Proof about L(U,V). Is this reasoning correct?

Question

If $U\cong U'$ and $V\cong V'$ then $L(U,V)\cong L(U',V')$
Proof:
Since $U\cong U'$ and $V\cong V'$ then $$dim_{\mathbb{K}}U=dim_{\mathbb{K}}U'=m$$ and $$dim_{\mathbb{K}}V=dim_{\mathbb{K}}V'=n.$$ Due to the above, it is necessary $$dim_{\mathbb{K}}L(U,V)=mn=dim_{\mathbb{K}}L(U',V')$$ then $$L(U,V)\cong L(U',V')$$
Is this reasoning correct? If not, any suggestions?

Answer 1

I assume $\mathbb{K}$ is a field and $U,U',V,V'$ are $\mathbb{K}$-vector spaces. In this case, your reasoning is exactly right, as dimension clasasifies finite-dimensional vector spaces up to ismorphism. Moreover, if you have explicit isomorphisms $\varphi \colon U \to U'$ and $\psi \colon V \to V'$, you can build an explicit isomorphism $\Theta \colon L(U,V) \to L(U',V')$ by $\Theta(f) = \psi \circ f \circ \varphi^{-1}$, for $f \in L(U,V)$.