3
$\begingroup$

Let $x, y$ be two complex vectors, $$\cos\Theta(x,y):=\operatorname{Re} \frac{y^*x}{\|x\|\|y\|} .$$ Then I want to prove that $$\Theta(x,y)\le \Theta(x,z)+\Theta(z,y) .$$

  • 2
    Did you look in his original paper?2011-05-18
  • 0
    How do you define $\Theta(x,y)$? Taking the inverse cosine mapping $[-1,1]$ to the interval $[0,\pi]$? This is one way to say that "the angle between two vectors defines a metric on the n-dimensional sphere." Seems pretty intuitive to me though.2011-05-18
  • 0
    Who says that this is due to Krein? This must have been known decades if not a hundred years earlier. Do you have a reference saying this explicitly?2011-05-18
  • 0
    Thanks for the reference. But as this is essentially a statement on $\mathbb{R}^3$, I'm still doubtful. (By the way: if you make references such as @Jonas T - include the full username and no space after the `@` - then we get notified).2011-05-19
  • 0
    I'm afraid the fact you attempt to quote involves the real part of the inner product in a Hilbert space. It is indeed obvious if everything involved is over the real numbers. $$ $$ http://rgmia.org/papers/v11e/IPI2ENApp.pdf2011-05-19
  • 0
    @Theo Buehler and @Jonas T the OP incorrectly described the contents of his reference, see my previous comment. I was unaware that a correctly placed at sign generates a notification to the named user.2011-05-19
  • 0
    @Will: Thank you for the notification and the link, now the question makes much more sense! There's one thing that you should know the @-notifications: only one user per comment gets notified, the first one, so Jonas T hasn't received a notification. [This link](http://meta.stackexchange.com/q/43019/155585) contains a detailed description.2011-05-19
  • 0
    @Jonas T: Will Jagy wrote a comment directed at you.2011-05-19
  • 0
    @Theo: Thanks! I've seen it.2011-05-19
  • 0
    Good to know about the notification system.2011-05-19

2 Answers 2

1

The inequality simply says that the "angle between $\vec x$ and $\vec z$ plus the angle between $\vec z$ and $\vec y$ is never smaller than the angle between $\vec x$ and $\vec y$". That's obvious in the Euclidean space. First, as ncmathsadist pointed out, the angles don't change if you normalize $\vec x,\vec y,\vec z$ to unit vectors.

Then the angles (in radians) between two vectors are measured just as distances measured along the surface of the unit sphere and the inequality says that there the crow-fly, straight distance between $x$ and $y$ can't be made shorter by inserting a detour to $z$. Well, I hope that it's not hard to see how to prove that the crow-fly distance is the shortest path between the two vectors on the sphere.

1

You should think about distances on the unit sphere.