Dot product

The dot product is the most common way to define an inner product between elements of $\R^n$ ($n$-dimensional vectors).

Definition
Let $\mathbf x=(x_1,\ldots,x_n)\in\R^n$ and $\mathbf y=(y_1,\ldots,y_n)\in\R^n$ . We define the dot product $\mathbf x\cdot\mathbf y$ between $\mathbf{x}$ and $\mathbf{y}$ by $\mathbf x\cdot\mathbf y=\sum_{i=1}^n x_iy_i=x_1y_1+\cdots+x_ny_n$

Note that some texts use the symbol $\langle\mathbf x,\mathbf y\rangle$ to denote the dot product between $\mathbf{x}$ and $\mathbf{y}$ , preserving the inner-product notation.

The dot product is one of three common types of multiplication compatible with vectors; the other being the cross product and scalar multiplication, the latter belonging to the vector space nature of $\R^n$ .

$\R^n$ as an inner-product space

We will now prove that the dot product $\cdot:\R^n\times\R^n\to\R$ turns $\R^n$ into an inner-product space. There are four statements to prove, namely, given any $\mathbf x,\mathbf y,\mathbf z\in\R^n$ and any scalar $\alpha \in \R$ , the following is true:

1. $\mathbf x\cdot\mathbf y=\mathbf y\cdot\mathbf x$
2. $(\alpha\mathbf x)\cdot\mathbf y=\alpha(\mathbf x\cdot\mathbf y)$
3. $(\mathbf x+\mathbf y)\cdot\mathbf z=\mathbf x\cdot\mathbf z+\mathbf y\cdot\mathbf z$
4. $\mathbf x\cdot\mathbf x\ge0$ with equality if and only if $\mathbf x=\mathbf 0$

Proof.

1. $\mathbf x\cdot\mathbf y=\sum_{i=1}^n x_iy_i=\sum_{i=1}^n y_ix_i=\mathbf y\cdot\mathbf x$
2. $(\alpha\mathbf x)\cdot\mathbf y=\sum_{i=1}^n(\alpha x_i)y_i=\alpha\sum_{i=1}^n x_iy_i=\alpha(\mathbf x\cdot\mathbf y)$ , where we in the second step factored out the $\alpha$
3. $(\mathbf x+\mathbf y)\cdot\mathbf z=\sum_{i=1}^n(x_i+y_i)z_i=\sum_{i=1}^n(x_iz_i+y_iz_i)=\sum_{i=1}^n x_iz_i+\sum_{i=1}^n y_iz_i=\mathbf x\cdot\mathbf z+\mathbf y\cdot\mathbf z$
4. $\mathbf x\cdot\mathbf x=\sum_{i=1}^n x_i^2=x_1^2+\cdots+x_n^2$ . But each $x_i^2\ge0$ ($i=1,\ldots,n$), so $\mathbf x\cdot\mathbf x\ge0$ , as required. Now suppose that $\mathbf x=\mathbf 0$ . Then clearly $\mathbf x\cdot\mathbf x=\sum_{i=1}^n 0^2=0$ . If $\mathbf x\cdot\mathbf x=0$ , then $x_1^2+\cdots+x_n^2=0$ . Suppose for the sake of contradiction that some $x_i\ne0$ . Then $x_i^2>0$ so that $\mathbf x\cdot\mathbf x\ne0$ . But this is a contradiction, so we must have $\mathbf x=\mathbf 0$

This completes the proof.

Euclidean norm and $\mathbf R^n$ as a metric space

Once we have defined the dot product between elements of Euclidean $n$-space, we may define a map $\|\cdot\|:\R^n\to\R$ , when applied to $\mathbf x \in \mathbb R^n$ is called the norm of $\mathbf{x}$ .

Definition
If $\mathbf x=(x_1,\ldots,x_n)\in\R^n$ , we define the norm of $\mathbf{x}$ , denoted by $\|\mathbf x\|$ , by $\|\mathbf x\|=\sqrt{\mathbf x\cdot\mathbf x}=\sqrt{\sum_{i=1}^n x_i^2}$

One can show that if $\mathbf x \in \mathbb R^n$ and $\mathbf y\in\R^n$ , then $\|\mathbf y-\mathbf x\|$ is a valid distance between $\mathbf{x}$ and $\mathbf{y}$ , and hence turns $\R^n$ into a metric space. In fact, this metric space is complete, meaning that every Cauchy sequence of elements in $\R^n$ converges to some point in $\R^n$ .

Angles between two elements

The dot product can be used to determine the angle between two elements: $\cos(\theta)=\frac{V\cdot W}{|V||W|}$

Orthogonality

Two elements in an inner-product space are said to be orthogonal if and only if their inner-product is 0. In $\R^n$ this translates to: $\mathbf{x}$ and $\mathbf{y}$ in $\R^n$ are orthogonal if and only if $\mathbf{x}\cdot\mathbf{y} = 0$ . Note that the zero vector is orthogonal to every vector.

See also

• Cross product
• Vector space
• Metric space
• Cauchy sequence