$\mathbb{R}^2$-linear but not $\mathbb{C}$-linear

Question

How can I construct a function that is linear in the plane, but not in the complex numbers (as specified by the Threads name)?

Answer 1

One example: $(x,y) \mapsto (x,2y)$

Answer 2

For instance $\operatorname{Re}\colon\mathbb{C}\longrightarrow\mathbb{C}$ has that property, as does the $\operatorname{Im}$ function.

Answer 3

The $\mathbb C$-linear functions are exactly those of the form $z \mapsto wz$ and so are represented in the canonical basis by matrices of the form $$ \pmatrix{ a & -b \\ b & a} $$ where $w=a+bi$. Not all $\mathbb R$-linear functions can be represented in this form.

Answer 4

I guess you want to know when maps $T:\mathbb C\to\mathbb C$ that are $\mathbb R$-linear are also $\mathbb C$-linear. First of all, $T$ is $\mathbb R$-linear if and only if $$T(z)=T(1)x+T(i)y=\lambda z+\mu\bar z,$$ where $\lambda=\frac{T(1)-iT(i)}{2}$ and $\mu=\frac{T(1)+iT(i)}{2}$.
If we are given any such $\mathbb R$-linear map, then it is $\mathbb C$-linear if and only if $T(i)=iT(1)$. It then follows immediately that $T(z)=T(1)z$. Of course, we could also interpret $T$ as a map $\mathbb R^2\to \mathbb R^2$ and represent it by a $2\times 2$-matrix $A$, say $$A=\begin{pmatrix}a&b\\ c&d \end{pmatrix}.$$ The condition of $\mathbb C$-linearity then translates to $c=-b$ and $d=a$. (Actually, this is the reason we have Cauchy-Riemann differential equations for holomorphic functions -- one way to define holomorphicity of a function is to say that is induces a $\mathbb C$-linear map on tangent spaces.)

Anyway, this should enable you to construct as many $\mathbb R$-linear functions that are not $\mathbb C$-linear as you wish. And I hope that you further understand a bit more the difference between those two notions.