Question about an application of the Rank-Nullity theorem

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 $$K \subset W$$,

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

• Consider the map from $f^{-1}(K)$ to $K$ induced by $f$. Sep 30, 2011 at 17:27
• Any vector space complement to $Ker(f)$ inside $f^{-1}(K)$ maps isomorphically to $K\cap Im(f)$. Sep 30, 2011 at 17:28
• 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. Sep 30, 2011 at 17:35
• @Kevin What if $f$ is singular? Does that imply $\dim(f^{-1}(K)) = \infty$? Oct 1, 2011 at 2:20
• @user7980 Your vector spaces in question may not even be finite dimensional. How can you apply Rank-nullity?
– user38268
Oct 4, 2011 at 23:22

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 $U \supseteq \ker f$.