# Dot Product with Self is Zero iff Zero Vector

## Theorem

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

Then:

$\mathbf u \cdot \mathbf u = 0 \iff \mathbf u = \mathbf 0$

## Proof 1

 $\ds \mathbf u \cdot \mathbf u$ $=$ $\ds 0$ $\ds \leadstoandfrom \ \$ $\ds \sum_{i \mathop = 1}^n u_i^2$ $=$ $\ds 0$ Definition of Dot Product $\ds \leadstoandfrom \ \$ $\ds \forall i: \,$ $\ds u_i$ $=$ $\ds 0$ $\ds \leadstoandfrom \ \$ $\ds \mathbf u$ $=$ $\ds \bszero$ Definition of Zero Vector

$\blacksquare$

## Proof 2

Let $\mathbf u \cdot \mathbf u = 0$.

Then:

 $\ds 0$ $=$ $\ds \norm {\mathbf u }^2 \cos \angle \mathbf u, \mathbf u$ Definition of Dot Product $\ds$ $=$ $\ds \norm {\mathbf u}^2 \cos 0$ $\ds$ $=$ $\ds \norm {\mathbf u}^2$

The only way for this to happen is if:

$\norm {\mathbf u} = 0$

which implies:

$\mathbf u = \mathbf 0$

Now suppose $\mathbf u = \mathbf 0$.

Then:

 $\ds \mathbf u \cdot \mathbf u$ $=$ $\ds \mathbf 0 \cdot \mathbf 0$ $\ds$ $=$ $\ds \norm {\mathbf 0}^2 \cos \angle \mathbf 0, \mathbf 0$ $\ds$ $=$ $\ds 0$

$\blacksquare$