Show that ||x|| is a norm on R^n

[Samuelb88]

Homework Statement

Show ||x|| = \sqrt{x \cdot x} is a norm on \mathbb{R}^n.

Homework Equations

Prop. 1. ||x|| = 0 IFF x=0.
2. \forall c \in \mathbb{R} ||cx|| = |c| \, ||x||.
3. ||x+y|| \leq ||x|| + ||y||.

Cauchy-Schwarz Inequality.

The Attempt at a Solution

Just want to check if I am showing this correctly. Note that I am using x in place of \vec{x} \in \mathbb{R}^n.

Let x \in \mathbb{R}^n. Suppose that ||x|| = \sqrt{x \cdot x}.

1. Suppose first that ||x||=0. Then 0 = \sqrt{x \cdot x} imples x \cdot x = 0. But x \cdot x = x_1^2 + \cdots + x_2^2 = 0 if and only if x_i =0 \forall x_i. Thus if ||x|| =0, then x=0. Suppose next that x=0. Then ||0|| = \sqrt{0^2 + \cdots + 0^2} = 0. Therefore if x=0, then ||x||=0.

2. Let c \in \mathbb{R}. Then ||cx|| = \sqrt{(cx_1)^2 + \cdots + (cx_n)^2} = c ||x||.

3. This follows from the Cauchy-Schwarz inequality which I have proven and will not show here.

 

[Charles49]

Everything looks correct.