I asked this question on Mathematics Stack Exchange, but got no answer:
Given two vector spaces $V$ and $W$ over a field $K$, what is the dimension of $\operatorname{Hom}_K(V,W)\ $?
To state the partial result I've been able to obtain, let me introduce the notation $$ \alpha:=\dim V,\quad\beta:=\dim W,\quad d(K,\alpha,\beta):=\dim\operatorname{Hom}_K(V,W), $$ $$ \quad\nu:=\operatorname{card}(\mathbb N),\quad\kappa:=\operatorname{card}(K). $$ We can assume $\alpha\ge\nu$.
By the Erdős-Kaplansky Theorem and the inequality $$ \dim\operatorname{Hom}_K(\oplus V_i,\oplus W_j)\le\dim\prod\operatorname{Hom}_K(V_i,W_j), $$ we can also assume $\alpha < \beta$, and we get $$ \kappa^\alpha\beta\le d(K,\alpha,\beta)\le\kappa^\beta $$ (for $\nu\le\alpha < \beta$). Indeed, $\kappa^\alpha\beta$ is the dimension of the space of finite rank linear maps from $V$ to $W$.
For the sake of completeness, let us add explicitly that $$ 1\le\beta\le\alpha\ge\nu $$ implies $$ d(K,\alpha,\beta)=\kappa^\alpha. $$
