$z\perp x_0\implies z\perp y_0$ for all $z$ in a dense subset implies $y_0$ is a multiple of $x_0$?

Question

Consider an inner product space $V$ and vectors $x_0, y_0\in V$. Let $E := \{z\in V : z\perp x_0\implies z\perp y_0\}$ be dense in $V$. Is it true that $y_0$ must be a scalar multiple of $x_0$?

If $x_0 = 0$, then $E = \{y_0\}^\perp$, whose denseness implies that $y_0 = 0$ (I can produce a proof if you want). Thus, w.l.o.g., assume that $\|y_0\| = 1$.

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I am able to prove that the answer is affirmative if $E = V$:

Take $z := y_0 - \langle y_0, x_0\rangle x_0$. Then $z\perp x_0$ and thus $z\perp y_0$ so that $\|y_0\|^2 - |\langle y_0, x_0\rangle|^2 = 0$ and thus $$ \|y_0\| = |\langle y_0, x_0\rangle|\le\|y_0\|\|x_0\| = \|x_0\|. $$ Hence, the inequality becomes an equality, and the claim follows.

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However, what about the case when we are only given the denseness of $E$?

Answer 1

Let $C$ be the orthogonal complement of $x_0$. Note that every $z\in V\setminus C$ satisfies $z\bot x_0\implies z\bot y_0$, because the antecedent $z\bot x_0$ is false. So your set $\{z: z\bot x_0\implies z\bot y_0\}$ includes $V\setminus C$ and is therefore dense, provided $V$ has dimension at least $2$.