Linear transformation on the vector space of complex numbers over the reals that isn't a linear transformation on $\mathbb{C}^1$. I am having a little trouble with the following question (that is from the Linear Transformations chapter in Hoffman's / Kunze Linear Algebra):
Let $\mathbb{V}$ be the set of all complex numbers regarded as a vector space over the field of real numbers (usual operations). Find a function from $\mathbb{V}$ into $\mathbb{V}$ which is a linear transformation on the above vector space, but which is not a linear transformation on $\mathbb{C}^1$, i.e., which is not complex linear.
$\mathbb{C}^1$ is the set of all complex numbers regarded as a vector space over the field of complex numbers right? If then, isn't all linear transformations in $\mathbb{V}$ a linear transformation on $\mathbb{C}^1$? (Because $\mathbb{R}\subset\mathbb{C}$)
I think I'm missing something fundamental here.
Any help is welcome. Thanks!
 A: The question is asking, in other words, for a function $f: \mathbb C \to \mathbb C$ which is $\mathbb R$-linear but not $\mathbb C$-linear. In other words, $f$ should satisfy $f(x+y)=f(x)+f(y)$ and $f(cx)=cf(x)$ for all $x,y\in\mathbb C$ and $c\in\mathbb R$, but it should not satisfy $f(cx)=cf(x)$ for all $c\in\mathbb C$.
A: As a vector space over $\mathbb{C}$, $\mathbb{V}$ is a 1-dimensional vector space (with basis = $\{1\}$), and all linear maps $T: \mathbb{V} \rightarrow \mathbb{V}$ are of the form $T(z) = z_0z$, for some constant $z_0 = a + ib$. Thus, for $z = x + i y$, 
$$
T(z) =T(x + iy) = (a + ib)(x+iy) = (ax-by) + i(ay+bx).
$$ 
As a vector space over $\mathbb{R}$, $\mathbb{V}$ is a 2-dimensional vector space, with basis $\{1, i\}$. The $\mathbb{C}$-linear map $T(z)$ above is also $\mathbb{R}$-linear, and its matrix representation, with respect to this basis, is 
$$
\begin{bmatrix}a & -b \\b&a\end{bmatrix} \;\;\;\; (*)
$$
So for your example, take any linear transformation with matrix representation that has a form different from $(*)$.
A: Define $T:\mathbb{C}\to \mathbb{C}$ by $T(z)=\bar{z}$, where $\overline{x+iy}=x-iy$. Then one can check that $T$ is $\mathbb{R}$-linear, but since
$$ T(i)=-i\neq i=i\cdot T(1) $$
it follows that $T$ is not $\mathbb{C}$-linear.
