# dim Hom(V,W) =?

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 $$d(K,\alpha,\beta)=\kappa^\alpha.$$

Using $\hom_K(V,W)=W^\alpha$ and
we get $\dim \hom_K(V,W)=(\#W)^\alpha=((\# K)\cdot\beta)^\alpha$.