The determinant of $T: \mathbb C^2\to \mathbb C^2$ as an $\mathbb R$-linear operator Suppose the determinant of a $\mathbb C$-linear transformation $T:\mathbb C^2\to \mathbb C^2$ is $a+bi$. I'm trying to prove that when $\mathbb C^2$ is identified with $\mathbb R^4$, the determinant of the $\mathbb R$-linear transformation $T:\mathbb R^4\to \mathbb R^4$ is $a^2+b^2$.
I started off with a lower dimensional case: a complex-linear operator $T': \mathbb C\to \mathbb C$ must be multiplication by a complex number $a+bi$, and if I write the matrix of $T'$ w.r.t. the basis $(1,i)$ of $\mathbb C$ over $\mathbb R$, then the determinant of the matrix is $a^2+b^2$.
In higher dimensional case, the complex-linear $T$ is multiplication by a $2\times 2$ matrix $$A=\begin{bmatrix}x_1+ix_2&z_1+iz_2\\y_1+iy_2&w_1+iw_2\end{bmatrix}.$$ The set $((1,0)^T,(i,0)^T,(0,1)^T,(0,i)^T)$ would be an $\mathbb R$-basis of $\mathbb C^2$. W.r.t. this basis, the matrix of $T$ is $$A'=\begin{bmatrix}x_1&-x_2&z_1&-z_2\\x_2&x_1&z_2&z_1\\y_1&-y_2&w_1&-w_2\\y_2&y_1&w_2&w_1\end{bmatrix}.$$
Am I supposed to compute the determinant of this last matrix and make sure it equals $a^2+b^2$ provided $\det A=a+bi$? It seems like a lot of calculations.
 A: Let do this in full generality. Consider a matrix $Z=A+iB$ of shape $n\times n$ with entries in $\Bbb C$, isolate then $A,B$ from it with entries in $\Bbb R$. Then $Z$ induces a linear map $\Bbb C^n\to\Bbb C^n$, thus it induces by using the forgetful functor "$!$" a map, 
$$
!Z:(\Bbb R^2)^n\to (\Bbb R^2)^n\ ,
$$
and thus a matrix, after fixing some identification $\Bbb R^{2n}\to\Bbb R^{2n}$, and we are considering the determinant of this map.
Using the basis generalized from the one in case $n=2$ in the posted question, we get the matrix
$$
!Z=
\begin{bmatrix} A & -B\\B&A \end{bmatrix} \ ,
$$
and we want to compute its determinant. (With respect to some/any basis).
Let $E$ (German notation, it is here better than the international notation $I$,) be the unit matrix of shape $n\times n$. We have the following equalities between block matrices:
$$
\begin{aligned}
\underbrace{\begin{bmatrix} A & -B\\B&A \end{bmatrix}}_{2\times 2}
\underbrace{\begin{bmatrix} E \\ -iE\end{bmatrix}}_{2\times 1}
&=
\underbrace{\begin{bmatrix} E \\ -iE\end{bmatrix}}_{2\times 1}
\underbrace{\begin{bmatrix} A +iB \end{bmatrix}}_{1\times 1}
\ ,
\\
\underbrace{\begin{bmatrix} A & -B\\B&A \end{bmatrix}}_{2\times 2}
\underbrace{\begin{bmatrix} E \\ iE\end{bmatrix}}_{2\times 1}
&=
\underbrace{\begin{bmatrix} E \\ iE\end{bmatrix}}_{2\times 1}
\underbrace{\begin{bmatrix} A -iB \end{bmatrix}}_{1\times 1}
\ ,
\qquad\text{(conjugated version)}\\
&
\qquad\text{so putting all together in one block matrix computation}\\
\underbrace{\begin{bmatrix} A & -B\\B&A \end{bmatrix}}_{2\times 2}
\underbrace{\begin{bmatrix} E & E\\ -iE & iE\end{bmatrix}}_{2\times 2}
&=
\underbrace{\begin{bmatrix} E & E\\ -iE & iE\end{bmatrix}}_{2\times 2}
\underbrace{\begin{bmatrix} A +iB & \\ & A -iB\end{bmatrix}}_{2\times 2}
\ ,
\\
&\qquad\text{so we have after base change with
$\begin{bmatrix} E & E\\ -iE & iE\end{bmatrix}$
the similitude}
\\
\begin{bmatrix} A & -B\\B&A \end{bmatrix}
&\sim
\begin{bmatrix} A +iB & \\ & A -iB\end{bmatrix}
\ ,\qquad\text{so}
\\
\det\begin{bmatrix} A & -B\\B&A \end{bmatrix}
&=
\det\begin{bmatrix} A +iB & \\ & A -iB\end{bmatrix}
\\
&=\det(A+iB)\cdot\det(A-iB)
\\
&=\det(A+iB)\cdot\overline{\det(A+iB)}
\\
&=|\ \det(A+iB)\ |^2
\ .
