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$dimV<\infty$, $f,g$ are nonzero linear functional on $V$ real vector space, $Ker(f)\subsetneq Ker(g)$ we need to pick out true statements

  1. $Ker(f)=Ker(g)$

  2. $ker(g)/ker(f)\cong\mathbb{R}^k, 1\le k<n$

  3. there exist a constant $c$ such that $g=cf$

In any case Kernel has to be of dimension $n-1$ assuming $dimV=n$, so $1$ and $3$ are true,and hence $2$ is wrong trivially., am I right?

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Because $\ker(f) \subsetneq \ker(g)$ is not possible when $f,g$ are nonzero linear functionals, then maybe you should change something in the hypothesis. – user26857 Oct 26 '12 at 18:33
up vote 1 down vote accepted

If $f$ and $g$ are non-zero, then indeed $\ker(f)$ and $\ker(g)$ are $n - 1$ dimensional. If also $\ker(f) \subset \ker(g)$, then you must have $ker(f) = ker(g)$ and so $\ker(f) \subsetneq \ker(g)$ is not possible... Anyway, if you assume that $ker(f) \subset \ker(g)$, then indeed $(1)$ and $(3)$ are true and the quotient $\ker(g) / \ker(f)$ is zero-dimensional and hence $(2)$ is wrong.

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How could assume $\ker(f) \subsetneq \ker(g)$ if this is not possible when $f,g$ are nonzero linear functionals? – user26857 Oct 26 '12 at 18:32

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