# The Unapologetic Mathematician

## Inner Products and Angles

We again consider a real vector space $V$ with an inner product. We’re going to use the Cauchy-Schwarz inequality to give geometric meaning to this structure.

First of all, we can rewrite the inequality as

$\displaystyle\frac{\langle v,w\rangle^2}{\langle v,v\rangle\langle w,w\rangle}\leq1$

Since the inner product is positive definite, we know that this quantity will be positive. And so we can take its square root to find

$\displaystyle-1\leq\frac{\lvert\langle v,w\rangle\rvert}{\langle v,v\rangle^{1/2}\langle w,w\rangle^{1/2}}\leq1$

This range is exactly that of the cosine function. Let’s consider the cosine restricted to the interval $\left[0,\pi\right]$, where it’s injective. Here we can define an inverse function, the “arccosine”. Using the geometric view on the cosine, the inverse takes a value between $-1$ and ${1}$ and considers the point with that $x$-coordinate on the upper half of the unit circle. The arccosine is then the angle made between the positive $x$-axis and the ray through this point, as a number between ${0}$ and $\pi$.

So let’s take this arccosine function and apply it to the value above. We define the angle $\theta$ between vectors $v$ and $w$ by

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

Some immediate consequences show that this definition makes sense. First of all, what’s the angle between $v$ and itself? We find

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

and so $\theta=0$. A vector makes no angle with itself. Secondly, what if we take two vectors from an orthonormal basis $\left\{e_i\right\}$? We calculate

$\displaystyle\cos(\theta_{ij})=\frac{\lvert\langle e_i,e_j\rangle\rvert}{\langle e_i,e_i\rangle^{1/2}\langle e_j,e_j\rangle^{1/2}}=\delta_{ij}$

If we pick the same vector twice, we already know we get $\theta_{ii}=0$, but if we pick two different vectors we find that $\cos(\theta_{ij})=0$, and thus $\theta_{ij}=\frac{\pi}{2}$. That is, two different vectors in an orthonormal basis are perpendicular, or “orthogonal”.

April 17, 2009