# determinant involving unitary hermitian matrices

Let $$S$$ be the set of complex $$N\times N$$ matrices that are both unitary and hermitian.

I have observed the following fact. For any pair of matrices $$A$$, $$B$$ from $$S$$, the value of $$\det(A+iB)$$ is an imaginary number if $$N \equiv 2 \text{ mod } 4$$.

Any ideas how to prove this?

All I could do was compute $$\det(A+iB)^*=\det(A^\dagger-iB^\dagger)=\det(A^{-1}-iB^{-1})=\det(A-iB),$$ but this did not help me much.

• Try putting $A=B=I$. – user647486 Apr 3 at 14:38
• @user647486 the fact does not hold for all dimensions, sorry – thedude Apr 3 at 14:50
• Take $A=\begin{pmatrix}1&0\\0&-1\end{pmatrix}$ and $B=I$. – user647486 Apr 3 at 15:38
• Without computing a thing it was always clear that the property is obviously false, since an analytic function that returns values on a line must be constant. – user647486 Apr 3 at 17:37

## 1 Answer

Consider$$[\det(A+iB)]^2=\det[(A+iB)^2]=\det[A^2+iAB+iBA-B^2]\tag1$$Now note that due to $$A$$ (similar for $$B$$) being Hermitian, $$A^\dagger=A$$. Due to unitarity, $$A^\dagger=A^{-1}$$. Combining, $$A=A^{-1}\implies A^2=\Bbb I$$ So $$(1)$$ becomes $$i^n\det(AB+BA)$$ If $$n\equiv2\mod4$$, then $$i^n=-1$$. So $$[\det(A+iB)]^2=-\det(AB+BA)\tag2$$ Thus, if we can show that $$\det(AB+BA)\ge0$$ for all $$A,B\in S$$, then your hypothesis holds. However, inspired by user647468's comment, we can take $$B=\Bbb I\in S$$, and $$A_{ij}=(-1)^{i+1}\delta_{ij}$$. Then $$\det(AB+BA)=2^n\prod_{i=1}^{n}(-1)^{i+1}=-2^n<0\tag3$$Hence the $$RHS$$ of $$(2)$$ can be positive, and so $$\det(A+iB)$$ is not necessarily imaginary.