Determinant of $(A+iB)(A-iB)$

Question

Let $A,B$ real $n \times n$ matrices such that $A+iB$ is invertible.

I want to prove (if true) that $|A+iB|\cdot|A-iB| >0$

Of course, this is the same as $\big| (A+iB)(A-iB) \big|>0$, so, multiplying them gives $C= A^2 + iBA -iAB + B^2$

Can we say anthing about the determinant of $C$?

Answer 1

It is true that $\det(A+iB)\det(A-iB) >0$ for your matrices. Here is one possible idea how to see and prove it. We know that $A$ and $B$ are real matrices. From this follows that $$ \overline{(A + iB)} = A - iB. $$ Now we would like to show that $\overline{\det (M)} = \det(\overline M)$. If this is true then we are almost finished because $$|A+iB|\cdot|A-iB| = z \overline z = |z|^2$$ for $z=\det(A+iB)$. Invertibiltiy will then be used to provide that $\det(A+iB)\neq 0$.

Now remains to show that $\overline{\det (M)} = \det(\overline M)$. This follows from the fact that determinant is a polynomial in matrix entries and $\overline{P(z_1,\cdots,z_n)} = P(\overline{z_1},\cdots, \overline{z_n})$ for all polynomials $P$ with real coefficients.

Answer 2

Conjugation commutes with addition and multiplication (i.e. $\overline{x\cdot y}=\overline x\cdot\overline y$ and $\overline{x+y}=\overline x+\overline y$), and taking the determinant is just a complicated set of additions and multiplications on the entries, so conjugation also commutes with taking the determinant. That implies the result you want.