Krdytkyu

Is it possible to give a nice characterisation of all couples $(x,y) in mathbb{R}^{2n}$ such that :
$$langle x,yrangle = langle x,xranglelangle y,yrangle?$$

$langle,rangle$ is the euclidian inner product on $mathbb{R}^n$.

Here are some thoughts :

• With the intuition : $langle x,yrangle = $ the length of the projection of $x$ onto $y times$ the length of $y$. It would mean that the above equality means :
  $$text{length projection of ... = the square of the length of }x times text{ the square of the length of }y$$
  But by C-S this is possible only if the length of $y$, or $x$ is less than $1$.




• The condition reminds me the independence of random variables, but I don’t know if it helps.

Thank you !

There doesn't seem to be a nice charcarterization but you can do the following: write $x=cy+z$ $,,$ (1) with $z$ orthogonal to $y$. [We can always do this]. Using the fact that $|cy+z|^{2}=c^{2}|y|^{2}+|z|^{2}$ we can see from the given equation that $c$ satisfies a quadratic equation. Solving this gives $c$ and these steps are all reversible so whenever $x$ and $y$ have this form we get $langle x,yrangle =langle x,xrangle langle y,yrangle$. I think this is the best characterization we can get. [ We will be forced to impose a condition like $|z|leq |y|$ because a real number $c$ satisfying (1) surely exists and the discriminant of the quadratic has to be non-negative].

The equation $langle x,yrangle = langle x,xranglelangle y,yrangle$ is trivially satisfied if $x=0$ or $y=0$. Therefore let us suppose that $x ne 0 ne y$. If $ phi$ is the angle between $x$ and $y$, then we have

$$ cos phi= frac{langle x,yrangle}{||x|| cdot ||y||}.$$

Then it is easy to see that

$$langle x,yrangle = langle x,xranglelangle y,yrangle iff langle x,yrangle = cos^2 phi.$$

The pair $(x,y)$ satisfies $langle x,yrangle = langle x,xranglelangle y,yrangle$ when $y=0$. When $yneq 0$ we may set $hat y = y/langle y,yrangle$ to obtain the equation $langle x,hat y rangle = langle x,xrangle$ which is equivalent to $x$ and $hat y-x$ being orthogonal. This should remind you of Thales's theorem, as the set of all such $x$ is the sphere with diameter the line segment from $0$ to $hat y$. That is, it is the set of all $x$ satisfying
$$
left| x - frac{hat y}{2}right| = left| frac{hat y}{2}right|
$$
or in the original variables
$$
left| x - frac{y}{2langle y,yrangle}right| = left| frac{y}{2langle y,yrangle}right| = frac{1}{2|y|}.
$$

In summary for $y=0$ every $x$ is possible, for each $yneq 0$ you have all $x$ on the sphere with diameter the line segment from $0$ to $y/langle y,yrangle$.