Prove that $\dim E=\dim F=1$

Question

$E$ and $F$ are two subspaces of the inner product space $H$. $E,F\neq\{0\}$. If there exists a constant $\alpha>0$ such that $$|\langle x,y\rangle|=\alpha\|x\|\|y\|$$ for all $x\in E$ and $y\in F$, how to prove that $\dim E=\dim F=1$?

Answer 1

I suppose that the base field is $\mathbb R$.

Take $e \in E$ and $f_1,f_2 \in F$ all having norm equal to one. Even if it means switching $f_i$ to $-f_i$, we can suppose that $\langle e,f_i\rangle \gt 0$.

We have

$$\langle e, f_1\rangle=\langle e, f_2\rangle=\alpha$$ and $$4 \alpha^2= \langle e, f_1+f_2\rangle^2 = \alpha^2 \Vert f_1 + f_2 \Vert^2 = 2 \alpha^2+2\alpha^2 \langle f_1, f_2\rangle^2.$$

Therefore $\vert \langle f_1, f_2\rangle \vert = 1$. Which means that whatever $f_1,f_2$ are and according to Cauchy-Schwarz equality case $f_1,f_2$ are colinear. We can conclude that $\dim F =1$ and also to $\dim E=1$ by symmetry as desired.