Please comment on the validity of my proof. Any tips and suggestions are appreciated.
Prove: If two finite-dimensional vector spaces are isomorphic, then they have the same dimension.
Let $B=${$b_1,…,b_n$} be a basis for a finite-dimensional vector space $V$. Let $W$ be a finite-dimensional vector space, and define $T: V→W$ as an isomorphism between them.
Let $x∈V$, then $T(x)=T(c_1b_1+…+c_nb_n)$ for some real scalars $c_1…c_n$ as $B$ is a basis for $V$. Also, because $T$ is an isomorphism, $T$ is a linear transformation, onto, and one-to-one.
Because $T$ is onto, for any $y∈W$, $y=T(x)=T(c_1b_1+…+c_nb_n)=c_1T(b_1)+…+c_nT(b_n)$. Thereofore, $span${$T(b_1),…T(b_n)$}=$W$.
Let $c_1T(b_1)+…+c_nT(b_n)=0$. Because $T$ is one-to-one and linear $c_1T(b_1)+…+c_nT(b_n)=T(c_1b_1+…+c_nb_n)=0$ implies that $c_1b_1+…+c_nb_n=0$. But because $b_1…b_n$ is a basis, it is linearly independent and so $c_1=…=c_n=0$. Therefore {$T(b_1),…T(b_n)$} is a linearly independent set.
So, {$T(b_1),…T(b_n)$} is a basis for $W$ and is clearly $n$ dimensional. So, $dim(V)=dim(W)$.
Best Answer
Your proof is correct. It is also written down very nicely and formally clean. I like it and do not have anything to add. I honestly think there isn't much to add.
Only suggestion I have is to replace onto and one to one with surjective and injective, as I think those are more widely understood. But then again, I'm not from a english speaking country so maybe I'm wrong about that and it's probably personal taste.