Geometric proof

We’ll be using this diagram of our vectors and , as well as the vector :

Using the law of cosines [2] on the triangle formed by the three vectors:

Since for any vector , we have (see Appendix), let’s rewrite this equation as:

But and the dot product obeys the distributive property (see Appendix). Therefore:

Projection proof

For this proof, we’ll assume the geometric definition is correct and will see how it leads to the component definition. We’ll begin by denoting vectors as the standard orthonormal basis for . For example, in 2D space, these basis vectors are and , shown in this diagram:

If we take an arbitrary and calculate its dot product with a basis vector, we can use the geometric definition:

where is the component of in the direction of . The diagram makes it easy to see why this is true from basic trigonometry, but in the more general case this is just a vector projection.

Now let’s represent vectors and as linear combinations of the basis vectors:

And calculate the dot product , beginning by rewriting with its linear combination of basis vectors representation:

Using the fact that the dot product distributes over linear combinations:

But earlier we’ve shown that . Therefore:

Which is the component definition .

Appendix A: Inner product space

A generalization of dot products in is the inner product, which is an operation meeting some specific requirements, defined on a vector space.

The inner product is denoted as , and must satisfy the following requirements for all vectors and scalars :

• Symmetry:
• Linearity in the first argument:
• Positive-definiteness: if then

For , we define the inner product operation in its component formulation as:

Let’s prove the requirements listed above for this operation; this is fairly straightforward, given the well-known properties of scalar multiplication and addition on :

Symmetry:

Linearity in the first argument:

Positive-definiteness:

Consider the components of vector . Clearly, . Since the vector is not the zero vector, at least one of its components is nonzero, and for that component . Therefore:

Now that we’ve proved all the inner product requirements on our operation , we can say that is an inner product space with this operation.

By meeting these requirements, it can be readily shown that our inner product operation has additional useful properties:

• if and only if

The third property is particularly helpful, because it means the inner product is bilinear, and thus is distributive over addition.

Note that these are shown for the component definition of dot product. It’s not too hard to prove distributivity for the geometric definition using the notion of projections and how they add up.

Norm

The norm of a vector in an inner product space is defined as . Therefore, the square of the norm is .

The norm is used to express the notion of magnitude, or length of a vector. If you think of a vector in Cartesian coordinates, the definition of the norm is a generalization of the Pythagorean theorem.

[1]	By this we mean , where each vector is an n-tuple of real numbers, with the usual mathematical operations making this a vector space.