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Notation

Consider an arbitrary vector space $(V, \oplus, \odot)$ over a field $F$ with

  • vector addition $\oplus : V \times V \to V$ and
  • scalar multiplication $\odot : F \times V \to V$,

both satisfying all the axioms defining a vector space.

Background

Let us fix the field $F$ and the set $V$. It is obvious that the vector addition does not have to be unique. Any binary operation $\oplus$ that makes $(V, \oplus)$ an Abelian group, would actually work. But I am not so sure if this is also true for (the) scalar multiplication.

Question

Let us fix the field $F$ and the Abelian group $(V, \oplus)$. Is the action of $F$ on $(V, \oplus)$, i.e., the scalar multiplication $\odot$ satisfying the axioms of a vector space, a unique operation?

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    $\begingroup$ You should be careful about changing the addition: If $(V,+,\cdot)$ is a vector space over a field $F$ and $\oplus \colon V \times V \to V$ is a binary operation such that $(V,\oplus)$ is an abelian group, then $(V,\oplus,\cdot)$ is not necessarily also a vector space! The scalar multiplication depends far more on the addition then the other way around. $\endgroup$
    – Jendrik Stelzner
    Dec 31, 2015 at 13:40
  • $\begingroup$ @JendrikStelzner: I am aware of this , which is why in the background section I fixed the field $F$ and the set $V$ underlying the Abelian group, but not yet $\odot$. But thanks to have emphasized this. $\endgroup$
    – Björn Friedrich
    Dec 31, 2015 at 13:46

1 Answer 1

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No. On any complex vector space $(V,+,\cdot)$ you can introduce a new scalar multiplication $*$ given by $z * v = \overline{z} \cdot v$ for all $z \in \mathbb{C}$ and $v \in V$.

More generally: If $(V,+,\cdot)$ is an $F$-vector space and $\phi \colon F \to F$ a field automorphism then $z * v = \phi(z) \cdot v$ defines a new scalar multiplication $*$.

PS: The scalar multiplication is unique if $F$ is a prime field, i.e. if $F = \mathbb{Q}$ or $F = \mathbb{F}_p$ with $p > 0$ prime. This follows because the action of $1 \in F$ is uniquely determined by the axioms of the scalar multiplicaton and each element in these fields is a multiple of $1$ (if $F = \mathbb{F}_p$) or can be written as a quotient of multiples of $1$ (if $F = \mathbb{Q}$).

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  • $\begingroup$ This is what confuses me, a vector space is defined for a set ($V$) relative to a field ($F$). If $F$ is not a field, for example the integers (not a field) then is this still a vector space?, can you have a scalar multiplication in which for example $v + v \neq 2\cdot v$? $\endgroup$
    – alfC
    May 9, 2018 at 19:33

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