# The determinant of a unitary matrix is 0

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Runrun Xu 2021 年 4 月 18 日 10:11

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?

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### 回答 (2 件)

Matt J 2021 年 4 月 18 日 11:59

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
[L,U,P]=lu(H);
tridet=@(M) exp(sum(log(diag(M))));
Determinant=real(det(P)*tridet(L)*tridet(U))
Determinant = -1.0000
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Matt J 2021 年 4 月 18 日 11:19

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
min(H*H.'-eye(4000),[],'all')
ans = -9.9920e-16
max(H*H.'-eye(4000),[],'all')
ans = 1.1102e-15
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Bruno Luong 2021 年 4 月 18 日 14:52
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

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