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A set V qualifies to be a vector space if it satisfies properties under addition and scalar multiplication over a field F. Let's say if a set V is given but the field F is not specified and it is given to check if set V is a vector space or not. Shall I consider F as both R (set of real numbers) and C (set of complex numbers) to check if V is a vector space or not OR anyone is sufficient??

novice
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  • Do you assume you know the addition structure on $V$? (btw, any $\mathbb C$-vector space is automatically also an $\mathbb R$-vector space.) – Kenta S Aug 08 '21 at 07:49
  • @KentaS yes addition structure is known. – novice Aug 08 '21 at 08:41
  • @KentaS I am new to Linear Algebra course. Can you suggest few books or video lectures that would be help me in developing basics concepts !! – novice Aug 08 '21 at 13:10
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    Sorry, it's been a while since I studied linear algebra, so I'm not super sure. Lax's Linear Algebra and Its Applications and Lang's Algebra are canonical references, though. Also try checking out this: https://math.stackexchange.com/questions/2377980/book-recommendations-for-linear-algebra – Kenta S Aug 08 '21 at 13:23

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I'm going to interpret the question as being:

Let $V$ be an abelian group. When does there exist a field $F$ and an $F$-action on $V$ (i.e., scalar multiplication) which makes $V$ an $F$-vector space?

The first observation is that any field $F$ is an extension of $\mathbb Q$ or a finite field $\mathbb F_p$, depending on the characteristic. Thus, it suffices to consider whether $V$ is a $\mathbb Q$-vector space or a $\mathbb F_p$-vector space.

I claim $V$ can be made an $\mathbb F_p$-vector space iff $V$ is $p$-torsion. (i.e., for any $v\in V$, we have $pv=0$.)

$\Rightarrow$: $pv:=v+\dots+v\ (p\text{ times})=(1+\dots+1)v=0v=0$.

$\Leftarrow$: Any element $a\in\mathbb F_p\cong\mathbb Z/p$ lifts to a positive integer $\tilde a>0\in\mathbb Z$. Let $a\cdot v:=\tilde a\cdot v$. This is well-defined since $V$ is $p$-torsion. (Note that $\tilde a\cdot v$ simply means $v+\dots+v$, adding together $\tilde a$ times.) It is easily checked that this actually does make $V$ an $\mathbb F_p$-vector space.

Next, I claim $V$ can be made a $\mathbb Q$-vector space iff $V$ is uniquely divisible (i.e., for any $v\in V$ and nonzero integer $n$, there exists a unique $w\in V$ such that $v=nw$).

$\Rightarrow$: trivial.

$\Leftarrow$: For each $p/q\in\mathbb Q$ and $v\in V$, let $w\in V$ be such that $v=qw$. Now, we define $p/q\cdot v:=pw$. Again, it is easy to check that this action is well-defined, and that this makes $V$ a $\mathbb Q$-vector space.

To summarize, the answer is:

$V$ must either be uniquely divisible, or there must exist a prime number $p$ such that $V$ is $p$-torsion.

Kenta S
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  • Addendum: don't confuse scalar multiplication $n\cdot v$ with $v+\dots+v\ (n\text{ times})$, which unfortunately is also denoted $n\cdot v$! – Kenta S Aug 08 '21 at 13:28
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You can’t hope that considering $\mathbb R$ and $\mathbb C$ would be sufficient.

For example, consider the finite fields $\mathbb F_p$ or $\mathbb F_p \times \mathbb F_p$ where $p$ is a prime number.

On those « sets » $V$, you can define a scalar multiplication $\mathbb F_p \times V \to V$… just take the multiplication of the field!

But there is no obvious scalar multiplication $\mathbb R \times V \to V$. Hence considering $\mathbb R $ or $\mathbb C$ as « natural » fields for vector spaces can’t be done.

Note: if you’re doing physics, then vector spaces are in most of the cases used with $\mathbb R $ or $\mathbb C$ fields.

mathcounterexamples.net
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