# Another aspect of Rank Nullity theorem of linear transformations

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.

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

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).