I came across a question asking how the proof that the transform $f(g) = g(x-a) + g(x+a)$ was Hermitian worked. I was a bit surprised that the proof was symbolic, because if you plot out the coefficients it's obvious.

By "plot out the coefficients" I mean "treat the point (x, y) as the flow from input=y to output=x". In the case of f$(g) = g(x-a) + g(x+a)$, the flows are from $x-a$ to $x$ and from $x+a$ to $x$. If we set $a$ to 2 w.l.o.g., then points matching $(x-2, x)$ and $(x+2, x)$ will have coefficient 1 while all other points will have coefficient 0. So we get two lines:


(The lines differing in color is an artifact of how I plotted the coefficients; they both represent coefficients equal to 1.)

This is just a continuous version of your typical finite matrix grid-of-numbers representation. Noting that the coefficients are not affected by transposing (mirroring about the x=-y axis), we see that the transform must be Hermitian.

The fact that the original proof was done symbolically made me wonder if it's common to plot the coefficients of linear transforms on $\mathbb{R}$ the way I'm doing. Are there gotchas? Are there better ways?

Other examples:

Phase (arg) of the Fourier transform's coefficients (contour plot):

arg Fourier transform

LaPlace transform coefficients (contour plot):

LaPlace transform

Craig Gidney
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  • Can you please elaborate what you mean by the term " plot out the coefficients " ? What coefficients are you talking about ? Thanks. – Srinivas K Apr 05 '15 at 19:38
  • @SrinivasK I've tried to clarify it. Each point on the plane represents a coefficient, and that coefficient is a flow from input to output. So if the coefficient for (1,2) is 5, then increasing the input by v at x=2 must increase the output by 5v at x=1. The coefficients are effectively a 2d function, so you just plot them. – Craig Gidney Apr 05 '15 at 19:45
  • Sorry to disturb you again. Since I'm new to maths, I'm not able to get what you mean by " flow from input to output " clearly. How are these related to the transform and the function ? – Srinivas K Apr 05 '15 at 19:51
  • @SrinivasK It's not standard terminology, so I'm not surprised it was confusing. Linear transforms have the property that each input value contributes to each output value linearly, and if you know all of these contributions (the "flows") then you know the whole transform. You can reverse-engineer the contribution coefficients e.g. by running `g(y) = {1 if y=5 else 0}` through the transform, at which point the output function `f(g)(x)` tells you the contribution from 5 to x at each x. – Craig Gidney Apr 05 '15 at 19:57

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