# Why matrix representation of convolution cannot explain the convolution theorem?

A record saying that Convolution Theorem is trivial since it is identical to the statement that convolution, as Toeplitz operator, has fourier eigenbasis and, therefore, is diagonal in it, has disappeared from Wikipedia.

The Convolution Theorem states that convolution of functions, h(t) and x(t), in time domain is equivalent to their multiplication in the frequency domain. That is, you convolve them, h*x, and take result into frequency domain. Result F(h*x) must be the same as multiplying their Fourier images, H and X: $F(h*x) = H \cdot X$, where H and X are fourier images, $H = F(h)$ and $X = F(x)$.

This was the definition where I used the letters h and x for the functions, instead of conventional f and g, because convolution h*x is often represented by h(x), where h is a convolution matrix derived from the first function, h, which is applied to the second function, x. Being Toeplitz, matrix h has eigenbasis F -- the same as apply to vector to take it into Fourier domain (see change of basis to see why base matrix product for base transform). Therefore, matrix h, translated into its eigenbasis, happens to be diagonal and H. That is, according to the Convolution Theorem, we must prove that

$$F(h \vec x) = H \vec X$$

But, there is nothing to prove. We just can show that

$$H \vec X = (FhF^{-1})(F\vec x) = F(F^{-1}F)h(F^{-1}F)\vec x = F(h \vec x)$$

or, the other way around

$$F(h \vec x) = F(F^{-1}F)h(F^{-1}F)\vec x = (FhF^{-1})(F\vec x) = H \vec X$$

just to exercise the beautiful algebra of relationships and diagonalization in F. We just need to keep in mind that $H = FhF^{-1}$ is diagonal (multiplication operator) because F is eigenbasis of h.

This discussion was classified as nonsense. But why? I find it amazing that Toeplitz (or convolution) has Fourier eigenbasis. Should this precious fact be hidden from the general population? Why should general population appreciate the integral-based proof rather than enjoy this fact? Can I say that Toeplitz operator = Convolution (operator)? Can I say that operator is a (continuous-case) generalization of matrix?

Is convolution theorem related with the Fourier eigenbasis? Can it be explained simpler, based on this fact?

• For future reference: It's spelled eigen. No h. – kahen Apr 30 '13 at 19:19
• Why $H = Fh$ and also = $FhF^{-1}$? – user10024395 Nov 9 '17 at 12:58
• This is actually exactly what I was looking for when first reading about convolution and Fourier/Laplace transforms - thanks! – MGwynne Mar 24 '18 at 15:06
• I also found math.stackexchange.com/questions/918345/… useful as well, particularly the linked blog. – MGwynne Mar 24 '18 at 15:33