If $X $ is an inner product space and if there exists $x \in X $ s.t. $\{x\}^{\perp}=0$. Is $X$ one dimensional?
The way I have written out this question a bit wrong because it was in my exam so I don't know word for word. I thought that $\{x\}^{\perp}=0$ then this means by theorem that $<x> =\{ ax : a\in \mathbb R \}$ is dense in $X$. So $X= \bar{< x >}$. So it is one dim?
Suppose $x=0$. Then, it is easy to show that $X=\{0\}$. Therefore $\mathrm{dim}X=0$.
Suppose $x\neq 0$. Assume WLG that $\|x\|=1$. Then, there is at least one vector $u\in X$, $u\neq x$. Take $u^\prime=u-\langle u,x\rangle x$. It is easy to show that $0=\langle u^\prime,x\rangle$. Thus, $u^\prime=0$. Therefore $u-\langle u,x\rangle x=0$. That is $u\in\mathrm{span}\{x\}$. This means $X=\mathrm{span}\{x\}$, and therefor $\mathrm{dim}X=1$.