## Dot product

Given two tuples $$u,v\in\mathbb{R}^n$$, the dot product of $$u$$ and $$v$$, denoted by $$u\cdot v$$, is defined to be the quantity $u_1v_1 + u_2v_2 + \cdots + u_nv_n.$

For example, if $$u = \begin{bmatrix} 1 \\2 \\3\end{bmatrix}$$ and $$v = \begin{bmatrix} 4 \\ 5 \\ 6\end{bmatrix}$$, then $u\cdot v = (1)(4)+ (2)(5) + (3)(6) = 4 + 10 + 18 = 32.$

(Notice that the dot notation is overloaded and is used for multiplication in a field as well as the dot product of two tuples. As confusing as it might seem, the context will make it clear which is intended.)

## Orthogonality

We say that $$u, v\in \mathbb{R}^n$$ are orthogonal if $$u\cdot v = 0$$.

When $$n = 2$$ or $$n = 3$$, if neither $$u$$ nor $$v$$ is the zero vector, then $$u \cdot v = 0$$ if and only if the arrows representing $$u$$ and $$v$$ form a right angle. In fact, a bit more is known.

Let $$\theta$$ denote the value of the smaller angle between the arrows representing $$u$$ and $$v$$. (We often refer to this angle simply as the angle between $$u$$ and $$v$$.) Then $\cos\theta = \displaystyle\frac{u\cdot v}{\sqrt{u \cdot u}\sqrt{v \cdot v}}.$ Clearly, when $$u \cdot v = 0$$, $$\theta = \frac{\pi}{2}$$ radians, the measure of a right angle.

One often writes $$\|u\|$$ for $$\sqrt{u\cdot u}$$. When $$n = 2$$ or $$n = 3$$, one can see from the Pythagorean Theorem that $$\|u\|$$ gives the length of the arrow representing $$u$$. For general $$n$$, $$\|u\|$$ is called the norm of $$\|u\|$$ and is the quantity $$\sqrt{u_1^2 + u_2^2 + \cdots + u_n^2}$$.

A vector having norm $$1$$ is called a unit vector.

Remark: One can in fact show that for all integers $$n \geq 1$$ and $$u,v\in\mathbb{R}^n$$, $|u\cdot v| \leq \|u\|\|v\|$ This inequality is known as the Cauchy-Schwarz Inequality. Hence, $$\displaystyle\frac{|u\cdot v|}{\|u\|\|v\|}\leq 1$$. With this, one can define the cosine of the angle between two vectors $$u,v\in\mathbb{R}^n$$ as $$\displaystyle\frac{u\cdot v}{\|u\|\|v\|}$$ even for $$n \geq 4$$. Clearly, one cannot visualize an angle beyond three dimensions. However, the terminology is retained as an analogue for higher dimensions.

### Examples

1. Let $$u = \begin{bmatrix} 2 \\ 0\end{bmatrix}$$ and $$v = \begin{bmatrix} 0 \\ -1 \end{bmatrix}$$. Clearly, the angle between $$u$$ and $$v$$, call it $$\theta$$, is $$\frac{\pi}{2}$$ radians. (Recall that we take the smaller angle.) Thus $$\cos \theta = 0.$$ One can also obtain this by applying the formula above: $\cos\theta = \frac{u\cdot v}{\|u\|\|v\|} = \frac{ (2)(0) + (0)(-1) } {\|u\| \|v\|} = 0.$

2. Let $$u = \begin{bmatrix} 3 \\ 3\end{bmatrix}$$ and $$v = \begin{bmatrix} 0 \\ 2 \end{bmatrix}$$. Clearly, the angle between $$u$$ and $$v$$, call it $$\theta$$, is $$\frac{\pi}{4}$$ radians. Thus $$\cos \theta = 1/\sqrt{2}.$$ One can also obtain this by applying the formula above: $\cos\theta = \frac{u\cdot v}{\|u\| \|v\|} = \frac{ (3)(0) + (3)(2) } {\sqrt{3^2+3^2}\sqrt{0^2+2^2}} = \frac{ 6 } {\sqrt{18}\sqrt{4}} = \frac{ 6 } {6\sqrt{2}} = \frac{ 1 } {\sqrt{2}}.$

## A bit of notation

If $$u,v\in\mathbb{R}^n$$, then $$u^\mathsf{T} v$$ technically is a $$1\times 1$$ matrix whose entry is $$u\cdot v$$. However, one often for convenience regards a $$1\times 1$$ matrix as the entry that it contains. In other words, one often writes $$u^\mathsf{T} v$$ and $$u \cdot v$$ interchangeably. The context usually makes it clear if $$u^\mathsf{T}v$$ refers to a number or a matrix.

## Exercise

For each of the following, find the value of $$u\cdot v$$. Provide a sketch in cases when the dimension is at most $$3$$.

1. $$u = \begin{bmatrix} 2 \\ 3\end{bmatrix}$$ and $$v = \begin{bmatrix} 4 \\ -1\end{bmatrix}$$.

2. $$u = \begin{bmatrix} 1 \\ 1 \\ -1 \end{bmatrix}$$ and $$v = \begin{bmatrix} 1 \\ -1 \\ 1 \end{bmatrix}$$.

3. $$u = \begin{bmatrix} 1 \\ 0 \\ 0 \\ 1\end{bmatrix}$$ and $$v = \begin{bmatrix} 0 \\ 1 \\ 1 \\ 0\end{bmatrix}$$.