What is the Precise Definition of a “Complex Vector Space”? I am studying linear algebra (as a second year) on my own using Axler’s, “Linear Algebra Done Right.”  
I have run into a definitional problem that I can’t get past.
Specifically, Axler (and Wolfram, and others) define a “complex vector space” as a vector space in which the field is the complex numbers.  According to this definition, the complex numbers over the real numbers are not a complex vector space, but the complex numbers over the complex numbers are a complex vector space.  This despite the fact that the two vector spaces are identical (or at least isomorphic).
I already see that many theorems concerning eigenvalues/vectors, adjoints, and spectral theory vary according to whether we are looking at complex or real vector spaces.  Thus, the definition of “complex vector space” is critical.
I’m sure there is an easy explanation, but I don’t see it.  Thanks.
 A: It might help to remember that a vector space is defined by more than just the set of vectors $V$. Although we frequently abbreviate it that way, a vector space really consists of four things, $(k, V, {}+{}, {}\cdot{})$: the base field $k$, the set of vectors $V$, the vector addition operation $+$, and the scalar multiplication ${}\cdot{}$. Keeping that in mind will make it easier to understand how $\mathbb C$ can sometimes be a complex vector space and sometimes not.
When we're thinking about $\mathbb C$ as a vector space over $\mathbb C$, we mean the vector space $(\mathbb C, \mathbb C, {}+{},{}\cdot{})$, where the addition is addition of complex numbers and scalar multiplication is multiplication of complex numbers.
When we're thinking about $\mathbb C$ as a vector space over $\mathbb R$, we mean the vector space $(\mathbb R, \mathbb C, {}+{},{}\cdot{})$,  where the addition is addition of complex numbers and scalar multiplication is multiplication of reals by complex numbers.
So these really are two different vector spaces, even though the set of vectors is the same in both. And one of them is a complex vector space, while the other is a real vector space.
A: I hope this gives some insight:
$\Bbb R$ as an $\Bbb R$-vector space
This is the real line you are used to. It has a basis consisting of one element, which we can choose to be $1$. You can view this as a vector with just one coefficient. Notice that any $r \in \Bbb R$ can be expressed in this basis as $r \cdot 1$. A vector is just a real number, and we have scalar multiplication that is just the usual multiplication.
$\Bbb C$ as an $\Bbb R$-vector space
This is the complex plane. We need two basis elements, for example $1$ and $i$. Now any vector in $\Bbb C$ can be expressed as $a + bi$ with $a,b \in \Bbb R$. If $r \in \Bbb R$ we have scalar multiplication $r(a + bi) = ra + rbi$.
$\Bbb C$ as an $\Bbb C$-vector space
Now we're back to being one dimensional. We only need one basis element, $1$ for example. Any element $a + bi \in \Bbb C$ can be expressed as $(a + bi) \cdot 1$. Compare this to the first example and convince yourself that these are essentially the same. Scalar multiplication happens with elements from $\Bbb C$.
Note that, for example, $\Bbb R$ as a $\Bbb C$-vector space does not make sense. We need to be able to multiply by scalars in a meaningful way. For general $z \in \Bbb C$, the product with a real number $r \in \Bbb R$ is a complex number $r \cdot z \in \Bbb C$.
A: In case the other answers are too abstract, here is a concrete example.
Treating $\mathbb C$ as a complex vector space of one dimension,
an example of a scalar is $a = 2 + i.$ An example of a vector is 
$\mathbf v = (1 + 2i).$
We can perform a scalar multiplication of a vector like so:
$$ a \mathbf v = (2 + i)(1 + 2i) = (5i), $$
where $(5i)$ is also a vector in this space.
Now if we take $\mathbb R^2$ as a two-dimensional vector space,
we could make the obvious correspondence between $\mathbb R^2$ and $\mathbb C$
such that $(1, 2)^T \in \mathbb R^2$ corresponds to $1 + 2i = \mathbb C.$
But $2+i$ is not a scalar in this vector space.
Instead we would have to treat $2 + i$ as another vector.
It is impossible to treat $(2 + i)(1 + 2i)$ as scalar multiplication in 
$\mathbb R^2$ taken as a two-dimensional vector space,
although in $\mathbb C$ as a one-dimensional vector space this is scalar multiplication.
So although $\mathbb R^2$ has some properties isomorphic with the corresponding properties of $\mathbb C,$ it's missing at least this one.
