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

Note that some texts use the symbol ${\displaystyle \langle\mathbf x,\mathbf y\rangle}$ to denote the dot product between ${\displaystyle \mathbf{x}}$ and ${\displaystyle \mathbf{y}}$ , 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 ${\displaystyle \R^n}$ .

${\displaystyle \R^n}$ as an inner-product space

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

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

Proof.

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

This completes the proof.

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

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

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

Angles between two elements

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

Orthogonality

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

See also

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