16
$\begingroup$

The nullspace of a linear functional that is not $\equiv 0$ is a linear subspace of codimension $1$.

I don't understand this statement on page 57, Functional Analysis(Pater Lax). Does it mean the dimension of nullspace of a linear functional is either zero or the dimension of the domain of the functional minus one, which I don't see why it's necessarily true.

Added: Thank you all for your valuable comments and answers. I didn't realize that it's wrong to interpret "The nullspace of a linear functional that is not $\equiv 0$" as the nullspace (of a linear functional) that is not $\equiv 0$ until I saw the answers.

$\endgroup$
4
  • 2
    $\begingroup$ The image of a functional has dimension $1$. $\endgroup$
    – Asaf Karagila
    Commented Mar 15, 2013 at 5:33
  • 1
    $\begingroup$ Do you know what codimension means for a subspace of an infinite-dimensional vector space? $\endgroup$ Commented Mar 15, 2013 at 5:36
  • 1
    $\begingroup$ @QiaochuYuan: No, I don't :-( $\endgroup$ Commented Mar 15, 2013 at 5:51
  • 4
    $\begingroup$ @Metta: it means the dimension of the quotient space. Now this is just an application of the first isomorphism theorem. $\endgroup$ Commented Mar 15, 2013 at 5:55

2 Answers 2

27
$\begingroup$

For simplicity, suppose $X$ is a vector space over $\mathbb{R}$, and $f$ a linear functional on $X$.

If $f=0$, then $\ker f = X$, so the codimension is zero.

If $f \neq 0$, then there is some $x_0$ such that $f(x_0) \neq 0$. Now consider the quotient space $Q={X}/{\ker f}$ (ie, two points $x_1,x_2$ are equivalent iff $f(x_1) = f(x_2)$, which basically 'flattens' $X$ 'down' to the values of $f$, ie $\mathbb{R}$).

Pick some $q \in Q$, then we must have $q = \{y\}+\ker f$ for some $y$. Note that by linearity we have $f(y+\frac{-f(y)}{f(x_0)}x_0) = 0$, hence $q = \frac{f(y)}{f(x_0)}(\{x_0\}+ \ker f)$. Hence $\{ \{x_0\}+ \ker f \}$ is a basis for $Q$, hence $\ker f$ has codimension one.

$\endgroup$
5
  • $\begingroup$ $f(y+\frac{-f(y)}{f(x_0)}x_0) = 0$ could you clarify why this is true? $\endgroup$
    – Killaspe
    Commented Mar 29, 2021 at 20:51
  • $\begingroup$ I understand that $f(0)=0$ I guess I am just confused why we are able to say that $y=\frac{f(y)}{f(x_0)}x_0$. $\endgroup$
    – Killaspe
    Commented Mar 29, 2021 at 20:59
  • 2
    $\begingroup$ @NewbieMather You can't. But since $y+\frac{-f(y)}{f(x_0)}x_0 \in \ker f$, you can write $y = \frac{f(y)}{f(x_0)}x_0 + k$, where $k \in \ker f$ and so you can write $q= \frac{f(y)}{f(x_0)} \{ x_0 \} + \ker f$. $\endgroup$
    – copper.hat
    Commented Mar 29, 2021 at 21:05
  • $\begingroup$ I guess I am not sure why the above implies that $y-\frac{f(y)}{f(x_0)}x_0\in \ker f$. $\endgroup$
    – Killaspe
    Commented Mar 29, 2021 at 21:08
  • $\begingroup$ @copper.hat Please take a look at this question. $\endgroup$ Commented Mar 29, 2021 at 21:09
4
$\begingroup$

If your linear functional is $f\colon V \to \mathbb R$ (or whatever field you're working over) it means the nullspace has dimension $\dim V$ if $f = 0$ and $\dim V - 1$ otherwise.

The reason why is the rank nullity theorem: $\dim\operatorname{rank}f + \dim\operatorname{nullity}f = \dim V$. The rank of $f$ is $0$ if $f = 0$. If $f \neq 0$ then the rank is $1$.

Edit: As Yuan points out the above argument only works in the finite dimensional case. For the infinite dimensional case take a basis of the nullspace and extend to a basis of $V$. You can add at most $1$ additional vector because if $f(v) = a$ and $f(w) = b$ then $f(bv - aw) = 0$.

$\endgroup$
2
  • 4
    $\begingroup$ The rank-nullity theorem doesn't apply if $\dim V$ is infinite, but the statement is still true in that case. $\endgroup$ Commented Mar 15, 2013 at 5:36
  • $\begingroup$ @QiaochuYuan: Thanks, good catch :) I was assuming finite dim because of the wording of the question. $\endgroup$
    – Jim
    Commented Mar 15, 2013 at 5:40

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .