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.


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.


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?

  • 1
    $\begingroup$ 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 '15 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 '15 at 13:46

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 $F = \mathbb{Q}$).

| cite | improve this answer | |
  • $\begingroup$ 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 '18 at 19:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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