Properties of Dot Product

Theorem
Let $\mathbf u, \mathbf v, \mathbf w$ be vectors in the vector space $\R^n$.

Let $c$ be a real scalar.

The dot product has the following properties:

Dot Product with Self is Non-Negative

 * $\mathbf u \cdot \mathbf u \ge 0$

Dot Product with Self is Zero iff Zero Vector

 * $\mathbf u \cdot \mathbf u = 0 \iff \mathbf u = \mathbf 0$

Dot Product Operator is Commutative

 * $\mathbf u \cdot \mathbf v = \mathbf v \cdot \mathbf u$

Dot Product Operator is Bilinear

 * $(\mathbf u + \mathbf v) \cdot \mathbf w = \mathbf u \cdot \mathbf w + \mathbf v \cdot \mathbf w$


 * $\left({ c \mathbf u }\right) \cdot \mathbf v = c (\mathbf u \cdot \mathbf v)$

Proofs
From the definition of dot product


 * $\displaystyle \mathbf a \cdot \mathbf b = \sum_{i \mathop = 1}^n a_i b_i$

Proof of (1)
$\displaystyle \mathbf u \cdot \mathbf u = \sum_{i \mathop = 1}^n u_i^2 \ge 0$.

Proof of (2)
Let $\mathbf u \cdot \mathbf u = 0$.

Then $\displaystyle \sum_{i \mathop = 1}^n u_i^2 = 0$ and so $\forall i: u_i = 0$.

Now suppose $\mathbf u = \mathbf 0$.

Then $\displaystyle \sum_{i \mathop = 1}^n u_i^2 = 0$ and so $\mathbf u \cdot \mathbf u = 0$.

Alternative Proofs
Because these properties are used to demonstrate the equivalence of the definitions of the dot product, it is necessary to derive them for both definitions.

From our Alternative Definition of Dot Product
 * $\displaystyle \mathbf a \cdot \mathbf b = \left\Vert{ \mathbf a }\right\Vert \left\Vert{ \mathbf b }\right\Vert \cos \angle \mathbf a, \mathbf b$

Alternative Proof of (2)
Let $u\cdot u = 0$.

The only way for this to happen is if $\left\Vert{ \mathbf u }\right\Vert = 0$, which implies $\mathbf u = \mathbf 0$.

Now suppose $\mathbf u = \mathbf 0$.