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Suppose we have a Hilbert space $X$ (or maybe just Banach even), and have the set of linear functionals $L(X,\mathbb{R})$. Further suppose two functionals $x, x'$ in this set have the same kernel, then are they same up to similarity?

Edit:

1.By similarity I mean same up to a scalar factor

  1. The setting is a hilbert space X, as the question came up as I was tryign to understand the proof of Riesz Representation theorem for hilbert spaces.

Motivation:

I got this idea when looking at $\mathbb{R^2}$ where equations of lines through origin can be identified with linear functionals of form $\alpha : (x,y) \to \alpha x + \beta y$, now zero set of such functionals is the perpendicular lines to the line corresponding to the linear functional. If we know the kernel line, we can then find the linear functional upto similarity. Now I am wondering how this idea gneeralizes.

Clemens Bartholdy
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  • Please define formally "$x$ and $x'$ are the same up to similarity" – azif00 Mar 13 '25 at 22:06
  • Yes; it follows from this: https://math.stackexchange.com/questions/709781/intersection-of-kernels-and-linear-dependence-of-functionals – Just dropped in Mar 13 '25 at 22:13
  • @Justdroppedin perhaps it answers the question but it is a different question. – Clemens Bartholdy Mar 13 '25 at 22:17
  • I don't understand the close votes. It's a good question that OP has put thought into. – littleO Mar 13 '25 at 22:20
  • @Anon001 fair enough! – Just dropped in Mar 13 '25 at 22:30
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    This is a duplicate: https://math.stackexchange.com/questions/60460/ker-t-subset-ker-s-rightarrow-s-rt-when-s-and-t-are-linear-functionals – azif00 Mar 13 '25 at 22:32
  • @littleO It's not clear whether the OP has tried to come up with a proof or not. There is also some potential ambiguity as to whether $L(X, \mathbb{R})$ should consist of the continuous linear functionals on $X$. Otherwise, the question appears to be a purely algebraic problem and it is not clear why the assumption that $X$ is a Hilbert or Banach space is even needed. – Dean Miller Mar 14 '25 at 04:22

1 Answers1

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As I mentioned in the comments, this follows in much greater generality, for vector spaces in general, from this.

Here is a short argument however: say $X$ is a linear space and $a,b:X\to\mathbb{K}$ are two functionals with the same kernel. If $a=0$, then $\ker(a)=X$ and so $\ker(b)=X$ implies that $b=0$ so there is nothing to show.

Otherwise, if $a\ne0$, let $x_0\in X$ be such that $a(x_0)=1$ (such $x_0$ exists by scaling a vector on which $a$ does not vanish). Then, for $x\in X$, we have that $x-a(x)x_0\in\ker(a)$, where $a(x)x_0$ should be interpreted as $x_0$ scaled by $a(x)$. Indeed, we have

$$a(x-a(x)x_0)=a(x)-a(x)a(x_0)=a(x)-a(x)=0,$$

so $x-a(x)x_0\in\ker(b)$, and so $b(x-a(x)x_0)=0$, so $b(x)=a(x)b(x_0)$. As this is true for all $x\in X$, we have that $b=\lambda a$, where $\lambda =b(x_0)$.

Just dropped in
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