Let $V$ and $W$ be finite-dimensional vector spaces over the field $\textbf{F}$. Prove that $V$ and $W$ are isomorphic if and only if $\dim V = \dim W$.
MY ATTEMPT
If $V$ and $W$ are isomorphic, there exists a bijective linear transformation $T:V\rightarrow W$. Since $T$ is bijective, it is invertible. Once invertible linear transformations are non-singular, it takes linear independent sets into linear independent sets.
More precisely, let $\{\alpha_1,\alpha_2,\ldots,\alpha_n\}$ be a basis for $V$ and let $\{\beta_1,\beta_2,\ldots,\beta_m\}$ be a basis for $W$. Then $\{T\alpha_1,T\alpha_2,\ldots,T\alpha_n\}$ is linear independent. Since $\{\beta_1,\beta_2,\ldots,\beta_m\}$ spans $W$, we conclude that $n\leq m$. The same reasoning applied to $T^{-1}$ implies that $m\leq n$. Thus $m = n$.
Now let us assume that $n = \dim V = \dim W$. Therefore there is a basis $\{\alpha_{1},\alpha_{2},\ldots,\alpha_{n}\}$ for $V$ and a basis $\{\beta_{1},\beta_{2},\ldots,\beta_{n}\}$ for $W$. Let us set the linear mapping $T\alpha_{j} = \beta_{j}$, which exists and is unique. Thus it is injective, since \begin{align*} T\alpha = T(a_{1}\alpha_{1} + a_{2}\alpha_{2} + \ldots + a_{n}\alpha_{n}) & = a_{1}T\alpha_{1} + a_{2}T\alpha_{2} + \ldots + a_{n}T\alpha_{n}\\\\ & = a_{1}\beta_1 + a_2 \beta_2 + \ldots a_n \beta_n = 0\\\\ &\Rightarrow a_1 = \ldots = a_n = 0 \Rightarrow \alpha = 0 \end{align*} Given that $\dim V = \dim W = \dim N + \dim R = \dim R$, $T$ is also surjective. Hence $T$ is a bijection.
Am I reasoning correctly? Any other approach would be welcome.