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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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What have you tried? If this is homework, tag it as such. –  lhf Sep 14 '12 at 1:50
    
This isn't homework. –  Parakee Sep 14 '12 at 1:52

3 Answers 3

up vote 3 down vote accepted

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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