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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).

t.b.
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resu
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    There isn't a "canonical" way. But there are a lot of ways, in general. For example, as in Florian's answer. – George Sep 08 '11 at 09:15
  • And without Choice? – PseudoNeo Sep 08 '11 at 12:56
  • @PSeudoNeo The questions, answers, and links starting at http://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 00:30

3 Answers3

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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).

Florian
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  • @anon: You can assign whatever nonzero values you please to each $|\lambda_b|$. For example, declare each to be $1$. – George Sep 08 '11 at 09:46
  • @Pierre: ...right. – anon Sep 08 '11 at 09:52
  • Well, are you assuming that the vector space is finite-dimensional? – Srivatsan Sep 08 '11 at 11:34
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    @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 08 '11 at 11:45
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    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 08 '11 at 12:28
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    As far as I can tell, this process gives the vector space the initial topology w.r.t. the linear functionals or in other words, the algebraic dual of the vector space is equal to the continuous dual. It turns out in most applications for infinite-dimensional vector spaces that this is not what you want. – kahen Sep 08 '11 at 14:04
  • Here we are aussuming each vector space has a basis. Can we define norm on each vector space without assuming AC? – Sushil Jun 10 '15 at 15:02
  • Why is it so that only finitely many of $\lambda_b$ are non-zero, why can't all of them be non-zero ? – ARROW Feb 03 '24 at 09:51
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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:

t.b.
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Tim van Beek
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    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 08 '11 at 11:58
  • Right, and thanks for adding the links to google books, BTW. – Tim van Beek Sep 08 '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 09 '11 at 02:38
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The first axiom for a norm space is N1: $\left\Vert \mathbf{x}\right\Vert =0\implies \mathbf{x}=\mathbf{0}$. Now, if we take this away and have the other two viz. homogenity and the triangle inequality, then we're left with what is called a seminorm. Any vector space over $\mathbb{R}$ or $\mathbb{C}$ can be converted into a seminorm space as follows: take any functional $f$. This gives rise to a seminorm such that $\left\Vert \mathbf{x% }\right\Vert =\left\vert f\left( \mathbf{x}\right) \right\vert $ where $% \left\vert .\right\vert $ is the absolute value of an element $f\left( \mathbf{x}\right) $ of the underlying field. You may want to look up for how to define absolute values over an integral domain/field in aglebra. Now, the answer to your question, a seminorm space can $N$ be converted to a norm space $N/W$ by taking a collection $W$ of all vectors $% \mathbf{v}$ such that $\left\Vert \mathbf{v}\right\Vert =0$.\ This is a subspace since it trivially satisfies the axioms of a vector space and, furthermore, this function is a seminorm, given the properties of the absolute value. The new norm so defined is $\left\Vert \mathbf{x}+W\right\Vert _{N/W}=\left\Vert \mathbf{x}% \right\Vert _{N}$ for all $x+W\in N/W$. Thus every vector space can be converted into a norm space. For finite fields, however, the only absolute value definable is the trivial absolute value, making the freshly made norm space uninteresting.

Coward
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