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:

\(\ds \mathbf u \cdot \mathbf u\)

\(=\)

\(\ds \mathbf 0 \cdot \mathbf 0\)

\(\ds \)

\(=\)

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

\(\ds \)

\(=\)

\(\ds 0\)

$\blacksquare$