Prove that a linear map is an isomorphism iff the image set of the basis of V is a basis of W. So, here's the result that I'm trying to prove:

Let $V$ and $W$ be vector spaces over $F$ and let $(v_1,v_2,...,v_n)$
  be a basis of $V$. Then, a linear map $f:V \to W$ is an isomorphism if
  and only if the list $(f(v_1),f(v_2),...,f(v_n))$ is a basis of $W$.


Proof Attempt:
Let $A = (v_1,v_2,....,v_n)$ and $B = (f(v_1),f(v_2),....,f(v_n))$. 
Then, we have to show that the list B is linearly independent and generates $W$. Consider the following linear combination:
$$\sum_{k=1}^{n}\alpha_k f(v_k) = 0\\
\implies \sum_{k=1}^{n} f(\alpha_k v_k) = 0\\
\implies f[\sum_{k=1}^{n} \alpha_k v_k] = 0\\
\implies \sum_{k=1}^{n} \alpha_k v_k = 0\\
\implies \forall k \in \{1,2,...,n\}: \alpha_k = 0$$
In the above computation, we have used the linearity of $f$ and the fact that $f(0) = 0$, alongside the bijectivity of $f$. This proves the linear independence of the list of vectors in $B$. 
Then, we need to show that it generates $W$. Clearly, $L(B) \subset W$. Let $w \in W$. Then, there exists a unique element $v \in V$ such that $f(v) = w$. So, we have:
$$w = f(v) = f[\sum_{k=1}^{n} \alpha_k v_k]\\
\implies w = \sum_{k=1}^{n} f(\alpha_k v_k)\\
\implies w = \sum_{k=1}^{n} \alpha_k f(v_k)\\
\implies w \in L(B).$$ 
This proves that $W \subset L(B)$ and, thus, proves that $L(B) = W$. This shows that the list $B$ is a basis of $W$. 
Now, suppose that B is a basis of W. Then, we have to show that $f:V \to W$ is bijective. Let $u_1,u_2 \in V$ such that:
$f(u_1) = f(u_2)$
Then, $u_1 = \sum_{k=1}^{n} \alpha_k v_k$ and $u_2 = \sum_{k=1}^{n} \beta_k v_k$. So, we have:
$f(u_1)-f(u_2) = f(u_1-u_2) = 0$
$\implies f[\sum_{k=1}^{n} (\alpha_k - \beta_k)v_k)] = 0$
$\implies \sum_{k=1}^{n} (\alpha_k-\beta_k)f(v_k) =0$
$\implies \forall k \in \{1,2,...,n\}: \alpha_k = \beta_k$
Hence, that proves that $u_1 = u_2$. This proves the injectivity of $f$. Then, consider $f(V)$. Clearly, $f(V) \subset W$. Let $w \in W$. Then, we have:
$$w = \sum_{k=1}^{n} \alpha_k f(v_k)\\
\implies w = \sum_{k=1}^{n} f(\alpha_k v_k)\\
\implies w = f[\sum_{k=1}^{n} \alpha_k v_k]\\
\implies w \in f(V)$$. 
This proves that $W \subset f(V)$ and, therefore, $f(V) = W$. That proves that f is surjective. Since f is injective and surjective, it follows that it is bijective and that proves that it is an isomorphism. 
I'd like some serious criticism on my proof above. Is it correct? Is my proof writing okay? How can I improve it? 
 A: The first part of the implication $(\Rightarrow)$ sounds good. As to its second part, you may approach it alternatively as follows. Let $\mathcal{B}_{V} = \{v_{1},v_{2},\ldots,v_{n}\}$ be a basis for $V$ and let $\mathcal{B}_{W} = \{w_{1},w_{2},\ldots,w_{m}\}$ be a basis for $W$. Since $f$ is an isomorphim, we conclude that $f$ is injective as well as $f^{-1}$.
Once injective linear mappings take LI sets into LI sets, we conclude $f(\mathcal{B}_{V}) = \{f(v_{1}),f(v_{2}),\ldots,f(v_{n})\}$ as well as $f^{-1}(\mathcal{B}_{W}) = \{f^{-1}(w_{1}),f^{-1}(w_{2}),\ldots,f^{-1}(w_{m})\}$ are LI. But $\mathcal{B}_{V}$ spans $V$ and $\mathcal{B}_{W}$ spans $W$, whence we conclude that $n\leq m$ and $m\leq n$, that is to say, $m = n$.
The result which supports this statement is:

Let $V$ be a finite dimensional vector space and suppose that $S = \{s_{1},s_{2},\ldots,s_{n}\}$ spans $V$. Then any set of linear independent vectors has no more than $n$ vectors.

Finally, once $f(\mathcal{B}_{V}) = \{f(v_{1}),f(v_{2}),\ldots,f(v_{n})\}$ is LI and $\dim W = n$, we conclude $f(\mathcal{B}_{V})$ is a basis for $W$ indeed.
As to the implication $(\Leftarrow)$, I think it suffices to show that $f$ is injective, since $\dim V = \dim W = n$ by assumption. Indeed, according to the  rank-nullity theorem, one has that
\begin{align*}
\dim V = \dim W = \dim f^{-1}(\{0\}) + \dim f(V)
\end{align*}
Thus, if $f$ is injective, we conclude that $\dim W = \dim f(V)$, and since $f(V)$ is a subspace of $W$, the surjectivity holds.
In order to prove it, it is enough to check that $f$ is nonsingular, that is to say, $f(x) = 0$ implies that $x = 0$. Indeed, this is the case
\begin{align*}
f(x) & = f(x_{1}v_{1} + x_{2}v_{2} + \ldots + x_{n}v_{n})\\\\
& = x_{1}f(v_{1}) + x_{2}f(v_{2}) + \ldots + x_{n}f(v_{n}) = 0\\\\
& \Rightarrow x_{1} = x_{2} = \ldots = x_{n} = 0 
\end{align*}
because $\{f(v_{1}),f(v_{2}),\ldots,f(v_{n})\}$ is LI, and we are done.
