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.

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    $\begingroup$ It depends on what basic definitions your start with. For example, high schools here usually define the standard dot product precisely as $$x\cdot y=||x||\,||y||\cos\theta$$ with $\;\theta=$ the angle between both vectors. This definition is rather poor for universities, say...but it is a starting point to work. $\endgroup$ – Timbuc Feb 3 '15 at 9:39
  • $\begingroup$ You second approach does not work because you should first show that the right hand side is $\le 1$ so it can be the cosine of something (which is what you do in the first approach) Instead, as @Timbuc say you could define the scalar product in that way, but it is not optimal for a lots of reasons (What the heck is an angle in a $17$-dimensional space? What is the angle between two functions in the hilbert space $L^2$? )These example show that the first approach is easy to generalize if one wants, the other not so much and it does not make much sense as long as you leave $\Bbb R^3$ $\endgroup$ – Ant Feb 3 '15 at 10:29

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.

  • $\begingroup$ How does this answer address the questions raised in the first post? $\endgroup$ – daw Feb 3 '15 at 10:37
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    $\begingroup$ I think it does answer what it must answer: first, it gives a proof of CSB inequality without any need of angles, cosines and stuff, from which then we can define cosine of angle between vectors and stuff. This clears both (1)-(2), and the other question "which approach is right", besides being addressed already in other comments/answers, depends heavily on the first, basic definitions given, and I address this with an example in my comment below the question. $\endgroup$ – Timbuc Feb 3 '15 at 10:40
  • $\begingroup$ Of course, another option is to downvote what you don't understand, @daw...:) Oh, well. $\endgroup$ – Timbuc Feb 3 '15 at 10:45

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 $1$.

  • $\begingroup$ This was unfortunately too long to be a comment. $\endgroup$ – Surb Feb 3 '15 at 9:49
  • $\begingroup$ Can you prove all the necessary prerequisites without Cauchy-Schwarz? After all, you use a lot of advanced techniques: compactness, multi-variate calculus, critical points. $\endgroup$ – daw Feb 3 '15 at 10:35
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    $\begingroup$ @daw If you assume that $\|x\|= \langle x,x\rangle$ and $\|\cdot\|$ is a norm, YES! However, the margin is too small to include the proof. The analysis lecture I had in first year shows all the required material to understand this proof without using the Cauchy-Schwartz inequality (under the assumptions cited above). $\endgroup$ – Surb Feb 3 '15 at 10:51

(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.

  • $\begingroup$ What do you mean exactly by: "The definition of the angle between two vectors in ℝn is based on the Cauchy-Schwarz."? $\endgroup$ – Surb Feb 3 '15 at 9:54
  • $\begingroup$ @Surb because of Cauchy-Schwarz, the value $a=\frac{\vec{u}\centerdot\vec{v}}{|\vec{u}||\vec{v}|}$ is always within $[-1,1]$, which makes it possible to define $a$ as the $\cos\theta$. $\endgroup$ – Vim Feb 4 '15 at 14:38
  • $\begingroup$ No... You can prove that $a\in [-1,1]$ without using Cauchy-Schwartz... $\endgroup$ – Surb Feb 4 '15 at 15:31
  • $\begingroup$ @Surb I'd like to learn how :P $\endgroup$ – Vim Feb 4 '15 at 15:33
  • $\begingroup$ See my answer in this question. Since you were the only one claiming that, I wrote it for this purpose. $\endgroup$ – Surb Feb 4 '15 at 15:37

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.


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