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Let $V$ be an infinite dimensional vector space over a field $\mathbb{F}$ such that its dimension over $\mathbb{F}$ is a cardinal $k$.

Is true that the dimension of $End(V)$ as $\mathbb{F}$ vector space is again $k$?

EDIT: egreg answer below proves that if $k \geq |\mathbb{F}|$ the statement is false. But what happens in other cases? In particular what if we set $\mathbb{F}=\mathbb{C}$ or $\mathbb{R}$?

Sabino Di Trani
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  • OP asks for infinite dimensions. And yeah, the claim is correct. – Levent Dec 11 '17 at 21:16
  • Is $k$ the dimension of $V$ over $\mathbb K$ ? – Tom-Tom Dec 11 '17 at 21:17
  • Related – Kenny Lau Dec 11 '17 at 21:19
  • What's $\Bbb{K}$? Is it a typo for $\Bbb{F}$? If so, then the claim is false: e.g., take $\Bbb{F}$ to be any finite or countable field and take $k = \omega$. Then if $V$ is an $\Bbb{F}$-vector space of dimension $k$, $V$ is countable, but $\mbox{End}(V)$ is uncountable (since the uncountably many functions from $k$ to itself induce uncountably many distinct endomorphisms of $V$). – Rob Arthan Dec 11 '17 at 21:34
  • Sorry but I'm not talking of the cardinality of $End(V)$, I'm looking at its dimension as $\mathbb{F}$-vector space. – Sabino Di Trani Dec 11 '17 at 22:38
  • I should have explained that in my example, the cardinality of $\mbox{End}(V)$ is the same as its dimension. I see egreg has written this up as an answer. – Rob Arthan Dec 12 '17 at 16:19

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Let's do a specific case, with $\mathbb{F}$ the rational field or a finite field. Then, for an infinite dimensional vector space $V$, we have $$ \dim V=|V| $$ because an element of $V$ is a finite linear combination of elements of a basis and the choice of the scalars is in a countable (or finite) set.

Note now that $\lvert\operatorname{End}(V)\rvert=|V^k|$, because any endomorphism of $V$ is defined by an arbitrary function from the basis to $V$. Thus $$ \dim\operatorname{End}(V)=\lvert\operatorname{End}(V)\rvert= k^k>k=\dim V $$ The same can be said whenever $k\ge|\mathbb{F}|$.

egreg
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