2
$\begingroup$

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)$?

$\endgroup$
5
  • 1
    $\begingroup$ Consider the map from $f^{-1}(K)$ to $K$ induced by $f$. $\endgroup$ Sep 30, 2011 at 17:27
  • 1
    $\begingroup$ Any vector space complement to $Ker(f)$ inside $f^{-1}(K)$ maps isomorphically to $K\cap Im(f)$. $\endgroup$ Sep 30, 2011 at 17:28
  • 2
    $\begingroup$ 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. $\endgroup$ Sep 30, 2011 at 17:35
  • $\begingroup$ @Kevin What if $f$ is singular? Does that imply $\dim(f^{-1}(K)) = \infty$? $\endgroup$
    – user7980
    Oct 1, 2011 at 2:20
  • 1
    $\begingroup$ @user7980 Your vector spaces in question may not even be finite dimensional. How can you apply Rank-nullity? $\endgroup$
    – user38268
    Oct 4, 2011 at 23:22

1 Answer 1

2
$\begingroup$

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 $U \supseteq \ker f$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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