\end{aligned}
$$
Note:
Structurally this means the following. We start with
$\require{AMScd}$
\begin{CD}
    !\Bbb C^n @>!Z>> !\Bbb C^n\\
    @V \cong V V @VV \cong V\\
    \Bbb R^{2n} @>> W> \Bbb R^{2n}
\end{CD}
where $W$ is a matrix that we write down when the choice of the base change matrix for the two same vertical isomorphisms is fixed. Here, $\det W$ does not depend on the two same $\cong$ arrows.
We need $\det W$. The idea is to extend once more the field of scalars, from $\Bbb R$ to $\Bbb C$! The determinant remains. (This is similar to the fact, that the determinant of a matrix with entries in $\Bbb Q$ is the same one, if we consider all entries to be in $\Bbb R$ or $\Bbb C$...)
So we formally tensor over $\Bbb R$ with $\Bbb C$. The functorially induced diagram is 
$\require{AMScd}$
\begin{CD}
    @. !\Bbb C^n\otimes_{\Bbb R}\Bbb C @>!Z\otimes\operatorname{id}>> !\Bbb C^n\otimes_{\Bbb R}\Bbb C\\
    @. @V \cong V V @VV \cong V\\
    \Bbb C^{2n} @= \Bbb R^{2n}\otimes_{\Bbb R}\Bbb C @>> W> \Bbb R^{2n}\otimes_{\Bbb R}\Bbb C @= \Bbb C^{2n}
\end{CD}
But now we are free to choose the matrix $W$ using the $\cong$ arrows with entries in $\Bbb C$. And it turns out that we can make the choice such that over $\Bbb C$:
$\require{AMScd}$
\begin{CD}
    !\Bbb C^n\otimes_{\Bbb R}\Bbb C @>!Z\otimes\operatorname{id}>> !\Bbb C^n\otimes_{\Bbb R}\Bbb C\\
    @V \cong V V @VV \cong V\\
    \Bbb C^{2n} @>> \begin{bmatrix} Z&\\&\bar Z\end{bmatrix}> \Bbb C^{2n}
\end{CD}
(Note: Such simple linear algebra "computations" may become structurally important, e.g. when studying Hodge structures... This is the only reason for the categorial overkill, that would be misplaced without this connection.)
A: Suppose $A \in \mathbb{C}^{n \times n}$. Define $\phi:\mathbb{R}^{2n} \to \mathbb{C}^n $ as 
$\phi((x_1,y_1,...,x_n,y_n)) = \sum_k (x_k+iy_k) e_k$. It is straightforward to see that $\phi$ is linear and invertible (the inverse being real linear).
Write the basis $e_1,...,e_{2n} $ of $\mathbb{R}^{2n}$ as $u_1,v_1,...,u_n,v_n$.
The corresponding real basis of $\mathbb{C}^n$ is $e_1 = \phi(u_1),ie_1= \phi(v_1),...,e_n=\phi(u_n),ie_n=\phi(v_n)$.
We want to compute $\det \tilde{A}$, where $\tilde{A} = \phi^{-1}\circ A \circ \phi$.
If $V$ is invertible, note (some expansion needed) that
$\det(\phi^{-1}\circ V^{-1}A V \circ \phi) = \det \tilde{A}$, hence we can choose $A$ to have whatever form suits, in this case
the Jordan normal form.
