Dot Product with Self is Non-Negative/Proof 2
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Theorem
Let $\mathbf u$ be a vector in the real Euclidean space $\R^n$.
Then:
- $\mathbf u \cdot \mathbf u \ge 0$
where $\cdot$ denotes the dot product operator.
Proof
\(\ds \mathbf u \cdot \mathbf u\) | \(=\) | \(\ds \norm {\mathbf u}^2\) | Dot Product of Vector with Itself | |||||||||||
\(\ds \) | \(\ge\) | \(\ds 0\) | Square of Real Number is Non-Negative |
$\blacksquare$