Q:

The determinant of a complex linear operator regarded as a real linear operator?

I was trying to solve the following question
Let $T: \mathbb{C}^{2} \rightarrow \mathbb{C}^{2}$ a linear operator with determinant a+bi. If we regard $\mathbb{C}^{2}$ as a vector space over $\mathbb{R}^{4}$ and $T$ as a real linear operator, then we have that $det(T)=a^{2}+b^{2}$ as a real linear operator.
Here are my thoughts:
I was viewing $\mathbb{C}^{2}$ as a real vector space with basis $(1,0), (i,0), (0,1), (0,i)$. This gives the linear isomorphism $\mathbb{R}^{4} \rightarrow \mathbb{C}^{2}$. 
Setting $z_{k}=a_{l} + b_{k}i$, we have the following $2$ by $2$ matrix for $T$, 
$\begin{bmatrix} z_{1} &z_{2} \\ z_{3} &z_{4} \end{bmatrix}$.
As a real operator I found the operator $T$ to have the matrix
$\begin{bmatrix}
a_{1} &-b_{1} &a_{2} &-b_{2}\\
b_{1} &a_{1} &b_{2} &a_{2} \\
a_{3} &-b_{3} &a_{4} &-b_{4}\\
b_{3} &a_{3} &b_{4} &a_{4}\\
\end{bmatrix}
$
Finding the determinant of the above matrix looks very messy to me, I was considering finding the determinant of the above matrix and collecting terms to prove the point but I may be working too hard.
Is there another way to do this?

A:

Here is another approach. Pick an ordered basis $\mathcal B_2=\{u,v\}$ of the complex linear space $\mathbb C^2$ under which the matrix of $T$ is in Jordan form. Then $\mathcal B_4=\{u,iu,v,iv\}$ is an ordered basis of the real linear space $\mathbb C^2$. So, if the matrix of $T$ under $\mathcal B_2$ is $\pmatrix{z_1&\ast\\ 0&z_2}$, with $z_j=a_j+ib_j\ (a_j,b_j\in\mathbb R)$, the matrix of $T$ under $B_4$ would be $\pmatrix{a_1&-b_1&\ast&\ast\\ b_1&a_1&\ast&\ast\\ 0&0&a_2&-b_2\\ 0&0&b_2&a_2}$. So, if the determinant of $T$ under $\mathcal B_2$ is $z_1z_2=a+bi$, the determinant of $T$ under $\mathcal B_4$ is $(a_1^2+b_1)^2(a_2^2+b_2)^2=|z_1z_2|^2=a^2+b^2$.

A:

We can write $T$ as a block matrix
$$
\begin{bmatrix}
A & B\\
C & D
\end{bmatrix}
=
\begin{bmatrix}
\pmatrix{
a_{1} &-b_{1}\\
b_{1} &a_{1}}&
\pmatrix{
a_{2} &-b_{2}\\
b_{2} &a_{2}}\\
\pmatrix{
a_{3} &-b_{3} \\
b_{3} &a_{3} }&
\pmatrix{
a_{4} &-b_{4}\\
b_{4} &a_{4}}\\
\end{bmatrix}
$$
Because $C$ and $D$ commute, we can apply the formula here to find
$$
\det\begin{bmatrix}
A & B\\
C & D
\end{bmatrix} = 
\det[AD - BC]
$$
this is a $2 \times 2$ determinant that you'll have no trouble computing.
Also, note that
$$
\pmatrix{a&-b\\b&a} \pmatrix{c&-d\\d&c} = 
\pmatrix{ac - bd & -(ad + bc)\\ad + bc & ac - bd}
$$
in analogy with the multiplication of complex numbers.

