I asked this [question](https://math.stackexchange.com/q/813334/660) 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](https://math.stackexchange.com/a/61537/660) 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
$$
d(K,\alpha,\beta)=\kappa^\alpha.
$$



dim Hom(V,W) =?
Using $\hom\_K(V,W)=W^\alpha$ and 


[Dimension of infinite product of vector spaces](https://mathoverflow.net/questions/49551/dimension-of-infinite-product-of-vector-spaces) 


we get $\dim \hom\_K(V,W)=(\#W)^\alpha=((\# K)\cdot\beta)^\alpha$.


