Show that $\det(AC) \ge 0$ for real matrices with $(A+iB)^{-1} = C+iD$. Let $H=A+Bi$ be a complex $n \times n$ invertible matrix where $A,B$ are real matrices, with inverse $H^{-1}=C+Di$ for $C,D$ real.
Prove $\det(AC)\geq 0$.
My attempt so far, $$I=HH^{-1}=(A+Bi)(C+Di)=AC-BD+(AD+BC)i$$
So $$AC-BD+(AD+BC)i=I$$
Solving by $AC$ and taking the determinant in both sides, I end up with $$
\det(AC)=\det(I-(AD+BC)I+BD)$$
Since  $AC-BD+(AD+BC)i$ is the identity then it is a real matrix so it must be true that $(AD+BC)=O$,
So finally I get $$\det(AC)=\det(I+BD).$$
But this is where I am stuck, is $\det(I+BD)$ non-negative?
Or did I take a completely wrong approach?
 A: We can assume that $A$ is non-singular. Then
$$
AD + BC = 0 \implies D = -A^{-1}BC
$$
and therefore
$$
I = AC-BD = AC+AA^{-1}BA^{-1}BC = A(I+X^2)C
$$
with $X= A^{-1}B$. We conclude that
$$
 1 = \det(AC) \cdot \det(1+X^2) \, .
$$
The second factor cannot be zero. The second factor is also not negative, see for example

*

*is $\det(A^2 + I)$ always non negative? or

*Polynomial with no real roots implies that $\det(P(A))\ge 0$.

It follows that $\det(AC) \ge 0$.

Remark: In addition to $\det(AC) \ge 0$ we also have $\det(-BD)\ge 0$. This can be obtained from the above result in various ways:

*

*Apply the above result to $I = (A+iB)(C+iD) = (-B+iA)(D-iC)$.

*Or write the matrix identities as
$$
 AC + (-B)D= I \, , \, AD = (-B)C \, .
$$
and use a symmetry argument.

*Or use that
$$
 \det(AC) \cdot \det(BD) = \det(AD) \cdot \det(BC) = -\det(AD)^2 \le 0 \, .
$$

With respect to your approach: It is correct that $\det(AC)=\det(I+BD)$, but that alone cannot give the desired conclusion. The other identity $AD+BC=0$ must be used as well.
