How axioms of inner product ensure that an instantiation/realization capture notion of angle correctly?

Question

Axiomatic definition of inner product can lead to various instantiations like euclidean inner product or complex inner product or weighted inner product etc.

Whatever the special case, we can be sure, $\langle x, x \rangle$ will be candidate for norm. We can show $\langle x, x \rangle$ will always satisfy axioms for a norm using 1. axioms of inner product and 2. Cauchy–Schwarz inequality. So I can see how axioms of inner product have intuition for length embedded in them.

I wonder, what about angle? How do axioms ensure that any instantiation captures the notion of angle correctly?

Answer 1

Note that

1. Any statement you can make involving angles (or lengths) which only involves finitely many vectors $v_1, \dots v_n$ in an inner product space $V$ at a time only involves the finite-dimensional subspace $\text{span}(v_1, \dots v_n)$, and

2. In an inner product space $V$, every $d$-dimensional subspace is isometric to Euclidean space $\mathbb{R}^d$, and this property characterizes inner product spaces among normed vector spaces by the polarization identity (in fact it suffices to take $d = 2$).

So, the inner product axioms characterize exactly the geometric facts about lengths and angles true in Euclidean spaces, but also allow us to generalize those facts to infinite dimensions.

Answer 2

Given an inner product space and two nonzero vectors, the two vectors span a two dimensional space which is isometrically isomorphic to the Euclidean plane. Using the axioms of inner product, the definition of angle using the inner product agrees with the definition of angle in the Euclidean plane.

Answer 3

A real inner product on a real vector space $V$ is a symmetric positive definite bilinear form $V \times V \to \mathbb{R}$.

You already accept that $|x| = \sqrt{\langle x, x\rangle}$ captures the notion of length.

Before getting to angles in general we should discuss right angles (orthogonality).

It should be geometrically reasonable that $v \perp w$ if and only if $|v - \alpha w|$ has $\alpha = 0$ as the unique minimizer.

$$ |v - \alpha w|^2 = |v|^2 + \alpha^2 |w|^2 +2\alpha\langle v, w\rangle $$

This is a quadratic in $\alpha$ which achieves a minimum value at $\alpha = -\frac{\langle v, w \rangle}{|w|^2}$. So it should be geometrically reasonable to define $v \perp w$ if and only if $\langle v, w\rangle = 0$.

Now that we have defined perpendicularity, it is reasonable to define angles through right triangle trigonometry.

The same calculation shows that $v$, $\alpha w$ and $v - \alpha w$ form a right triangle when $\alpha = -\frac{\langle v, w \rangle}{|w|^2}$ with $v$ as the hypotenuse. Assume for the moment that $\langle v, w \rangle > 0$ so that I can ignore sign issues. So it would be reasonable to define

$$\cos(\theta) = \frac{|\alpha w|}{|v|} = \frac{\langle v, w\rangle}{|v||w|}$$

The Cauchy-Schwarz inequality justifies that this equation has a solution (the RHS is between $-1$ and $1$).