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We said in class that all complex linear maps have the form $\rm\:x\to (a+bi)\, x\,$.

It isn't immediately clear to me why all complex linear maps have this form.

What about a transformation such that $x\to \left[ \begin{array}{c} 1& 2 \\ 6 & 5 \end{array} \right] x$ what is wrong with this transformation going from $C $ to $C$?

If I can see $C$ as equivalent to $R^2$ (as claimed in class) surely my map must be ok, what am I brutally misunderstanding?

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    It's not linear in the complex sense. By _definition_, a function $f$ from $\mathbb{C}$ to $\mathbb{C}$ which is _linear_ over $\mathbb{C}$ must be such that, for any $c \in \mathbb{C}$, $f(cz) = cf(z)$ – 2017-02-25
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    The matrix transforms $C$ to $C^2$. The challenged claim is about transformations from $C$ to $C$. – 2017-02-25
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    Thus, the function you proposed is an $\mathbb{R}$-linear transformation from $\mathbb{R^2}$ to $\mathbb{R^2}$, but not a $\mathbb{C}$-linear transformation from $\mathbb{C}$ to $\mathbb{C}$. – 2017-02-25

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It depends on how you consider $\mathbb C$ to be a vector space.

If $\mathbb C$ is a vector space over $\mathbb C$ (and scalar multiplication is the ordinary multiplication of complex numbers), then the only linear maps are given by multiplication by a complex constant. This is true whenever a field is considered to be a vector space over itself: if $f$ is a linear function then $$ f(x) = f(x\cdot 1) = x\cdot f(1) = f(1)x $$ so $f(1)$ must be a constant that $f$ multiplies everything by.

On the other hand, if you view $\mathbb C$ as a real vector space -- that is, you consider $\mathbb R$ to be the scalar field -- then the conditions for being linear puts fewer demands on $f$ because there are fewer scalars that it needs to be compatible with. In that case there are more functions that qualify as linear, such as the one you suggest.

In order to prevent such confusion, one sometimes speaks about "$\mathbb C$-linear" or "$\mathbb R$-linear" functions when it is not otherwise clear what the scalar field is.

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    Thanks, the next question that comes to mind is why is it more useful or natural to view the linear transformations as $\mathbb C$-linear as opposed to $\mathbb R$-linear? – 2017-02-25
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    Whether it's useful or natural depends on context: whether there's some reason why the only transformations to consider for a particular purpose should be the complex-linear ones. – 2017-02-25