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I think you get a basis of an n-dimensional vector space by the definition of dimension (if e.g. the dimension is the maximum size of a set of linearly independ

I think you get a basis of an n-dimensional vector space by the definition of dimension (if e.g. the dimension is the maximum size of a set of linearly independent vectors). It's a bit trickier when the dimension is infinite, but again most definitions of dimension require there to be a basis of a particular size, the difficult part is proving that this makes sense (i.e. the dimension is unique and defined).

zodiac 9y ago
Yes, it is easy to prove that "a basis exists" (almost by definition depending on how exactly you define dimension) and of course it's legal to use it in proofs, but I think proofs that avoid using this fact directly are more elegant
- Bromskloss 9y ago
  > almost by definition depending on how exactly you define dimension What definition of _dimension_ is there that does not rely upon the existence of a base?
  - zodiac 9y ago
    Does "maximum size of a linearly independent set" count? You would have to prove something about spanning sets to get to the existence of a basis
  - contravariant 9y ago
    You could define the dimension to be the supremum of the cardinality of all linearly independent sets. In the infinite case this does not trivially give you a basis as 1) the supremum could be strictly larger than the cardinality of all linearly independent sets and 2) adding an extra vector to an infinite linearly independent set doesn't increase it's cardinality, hence there is no reason for the basis to span the entire space. You could also take the infimum of all sets that span the entire space, but you run into similar problems.
Sniffnoy 9y ago
I mean, in order to define dimension, we have to know that bases exist (and all have the same cardinality). Which means that before I can use the notion of "dimension", I have to know about bases. I mean I suppose you could just restrict the definition of dimension to vector spaces that have bases, and then you wouldn't have to. I guess that's what you're implicitly suggesting, that dimension would just not apply to vector spaces that don't have bases. That would make sense. But I'm used to thinking of dimension as, well, a function of vector spaces, not a partial function, so as I was thinking of it, you have to prove they all have bases before you can use it!
cgmg 9y ago
> It's a bit trickier when the dimension is infinite For those who are wondering, 'every vector space has a basis' is equivalent to the Axiom of Choice.

zodiac 9y ago

Yes, it is easy to prove that "a basis exists" (almost by definition depending on how exactly you define dimension) and of course it's legal to use it in proofs, but I think proofs that avoid using this fact directly are more elegant

Bromskloss 9y ago
> almost by definition depending on how exactly you define dimension What definition of _dimension_ is there that does not rely upon the existence of a base?
- zodiac 9y ago
  Does "maximum size of a linearly independent set" count? You would have to prove something about spanning sets to get to the existence of a basis
- contravariant 9y ago
  You could define the dimension to be the supremum of the cardinality of all linearly independent sets. In the infinite case this does not trivially give you a basis as 1) the supremum could be strictly larger than the cardinality of all linearly independent sets and 2) adding an extra vector to an infinite linearly independent set doesn't increase it's cardinality, hence there is no reason for the basis to span the entire space. You could also take the infimum of all sets that span the entire space, but you run into similar problems.

Sniffnoy 9y ago

I mean, in order to define dimension, we have to know that bases exist (and all have the same cardinality). Which means that before I can use the notion of "dimension", I have to know about bases. I mean I suppose you could just restrict the definition of dimension to vector spaces that have bases, and then you wouldn't have to. I guess that's what you're implicitly suggesting, that dimension would just not apply to vector spaces that don't have bases. That would make sense. But I'm used to thinking of dimension as, well, a function of vector spaces, not a partial function, so as I was thinking of it, you have to prove they all have bases before you can use it!

cgmg 9y ago

> It's a bit trickier when the dimension is infinite For those who are wondering, 'every vector space has a basis' is equivalent to the Axiom of Choice.