Dot Product with Self is Zero iff Zero Vector/Proof 2
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Theorem
- $\mathbf u \cdot \mathbf u = 0 \iff \mathbf u = \mathbf 0$
Proof
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:
\(\ds \mathbf u \cdot \mathbf u\) | \(=\) | \(\ds \mathbf 0 \cdot \mathbf 0\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \norm {\mathbf 0}^2 \cos \angle \mathbf 0, \mathbf 0\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 0\) |
$\blacksquare$