The Unapologetic Mathematician

Mathematics for the interested outsider

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 Posted by | Algebra, Geometry, Linear Algebra | 15 Comments

   

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