Given two vector spaces $V,W$ over the field $\mathbb F$ with finite dimension,show that they are isomorphic if and only if $\text{dim(V)}=\text{dim(W)}$.
$\Longrightarrow$
This this direction looks pretty straightforward to me,since if the two vector spaces are isomorphic then it means that an isomorphic exists between them,which is a bijective homomorphism,and we know that if there exist a bijection between two sets then their cardinality is the same and if they are finite then $$\text{dim(V)}=\left|V\right|=\left|W\right|=\text{dim(W)}$$
$\Longleftarrow$
One can use the fact that every finite dimension vector space $V$ with $\text{dim (V)}=n$ is isomorphic to $\mathbb F^n$ and use the fact that the composition of isomorphism is another isomorphism.
I'm not sure if my proof is right,it would be nice if someone checks that,thanks.
Best Answer
In $\Longrightarrow$, $\text{dim(V)}=\left|V\right|=\left|W\right|=\text{dim(W)}$ is false since dimension of a finite dimensional vector space is the size of a basis for that vector space, not the cardinality (or size) of the vector space itself.
To prove this part, first define $B_V = \{v_1,...,v_n\}$ as a basis for $V$ with $\dim(V) = n$. Then, since $V$ and $W$ are isomorphic, there exists a linear transformation $T:V \to W$ such that $T$ is a bijection. Then, since $T$ is a linear transformation, it is uniquely determined by where it sends $v_1,...,v_n$ to. Now, you can show that $B_W = \{T(v_1),...,T(v_n)\}$ is a basis for $W$ by using the assumption that $T$ is bijective (injectivity will be useful for showing linear independence of $B_W$ and surjectivity will be useful for showing $\text{Span}(B_W) = W$).
For $\Longleftarrow$, I think your second argument is fine.