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Dot Product: Equivalence of Definitions

We have operated with the dot product as an operation with two different definitions. In this section, we will show, in multiple different ways, that these definitions are equivalent.

Table of Contents​

The Definitions to Equate​

The dot product of two vectors and can be defined in two ways, as we have previously seen:

We will show that these two definitions are equivalent. That is:

By the Cosine Rule​

Brief Overview

This proof assumes the component definition of the dot product () and uses the cosine rule to derive the geometric definition of the dot product ().

Additionally, it uses the corollary that the square of the magnitude of a vector is the dot product of the vector with itself; .

For any triangle with sides , , and , and angles , , and , the cosine rule states:

We can apply this rule to the triangle formed by the vectors and , and their difference . Below is a diagram of the triangle:

If we let , , and , we can apply the cosine rule to the triangle:

Recall that the square of the magnitude of a vector is the dot product of the vector with itself:

Substituting this back into the cosine rule equation, we get:

Finally, we can cancel out the and terms to get the equivalence of the dot product definitions:

Hence, the equivalence is proven.

By Vector Decomposition​

Brief Overview

This proof assumes the geometric definition of the dot product () and uses the decomposition of vectors into their components to derive the component definition of the dot product (). Along the way, it uses various properties of the dot product, such as its distributivity over vector addition and scalar multiplication.

Let and . As we know, these vectors are in 3D space, but the proof can be generalized to any dimension.

We can express the dot product of and as:

By the distributivity of the dot product over vector addition, we can expand this expression into the sum of the dot products of the individual components. This is analogous to how we expand the product of two binomials; .

Notice that in the product of the two binomials, each term is the product of one term from the first binomial and one term from the second binomial. We get every possible combination of terms from the two binomials. This is the same for trinomials; .

Applying this to the dot product of and , we get a similar expansion:

Recall from the previous section on the corollaries of the dot product that the dot product of two basis vectors is 1 if they are the same and 0 if they are different. As such, all the terms in the expansion where the basis vectors are different will evaluate to 0, and the terms where the basis vectors are the same will evaluate to 1:

Since this proof starts with the geometric definition of the dot product and derives the component definition, it shows that the two definitions are equivalent:

Hence, the equivalence is proven.

As mentioned earlier, this proof can be generalized to any dimension using the same principles. The key idea is to expand the dot product of two vectors into the sum of the dot products of their components, and then use the properties of the dot product to simplify the expression.

There are a few interesting things to note about this proof:

  1. Notice that this is independent of the cosine rule. This means that we can actually derive the cosine rule from the dot product, rather than the other way around.

  2. This proof gives you an explicit formula to get the magnitude of a vector. In the previous proof, we just assumed that the magnitude squared gives you . But this proof shows this independently.

    Not only is this also a proof of the Pythagorean theorem, but this proof is also especially important when we consider vector spaces without an orthonormal basis. In such cases, the magnitude is not just , but something else. These situations are more common than one would think, especially in fields such as relativity and quantum mechanics. With that in mind, we say that the magnitude, or norm, of a vector is actually defined via the dot product.

Dot Product as a Linear Transformation​

Brief Overview

This proof assumes we don't know the geometric or component definitions of the dot product. Instead, it treats the act of projection as a linear transformation and shows that the computation of this transformation is equivalent to the component definition of the dot product.

This proof originates from 3b1b's series on linear algebra. It's a very interesting way to think about the dot product.

Imagine that we do not know anything about the dot product, only that it does some kind of projection. Let be some unit vector that we want to project onto. Essentially, we are turning this two-dimensional onto a line defined by . Hence, this is a transformation from to .

(Note that this line is just a visual representation of the projection; it does not actually "sit" in this 2D space.)

Recall that linear transformations are defined by how they act on the basis vectors. Since we want to project it onto a line, each column in the matrix representing the transformation would only have one entry. Call this matrix :

Recall that each column in the matrix represents the result of applying the transformation to the basis vectors. To find these columns, we can zoom in on the diagram above. Below is the projection of onto the line:

Here's the key part: the projection of onto the line, , is exactly the same as the projection of onto . We can see this visually by creating a line of symmetry between the two vectors:

But the projection of onto is just the -component of :

Thus, the projection of onto is the same as the projection of onto , which is . The same applies for the projection of onto , which is . Thus, the matrix is:

Notice what happened here: we have a transformation that takes a vector and projects it onto a line. This transformation is represented by a matrix, and the matrix is the dot product of the vector with the basis vectors. But notice that this transformation is exactly the same as what the dot product does.

Now, we can apply this transformation to any arbitrary vector .

This is the same as the component definition of the dot product.

Hence, the equivalence is proven.

Note: What we have derived here has a name: covectors, or dual vectors. Essentially, when a linear transformation maps to , it can be represented as a row vector. Then, if we "tip" this row vector over, we can express this transformation as a dot product. This vector that corresponds to the linear transformation is precisely what we call a covector.

Non-Orthonormal Vectors (Optional Section)​

The component definition of the dot product, as shown in the second proof, is only valid for orthonormal vectors. However, the geometric definition of the dot product, as shown in the first proof, is valid for any pair of vectors because it is independent of the basis.

We can try to extend the component definition to non-orthonormal vectors. Let and . The dot product of and is, by distributing everything:

Notice that each row has a constant index for (red) and a varying index for (blue). We can group the terms by the the index by a product between a row vector and a column vector:

We can further group the terms by the index by a product between a matrix and a column vector:

This is the most general form of the dot product, and it is valid for any pair of vectors, orthonormal or not. The matrix in the middle is called the metric tensor matrix. It is a matrix that encodes the information about the dot product of the basis vectors. If we denote this matrix as , then the dot product of and can be written as:

It is often convenient to refer to a specific part of the metric tensor matrix. The element in the th row and th column of the metric tensor matrix is denoted as :

Then, we can write the dot product as a sum over the components of the vectors:

In the case of orthonormal vectors, we simply have , where is the Kronecker delta. This follows from the fact that the dot product of two basis vectors is 1 if they are the same and 0 if they are different. The matrix is then the identity matrix, and the dot product simplifies to .

Summary and Next Steps​

In this section, we have shown that the two definitions of the dot product are equivalent. This is a very important result, as it shows that the dot product is a very fundamental operation in linear algebra.

Here are the key points to remember:

  • The geometric definition of the dot product, , is equivalent to the component definition, .

  • The equivalence can be shown in multiple ways, such as using the cosine rule, vector decomposition, or treating the dot product as a linear transformation.

  • The component definition of the dot product can be extended to non-orthonormal vectors using the metric tensor matrix:

In the next section, we will end our introduction to the dot product by exploring some of its real-world applications.