- #1

Samuelb88

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## Homework Statement

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

## Homework Equations

**Prop.**1. [itex]||x|| = 0[/itex] IFF [itex]x=0[/itex].

2. [itex]\forall c \in \mathbb{R}[/itex] [itex]||cx|| = |c| \, ||x||[/itex].

3. [itex]||x+y|| \leq ||x|| + ||y||[/itex].

Cauchy-Schwarz Inequality.

## The Attempt at a Solution

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

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

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

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

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