Let $V = \{(a_1,a_2,\ldots,a_n): a_i \in \mathbb{R}\ \text{ for } i = 1,2,\ldots,n\};$ So $V$ is a vector space over $\mathbb{R}$. Is $V$ a vector space over the field of complex numbers with the operations of coordinatewise addition and multiplication?

I don't need an actual answer for this, as I just want to confirm that I understand what the question is asking. Paraphrasing: If $V = \mathbb{R}^n$ and $\mathbb{F} = \mathbb{R}$, then we know that $V$ is a vector space. If $\mathbb{F} = \mathbb{C}$, then is $V$ still a vector space (Is the set of real vectors $v \in \mathbb{R}^n$ over the field of complex scalars a vector space)?

  • 2
    $\begingroup$ How would $i\cdot v$ be defined? You can make every $\mathbb{R}$-vector space of even dimension into a $\mathbb{C}$-vector space, but usually, there is no "natural" multiplication with nonreal scalars defined. $\endgroup$ – Daniel Fischer Jan 10 '14 at 20:37
  • $\begingroup$ Would $iv = i(v_1,v_2,\ldots,v_n) = (iv_1, iv_2,\ldots,iv_n)$? So in this case $iv \not\in V$. $\endgroup$ – St Vincent Jan 10 '14 at 20:41

One of the axioms of a vector space is that $\lambda \vec{v} \in V$ for all $\lambda \in \mathbb{F}$ and $\vec{v} \in V$.

Let $V=\mathbb{R}$ and $\mathbb{F}=\mathbb{C}$. If $\vec{v}=1$ and $\lambda=\operatorname{i}$ then $\lambda \vec{v} = \operatorname{i}$ and that does not belong to $\mathbb{R}$.

In such a case, the pair $(\mathbb{R},\mathbb{F})$ fails to be closed under scalar multiplication.

  • $\begingroup$ Is the $\mathbb{K}$ supposed to be $\mathbb{F}$? Or is it just a different notation for a field? $\endgroup$ – St Vincent Jan 10 '14 at 20:44
  • $\begingroup$ My earlier comment was based on a misreading of the question, so I removed it. $\endgroup$ – Keenan Kidwell Jan 10 '14 at 20:53
  • $\begingroup$ @StVincent You're right: the $\mathbb{K}$ should be an $\mathbb{F}$. A common notation for a field is the letter $k$. $\endgroup$ – Fly by Night Jan 10 '14 at 21:49
  • $\begingroup$ @KeenanKidwell I didn't see your earlier comment, so there's no harm done :-) $\endgroup$ – Fly by Night Jan 10 '14 at 21:53

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.