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
A: As another way of looking at this, you may remember $\mathbb{R}$ linear maps only having linear factors of what you're putting in? This also applies in the complex setting, but we also can't have complex conjugation showing up (messes up pulling out constants). In your example, we easily see that (noting $x = (z + \bar z)/2$ and $ y = ( z - \bar z)/2i$)).
$$ T(z) = z + \bar z - \frac{z - \bar z}{2i} + i \left (\frac{z + \bar z }{2} - 3 \frac{z - \bar z}{2i} \right ) =\left( \frac{-1+5i}{2} \right )z + \left (\frac{1 + 5i}{2} \right) \bar z$$ 
Since $\bar z$ shows up we see this will never allow $T( az) = a T(z)$ with $a \in \mathbb{C}$.
