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?

  • $\begingroup$ And if $N\equiv 0 \bmod 4$, then the determinant obviously need not be imaginary. $\endgroup$ Apr 3 '19 at 19:12
  • $\begingroup$ @ChristianRemling I wouldn't mind learning why that is obvious... $\endgroup$
    – thedude
    Apr 3 '19 at 19:15
  • $\begingroup$ The diagonal matrix $\textrm{diag}(1+i,1+i,-1-i,-1-i)$ has determinant $-4$, and then in general build the matrix from such blocks. $\endgroup$ Apr 3 '19 at 19:18
  • 2
    $\begingroup$ Or, more succinctly perhaps, with $B=A$, we have $\det (A+iB)=(1+i)^N\det A$. $\endgroup$ Apr 3 '19 at 19:25
  • $\begingroup$ A guess: rewrite as $\det(D + i U^*BU)$ expand as a series in $i$, and collect $\endgroup$ Apr 3 '19 at 22:05

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 $A=\begin{pmatrix}0&I\\I& 0\end{pmatrix}$.


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