If $M=\{ x: Lx=0\}$, where $L$ is continuous linear functional on $H$ (Hilbert Space). Prove that $M^\perp$ is vector space of one-dimensional unless $M= H$.
I know $M$ is closed so that $M^\perp$ is complement of $M$ union $\{0\}$. But i am unable to see how it is span by only one basis.
Could anyone please help me.
Suppose $u,v\ne 0$ and $u,v\perp M$. By this, it follows that $Lu\ne 0$ and $Lv\ne 0$.
By norming them: $u_1:=\displaystyle\frac1{Lu}u,\ \ v_1:=\frac1{Lv}v$, we get $Lu_1=1=Lv_1$.
But then $u_1-v_1\in M$, however $u_1-v_1$ is still $\perp M$, leading to $u_1=v_1$.