Suppose I have some $N \times N$ complex matrix $A$, that commutes with some antiunitary operator $U$ that satisfies $U^2 =-1$.
It can be shown that $\det(A)\ge 0$ , because for every eigenvector $v$ with eigenvalue $\lambda$ there will also exist an orthogonal eigenvector $Uv$ with $\lambda^*$.
it seems that this symmetry provides some limitations as to the form of $A$, but can I use this symmetry to efficiently calculate this determinant on a computer?
So far I tried the following: Create a set of orthonormal vectors that are labeled as $v_n$ , $U v_n$, with $n=1...N/2$ (suppose $N$ is even). Then you find $(Uv_i, A Uv_j) =(v_i,A v_j)^* $ , and $(U v_i , A v_j) = -(v_i,A U v_j)^*$. So essentially by knowing half of the matrix elements you know them all. But still, how does this help to numerically calculate the determinant?