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In group or ring theory we have the first isomorphism theorem which gives an isomorphism between two structures involving kernel and image of the homomorphism. The Rank Nullity theorem also has a kind of similarly if we write it like this: $$\operatorname{dim}(V)-\operatorname{dim}(\operatorname{ker}T)=\operatorname{dim}(\operatorname{Im}T)$$ So my question is, does this theorem also indicates some kind of isomorphism between two spaces? Because we know same dimensional spaces over same field are isomorphic.

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  • $\begingroup$ Of course it does. Where does the rank nullity theorem come from to begin with? From the 1st isomorphism theorem for vector spaces. $\endgroup$ Commented Jul 2, 2020 at 15:30

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Yes, the first isomorphism theorem also holds for vector spaces, and is even more general than the statement of the rank-nullity theorem you gave!

Theorem (First Isomorphism Theorem). Let $f : V \to W$ be any linear map between two vector spaces. Then $f$ induces an isomorphism $V/ \ker(f) \to \operatorname{img}(f)$.

You should definitely try to prove this! This is more general than the rank-nullity statement you gave because this holds for any vector spaces whatsoever (not just the finite-dimensional ones, as in your statement)! Also, the exact sequence $0 \to \ker(f) \to V \to \operatorname{img}(f) \to 0$ tells us that $\dim \ker(f) + \dim \operatorname{img}(f) = \dim V$, even when all three of these quantities are infinite cardinals (so we cannot subtract).

Indeed, the first isomorphism theorem for vector spaces is the "best" way to prove the rank-nullity theorem.

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  • $\begingroup$ So can I say Rank Nullity theorem is a consequence of first isomorphism theorem for finite dimensional case? $\endgroup$
    – user598858
    Commented Jul 2, 2020 at 15:51
  • $\begingroup$ Yes! (character limit) $\endgroup$ Commented Jul 2, 2020 at 15:56

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