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I already asked this question here but there is no response. I'd like to ask this question in signal processing domain.

It is well-known that for a real symmetric matrix $L$ (here, graph Laplacian) one can write the eigenvalue decomposition as

$$ L = U \Lambda U^{\mathsf T}, $$ where $U$ is a real eigenvector matrix. Moreover, in graph signal processing papers, including the great paper by Shuman et al. (cf. page 4), the adjoint (complex conjugate) of $U$ is used to define the graph Fourier transform $\mathcal{F}_{G}$ as $$ \hat{x} = \mathcal{F}_{G} x = U^{*}x, $$ where $x$ is the signal in vector form and $U^{*}$ is the complex conjugate of $U$.

I am curious to know is there any specific reason for using the notation of complex conjugate since $U$ is real?

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You terminology is something I am not used to.

$$ \hat{x} = \mathcal{F}_{G} x $$

If by $\hat{x}$ you mean the DFT (complex except very special cases), $x$ is your input (real or complex), and $\mathcal{F}_{G}$ is the matrix composed of the sinusoidal basis vectors, then your assertion that $U$ is real is dubious. $\mathcal{F}_{G}$ is most certainly not. With the proper normalizattion:

$$ \mathcal{F}_{G}^{-1} = \mathcal{F}_{G}^{*} $$

Now, for a known $x$ where $\hat{x}$ is known to be complex, how can $U$ be real?

Since they are using the complex conjugate, I would say that U is expected to also be capable of being complex. I have not read the paper so I don't know where you picked up that assumption.

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  • $\begingroup$ $\hat{x}$ is the graph Fourier transform of $x$ not DFT! $\endgroup$ – Amin Jul 20 at 3:55
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    $\begingroup$ @Amin Okay, I guess I have something else to study. $\endgroup$ – Cedron Dawg Jul 20 at 4:05
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    $\begingroup$ @Amin It seems you are not alone in asking that. Check out the comments: johndcook.com/blog/2016/02/09/… Maybe you should pose your question there, but the other guy didn't get an answer. The only papers I could find were behind the IEEE paywall and, sorry, I won't support them. $\endgroup$ – Cedron Dawg Jul 20 at 4:16
  • $\begingroup$ It is ambiguous for me why we need complex conjugate there. $\endgroup$ – Amin Jul 20 at 4:31

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