I was reading a book of linear algebra and started the chapter of isomorphisms. The book describes them as linear maps that are injective and surjective. Then I realised that that condition is the same as invertibility of the matrix of the linear map, but I don't know how to prove it. So, my idea is:
Given a linear map $f: E\to F$ (with $dim(F) = dim(E)$ and basis $B$ of $E$ and $B'$ of $F$, I want to proof that $f$ is an isomorphism if and only if it is invertible. That is, I want to prove they are equivalent. In some way I kind of "feel" it, but don't know how to proceed.
Here is my attempt for the right-to-left implication: we assume $f$ is invertible. Then we want to show it is injective and surjective. Let $u,v \in E$, and $f(u) = f(v)$, then $u = f^{-1}(f(u)) = f^{-1}(f(v)) = v$, therefore $f$ is injective. To prove it is surjective we want to prove that $range(f) = F$. Let $w \in F$, then take $v = f^{-1}(w)$, the existance and uniqueness of this element is guaranteed by the invertibility of the inverse of $f$. Then $f(v) = f(f^{-1}(w)) = w$. So $range(f) = F$.
However, for the other direction I don't know how to do it. Can somebody help me?
