What does it mean to treat $\mathbb{C}$ as a real vector space? My textbook (“Linear Algebra Done Wrong” by Sergei Treil) says (on page 27 of the PDF file):

The set $\mathbb{C}$ of complex numbers can be canonically identified with the space $\mathbb{R}^2$ by treating each $z = x+iy$ as a column $(x,y)^T \in \mathbb{R}^2$.

Does this just mean that you can either represent a complex number $a+ib$ as $(a,b)$, a column in $\mathbb{R}^2$, or alternative as just $(a+ib)$, a one dimensional vector in $\mathbb{C}^1$? Does it have any other implications? For example, are the scalars in both cases still complex numbers?
The text later writes:

Treating $\mathbb{C}$ as the real vector space $\mathbb{R}^2$…

Is this referring to the same canonical identification as above? The way its written confuses me and makes me think because we're treating $\mathbb{C}$ as a real vector space the scalars can no longer be complex numbers, but is this wrong?
Finally, it defines a linear transformation and asks the question:

Show that this transformation is not a linear transformation in the complex vectors space $\mathbb{C}$, but if we treat $\mathbb{C}$ as the real vector space $\mathbb{R^2}$ then it is a linear transformation there (i.e., that it is a real linear but not complex linear transformation.

I think I'm understanding this wrong. If treating the vector spaces as real or complex simply means representing them differently (as a tuple vs a 1d vector), shouldn't this not change the linearity of any transformation?
 A: Whenever we say that we are treating a complex vector space $V$ as a real vector space, then the only scalars that we use are real numbers. So, $\mathbb C$, which is a one-dimensional complex vector space, can also be seen as a two-dimensional real vector space.
A: A vector space is always over some field of scalars. A real vector space takes this field to be $\Bbb R$; a complex vector space takes this field to be $\Bbb C$. You may treat $\Bbb C$ as $1$-dimensional over $\Bbb C$, with the scalars being complex, or as $2$-dimensional over $\Bbb R$, with the scalars being real. (In fact, you may treat $\Bbb C^n$ as $n$-dimensional over $\Bbb C$, with the scalars being complex, or as $2n$-dimensional over $\Bbb R$, with the scalars being real.)
A transformation on $\Bbb C$ might be linear only in the real case. Take complex conjugation: over $\Bbb R$ it's just the matrix $\left(\begin{array}{cc}
1 & 0\\
0 & -1
\end{array}\right)$, but over $\Bbb C$ it's not linear at all, because there's no $k\in\Bbb C$ for which $z^\ast$ is identically $kz$.
