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I know that if $U$ and $V$ are vector spaces over a field $\mathbb{F}$ then $\dim_{\mathbb{F}}\mathcal{L}(U,V)\geq \dim_{\mathbb{F}}U\cdot \dim_{\mathbb{F}}V$. I am looking for an example of two infinite-dimensional vector spaces such that $\dim_{\mathbb{F}}\mathcal{L}(U,V)> \dim_{\mathbb{F}}U\cdot \dim_{\mathbb{F}}V$. I've thought a lot, but I've failed!

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If both U and V have countable dimension, then the space of linear maps between them has uncountable dimension, so you are done. –  Tobias Kildetoft Jan 23 '12 at 22:06
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up vote 9 down vote accepted

Consider $F=\mathbb Q $ and $U=\mathbb Q[X] $ (the polynomial ring, seen just as a vector space over $\mathbb Q$) .

Then $U^*=$ the dual space of $U$ is identified with $\mathbb Q^{\mathbb N}$, the vector space of sequences of rationals.
The dimension of $U^*\simeq \mathbb Q^{\mathbb N}$ is already $ {\aleph _0}^{\aleph _0}=2^{\aleph _0}=\mathfrak c$, the continuum.

Now consider the vector space $\mathcal L(U,V)$, where $V \:$ is an arbitrary infinite dimensional vector space. We have:
$\dim \: \mathcal L(U,V)\geq \dim \: \mathcal L(U,\mathbb Q)=\dim\: U^*={\aleph _0}^{\aleph _0}=\mathfrak c $ so that you will have your strict inequality $\dim \: \mathcal L(U,V)\gt \dim(U)\cdot \dim(V) $ if $V$ has dimension $\aleph_0$ too .

A weird equality
Since the key point was to calculate the dimension of a dual vector space, let me give an unexpected formula for that dimension.
Given an arbitrary field $k$ (it may be finite!) and an arbitrary infinite dimensional $k$-vector space $U$, we have the formula (where of course $\dim$ means dimension over $k$) $$\dim(U^*) =\operatorname{card} ( U^*)=\operatorname{card}(k)^{\dim(U)}$$
The weird first equality is due to Erdős-Kaplansky.
I call it weird because we expect the dimension of a vector space, the cardinality of a basis, to be much smaller than that of the vector space itself.
Indeed, in accordance with our expectation the dimension $\aleph_0$ of $\mathbb R[X]$ is much smaller than the cardinality $2^{\aleph_0}=\mathfrak c$ of $\mathbb R[X]$.
Fine, but then Erdős-Kaplansky tell us that there does not exist a real vector space whose dual is isomorphic to $\mathbb R[X]$ !

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He wanted both to be infinite dimensional, so you should take them both to be $U$. –  Tobias Kildetoft Jan 23 '12 at 22:19
    
Of course you can embed $U^*$ in $L(U,V)$ by $\phi \mapsto v \phi$ for any nonzero $v \in V$. –  Robert Israel Jan 23 '12 at 22:36
    
Dear Tobias and Robert, you are both absolutely right: apologies for my misreading. Thanks a lot for your attention and suggestions. I have written an edit, in order to answer the actual question! –  Georges Elencwajg Jan 23 '12 at 23:48
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