Determinant of a linear transformation T over different fields. Let $V = \mathbb{C}^2$. Let $T: V\to V$ denote a $\mathbb{C}$-linear transformation with determinant $a + bi$, $a,b\in \mathbb{R}$. Prove that if we consider $V$ as a 4-dimensional real vector space, then the determinant of $T$ as an $\mathbb{R}$-linear transformation is $a^2 + b^2$.
I don't think I am supposed to do this computationally. I would like to use a convenient formula for the determinant, e.g., as a product of eigenvalues, but I don't think I am allowed to in this case (I don't think it applies to all linear transformations $T$). I would appreciate any direction. Thanks,
 A: I will prove this for arbitrary complex space (since $\dim V = 2$ is not really necessary here).
There is a matrix-based proof that goes in the following way. At first you take a $\mathbb{C}$-basis 
$E = \{e_1, \dots, e_n\}$ in $V$ and after that you have an $\mathbb{R}$-basis $E_r = \{e_1, \dots, e_n, ie_1, \dots, ie_n\} = E \bigcup iE$. Then as soon as operator $T:V \rightarrow V$ is $\mathbb{C}$-linear you have some structure in matrix $T(E_r)$ of your operator in basis $E_r$. It appears that
$$
T(E_r) =  \begin{bmatrix}
Re T(E) & - Im T(E)\\\
Im T(E) & Re T(E)
\end{bmatrix}
$$
where $T(E)$ is the complex matrix of operator $T$ in original basis $E$. Now we apply some basic linear algebra:
$$
2^{2n}\det T(E_r) = \det 
\begin{bmatrix}
I & iI\\\
I & -iI
\end{bmatrix}
\begin{bmatrix}
Re T(E) & - Im T(E)\\\
Im T(E) & Re T(E)
\end{bmatrix}
\begin{bmatrix}
I & I\\\
-iI & iI
\end{bmatrix}
$$
Here we used one simple fact:
$$
\begin{bmatrix}
I & iI\\\
I & -iI
\end{bmatrix}
\begin{bmatrix}
I & I\\\
-iI & iI
\end{bmatrix} = \begin{bmatrix}
2I & 0\\\
0 & 2I
\end{bmatrix}
$$
And therefore by computing block multiplication we can get
$$
2^{2n}\det T(E_r) = \det 
\begin{bmatrix}
2T(E) & 0\\\
0 & 2\overline{T(E)}
\end{bmatrix} = 2^{2n} \det T(E) \det \overline{T(E)} = 2^{2n} |\det T(E)|^2
$$ 
So, we obtained equality$\det T(E_r) = |\det T(E)|^2$.
