The determinant of a unitary matrix is 0

I was trying the calculate the determinant of the eigenvector matrix (let me call it U) of a Hermitian matrix (a Hamiltonian matrix H in a physical problem). As U should be a unitary matrix, its determinant should have modulus 1.

When I was doing the numerical calculation, I noticed that when the system size (or the size of the Hamiltonian matrix H) is relatively small, I get the correct value of det(U), however, when I enlarge the system size (e.g., H is of the order ), I found that matlab gives me . However, when I try det(U.'), it gives me 1.

So I just wonder what could be the reason for this strange result (issue) and how can I resolve it?

Answer 1

[Matt J]

Matlab's det involves taking the product of long matrix diagonals. This can overflow or underflow very easily. The following might be a more stable determinant calculation than what det() does:

[H,~]=qr(rand(4000));
det(H)

ans = 0

Instead:

[L,U,P]=lu(H);

tridet=@(M) exp(sum(log(diag(M))));

Determinant=real(det(P)*tridet(L)*tridet(U))

Determinant = -1.0000

Answer 2

[Matt J]

Determinant calculations for large matrices are numerically delicate (which is why they're often avoided). There are other/better ways you can verify the unitariness of H. Example,

[H,~]=qr(rand(4000));

det(H)

ans = 0

det(H.')

ans = 0

Instead, we can do:

min(H*H.'-eye(4000),[],'all')

ans = -9.9920e-16

max(H*H.'-eye(4000),[],'all')

ans = 1.1102e-15

[Bruno Luong]

MATLAB DET uses LU decomposition, which is far from the best to compute the determinant. IMO if they would use QR (with permutation) or SVD it would be more stable and accurate.

But who cares? DET is mostly useless for any pratical application regardless the algorithm. It can be useful for some very small dimensions (2, 3, up to 5). Beside that it just creates more problems than solving problem.

[Paul]

@Bruno Luong

I understand that prod(svd(A)) = abs(det(A)). How can sign(det(A)) be determined? Or am I on the wrong track altogether with using SVD to compute DET?

[Bruno Luong]

One can compute the seterminant of U and V. See my answer in this thread (I also the one who asks the question and you have participated actively).

Second method! In could imagin within SVD calculation (usually based on qr-algorithm ast some step) the determinant of U and V can be computed internally. It is just need to be reported back. TMW don't want to investigate such