Another aspect of Rank Nullity theorem of linear transformations

Question

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.

Answer 1

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.