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Context: I was reading the following set of notes (page 83): here and it says:

Thus, the Fourier transform of signal (or function) $ \mathbf{f} \in R^{|V|} $ on a graph can be computed as $$ \mathbf{s} = \mathbf{U}^T \mathbf{f} $$

Question: What happens if each node has multiple 'signals'? Are the Fourier transforms on each signal independent of one another?

Attempt: I assume that each signal is denoted as a column vector, and thus multiple signals may be written as a matrix $\mathbf{F} = [\mathbf{f_1}, \mathbf{f_2}, ..., \mathbf{f_n}]$ (where $ \mathbf{f_i} \in R^{|V|} $) for a graph with $n$ signals. Thus, the graph Fourier transform would be $$ \mathbf{S} = \mathbf{U}^T \mathbf{F} $$ and thus each of the Fourier transforms would be independent of one another. Is this the correct way to think about this.

Many thanks in advance!

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  • $\begingroup$ Single node can't have multiple signals they have $\mathbb{R}^n$ dimensional feature vector, so in this context each node have $n$ number of channel. The signal is define over the entire graph. Take the first channel of each node, you'll get a graph signal, then take second channel you'll get another signal(Think of RGB image where we have 3 channel). Now when we perform GFT, we perform it over individual graph signal. $\endgroup$
    – Swakshar Deb
    Commented Sep 10, 2021 at 7:54
  • $\begingroup$ Thanks @SwaksharDeb! Okay cool, yes I was thinking exactly along the lines of RGB (but didn't know how to word it). Okay so each nodes can have $n$-dimensional feature vectors and when taking the GFT, we just consider one layer at a time (over the whole graph).? That makes more sense now $\endgroup$
    – Rocky the Owl
    Commented Sep 11, 2021 at 3:29

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