Let $g: L \to M$ a linear transforming. $M, L$ finite dimensional. $g^{\ast} : M^{\ast} \to L^{\ast}$ How do I construct an isomorphism between $ \ker g^{\ast}$ and $coker~ g$? I really don't know what choose to build this isomorphism.

Thanks in advance.

  • $\begingroup$ I am sorry, I will edit. $\endgroup$ Feb 9, 2016 at 19:50
  • $\begingroup$ @DietrichBurde I did it. $\endgroup$ Feb 9, 2016 at 19:51
  • $\begingroup$ The result is a corollary to the fundamental theorem of linear algebra. We have $coker(g)\cong \ker (g^*)$ and $im(g^*)\cong coim (g)$. $\endgroup$ Feb 9, 2016 at 19:57
  • $\begingroup$ @DietrichBurde how to adaptat to this particular case? $\endgroup$ Feb 9, 2016 at 20:14

1 Answer 1


Start by constructing a pairing between the two spaces, and then show that it's non-degenerate. I think you need to assume the vector spaces are finite-dimensional for the claim to hold.

  • $\begingroup$ I don't know what means construct a pairing between these two spaces. I corrected about the dimension that is required be finite. $\endgroup$ Feb 9, 2016 at 19:56
  • $\begingroup$ Show how to interpret each element of $\ker(g^*)$ as a linear functional on $\mathrm{coker}(g)$. $\endgroup$ Feb 9, 2016 at 19:57
  • $\begingroup$ I will try, thanks! $\endgroup$ Feb 9, 2016 at 19:59
  • $\begingroup$ thank you! I I have followed your hint. $\endgroup$ Feb 23, 2016 at 0:11

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