# Let $f$ be bounder linear functional on Hilbert space $H$ , then dimensional of orthogonal complement of null space is $1$

Let $$f$$ be bounded linear functional (non trivial) on Hilbert Space $$H$$ . Let $$null(f)$$ be null space of $$f$$ in $$H$$. Then how to prove that dimension($$null(f)^⊥$$) is $$1$$.

• Maybe this can help math.stackexchange.com/questions/330948/… – user326159 Apr 21 '20 at 12:00
• @user326159 are co dimension and dimension of orthogonal complement same thing ? – user776591 Apr 21 '20 at 12:03
• In this case, since the kernel is closed, it is – user326159 Apr 21 '20 at 12:31

## 1 Answer

Let $$x$$ be a fixed non-zero vector in $$ker(f)^{\perp}$$. Any other vector $$y$$ in $$ker(f)^{\perp}$$ can be written as $$y=z+cx$$ where $$z=y-cx$$ and $$c$$ is chosen such that $$f(z)=f(y)-cf(x)=0$$ (or $$c=-\frac {f(y)}{f(x)}$$). Now $$z=y-cx\in ker(f)^{\perp}$$ because $$x, y \in ker(f)^{\perp}$$. But $$z \in ker (f)$$. Hence $$z=0$$ and we get $$y=cx$$. Thus evey vector $$y$$ in $$ker(f)^{\perp}$$ is a scalar multiple of $$x$$.

• are dimension of orthogonal complement and co dimension same thing ? – user776591 Apr 21 '20 at 12:14
• @Pollock The kernel $M$ of $f$ is a closed subspace and this implies $H=M\oplus M^{\perp}$. Hence the codimension of $M$ is same as dimension of $M^{\perp}$. – Kavi Rama Murthy Apr 21 '20 at 12:17