Characterising $\langle x,y\rangle = \langle x,x\rangle \cdot \langle y,y\rangle$

Question

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

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

Here are some thoughts :

• With the intuition : $\langle x,y\rangle = $ 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 !

Answer 1

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 $\langle x,y\rangle =\langle x,x\rangle \langle y,y\rangle$. 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].

Answer 2

The equation $\langle x,y\rangle = \langle x,x\rangle\langle y,y\rangle$ 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

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

Then it is easy to see that

$$\langle x,y\rangle = \langle x,x\rangle\langle y,y\rangle \iff \langle x,y\rangle = \cos^2 \phi.$$

Answer 3

The pair $(x,y)$ satisfies $\langle x,y\rangle = \langle x,x\rangle\langle y,y\rangle$ when $y=0$. When $y\neq 0$ we may set $\hat y = y/\langle y,y\rangle$ to obtain the equation $\langle x,\hat y \rangle = \langle x,x\rangle$ 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 $$ \left\| x - \frac{y}{2\langle y,y\rangle}\right\| = \left\| \frac{y}{2\langle y,y\rangle}\right\| = \frac{1}{2\|y\|}. $$

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