The Unapologetic Mathematician

Mathematics for the interested outsider

Inner Products and Lengths

We’re still looking at a real vector space V with an inner product. We used the Cauchy-Schwarz inequality to define a notion of angle between two vectors.

\displaystyle\cos(\theta)=\frac{\lvert\langle v,w\rangle\rvert}{\langle v,v\rangle^{1/2}\langle w,w\rangle^{1/2}}

Let’s take a closer look at those terms in the diagonal. What happens when we compute \langle v,v\rangle? Well, if we’ve got an orthonormal basis around and components v^ie_i, we can write

\displaystyle\langle v,v\rangle=\sum\limits_{i=1}^d\left(v^i\right)^2

The v^i are distances we travel in each of the mutually-orthogonal directions given by the vectors e_i. 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 v, and so the quantities in the denominator above were the lengths of v and w, 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 v as an arrow pointing from the origin (the zero vector) to the point at its tip, we should think of the norm \lVert v\rVert as the distance between these two points. Similarly, the distance between the tips of v and w should be the length of the displacement vector v-w 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 d(v,w) as the norm of v-w.

Here are some axioms: A function from V to \mathbb{R} is a norm, written \lVert v\rVert, if

  • For all vectors v and scalars c, we have \lVert cv\rVert=\lvert c\rvert\lVert v\rVert.
  • For all vectors v and w, we have \lVert v+w\rVert\leq\lVert v\rVert+\lVert w\rVert.
  • The norm \lVert v\rVert is zero if and only if the vector v 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

\displaystyle0=\lVert0\rVert=\lVert v-v\rVert\leq\lVert v\rVert+\lVert -v\rVert=\lVert v\rVert+\lvert-1\rvert\lVert v\rVert=2\lVert v\rVert

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

\displaystyle\sqrt{\langle cv,cv\rangle}=\sqrt{c^2\langle v,v\rangle}=\sqrt{c^2}\sqrt{\langle v,v\rangle}=\lvert c\rvert\sqrt{\langle v,v\rangle}

as we’d hope. Finally, let’s check the triangle inequality. We’ll start with

\displaystyle\begin{aligned}\lVert v+w\rVert^2&=\langle v+w,v+w\rangle\\&=\langle v,v\rangle+2\langle v,w\rangle+\langle w,w\rangle\\&\leq\lVert v\rVert^2+2\lvert\langle v,w\rangle\rvert+\lVert w\rVert^2\\&\leq\lVert v\rVert^2+2\lVert v\rVert\lVert w\rVert+\lVert w\rVert^2\\&=\left(\lVert v\rVert+\lVert w\rVert\right)^2\end{aligned}

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.

April 21, 2009 - Posted by | Algebra, Geometry, Linear Algebra


  1. So, what happens over a finite field of characteristic 2? You no longer have the same proof of non-negative norms!




    What would I even mean by non-negative in characteristic 2?


    Never mind me.

    Comment by Mikael Vejdemo Johansson | April 21, 2009 | Reply

  2. I don’t know, Mikael. What happens in a real vector space over a finite field at all?

    Comment by John Armstrong | April 21, 2009 | Reply

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