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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 $1$.

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

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  • $\begingroup$ are dimension of orthogonal complement and co dimension same thing ? $\endgroup$ – user776591 Apr 21 '20 at 12:14
  • $\begingroup$ @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 $M^{\perp}$. $\endgroup$ – Kavi Rama Murthy Apr 21 '20 at 12:17

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