$$\newcommand{L}{\| #1 \|}\newcommand{VL}{\L{ \vec{#1} }}\newcommand{R}{\operatorname{Re}\,(#1)}\newcommand{I}{\operatorname{Im}\, (#1)}$$

# Vectors and dot products¶

A vector is an ordered sequence of values:

$\begin{split}\vec{v} = [ v_1, v_2, \cdots v_n ] \\\end{split}$

## Videos¶

See these Khan academy videos for nice introductions to vector dot products:

## Vector scaling¶

A vector can be scaled by a scalar $$c$$:

$c \vec{v} \triangleq [ c v_1, c v_2, \cdots c v_n ]$

Say we have two vectors containing $$n$$ values:

$\begin{split}\vec{v} = [ v_1, v_2, \cdots v_n ] \\ \vec{w} = [ w_1, w_2, \cdots w_n ]\end{split}$

Vector addition gives a new vector with $$n$$ values:

$\vec{v} + \vec{w} \triangleq [ v_1 + w_1, v_2 + w_2, \cdots v_n + w_n ]$

Vector addition is commutative because $$v_i + w_i = w_i + v_i$$:

$\vec{v} + \vec{w} = \vec{w} + \vec{v}$

## Vector dot product¶

The vector dot product is:

$\vec{v} \cdot \vec{w} \triangleq \Sigma_{i=1}^n v_i w_i$

## Vector length¶

We write the length of a vector $$\vec{v}$$ as $$\VL{v}$$:

$\VL{v} \triangleq \sqrt{ \Sigma v_i^2 }$

This is a generalization of Pythagoras’ theorem to $$n$$ dimensions. For example, the length of a two dimensional vector $$[ x, y ]$$ is the length of the hypotenuse of the right-angle triangle formed by the points $$(x, 0), (0, y), (x, y)$$. This length is $$\sqrt{x^2 + y^2}$$. For a point in three dimensions $${x, y, z}$$, consider the right-angle triangle formed by $$(x, y, 0), (0, 0, z), (x, y, z)$$. The hypotenuse is length $$\sqrt{\L{ [ x, y ] }^2 + z^2} = \sqrt{ x^2 + y^2 + z^2 }$$.

From the definition of vector length and the dot product, the square root of the dot product of the vector with itself gives the vector length:

$\VL{v} = \sqrt{ \vec{v} \cdot \vec{v} }$

## Properties of dot products¶

We will use the results from Some algebra with summation.

### Commutative¶

$\vec{v} \cdot \vec{w} = \vec{w} \cdot \vec{v}$

because $$v_i w_i = w_i v_i$$.

$\vec{v} \cdot (\vec{w} + \vec{x}) = \vec{v} \cdot \vec{w} + \vec{v} \cdot \vec{x}$

because:

$\begin{split}\vec{v} \cdot (\vec{w} + \vec{x}) = \\ \Sigma{ v_i ( w_i + x_i) } = \\ \Sigma{ (v_i + w_i) } + \Sigma{ (v_i + x_i) } = \\ \vec{v} \cdot \vec{w} + \vec{v} \cdot \vec{x}\end{split}$

### Scalar multiplication¶

Say we have two scalars, $$c$$ and $$d$$:

$(c \vec{v}) \cdot (d \vec{w}) = c d ( \vec{v} \cdot \vec{w} )$

because:

$\begin{split}(c \vec{v}) \cdot (d \vec{w}) = \\ \Sigma{ c v_i d w_i } = \\ c d \Sigma{ v_i w_i }\end{split}$

From the properties of distribution over addition and scalar multiplication:

$\vec{v} \cdot (c \vec{w} + \vec{x}) = c (\vec{v} \cdot \vec{w}) + (\vec{v} \cdot \vec{x})$

## Unit vector¶

A unit vector is any vector with length 1.

To make a corresponding unit vector from any vector $$\vec{v}$$, divide by $$\VL{v}$$:

$\vec{u} = \frac{1}{ \VL{v} } \vec{v}$

Let $$g \triangleq \frac{1}{\VL{v}}$$. Then:

$\begin{split}\L{ g \vec{v} }^2 = \\ ( g \vec{v} ) \cdot ( g \vec{v} ) = \\ g^2 \VL{v}^2 = 1\end{split}$

## If two vectors are perpendicular, their dot product is 0¶

I based this proof on that in Gilbert Strang’s “Introduction to Linear Algebra” 4th edition, page 14.

Consider the triangle formed by the two vectors $$\vec{v}$$ and $$\vec{w}$$. The lengths of the sides of the triangle are $$\VL{v}, \VL{w}, \L{\vec{v} - \vec{w}}$$. When $$\vec{v}$$ and $$\vec{w}$$ are perpendicular, this is a right-angled triangle with hypotenuse length $$\L{\vec{v} - \vec{w}}$$. In this situation, by Pythagoras:

$\VL{v}^2 + \VL{w}^2 = \L{\vec{v} - \vec{w}}^2$

Write the left hand side as:

$\VL{v}^2 + \VL{w}^2 = v_1^2 + v_2^2 + \cdots v_n^2 + w_1^2 + w_2^2 + \cdots w_n^2$

Write the right hand side as:

$\L{\vec{v} - \vec{w}}^2 = (v_1^2 - 2v_1 w_1 + w_1^2) + (v_2^2 - 2v_2 w_2 + w_2^2) + \cdots (v_n^2 - 2v_n w_1 + w_n^2)$

The $$v_i^2$$ and $$w_i^2$$ terms on left and right cancel, so:

$\begin{split}\VL{v}^2 + \VL{w}^2 = \L{\vec{v} - \vec{w}}^2 \implies \\ 0 = 2(v_1 w_1 + v_2 w_2 + \cdots v_n w_n) \implies \\ 0 = \vec{v} \cdot \vec{w}\end{split}$

By the converse of Pythagoras’ theorem, if $$\VL{v}^2 + \VL{w}^2 \ne \L{\vec{v} - \vec{w}}^2$$ then vectors $$\vec{v}$$ and $$\vec{w}$$ do not form a right angle and are not perpendicular.