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

Dimension of infinite product of vector spaces

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

  • $\begingroup$ Dear Fernando: Thanks for you answer. I especially enjoyed the answer mathoverflow.net/a/49572/461 of Todd Trimble's to the question you linked to! $\endgroup$ May 30 '14 at 9:15
  • $\begingroup$ @Pierre-YvesGaillard you're welcome. I think, however, that my answer has little value, it is just a corollary of Todd's. $\endgroup$ May 30 '14 at 14:38
  • $\begingroup$ To me, it was invaluable! (Also I now realize - thanks to you, Todd and François - how easy it was to answer my question using the Erdős-Kaplansky Theorem, and how silly my approach was...) $\endgroup$ May 30 '14 at 14:56

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