I'm trying to prove that two vector spaces $V$ and $W$, where $\dim(V) = \dim(W) = n$ and $n$ is finite, will be isomorphic to each other. I've constructed the proof below and think I have convinced myself that it is correct. However, I'm wondering if there's a simpler way or if there is a fault in my logic.
Let $\{v_1, \cdots, v_n\}$ be a basis for $V$ and $\{w_1, \cdots, w_n\}$ be a basis for $W$. We must find an isomorphism/bijection between $V$ and $W$ for the two spaces to be isomorphic.
Let $T : V \to W$ be a linear transformation such that $T(v_i) = w_i$. It is trivial to show that this is a linear transformation (satisfies homogeneity and additivity).
We must show that $T$ is an injection and surjection in order to be an isomorphism.
Prove $T$ is an injection:
Any $x : V$ can be represented as the linear combination of a basis for $V$: $x = c_1v_1 + \cdots + c_nv_n$. Thus:
$$T(x) = c_1T(v_1) + \cdots + c_nT(v_n) = c_1w_1 + \cdots + c_nw_n = y$$
We know that $x : V$ and $y : W$ can only equal $0$ if $c_j = 0$ for $1 \leq j \leq n$, since they are the coordinates for the linear combination of a basis. This means that $\ker(T) = \{0\}$. Thus, $T$ is injective.
Prove $T$ is a surjection:
$\dim(V) = \dim(W) = \text{rank}(T) + \text{nullity}(T) = \text{rank}(T)$
This tells us that $\dim(W) = \text{rank}(T)$. Thus, $T$ must be surjective since all vectors in the output space are in the range of $T$.
This shows $T$ is an injection and thus $\dim(V) = \dim(W) \Rightarrow V$ and $W$ are isomoprhic.
This is from an exercise out of 8.3 of Anton and Rorre's 11th edition Elementary Linear Algebra.