Q:

Real Linear vs. Complex Linear

I recently started a new math course and got hung up on a particular problem from the book "Linear Algebra Done Wrong". Specifically, problem 1.3.6 (c). I am an engineer, and I believe I simply lack terminology/definition to solve the problem. Again, it is part (c) of the problem:
The set $\mathbb C$ of complex numbers can be canonically identied with the space $\mathbb R^2$
by treating each ($z = x + iy$) of $\mathbb C$ as a column $(x , y)^T$ of $\mathbb R^2$.
Define $T(x+iy) = 2x-y+i(x-3y)$. 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 $T$ is a real linear but not a complex linear transformation). Find the matrix of the real liner transformation $T$.
It appears to me that:
$T(au+bv) = aT(u)+bT(v)$
I can't see why it is not linear on $\mathbb C$ but it is on $\mathbb R^2$. However, I recognize that I don't understand how this changes the outcome. Any ideas?
Thanks

A:

For $\mathbb{C}$ as $\mathbb{R}^2$:
$T(v_1 + v_2) = T(x_1 + x_2 + i(y_1 + y_2)) = ... = T(v_1) + T(v_2)$
(This is easy to check)
Similarly, $T(\alpha v) = T(\alpha x + i \alpha y) = ... =\alpha T(v)$ is also easy to check.
This proves the linearity of T when $\mathbb{C}$ is viewed as $\mathbb{R}^2$.
Calculating Matrix: 
$T(\begin{bmatrix}
    1\\
    0
\end{bmatrix}) = \begin{bmatrix}
    2\\
    1
\end{bmatrix}$
$T(\begin{bmatrix}
    0\\
    1
\end{bmatrix}) = \begin{bmatrix}
    -1\\
    -3
\end{bmatrix}$
Matrisx of T is $\begin{bmatrix}
    2 & -1\\
    1 & -3
\end{bmatrix}$

Now for $\mathbb{C}$ as complex plane:
Take $\alpha = a + ib, v = = x + iy$
$T(\alpha v) = T(ax-by + i(ay + bx)) \neq \alpha T(v) $
Very easy to check. Use the definition of T
Hence, T is not Complex linear

