I've had a question on a problem sheet which I couldn't do at the time and am now struggling to understand the answer I've been given. The question is about proving that a map between two vector spaces is a linear Isomorphism. We can prove that the map is linear quite easily. We then need to prove that it's isomorphic.
The answer gives that the way to do this is to prove that the map is Bijective (which I understand), but it then jumps straight from the fact that having proven the map is simply injective, we can see that it is a linear isomorphism by the rank-nullity theorem.
Is this the case with all linear injective maps? Is there some point that I'm missing about the linear map being injective that makes it automatically bijective?
Many Thanks
Best Answer
If you have a linear map $f : K^n \longrightarrow K^n$, the rank-nullity theorem tells you that the following are equivalent:
$f$ is injective, i.e. $\ker f = 0$
$f$ is surjective, i. e. $Im f = K^n$
infact $$n = \dim Im(f) + \dim \ker f$$ implies that $$\dim \ker f = 0 \Leftrightarrow \dim Im f = n$$