# Can antiunitary symmetry be used to calculate determinant of a matrix

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?

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Correct me if I am wrong, but given any $A$ with a full eigen-basis, could not $U$ be built from that by taking the same eigenvectors and replacing all eigenvalues with $\pm\sqrt{-1}$? That does not limit $A$ very much. – adam W Jan 21 '13 at 15:47
@adam: You seem to be conceiving of $U$ as a linear operator. An antiunitary operator is antilinear. – joriki Jan 21 '13 at 15:52
@joriki yes I am unfamiliar with it, and went solely on the facts in the question of $U^2 = -I$ and $AU=UA$. (Not $U^2=-1$ since assuming $-1 = -I$ was the intended.) – adam W Jan 21 '13 at 16:03
@adamW the antiunitarity is important, and it does limit the form of A quite a lot. But you're right, I did mean $U^2 = -I$ (I is the unity operator) – jjj Jan 21 '13 at 16:38