}^2\\ & =⟨v+tw,v+tw⟩\\ & =⟨v,v⟩+2t⟨v,w⟩+t^2⟨w,w⟩\\ & =||v|{|}^2+2t⟨v,w⟩+t^2||w|{|}^2\end{array}$$

As $|w|\ne 0$ then it is a quadratic in $t$ which is non-negative so discriminant is non-negative
Hence

$$\text{discriminant}=4⟨u,w{⟩}^2-4||w|{|}^2||v|{|}^2\le 0$$

And then the Cauchy-Schwarz inequality follows
For equality then the discriminant has to be zero there is a repeated root $t=t_0$
Then $||v+t_{0}w||$ and hence $v+t_{0}w=0_V$
Therefore $v,w$ are linearly independent

Properties of Norm

Let $V$ be a inner product space
For $v,w\in V$ and $\alpha \in \mathbb{R}$

1. $||v||\ge 0$ and $||v||=0⟺v=0_V$
2. $||\alpha v||=|a|||v||$
3. $||v+w||\le ||v||+||w||$ (also known as the triangle inequality)

Proof - (3)

$$\begin{array}{rl}||v+w|{|}^2=⟨v+w,v+w⟩& \\ & =⟨v,v⟩+2⟨v,w⟩+⟨w,w⟩\\ & \le ||v|{|}^2+2||v||||w||+||w|{|}^2\\ & =(||v||+||w||{)}^2\end{array}$$

Hence the triangle inequality follows