1
$\begingroup$

Let $E,F$ be two vector spaces over a field $k$. $L(E,F)$ is the $k$-vector space of linear maps $E \rightarrow F$.

In ZFC, is there a functionnal relation between $dim(L(E,F))$ and $(dim(E),dim(F),|E|,|F|)$, and if so, what is it?

I know the answer in the case when $E$ and $F$ are finite dimensional.

If not, I also know that $Max(|k|,dim(L(E,F))) = |F|^{dim(E)}$ but I do not think that $dim(L(E,F)) \geq |k|$ holds for every $E,F,k$. (edit: I thought I had a kind of counter example but it turns out it was quite false!)

However, I wonder if there is a "better" relation when $|F|^{dim(E)} = |k|$.

Improve this question
$\endgroup$
3
  • $\begingroup$ As you know, for $E=F=k$ the space $L(E,F)$ is one-dimensional. In general, if $B$ and $C$ are bases of $E$ and $F$ I think that you have quite naturally $L(E,F)\cong F^B \cong \lbrace \phi:B\times C\to k:$ for all $b\in B$ only finitely many $c\in C$ satisfy $\phi(b,c) \neq 0\rbrace$. My feeling is that the dimension of this space should only depend on $B$ and $C$. $\endgroup$
    – Jochen Wengenroth
    Commented Feb 26, 2014 at 16:43
  • $\begingroup$ This is my feeling too, though I think I would have more trouble dealing with the last of your isomorphic spaces than with the other two. Unfortunately is difficult to find examples or counter examples as I know the dimension of no $L(E,F)$ with $dim(E),dim(F) \geq \aleph_0$. $\endgroup$
    – nombre
    Commented Feb 26, 2014 at 18:22
  • $\begingroup$ Bourbaki's exercises have related material (Erdos-Kaplansky). For each basis vector in the domain, the image is an arbitrary vector of the codomain, so you have a direct product of copies of the codomain, one for each basis vector of the domain ... $\endgroup$
    – user46855
    Commented Mar 1, 2014 at 4:34

0

Reset to default

You must log in to answer this question.

Browse other questions tagged .