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.