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Is there a relation between the cardinality of the basis of a vector space $V$ over $k$ and the cardinality of the basis of $\operatorname{End}{V}$, the set of $k$-linear endormophisms of $V$, over $k$? Please give me some hints, thanks.

(I know that if $\dim{V}$ is finite then $\operatorname{End}{V}$ has dimension $(\dim{V})^2$. But the standard basis used here, viz the 'Kronecker delta' linear maps, doesn't work if $\dim{V}$ is infinite because for example the identity linear map from $k^\mathbb{N}$ to itself is not a finite linear combination of these Kronecker delta maps...)

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Let $V$ be a $\mathbb K$-vector space, $char(\mathbb K)=0$, such that $dimV=\nu$ and $|\mathbb K|=\kappa$, with $\nu$ and $\kappa$ cardinals. I prove that, if $\nu$ is infinite, if $\kappa^\nu > \kappa$ then $dimEndV=\kappa^\nu$. In the case were $\kappa^\nu = \kappa$ i'm not able to prove it, but i think that $dimEndV$ is still $\kappa^\nu =\kappa$.

Let $\mathcal{B}$ be a base of $V$, $|\mathcal{B}|=\nu$. I use that if $V$ is a $\mathbb K$-vector space and $dimV$ is infinite then $|V|=\kappa\cdot\nu$. Infact we have a bijection between $V$ and $$\mathop{\cup}_{n\in\mathbb{N}}\left([\mathcal{B}]^n\times\,(\mathbb K^*)^n\right),$$

where $[\mathcal{B}]^n = \{U\subseteq\mathcal{B} : |U|=n\}$ and $(\mathbb K^*)^n = (\mathbb K^*)\times...\times(\mathbb K^*)$ $n$ times.

We have that $|EndV|=|Fun(\mathcal{B},V)|=|V|^\nu=(\kappa\cdot\nu)^\nu=\kappa^\nu\cdot\nu^\nu=\kappa^\nu\cdot2^\nu=\kappa^\nu$, so, because $\kappa^\nu > \kappa$ we have that $\kappa^\nu=|EndV|=dimEndV\cdot|\mathbb K|=dimEndV\cdot\kappa$ if and only if $dimEndV=\kappa^\nu$.

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