Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
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.

share|improve this answer
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 $1$.

share|improve this answer

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.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.