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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\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\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

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