# Dot Product of Vector with Itself

## Theorem

Let $\mathbf u$ be a vector in the real vector space $\R^n$.

Then:

$\mathbf u \cdot \mathbf u = \norm {\mathbf u}^2$

where $\norm {\mathbf u}$ is the length of $\mathbf u$.

## Proof 1

Let $\mathbf u = \tuple {u_1, u_2, \ldots, u_n}$.

Then:

 $\ds \mathbf u \cdot \mathbf u$ $=$ $\ds u_1 u_1 + u_2 u_2 + \cdots + u_n u_n$ Definition of Dot Product $\ds$ $=$ $\ds u_1^2 + u_2^2 + \cdots + u_n^2$ $\ds$ $=$ $\ds \paren {\sqrt {\sum_{i \mathop = 1}^n u_i^2} }^2$ $\ds$ $=$ $\ds \norm {\mathbf u}^2$ Definition of Vector Length in $\R^n$

$\blacksquare$

## Proof 2

 $\ds \mathbf u \cdot \mathbf u$ $=$ $\ds \norm {\mathbf u} \norm {\mathbf u} \cos \angle \mathbf u, \mathbf u$ Cosine Formula for Dot Product $\ds$ $=$ $\ds \norm {\mathbf u}^2 \cos 0$ since the angle between a vector and itself is $0$ $\ds$ $=$ $\ds \norm {\mathbf u}^2$ Cosine of Zero is One

$\blacksquare$

## Also presented as

This can also be seen presented as:

$\norm {\mathbf u} = \paren {\mathbf u \cdot \mathbf u}^{1/2}$

or:

$\norm {\mathbf u} = u = \sqrt {\mathbf u \cdot \mathbf u}$

and so on.