Note that
\begin{eqnarray}
\tilde{A}(\alpha u_k+\beta v_k) &=& \phi^{-1}(A (\alpha u_k+i\beta v_k)) \\
&=& \phi^{-1}(\alpha \operatorname{re}(A e_k) - \beta \operatorname{im}(A e_k)+ i [ \alpha \operatorname{im}(A e_k) + \beta \operatorname{re}(A e_k) ] ) \\
&=& \phi^{-1}( \sum_i (\alpha \operatorname{re}[A]_{ik} - \beta \operatorname{im}[A]_{ik}) e_i + (\alpha \operatorname{im}[A]_{ik} + \beta \operatorname{re}[A]_{ik} ) i e_i) \\
&=& \sum_i (\alpha \operatorname{re}[A]_{ik} - \beta \operatorname{im}[A]_{ik}) u_i + (\alpha \operatorname{im}[A]_{ik} + \beta \operatorname{re}[A]_{ik} ) v_i
\end{eqnarray}
In particular, the representation of the map from the subspace spanned by $u_k,v_k$ to the coordinates of $u_i,v_i$ is given by the block matrix
${\bf \tilde A}_{ik} = \begin{bmatrix} \operatorname{re}[A]_{ik} & -\operatorname{im}[A]_{ik} \\
\operatorname{im}[A]_{ik} & \operatorname{re}[A]_{ik}
\end{bmatrix}$
Hence, if $A$ is in Jordan normal form, then $\tilde{A}$ is an upper triangular
block matrix, with each block being a $2 \times 2$ real matrix of the above form. Furthermore, the diagonal blocks are of the form
$\begin{bmatrix} \operatorname{re} \lambda_k & -\operatorname{im} \lambda_k \\
\operatorname{im} \lambda_k & \operatorname{re} \lambda_k
\end{bmatrix}$, where $\lambda_k$ are the eigenvalues of $A$.
Hence $\det {\tilde A} = \prod_k  \det \begin{bmatrix} \operatorname{re} \lambda_k & -\operatorname{im} \lambda_k \\
\operatorname{im} \lambda_k & \operatorname{re} \lambda_k
\end{bmatrix}= \prod_k |\lambda_k|^2$.
Since $\det A = \prod_k \lambda_k$, we have the desired result.
A: Here is a terrible and lazy proof. If $W$ is a complex vector space, write $W^!$ for $W$ considered as a real vector space.
Given an element $\lambda$ of $\mathbb{C}^\times$, pick any $T$ with that element as its determinant (clearly such a $T$ exists), and consider the determinant of the linear transformation $T$ acting on the underlying real vector space $V^!$. I claim that this new determinant does not depend on the choice of $T$, but only on the element $\lambda$. Indeed, the map induced by $T$ on $\bigwedge^4 V^!$ is the same as the map induced by $\lambda$ on $\bigwedge^2 (\bigwedge^2 V)^!$ under the canonical isomorphism between the latter and $\bigwedge^4 V^!$.
It follows that we may check the claim on transformations of the form
$$
\begin{pmatrix} \lambda & 0 \\ 0 & 1\\
\end{pmatrix}
$$
for which it reduces to the case you already checked.
A: Assume w.l.o.g. that $A$ is in Jordan normal form, specifically, that it is upper triangular: $$A = \begin{bmatrix} \lambda_1 & * \\ 0 & \lambda_2 \end{bmatrix}$$ so that we have $\det A = \lambda_1\lambda_2 = a+ib$. Its eigenvalues can, of course, be complex: $\lambda_j=x_j+iy_j$, $x,y\in\mathbb R$. The corresponding real matrix is then block upper-triangular: $$A' = \begin{bmatrix} C_1 & * \\ 0 & C_2 \end{bmatrix}$$ with $$C_j = \left[\begin{array}{lr} x_j & -y_j \\ y_j & x_j \end{array}\right], \det C_j = \lambda_j\overline\lambda_j.$$ We then have $$\det A' = \det C_1 \det C_2 = \lambda_1 \overline\lambda_1 \lambda_2 \overline\lambda_2 = (\det A) (\overline{\det A}) = (a+ib)(a-ib) = a^2+b^2.$$
