This started with a question in the book Linear Algebra by S. Friedberg, A. Insel and L. Spence in which they define a vector space $\text{V} = \{(a_1, a_2, ...,a_n):a_i \in \mathbb{C}\}$, so a vector space over $\mathbb{C}$, and ask if it is a vector space over the field of real numbers.
I'm inclined to say this is false because $a_i$ could be any complex number, and multiplying by a scalar from $\mathbb{R}$ could still keep it in $\mathbb{C}$ right? In other words, if $\text{V}$ is defined over $\mathbb{C}$ and an element of $\text{V}$ with complex components is multiplied by a real scalar, its components could still be complex and non-real. In which case it is not an element in the field of $\mathbb{R}$.
The problem is that I've checked community solutions of the book online which say that it is in fact a vector field over the field of $\mathbb{R}$ and also consulted Apostol's Calculus: Vol 2, which says "Let $\text{V}= \mathbb{C}$, the set of all complex numbers, and define $ax$ to be multiplication of the complex number $x$ by the real number $a$. Even though the elements of $\text{V}$ are complex numbers, this is a real linear space because the scalars are real."
So I'm clearly wrong, but I fail to see why. How can a space be over a subset of a certain field by only limiting its scalar multiplication to that subset?