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Is there any vector space which is easy to proven to be a vector space but whose dimension is nontrivial to determine?

I want such a case as an exercise for the students.

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    $\begingroup$ te solutions of a linear system $Ax = 0$, where the rank of $A$ is not trivial to determine ? $\endgroup$ Nov 7 '15 at 10:41
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    $\begingroup$ Solution space of a linear differential equation? $\endgroup$
    – user99914
    Nov 7 '15 at 10:50
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    $\begingroup$ The span, over the rationals, of the zeros of the zeta function in the critical strip but off the half-line. $\endgroup$ Nov 7 '15 at 11:42
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    $\begingroup$ Not sure why this is getting close votes--the context is quite clear from the final sentence of the question. $\endgroup$ Nov 10 '15 at 5:33
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I think that the dual space of a finite-dimensional vector space is a good example for that.

What qualifies as a non-trivial method to determine the dimension? As @Tlön Uqbar Orbis Tertius mentioned in his comment, this can mean 'not easy to calculate the dimension by hand', in the case of a vector space given by ker($A$), but then the dimension is determined by rank $A$, even if it's hard to compute.

Another way to interpret the question is a vector space where you know that it's a vector space (because it's a subspace of a known vector space), but you don't immediately see the dimension.

Finite-dimensional spaces of functions come to mind here, and as a variant of the non-full-rank matrix, you could take a space spanned by some polynomials which are not linearly independent.

But among the sub-spaces of functions there is one lovely concept where you actually need to do some work to get the dimension: the dual space of a finite-dimensional space. The proof is not too long (Prove that vector space and dual space have same dimension).

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  • $\begingroup$ Thanks! This is really a good example. Personally I indeed struggled to prove this simple yet important fact. $\endgroup$
    – kaiser
    Nov 7 '15 at 11:10

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