$a = [ a_1, a_2, \ldots, a_n ]$

$b = [ b_1, b_2, \ldots, b_n ]$

$a \cdot b = \langle a, b \rangle = \sum_i a_i b_i$

$(x,y) \cdot e_1 = 1 x + 0 y = x$

$(x,y) \cdot e_2 = 0 x + 1 y = y$

This relies on $|e_1| = |e_2| = 1$.

• $c u \cdot v = c (u \cdot v )$
• $u \cdot v = v \cdot u$ (only for real numbers ... not complex)
• $u \neq 0 \implies u \cdot u > 0$
• $(u + v) \cdot w = u \cdot w + v \cdot w$

With complex numbers

For rotation $R$, let $u' = R u$ and $v' = R v$:

Note that the dot product $a \cdot b$ is the same as the matrix product $a^\mathsf{T} b$.

$\begin{array}[t]{rl} R u \cdot R v & = (R u)^\mathsf{T} (R v) \\ & = (u^\mathsf{T} R^\mathsf{T}) (R v) \\ & = u^\mathsf{T} (R^\mathsf{T} R) v \\ & = u^\mathsf{T} v \\ & = u \cdot v \end{array}$.

This works because $R$ is a rotation matrix and $R^\mathsf{T} = R^{-1}$.

1. $v \cdot \frac{u}{|u|}$ is the length of the projection of $v$ onto the line of $u$.
2. $(v \cdot \frac{u}{|u|}) \frac{u}{|u|}$ is the vector in the direction of $u$ with length equal to this projection.

   Note that $|u| |u| = u \cdot u$.

   Therefore, ${v \cdot u \over u \cdot u} u$ is the vector that is the projection of $v$ onto the line of $u$, and is the closest we can get to $v$ using only a multiple of $u$.

3. If $\theta$ is the angle between $u$ and $v$, then

   $\begin{array}{rl} \cos \theta & = {\text{adjacent} \over \text{hypotenuse}} \\ & = {v \cdot \frac{u}{|u|} \over |v|} \end{array}$
   

   So $u \cdot v = |u| |v| \cos\theta$.

We can write

$\begin{array}{rrrrrrrrrrrrr} f = [ & 0.8 & 2.1 & 1.4 & 1.0 & 1.5 & 1.9 & 1.6 & 1.6 & 2.0 & 1.7 & 0.9 & ] \\ g = [ & 0.0 & 1.0 & 1.5 & 1.0 & 0.0 & -1.0 & -1.5 & -1.0 & 0.0 & 1.0 & 1.5 & ] \end{array}$

Then the length of the projection of $f$ onto $g$ is $$f \cdot \frac{g}{|g|} = {f \cdot g \over |g|}$$

and the closest we can approximate $f$ with a multiple of $g$ is $$\begin{array}[t]{rl} (f \cdot \frac{g}{|g|}) {g \over |g|} & = ({f \cdot g \over |g|^2 }) g \\ & = ({f \cdot g \over g \cdot g}) g \end{array}$$

For a sampled function, we compute $${ f \cdot g \over g \cdot g } = { \sum_i f_i g_i \over \sum_i g_i^2 }$$

As the spacing between samples goes to zero, the sampled function becomes continuous and sums become integrals: $${ f \cdot g \over g \cdot g } = { \int f(x) g(x) dx \over \int g(x)^2 dx }$$