Proof of basic Inner Product Property

Question

I am searching for a proof of the following property for two vectors x and y in $\mathbb R^n$

$<x,y> = ||x||*||y|| \implies ||x||= \lambda*||y||$ for some $\lambda\in\mathbb R$

with $<,>$ being the inner product and $||.||$ the norm function.

I tried to prove it in $\mathbb{R^2}$ with $x=(x_{1},x_{2})$ and $y=(y_{1},y_{2})$. From the hypothesis I deduce that $$ <x,y> = ||x||*||y|| \implies x_{1}y_{1}+x_{2}y_{2}=\sqrt{x_{1}^{2}y_{1}^{2}+x_{1}^{2}y_{1}^{2}+x_{2}^{2}y_{1}^{2}+x_{2}^{2}y_{2}^{2}} $$

which implies that $x_{1}y_{2}+x_{2}y_{1}=0$

If my deduction is correct so far, how do we get the conclusion?

Answer 1

Assume that $y \not = 0$ and use orthogonal projection. Specifically, if you take $\lambda = \dfrac{\langle x,y \rangle}{\langle y,y \rangle}$ and $z = x - \lambda y$ you can compute rather easily that $x = \lambda y + z$ and $\langle y,z \rangle = 0$.

It follows (using the stated hypothesis) that $$\|x\| \|y\| = \langle x,y \rangle = \lambda \|y\|^2$$ so you get $\|x\| = \lambda \|y\|$. Now apply the Pythagorean theorem: since $y \perp z$ you have $$\|x\|^2 = \lambda^2 \|y\|^2 + \|z\|^2$$ which forces $z = 0$. Thus $x = \lambda y$.

Answer 2

By dividing through by $\|x\|, \|y\|$ we may assume that $x,y$ are unit vectors.

Note that $\|x-y\|^2 = \|x\|^2+\|y^2\| - 2 \langle x , y \rangle = \|x\|^2+\|y^2\| - 2\|x\| \|y\|= (\|x\|-\|y\|)^2 = 0$.

Since $x=y$, it follows that the original points lie on the same line.