Cauchy-Schwarz inequality and angle between vectors

Question

Let $\langle x,y\rangle=x\cdot y$ be the standard dot product on $\mathbb{R}^n$. By the Cauchy-Schwarz inequality, for $x,y$ non-zero, we have $$-1 \leq \frac{\langle x,y\rangle}{\|x\|\|y\|}\leq 1.$$

Thus there exists unique $\theta$ such that $\cos\theta$ is equal to the middle quantity in the above inequality. This $\theta$ is called as angle between $x$ and $y$.

However, while looking at proofs of Cauchy-Schwarz inequality, I came across one proof where it proceeds as follows:

since $\cos\theta=\frac{\langle x,y\rangle}{\|x\|\|y\|}$, taking modulus, we obtain $\left|\langle x,y\rangle\right|\leq \|x\|\|y\|$, which proves Cauchy-Schwarz inequality.

Question: Which of the two approaches is correct?

(1) Prove Cauchy-Schwarz inequality first and using it define angle between two vectors?

(2) Define angle between two (non-zero) vectors by $\cos\theta=\frac{\langle x,y\rangle}{\|x\|\|y\|}$ and use it to prove Cauchy-Schwarz inequality.

Answer 1

(2) is not correct. The definition of the angle between two vectors in $\mathbb R^n$ is based on the Cauchy-Schwarz.　you just cannot use it to prove what it's based on.

Answer 2

Here is a proof that $$\frac{\langle x,y\rangle}{\|x\|\|y\|} \in [-1,1]$$ which is not using the Cauchy-Schwarz inequality. Basically, the idea is to see this as the variational characterization of the singular values of the identity matrix, i.e. its Rayleigh quotient. Let $$f(x,y)=\frac{\langle x,y \rangle}{\|x\|\|y\|}.$$ Then, for every $x,y\neq 0$, we have $f(x,y)=f\left(\frac{x}{\|x\|},\frac{y}{\|y\|}\right)$. Since $f$ is smooth on $S=\{(x,y)\mid \|x\|=\|y\|=1\}$ it reaches its maximum and minimum values on $S$. All critical points of $f$ in $S$ must satisfy $$\nabla f(x,y)=0 \iff \left\{\begin{array}{l l} x=\langle x,y\rangle y \\ y=\langle x,y\rangle x\end{array}\right. \iff \left\{\begin{array}{l l} x=(\langle x,y\rangle)^2 x \\ y=(\langle x,y\rangle)^2 y\end{array}\right. \iff \langle x,y\rangle\in \{-1,1\}$$ and the associated critical value is $\langle x,y\rangle$. And thus the minimum of $f$ is $-1$ and its maximum is $1$.

Answer 3

I think at least in $\mathbb{R}^3$ you can take $\langle x,y\rangle = \lVert a\rVert \lVert b\rVert\cos \theta$ as the definition, then $$-1 \le \frac{\langle x,y\rangle}{\lVert a\rVert \lVert b\rVert}\le 1$$ all right.

But then you need to show that it's linear (which is difficult). So that you have the analytic form of the product which appears in the Cauchy-Schwarz inequality.

Answer 4

Take a parameter $\;a\in\Bbb R\;$, and since we're in a real linear space we have for any two vectors $\;x,y\;$ :

$$0\stackrel{\text{axiom!}}\le\langle ax+y\,,\,ax+y\rangle=||x||^2a^2+2\langle x,y\rangle a+||y||^2$$

Thus, the above is a non-negative quadratic in $\;a\;$ and thus its discrimimnant is non-positive (i.e. the corresponding parabola meets the $\;x$ - axis at most in one point):

$$\Delta=(2\langle x,y\rangle)^2-4||x||^2||y||^2\le0\implies|\langle x,y\rangle|\le||x||\,||y||$$

which is the Cauchy-Schwarz-Buniakovski inequality, without trigonometry and stuff, only very basic algebra.

Even now in graduate school, I find the above the easiest, most elegant and basic proof of this inequality in the real case.