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Let $V_1$, $V_2$, $V_3$, $V_4$ be linear spaces over the same field K

$$f_1: V_1 \to V_2 , f_2: V_2 \to V_3 , f_3: V_3 \to V_4 , f_4: V_4 \to V_1$$

with

$$\mbox{im}(f_1) = \ker(f_2) , \mbox{im}(f_2) = \ker(f_3) , \mbox{im}(f_3) = \ker(f_4) , \mbox{im}(f_4) = \ker(f_1)$$

Show that:

$$\dim(V_1) - \dim(V_2) + \dim(V_3) - \dim(V_4) = 0$$

I know I have to show it by applying dimension rules, but I have no idea how to practically do it.

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    $\begingroup$ this is why it is nice to know what an exact sequence is ;) $\endgroup$ Commented Nov 16, 2013 at 19:18

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Well, use the dimension (rank-nullity) theorem, which shows that:

$$\dim(V_{1}) - \dim(V_{2}) + \dim(V_{3}) - \dim(V_{4}) = $$$$(im(f_{1})+\ker(f_{1}))-(im(f_{2})+\ker(f_{2})) + (im(f_{3}) + \ker(f_{3})) - (im(f_{4})+\ker(f_{4}))$$$$ = im(f_{1}) + im(f_{4}) - im(f_{2}) - im(f_{1}) + im(f_{3}) + im(f_{2}) - im(f_{4}) - im(f_{3}) = 0$$

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  • $\begingroup$ thanks, unfortunately I did not know this theorem. $\endgroup$
    – sj134
    Commented Nov 16, 2013 at 19:44
  • $\begingroup$ @sj134 If you've seen the first isomorphism theorem for groups, then the rank-nullity theorem is just a corollary when restricted to finite dimensional vector spaces. $\endgroup$
    – Dan Rust
    Commented Nov 16, 2013 at 19:57
  • $\begingroup$ The rank nullity theorem holds of course also for infinite dimensional vector spaces. $\endgroup$ Commented Nov 16, 2013 at 20:49

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