# Vector spaces with complex field as scalar. [duplicate]

Sorry for stating the question informally. If we have a vector space whose scalars are the field $\mathbb{R}$, if we change the field to be $\mathbb{C}$ and "adapt" the addition and scalar multiplication operations, does it necessarily yield a new vector space?

Are there theorems developed in this manner or similar tests?

• By a quick search, I believe this is asking the same question, and the answer provided a straightforward counter-example. math.stackexchange.com/questions/634104/… – islamfaisal Apr 10 '15 at 13:45
• In particular, if the field is $\mathbb R$ itself (1 dimensional), what do you wan to get when you "adapt" it to $\mathbb C$? – GEdgar Apr 10 '15 at 13:48
• @islamfaisal it seems you have found out that what you wanted to know was already addressed in a previously asked question, making yours a duplicate. Following the site policy, I therefore flag as duplicate. – Hrodelbert Apr 10 '15 at 13:52

## 2 Answers

There is a general construction, informally stated: convert the scalars from $\mathbb R$ to $\mathbb C$.
In fancy language: if $V$ is a real vector space, then $$V_1 = V \otimes_{\mathbb R} \mathbb C$$ is the corresponding complex vector space.

You cannot do it in general because a finite-dimensional vector space over $\mathbb C$ has even dimension over $\mathbb R$. Of course, not every vector space over $\mathbb R$ has even dimension.

On the other hand, you can always complexify a real vector space $V$:

Take $V \times V$ and let $a+bi$ act on $(v,w)$ as if $(v,w)=v+wi$, following the natural rule.

This construction works even if the dimension of $V$ is infinite.