Dot product: Component vs. Geometric definition



The goal of this post is to answer a simple question: why are the following two definitions of the vector dot product in Euclidean space [1] equivalent for vectors \vec{a} and \vec{b}:

  • Component definition: \vec{a}\cdot\vec{b}=\sum_{i=1}^{n}a_i b_i
  • Geometric definition: \vec{a}\cdot\vec{b}=|\vec{a}||\vec{b}|cos(\theta), where |\vec{a}| is the magnitude of \vec{a} and \theta is the angle between the vectors’ directions

Here’s a graphical depiction of our vectors (focusing on \mathbb{R}^2 for clarity, though this applies to any-dimensional vectors). It shows both the components of the vectors and the angle between them. The length of the arrow for \vec{a} is |\vec{a}|.

Vectors a and b in the cartesian plane

We’ll show two proofs of the equivalence here, the geometric proof and the projection proof. The Appendix describes some properties of dot products that facilitate these proofs.

Geometric proof

We’ll be using this diagram of our vectors \vec{a} and \vec{b}, as well as the vector \vec{c}=\vec{a}-\vec{b}:

Vectors a, b and c in the cartesian plane

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

\[|\vec{c}|^2=|\vec{a}|^2+|\vec{b}|^2-2|\vec{a}||\vec{b}|cos(\theta)\]

Since for any vector \vec{a}, we have \vec{a}\cdot\vec{a}=|\vec{a}|^2 (see Appendix), let’s rewrite this equation as:

\[\vec{c}\cdot\vec{c}=\vec{a}\cdot\vec{a}+\vec{b}\cdot\vec{b}-2|\vec{a}||\vec{b}|cos(\theta)\]

But \vec{c}=\vec{a}-\vec{b} and the dot product obeys the distributive property (see Appendix). Therefore:

\[\begin{aligned} (\vec{a}-\vec{b})\cdot(\vec{a}-\vec{b})&=\vec{a}\cdot\vec{a}+\vec{b}\cdot\vec{b}-2|\vec{a}||\vec{b}|cos(\theta)\\ \vec{a}\cdot\vec{a}-2\vec{a}\cdot\vec{b}+\vec{b}\cdot\vec{b}&=\vec{a}\cdot\vec{a}+\vec{b}\cdot\vec{b}-2|\vec{a}||\vec{b}|cos(\theta)\\ -2\vec{a}\cdot\vec{b}&=-2|\vec{a}||\vec{b}|cos(\theta)\\ \vec{a}\cdot\vec{b}&=|\vec{a}||\vec{b}|cos(\theta) \end{aligned}\]

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 \vec{e}_1,\vec{e}_2\dots\vec{e}_n as the standard orthonormal basis for \mathbb{R}^n. For example, in 2D space, these basis vectors are \vec{e}_1=[1\ 0] and \vec{e}_2=[0\ 1], shown in this diagram:

Projection of vector a on basis vectors

If we take an arbitrary \vec{a}\in\mathbb{R}^n and calculate its dot product with a basis vector, we can use the geometric definition:

\[\vec{a}\cdot\vec{e}_i=|\vec{a}||\vec{e}_i|cos(\theta_i)=|\vec{a}|cos(\theta_i)=a_i\]

where a_i is the component of \vec{a} in the direction of \vec{e}_i. 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 \vec{a} and \vec{b} as linear combinations of the basis vectors:

\[\begin{aligned} \vec{a}&=\sum_{i=1}^{n}a_i\vec{e}_i\\ \vec{b}&=\sum_{i=1}^{n}b_i\vec{e}_i\\ \end{aligned}\]

And calculate the dot product \vec{a}\cdot\vec{b}, beginning by rewriting \vec{b} with its linear combination of basis vectors representation:

\[\vec{a}\cdot\vec{b}=\vec{a}\cdot\sum_{i=1}^{n}b_i\vec{e}_i\]

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

\[\vec{a}\cdot\vec{b}=\sum_{i=1}^{n}b_i(\vec{a}\cdot\vec{e}_i)\]

But earlier we’ve shown that \vec{a}\cdot\vec{e}_i=a_i. Therefore:

\[\vec{a}\cdot\vec{b}=\sum_{i=1}^{n}b_i a_i=\sum_{i=1}^{n}a_i b_i\]

Which is the component definition \blacksquare.

Appendix A: Inner product space

A generalization of dot products in \mathbb{R}^n is the inner product, which is an operation meeting some specific requirements, defined on a vector space.

The inner product is denoted as \langle x,y\rangle:\mathbb{R}^n\times\mathbb{R}^n\to\mathbb{R}, and must satisfy the following requirements for all vectors x,y,z\in\mathbb{R}^n and scalars a,b\in\mathbb{R}:

  • Symmetry: \langle x,y\rangle=\langle y,x\rangle
  • Linearity in the first argument: \langle ax+by,z\rangle=a\langle x,z\rangle+b\langle y,z\rangle
  • Positive-definiteness: if x\ne 0 then \langle x,x\rangle>0

For \mathbb{R}^n, we define the inner product operation in its component formulation as:

\[\langle x,y\rangle=\sum_{i=1}^{n}x_i\cdot y_i\]

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 \mathbb{R}:

Symmetry:

\[\langle x,y\rangle=\sum_{i=1}^{n}x_i\cdot y_i=\sum_{i=1}^{n}y_i\cdot x_i=\langle y,x\rangle\]

Linearity in the first argument:

\[\begin{aligned} \langle ax+by,z\rangle&=\sum_{i=1}^{n}(ax+by)_i\cdot z_i\\ &=\sum_{i=1}^{n}a x_i\cdot z_i+b y_i\cdot z_i\\ &=a\sum_{i=1}^{n}x_i\cdot z_i+b\sum_{i=1}^{n}y_i\cdot z_i=a\langle x,z\rangle+b\langle y,z\rangle \end{aligned}\]

Positive-definiteness:

Consider the components x_i of vector x. Clearly, \forall i\quad x_i\cdot x_i=x_i^2\ge 0. Since the vector x is not the zero vector, at least one of its components x_i is nonzero, and for that component x_i\cdot x_i>0. Therefore:

\[\langle x,x\rangle=\sum_{i=1}^{n}x_i\cdot x_i>0\]

Now that we’ve proved all the inner product requirements on our operation \langle x,y\rangle, we can say that \mathbb{R}^n 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:

  • \langle x,0\rangle=\langle 0,x\rangle=0
  • \langle x,x\rangle=0 if and only if x=0
  • \langle x,ay+bz\rangle=a\langle x,y\rangle+b\langle x,z\rangle
  • \langle x+y,x+y\rangle=\langle x,x\rangle+2\langle x,y\rangle+\langle y,y\rangle

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 x in an inner product space is defined as |x|=\sqrt{\langle x,x\rangle}. Therefore, the square of the norm is |x|^2=\langle x,x\rangle.

The norm is used to express the notion of magnitude, or length of a vector. If you think of a vector x\in\mathbb{R}^n in Cartesian coordinates, the definition of the norm is a generalization of the Pythagorean theorem.

[1]By this we mean \mathbb{R}^n, where each vector is an n-tuple of real numbers, with the usual mathematical operations making this a vector space.
[2]Which is a very fundamental theorem in geometry; it is (or rather, its non-trigonometric version) is proven from the basic Euclidean axioms in The Elements.

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