If $f:X \rightarrow \Bbb{K}$ is a linear functional,then $\textbf{Ker}(f)$ is a maximal subspace of $X$? Let $f$ be a linear functional on vector space $X$.
$K$ is a Field which can be $\Bbb{R}$ or $\Bbb{C}$.
Let $Z(f)$ denote the kernel of $f$
The Maximal subspace of $f$ :
$Z$ is said to be a maximal subspace of $X$ if for any subspace $Z_{1}$ with $Z \subset Z_{1} \subset X$ then either $Z = Z_{1}$ or $X = Z_{1}$
definition 2 :
$Z$ is a maximal subspace of $X$ $\textbf{iff}$ $\textbf{span}(Z \cup \{a\}) = X$ for any $a \in X\setminus Z$.
For $f$ a linear functional on $X$. Now I was thinking to prove that $Z(f)$ is a maximal subspace of $X$.
For proving I thought of this- Suppose $Z \neq Z_{1}$ then we need to show that $Z_{1} = X$,but how do I proceed with this?
$\textbf{EDIT}:$
Also as pointed out in the comments whether $X$ is finite dimensional or infinite dimensional? So it would be interesting to see if those two case go similarly or are there any different treatment or consequences in the proof of both the cases?
Any help is great.
 A: Nina Simone gave already the way for the proof which is independent of the dimension of $X$. I just improve the argument.

$Z$ is said to be a maximal subspace of $X$ if for any subspace $Z_1$ with $Z\subset Z_1\subset X$ then either $Z_1=Z$ or $Z_1=X$.

It is a little bit more comfortable to use

$Z$ is said to be a maximal subspace of $X$ if for any subspace $Z_1$ with $Z\subsetneq Z_1\subseteq X$ holds $Z_1=X$.

Let us consider $z\in Z_1\setminus Z\neq \emptyset$. We get $f(z)\neq 0$ since $z\notin Z$. For each $x\in X$ we can write
$$
x=\underbrace{\frac{f(x)}{f(z)}z}_{\in Z_1}+\underbrace{\left(x-\frac{f(x)}{f(z)}z\right)}_{\in Z}
$$
since $f\left(x-\frac{f(x)}{f(z)}z\right)=0$. From $Z\subseteq Z_1$ we conclude $x\in Z_1$. Since $x\in X$ was arbitrary we conclude $X\subseteq Z_1$ and therefore $Z_1=X$.
A: The existing answers are completely fine. Nonetheless, I would like to add one that demonstrate the power of normalization.
Let $f$ be a linear functional and $Z$ a subspace of $X$ such that $\ker(f)\subsetneq Z$. We would like to claim that $Z = X$. We pick $z\in Z\backslash X$ with $f(z)\neq 0$ and pick $x\in X$ with $f(x)\neq 0$; we want to show $x\in Z$.
Now we split things into cases:
We first consider $f(z) = f(x) = 1$. Then it follows clearly by linearity that $z-x\in \ker(f)$. It follows clearly that $x \in \langle\{z\}\rangle\oplus \ker(f)\subset Z$.
The general case, as is demonstrated by other answers, is just an application of the first case by replacing $x$ with $x/f(x)$ and $z$ with $z/f(z)$.
