# Kernel of Linear Functionals

Problem: Prove that for all non zero linear functionials $f:M\to\mathbb{K}$ where $M$ is a vector space over field $\mathbb{K}$, subspace $(f^{-1}(0))$ is of co-dimension one.

Could someone solve this for me?

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The following is a proof in the finite dimensional case: The dimension of the image of $f$ is 1 because $\textrm{im} f$ is a subspace of $\Bbb{K}$ that has dimension 1 over itself. Since $\textrm{im} f \neq 0$ it must be the whole of $\Bbb{K}$. By rank nullity,

$$\begin{eqnarray*} 1 &=& \dim \textrm{im} f \\ &=& \dim_\Bbb{K} M- \dim \ker f\end{eqnarray*}$$

showing that $\ker f$ has codimension 1.

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As long as $f$ is not the zero functional. – lhf Sep 14 '12 at 1:54
@lhf That was implicit when I wrote $\textrm{im} f \neq 0$. – user38268 Sep 14 '12 at 1:56

Take $u \in M$ so that $f(u) = 1$. Then for any $v \in M$ you can write $v = f(v) u + w$ where $w \in f^{-1}(0)$ since $f(w) = f(v - f(v) u) = 0$. That says $M = {\mathbb K} u + f^{-1}(0)$, so $f^{-1}(0)$ has codimension $1$.

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If $V$ and $W$ are vector spaces and $T:V\rightarrow W$ is linear, the isomorphism theorem says that $${V\over \ker T} \cong {\rm Im}(T)$$ In the case that $W$ is the base field and $T\not= 0$, $T$ must be onto, so the dimension of $V/{\ker(T)}$ is 1. This does not depend upon finite-dimensionality.

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First, we show that $Ker(f)$ is a hyperplane in $M$. Let $W_{1} \subset M$ a subspace, it's suffice shows that if $Ker(f) \neq W_{1}$ then $W_{1}=M$. Since $f \neq 0$ we have $Ker(f) \neq M$.

Suppose $Ker(f) \neq W_{1}$ and choose $v_{0} \in W_{1} \setminus Ker(f)$. Let $v \in M$ and taking $u= v - \frac{f(v)}{f(v_{0})}v_{0}$ we have that $u \in Ker(f) \subset W_{1}$, therefore $v \in W_{1}$, thus $W_{1}= M$.

Thereby, since $Ker(f)$ is a hyperplane in $M$, it's the maximal subspace in M. Write $W' = Lin(x_{0} \cup ker(f))$. Since that $W \subset W'$, we have that $W' = M$. Then,

$M = Lin(x_{0}) + ker(f)$

So, $ker(f)$ has codimesion 1

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@G.Sassatelli I edited the answer, thanks for the advice. But, i thought it was a well-know fact. – BBVM Jan 26 at 4:31
It is a well known fact. But, by definition, "$W$ is a maximal proper subspace $V$" means: "there exists $x\in V\setminus W$ and for all $y\notin W$, $W+\Bbb Ky=V$". On the other hand "the codimension of $W$ in $V$ is $1$" means: "there exists $x\notin W$ such that $W+\Bbb Kx=V$". If you have $\ker f$ to be a maximal proper subspace as a given, then your proof reads like: "Let's prove that there exists $x\notin \ker f$ such that $\Bbb Kx +\ker f=V$ knowing that there exists $x\notin \ker f$ and that for all $y\notin \ker f$, $V=\Bbb Ky+\ker f$". – G. Sassatelli Jan 26 at 5:56
There's something off (a typo, perhaps) in your first statement: given a (hyperplane) maximal proper subspace $W$ in a vector space $M$, if $\dim M\ge 2$ there always exists a subspace $W_1\neq W$ such that $W_1\subsetneqq M$. Of course, if $W$ is a hyperplane, every subspace $W_1\supsetneqq W$ must be $M$. But, as it's written, it's false. – G. Sassatelli Jan 26 at 6:45