A homework question asks me whether or not the matrix $W_N$ from the matrix representation of the Direct Fourier Transform is a Hermitian operator. From what I understand an Hermitian operator does not change when transposed and conjugated? What bothers me is that the exponentials lose their negative exponent.

  • $\begingroup$ Seems OK. The DFT is (up to a scalar) unitary, not Hermitian (except for $n \leq 2$). $\endgroup$
    – WimC
    Oct 20, 2013 at 14:29
  • $\begingroup$ Thanks for your comment. Hmm well glad that Im doing it right then. :) I can rearrange it so I can answer my own question you reckon? $\endgroup$
    – Leo
    Oct 20, 2013 at 14:33

1 Answer 1


Turns out I thought I did it wrong, but I didnt. This aswer was originally in my question. I coulndt find this on math.stackexchange so I will answer my own question so other people can find it. I did the following:

This is the definition of the element $w_3$

$w_3 = e^{-j \frac{2 \pi}{3} }$

This is how conjugating exponentials works

$\overline{w_3} = \overline{e^{-j \frac{2 \pi}{3} }} = e^{\overline{-j \frac{2 \pi}{3} }} = e^{j \frac{2 \pi}{3} }$

This is the consequence of $2 \pi$ periodicity

$e^{-j \frac{2 \pi}{3} 4} = e^{-j \frac{8 \pi}{3} } = e^{-j (\frac{8 \pi}{3} -2 \pi)} = e^{-j \frac{2 \pi}{3} }$

So then I try and apply what I know to the matrix $W_3$

$W_3 = \left[ \begin{array} {cc}1&1&1\\1&w_{3}^{1}&w_{3}^{2}\\1&w_{3}^{2}&w_{3}^{4}\end{array} \right]$

$W_3 = \left[ \begin{array} {cc}1&1&1\\1& e^{-j \frac{2 \pi}{3} }& e^{-j \frac{2 \pi}{3} 2 }\\1& e^{-j \frac{2 \pi}{3} 2 }& e^{-j \frac{2 \pi}{3} 4 }\end{array} \right]$

$W_{3}^{T} = W_3$

$W_{3}^{H} = W_{3}^{T*} = W_{3}^{*} = \left[ \begin{array} {cc}1&1&1\\1& e^{j \frac{2 \pi}{3} }& e^{j \frac{4 \pi}{3} }\\1& e^{j \frac{4 \pi}{3} }& e^{j \frac{2 \pi}{3} }\end{array} \right] \neq W_3$

So then W would not be a Hermitian operator.


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