Properties of Dot Product

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

Let $c$ be a real scalar.

The dot product has the following properties:


 * 1) $\vec u \cdot \vec u \ge 0$.
 * 2) $\vec u \cdot \vec u = 0 \iff \vec u = \vec 0$.
 * 3) $\vec u \cdot \vec v = \vec v \cdot \vec u$.
 * 4) $(\vec u + \vec v) \cdot \vec w = \vec u \cdot \vec w + \vec v \cdot \vec w$.
 * 5) $\left({ c \vec u }\right) \cdot \vec v = c (\vec u \cdot \vec v)$.

Proofs
From the definition of dot product


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

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

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

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

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

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

Alternate 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
 * $\displaystyle \vec a \cdot \vec b = \left\Vert{ \vec a }\right\Vert \left\Vert{ \vec b }\right\Vert \cos \angle \vec a, \vec b$

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

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

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