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Let $V$ be a (possibly infinite dimensional) vector space, and $U$ a non-zero subspace of $V$ such that $U\neq V$. Also, let $F = \{f_i\}$ be the family of linear functionals on $V$ such that $U \subseteq \text{ker}(f_i)$. Then,

  1. Is there always a non-zero function in $F$? And one for which $\ker{(f)}=U$?

  2. Must all $f_i$ be proportional?

My attempt:

Question 1:

Since $U\neq V$, $\text{dim}(U) < \text{dim}(V)$. Hence, let $\{u_i\}$ be a basis of $U$ and $\{v_i\}$ a set of vectors such that $v_i$ are orthogonal to $U$ and $V = U + \text{span}(\{v_i\})$. Then, we can always choose a functional such that $f(u_i)=0,\ f(v_i) \neq 0$. (Inspiration taken from here).

Nevertheless, the codimension of $\text{Ker}(f)$ must be $1$ for it to be a non-zero linear form. Does this mean that we can only set $f(v_i)\neq 0$ for a single vector $v_i$? I have problems understanding why this should be the case. And under what conditions can one guarantee the existence of a non-zero function $f$ such that $\ker{(f)}=U$?

Question 2:

Well, this depends on the answer to 1., but if there were a function such that $\text{ker}(f)=U$, by using this answer one could prove that all $f_i$ must be proportional to $f$, hence being proportional between each other. My intuition tells me that if such function didn't exists, $f_i$ don't need to be proportional since they would be acting differently in differrnt subspaces of $V$.

Thank you in advance

user3141592
  • 1,995
  • For question 1, there is no notion of orthogonality, and dimension and basis are of no help since infinite-dimensional spaces are allowed. The answer to 1.1 is yes because $U\ne V$. The answer to 1.2 is: no except if codim(U)=1, which requires that all the $f_i$s are proportional (have the same kernel). The answer to 2 is indeed no. – Anne Bauval Feb 25 '24 at 10:55
  • @AnneBauval Thank you for your comment! Then, when can we ensure that $\text{codim}(U)=1$ for infinite dimensional spaces? – user3141592 Feb 25 '24 at 13:03
  • Exactly when $U$ is the kernel of some nonzero linear form. In other words: a hyperplane. – Anne Bauval Feb 25 '24 at 13:04
  • To me, that is unfortunately a not very useful way to define it. Isn't there any other description of an hyperplane or some conditions a subspace must fulfill in order to be one? – user3141592 Feb 25 '24 at 13:38
  • This is also equivalent to: $\exists v\ne0$ (or equivalently: $\forall v\notin U$), $U\oplus kv=V$ (where $k$ is the field of scalars), and also equivalent to $\dim(V/U)=1$, where $V/U$ is the quotient vector space. – Anne Bauval Feb 25 '24 at 14:10

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