Let’s take a closer look at those terms in the diagonal. What happens when we compute ? Well, if we’ve got an orthonormal basis around and components , we can write
The are distances we travel in each of the mutually-orthogonal directions given by the vectors . But then this formula looks a lot like the Pythagorean theorem about calculating the square of the resulting distance. It may make sense to define this as the square of the length of , and so the quantities in the denominator above were the lengths of and , respectively.
Let’s be a little more formal. We want to define something called a “norm”, which is a notion of length on a vector space. If we think of a vector as an arrow pointing from the origin (the zero vector) to the point at its tip, we should think of the norm as the distance between these two points. Similarly, the distance between the tips of and should be the length of the displacement vector which points from one to the other. But a notion of distance is captured in the idea of a metric! So whatever a norm is, it should give rise to a metric by defining the distance as the norm of .
Here are some axioms: A function from to is a norm, written , if
- For all vectors and scalars , we have .
- For all vectors and , we have .
- The norm is zero if and only if the vector is the zero vector.
The first of these is eminently sensible, stating that multiplying a vector by a scalar should multiply the length of the vector by the size (absolute value) of the scalar. The second is essentially the triangle inequality in a different guise, and the third says that nonzero vectors have nonzero lengths.
Putting these axioms together we can work out
And thus every vector’s norm is nonnegative. From here it’s straightforward to check the conditions in the definition of a metric.
All this is well and good, but does an inner product give rise to a norm? Well, the third condition is direct from the definiteness of the inner product. For the first condition, let’s check
as we’d hope. Finally, let’s check the triangle inequality. We’ll start with
where the second inequality uses the Cauchy-Schwarz inequality. Taking square roots (which preserves order) gives us the triangle inequality, and thus verifies that we do indeed get a norm, and a notion of length.