Dot Product with Self is Non-Negative

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 1

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

\(=\)

\(\ds \sum_{i \mathop = 1}^n {u_i}^2\)

Definition of Dot Product

\(\ds \)

\(\ge\)

\(\ds 0\)

as $u_i \in \R$ it follows that ${u_i}^2 \ge 0$

$\blacksquare$

Proof 2

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

\(=\)

\(\ds \norm {\mathbf u}^2\)

Dot Product of Vector with Itself

\(\ds \)

\(\ge\)

\(\ds 0\)

Square of Real Number is Non-Negative

$\blacksquare$