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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)?

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    $\begingroup$ 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
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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.

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  • $\begingroup$ 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

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