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.

  • 2
    $\begingroup$ Try putting $A=B=I$. $\endgroup$ – user647486 Apr 3 at 14:38
  • $\begingroup$ @user647486 the fact does not hold for all dimensions, sorry $\endgroup$ – thedude Apr 3 at 14:50
  • $\begingroup$ Take $A=\begin{pmatrix}1&0\\0&-1\end{pmatrix}$ and $B=I$. $\endgroup$ – user647486 Apr 3 at 15:38
  • $\begingroup$ 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. $\endgroup$ – user647486 Apr 3 at 17:37

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.