# Dot Product Operator is Commutative

## Theorem

Let $\mathbf u, \mathbf v$ be vectors in the real Euclidean space $\R^n$.

Then:

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

## Proof 1

 $\ds \mathbf u \cdot \mathbf v$ $=$ $\ds \sum_{i \mathop = 1}^n u_i v_i$ Definition of Dot Product $\ds$ $=$ $\ds \sum_{i \mathop = 1}^n v_i u_i$ Real Multiplication is Commutative $\ds$ $=$ $\ds \mathbf v \cdot \mathbf u$ Definition of Dot Product

$\blacksquare$

## Proof 2

 $\ds \mathbf u \cdot \mathbf v$ $=$ $\ds \norm {\mathbf u} \norm {\mathbf v} \cos \angle \mathbf u, \mathbf v$ Definition of Dot Product $\ds$ $=$ $\ds \norm {\mathbf v} \norm {\mathbf u} \cos \angle \mathbf u, \mathbf v$ Real Multiplication is Commutative $\ds$ $=$ $\ds \norm {\mathbf v} \norm {\mathbf u} \cos \angle \mathbf v, \mathbf u$ Cosine Function is Even $\ds$ $=$ $\ds \mathbf v \cdot \mathbf u$ Definition of Dot Product

$\blacksquare$

## Examples

### Example: $\paren {2 + 5 i} \circ \paren {3 - i} = \paren {3 - i} \circ \paren {2 + 5 i}$

#### Example: $\paren {2 + 5 i} \circ \paren {3 - i}$

Let:

$z_1 = 2 + 5 i$
$z_2 = 3 - i$

Then:

$z_1 \circ z_2 = 1$

where $\circ$ denotes (complex) dot product.

#### Example: $\paren {3 - i} \circ \paren {2 + 5 i}$

Let:

$z_1 = 3 - i$
$z_1 = 2 + 5 i$

Then:

$z_1 \circ z_2 = 1$

where $\circ$ denotes (complex) dot product.

As can be seen:

$\paren {2 + 5 i} \circ \paren {3 - i} = \paren {3 - i} \circ \paren {2 + 5 i}$

$\blacksquare$