Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Is there a canonical way to define on any vector space over $\mathbb{K}=\mathbb{R},\mathbb{C}$ a norm ? (Or, if there isn't, can someone give me an example of a vector space over $\mathbb{K}$ that is not normable ?)

I have now looked through several books on the subject but nowhere is something like this mentioned and I also wasn't able to find a way to construct such norm (or to find a counterexample).

share|improve this question
6  
There isn't a "canonical" way. But there are a lot of ways, in general. For example, as in Florian's answer. –  George Sep 8 '11 at 9:15
    
And without Choice? –  PseudoNeo Sep 8 '11 at 12:56
    
@PSeudoNeo The questions, answers, and links starting at math.stackexchange.com/questions/207990/vector-spaces-and-ac seem to indicate that under AC, every vector space has a Hamel basis, and without AC, there exists a vector space without a Hamel basis –  Stefan Smith Mar 10 '13 at 0:30

2 Answers 2

up vote 18 down vote accepted

Pick a basis $B$ (in the algebraic sense, also known as a Hamel basis), so any vector can be uniquely written as $\sum_{b\in B}\lambda_b b$, with only finitely many of the $\lambda_b$ being nonzero. Define for instance $$\left \|\sum_{b\in B}\lambda_b b\right \| := \max _{b\in B} |\lambda_b|$$ (another possibility would be $\sum_{b\in B} |\lambda_b|$ instead of taking the maximum).

share|improve this answer
    
@anon: You can assign whatever nonzero values you please to each $|\lambda_b|$. For example, declare each to be $1$. –  George Sep 8 '11 at 9:46
    
@Pierre: ...right. –  anon Sep 8 '11 at 9:52
    
Well, are you assuming that the vector space is finite-dimensional? –  Srivatsan Sep 8 '11 at 11:34
7  
@Srivatsan: No, this works for every vector space, finite-dimensional or not. Note that every vector is (uniquely) a finite linear combination of basis vectors, so this quantity is always finite (and thus a norm). –  t.b. Sep 8 '11 at 11:45
3  
When choosing a "basis" in the infinite-dimensional case, perhapas it is best to say "Hamel basis" so everyone knows (or can look up) what you mean. –  GEdgar Sep 8 '11 at 12:28

Try books on the topic of "topological vector spaces": It is a theorem that every finite dimensional real or complex vector space has a norm, and that all norms are equivalent.

Correspondingly, there are infinite dimensional topological vector spaces that don't have a norm that induces the topology.

Canonical literature:

share|improve this answer
5  
Maybe it would be worth emphasizing "infinite dimensional topological vector spaces that don't have a norm that induces the topology" since the question itself doesn't involve a topology, a priori. So yours and Florian's are equally valid answers, I guess. –  t.b. Sep 8 '11 at 11:58
    
Right, and thanks for adding the links to google books, BTW. –  Tim van Beek Sep 8 '11 at 12:55
    
It is also a theorem that every finite-dimensional real or complex vector space has exactly one Hausdorff topology that makes it a nondiscrete topological vector space (meaning vector addition and scalar multiplication are continuous and the topology is not the discrete topology). This is harder than the theorem that all vector space norms on a finite-dimensional real or complex vector space define the same topology on the vector space, since a priori maybe a finite-dimensional space has a nondiscrete Hausdorff topology not coming from a norm. In fact it doesn't. –  KCd Sep 9 '11 at 2:38

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.