Determinant involving traceless unitary hermitian matrices

Question

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

A friend asked me the following question, motivated by a problem in condensed matter physics:

Is it true that for every two matrices $A$, $B$ in $S$, the value of $\det(A+iB)$ is always imaginary?

Well, if a matrix is unitary and hermitian, it can only have $\pm 1$ as eigenvalues. To be traceless, $N$ must be even.

I ran a computer experiment. I wrote $A=UDU^\dagger$ and $B=VDV^\dagger$ where $D={\rm diag}(1^M,(-1)^M)$ and $U$, $V$ are random unitary matrices of dimension $N=2M$.

The result is that $\det(A+iB)$ seems to be indeed always imaginary, if $N\equiv 2 \text{ mod } 4$.

Any ideas how this could be proved?

Answer 1

Yes. It's real when $N\equiv 0 \text{ mod } 4$ and imaginary when $N\equiv 2\text{ mod } 4$.

The square of the determinant is $\det(A+iB)^2=\det(1-1+i(AB+BA))=i^N\det(AB+BA)$, so for either parity of $N/2$ we need to show the Hermitian matrix $AB+BA$ has nonnegative determinant. Up to unitary transformation, $B=D$ and $A=\begin{pmatrix}X&Y\\Y^\dagger& -Z\end{pmatrix}$. $A^2=1$ implies $XY=YZ$, so if $Y$ is invertible then $\det X=\det Z$ and $AB+BA=\begin{pmatrix}2X&0\\0& 2Z\end{pmatrix}$ has nonnegative determinant.

Let $V$ be the variety of all such $A$. By a perturbation argument, it's enough to show that for a dense subset of $V$, the determinant of $Y$ is nonzero. The unitary group is (Zariski) irreducible and $V$ is its image under an algebraic map, so it too is irreducible, which means that it is enough to find a single $A$ with invertible $Y$. Take $A=\begin{pmatrix}0&I\\I& 0\end{pmatrix}$.