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.
Best Answer
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.