Decomposing linear mapping between $z$ and $\overline{z}$

Question

I found this exercise. It seems it's false, but I think adding some hypothesis or changing it in some way may make it true.

Let $T:\mathbb{C}\to\mathbb{C}$ be a $\mathbb{R}-$linear mapping. Prove there exist unique $\alpha,\beta\in\mathbb{C}$ such that for any $z\in\mathbb{C}$ we have $$T(z)=\alpha z+\beta \overline{z}.$$

It's false, since if we take $T=\begin{bmatrix} a & b\\ c & d\end{bmatrix}$ and $z=1$ then $T(z)=a+ic=(a+ic)\cdot 1$, so we can take $\alpha=a+ic,\beta=0$. But also, $\overline{1}=1$ so we can take $\alpha=0,\beta={a+ic}$. Or $\alpha=a,\beta=ic$, so uniqueness isn't guaranteed.

Now consider $T=\begin{bmatrix} 0 & 1\\ 1 & 0\end{bmatrix}$ and $z=x+iy$. Then $T(z)=y+ix$. The existence of such $\alpha$ and $\beta$ means $y=(\alpha+\beta)x$ and $x=(\alpha-\beta)y$. If we suppose both $x$ and $y$ are nonzero, we get $\frac{y}{x}=\alpha+\beta$ and $\frac{x}{y}=\alpha-\beta$. Then we get $\alpha=\frac{1}{2}(\frac{y}{x}+\frac{x}{y}),\beta=\frac{1}{2}(\frac{y}{x}-\frac{x}{y})$, and both of them will depend on $x$ and $y$, so even in this case we cannot guarantee the existence of the constants $\alpha,\beta$.

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The purpose of the exercise is to guarantee that given a real differentiable map $f:\mathbb{C}\to\mathbb{C}$, and a point $z_0\in \mathbb{C}$ there exist unique $\alpha,\beta\in\mathbb{C}$, which will be called $$\alpha=\frac{\partial f}{\partial z}(z_0),\beta=\frac{\partial f}{\partial\overline{z}}(z_0)$$ such that for any $dz=dx+idy\in\mathbb{C}$ we have $$J_f|_{z_0}(dz)=\alpha dz+\beta \overline{dz}= \frac{\partial f}{\partial z}(z_0)dz+\frac{\partial f}{\partial\overline{z}}(z_0) \overline{dz}.$$

But as I said, if the exercise is false, we cannot always guarantee that. So these are my questions: Am I misunderstanding something here? If not, under which changes can this become true?

Answer 1

Hint: by $\mathbb{R}$-linearity:

$$ \begin{align} T(z) = T\big(\operatorname{Re}(z) \cdot 1+\operatorname{Im}(z)\cdot i\big) &= \operatorname{Re}(z) \cdot T(1) + \operatorname{Im}(z) \cdot T(i) \\[5px] &= T(1)\cdot \frac{z+\bar z}{2} + T(i) \cdot \frac{z-\bar z }{2i} = \;\ldots \end{align} $$

Answer 2

You're misunderstanding something. Your first step, "Let $T=\begin{bmatrix} a & b\\ c & d\end{bmatrix}$" doesn't make sense, because $T$ is supposed to be a transformation, not a matrix. Assuming that you meant "the fractional linear transformation associated to this matrix", it makes a little sense...but those transformations are not generally $\Bbb R$-linear, so they doesn't make very good examples to work from.

I think you need to be a little more careful in reading what the problem asks, and in expressing your answer in ways that are mathematically precise.

Answer 3

For $T(z) = y + ix = 0(x + iy) + i(x - iy)$, so $\alpha = 0$ and $\beta = i$. I think the exercise is correct. Indeed, $T$ is well defined by $T(1)$ and $T(i)$, because $T(z= x+iy) = xT(1) + yT(i)$. Let $T(1) = a +ib$ and $T(i) = c+id$ then $T(z) = ax+ibx + cy +idy = a(x + iy) + ib(x - iy) + y((c-b)+i(d-a)) = \alpha z + \beta \bar{z}$ We use that $y = \frac{z - \bar{z}}{2i}$.