Hello,

I am trying to prove the following:

Let $I: \mathbb{C} \to \mathbb{C}$ be the identity and $\bar{I}:\mathbb{C} \to \mathbb{C}$ be conjugation. If $T: \mathbb{C} \to \mathbb{C}$ is real linear then there exists unique $w_1,w_2\in \mathbb{C}$ such that $T=w_1I+w_2 \bar{I}$.

The definition I have of real linear is: $T:\mathbb{C} \to \mathbb{C}$ is called real linear if $T(\lambda z + \gamma w)=\lambda T z + \gamma T w$ for all vectors and all real scalars.

I have also that $T$ is real linear if and only if $T \sim M$ with $M=\begin{pmatrix}\alpha & \beta \\ \gamma & \delta \end{pmatrix}$ and $T(x+iy)=\alpha x + \beta y + i(\gamma x + \delta y)$

So I wrote out $T=w_1I+w_2 \bar{I}$ and I'm having trouble believing this to be true at all as I would need, $$Tz\sim Mz = \begin{pmatrix}\alpha & \beta \\ \gamma & \delta \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} =\begin{pmatrix}w_1+w_2 & 0 \\ 0 & w_1-w_2\end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix}=(w_1I+w_2 \bar{I})z$$where $\mathbb{C} \ni z \sim \begin{pmatrix} x \\ y \end{pmatrix}$

This should imply that $w_1+w_2=\alpha$, $0=\beta=\gamma$ and $w_1-w_2=\delta$.

There has to be something really wrong with how I'm working this. Is it something to do with the one matrix having complex entries and the other real?