Is $T : \mathbb{C} \rightarrow \mathbb{C}: a + bi \mapsto \overline{a + bi} = a - bi$ a linear image? So a question in my handbook for linear algebra is the following; Is the linear map T
$$T :\mathbb{C} \rightarrow \mathbb{C}: a + bi \mapsto \overline{a + bi} = a - bi$$
an linear image if:

*

*$\mathbb{C}$ is interpreted as a real vector space

*$\mathbb{C}$ is interpreted as a complex vector space

I know how I would solve the question itself (I would check the axiom of linear image: if $L: V \rightarrow W$ is a linear map then $L(\lambda_1v_1 + \lambda_2v_2) = \lambda_1L(v_1) + \lambda_2L(v_2)$) but I'm stuck at what they mean with the interpretation.
 A: Assuming that you meant “linear map”, $T$ is a linear map if you see $\Bbb C$ as a real vector space, since, if $z,w\in\Bbb C$ and $\alpha,beta\in\Bbb R$, we have\begin{align}T(\alpha z+\beta w)&=\overline{\alpha z+\beta w}\\&=\alpha\overline z+\beta\overline w\text{ (because $\alpha,\beta\in\Bbb R$)}\\&=\alpha T(z)+\beta T(w).\end{align}
But $T$ is not a linear map if you see $\Bbb C$ as a complex vector space, because for instance, $T(i\times i)\ne i\times T(i).$
A: Hint.

*

*$T$ is a $\mathbb{R}$-linear map.

*$T$ is not a $\mathbb{C}$-linear map.


Notes.

*

*$\mathbb{C}$ is one-dimensional vector space over the filed $\mathbb{C}$   where $\alpha=(1)$ is a basis.

*$\mathbb{C}$ is two-dimensional vector space over the filed
$\mathbb{R}$ where $\beta=(1,i)$ is an (ordered) basis.

When you talk about "linear maps", you need to know the underlying vector spaces (over what field). The field determines what scalar $\lambda$ you need to check for
$$
T(\lambda_1v_1+\lambda_2v_2)=\lambda_1T(v_1)+\lambda_2T(v_2)\tag{1}
$$
If $T$ is $\mathbb{C}$-linear, it means that (1) is true
for all complex numbers $\lambda_1$ and $\lambda_2$.
If $T$ is $\mathbb{R}$-linear, it means that (1) is true
for all real numbers $\lambda_1$ and $\lambda_2$.
In your example, try to see what is $T(iz)$.
