# Why the nontrivial nullspace of a functional has codimension 1?

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.

• The image of a functional has dimension $1$. – Asaf Karagila Mar 15 '13 at 5:33
• Do you know what codimension means for a subspace of an infinite-dimensional vector space? – Qiaochu Yuan Mar 15 '13 at 5:36
• @QiaochuYuan: No, I don't :-( – Metta World Peace Mar 15 '13 at 5:51
• @Metta: it means the dimension of the quotient space. Now this is just an application of the first isomorphism theorem. – Qiaochu Yuan Mar 15 '13 at 5:55

## 2 Answers

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.

• $f(y+\frac{-f(y)}{f(x_0)}x_0) = 0$ could you clarify why this is true? – NewbieMather Mar 29 at 20:51
• I understand that $f(0)=0$ I guess I am just confused why we are able to say that $y=\frac{f(y)}{f(x_0)}x_0$. – NewbieMather Mar 29 at 20:59
• @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 $q= \frac{f(y)}{f(x_0)} \{ x_0 \} + \ker f$. – copper.hat Mar 29 at 21:05
• I guess I am not sure why the above implies that $y-\frac{f(y)}{f(x_0)}x_0\in \ker f$. – NewbieMather Mar 29 at 21:08
• @copper.hat Please take a look at this question. – José Carlos Santos Mar 29 at 21:09

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 $f(bv - aw) = 0$.

• The rank-nullity theorem doesn't apply if $\dim V$ is infinite, but the statement is still true in that case. – Qiaochu Yuan Mar 15 '13 at 5:36
• @QiaochuYuan: Thanks, good catch :) I was assuming finite dim because of the wording of the question. – Jim Mar 15 '13 at 5:40