According to the conjugate symmetry property of Fourier transform, shouldn't the following command not return 1 (=true):


However it does not (returns 0).

Is it due to some rounding error? Something else? Otherwise, how would you check that it is (avoiding to write a painful code with for loops etc)?

  • $\begingroup$ I assume x is 3D (R, G, B channels imported?). So is fft2 really doing a 3D FFT or, just 3 2D FFTs? $\endgroup$
    – M529
    Aug 9, 2019 at 20:39

1 Answer 1


In the 2D (continuous) Fourier domain, the Hermitian symmetry (for functions) writes (conjugate symmetric with respect to the origin):

$$F(-u,-v) = \overline{F}(u,v)\,.$$

For matrix $A_{i,j}$, we should have:

$$ a_{i,j} = \overline{a}_{j,i}$$

which means that the diagonal is real. As you can see, Hermitian matrices and Hermitian functions are slightly different. There is something in the diagonal that needs a center. But diagonals of matrices are real.

Plus, the FFT output matrix is not straightforwardly organized like the (continuous) Fourier transform. Friday night here, I am coming back later on FFT symmetries.


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