# Dot Product with Self is Zero iff Zero Vector

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## 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:

\(\ds \mathbf u \cdot \mathbf u\) | \(=\) | \(\ds \mathbf 0 \cdot \mathbf 0\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds \norm {\mathbf 0}^2 \cos \angle \mathbf 0, \mathbf 0\) | ||||||||||||

\(\ds \) | \(=\) | \(\ds 0\) |

$\blacksquare$