Let $V$ be a finite-dimensional vector space over field $F$, with $\mathrm{dim}(V)=n\in \Bbb{N}$. Then $V\cong W$$\iff$ $\mathrm{dim}(V)=\mathrm{dim}(W)$.
My attempt: $(\Leftarrow)$ Suppose $\mathrm{dim}(V)=\mathrm{dim}(W)$. Approach(1): Since $\mathrm{dim}(V)=\mathrm{dim}(W)=n$, we have $V\cong F^n$ and $W\cong F^n$, by this result. It’s easy to check $($class of vector space $,\cong )$ is an equivalence relation. By symmetry property, $W\cong F^n$ $\Rightarrow$ $F^{n}\cong W$. By transitive property, $V\cong F^n$ and $F^n\cong W$ $\Rightarrow$ $V\cong W$. Hence $V\cong W$. Approach(2): Since $\mathrm{dim}(V)=\mathrm{dim}(W)=n$, we have $\exists B_V=\{\alpha_1,…\alpha_n\}$ basis of $V$, $\exists B_W=\{\beta_1,…,\beta_n\}$ basis of $W$. By theorem 1 section 3.1, $\exists!$ $T\in L(V,W)$ such that $T(\alpha_j)=\beta_j$, $\forall j\in J_n$. So $\{T(\alpha_1),…,T(\alpha_n)\}$ $=\{\beta_1,…,\beta_n\}$ is basis of $W$. By theorem 9 section 3.2, $T$ is bijective. Thus $T$ is isomorphism. Hence $V\cong W$.
$(\Rightarrow)$ Suppose $V\cong W$. Then $\exists T\in L(V,W)$ such that $T$ is bijective. So $N_T=\{0_V\}$ and $R_T=W$. By rank nullity theorem, $\mathrm{dim}(V)$ $= \mathrm{dim}(N_T)+ \mathrm{dim}(R_T)$ $= \mathrm{dim}(W)$. Hence $\mathrm{dim}(V)= \mathrm{dim}(W)$. Is my proof correct?
In proof of $(\Rightarrow)$, we implicitly use $T$ is a linear map, because without linear condition we can’t conclude $N_T=\{0_V\}$. Am I right?