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 John Armstrong | Algebra, Geometry, Linear Algebra

18 Comments »

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

….

Wait

….

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

….

Never mind me.

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