say I have a linear space $V$ over $\Bbb C$ and a linear transformation $T:V \to V$ such that $T=A+iB$ where $A,B \in \Bbb R^{n \times n}$
I proved already that $T_\Bbb R = \begin{pmatrix} A & -B \\ B & A \end{pmatrix}$ (which mean $T$ reduced to $\Bbb R$ when you look at $V$ as a linear space over $\Bbb R$).
I need to prove that $\det(T_\Bbb R) = |\det(T)|^2$. can someone help please?
From the answer to your other question here, we know that $$ \det(T_\mathbb{R})=\det(A+iB)\det(A-iB). $$ Now $T=A+iB$ can be seen as a complex matrix with real part $A$ and imaginary part $B$. Denote $\overline{T}=A-iB$.
You just have to convince yourself that $\det \overline{T}=\overline{\det T}$. You can see it directly from the Leibniz formula.
Therefore $$ \det(T_\mathbb{R})=\det T\cdot \det \overline{T}=\det T\cdot \overline{\det T}=|\det T|^2. $$