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

Vectors and dot products

A vector is an ordered sequence of values:

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


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

Vector scaling

A vector can be scaled by a scalar \(c\):

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

Vector addition

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.


\[\vec{v} \cdot \vec{w} = \vec{w} \cdot \vec{v}\]

because \(v_i w_i = w_i v_i\).

Distributive over vector addition

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


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

Scalar multiplication

Say we have two scalars, \(c\) and \(d\):

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


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

From the properties of distribution over addition and scalar multiplication:

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

See: properties of dot products.

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.