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Let $V,W$ be two vector spaces over a field $F$ and let $f:V \to W$ be a linear mapping.

I have going over some facts about vector spaces (not necessarily finite dimensional) and I have been trying to apply the rank nullity theorem to derive a simple lemma. I have not been able to complete the proof and was wondering if I need to use something other then $\ker f + \mathrm{Im}\; f = \dim V$.

How do we show that for ever vector subspace $K \subset W$,

$ \dim(f^{-1}(K)) = \dim(K \cap \mathrm{Im}\; f) + \dim( \mathrm{Ker}\; f)$?

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Consider the map from $f^{-1}(K)$ to $K$ induced by $f$. –  Pierre-Yves Gaillard Sep 30 '11 at 17:27
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Any vector space complement to $Ker(f)$ inside $f^{-1}(K)$ maps isomorphically to $K\cap Im(f)$. –  Kevin Sep 30 '11 at 17:28
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You don't need to title all your questions "Question about". A reader already knows they are questions about something; otherwise they wouldn't appear here. –  Henning Makholm Sep 30 '11 at 17:35
    
@Kevin What if $f$ is singular? Does that imply $\dim(f^{-1}(K)) = \infty$? –  user7980 Oct 1 '11 at 2:20
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@user7980 Your vector spaces in question may not even be finite dimensional. How can you apply Rank-nullity? –  user38268 Oct 4 '11 at 23:22

1 Answer 1

up vote 1 down vote accepted

Let $U=f^{-1}(K)$ and consider $g=f|U$, the restriction of $f$ to $U$. Then $g:U \to W$ is a linear map and the rank nullity theorem gives $\dim U = \dim \mathrm{im} g + \dim \ker g$. Now $\mathrm{im} g = g(U) = f(U) = K \cap \mathrm{im} f$ and $\ker g = \ker f \cap U = \ker f$, since $U \supseteq \ker f$.

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