Does every $\mathbb{R},\mathbb{C}$ vector space have a norm?
By Sophia Hammond
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).
$\endgroup$ 33 Answers
$\begingroup$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).
$\endgroup$ 7 $\begingroup$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:
François Treves: "Topological Vector Spaces, Distributions and Kernels"
H.H. Schaefer, M.P. Wolff : "Topological Vector Spaces"
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.
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