In the case of linear time invariant filtering I know of the famous result that the circulant matrices always have an eigensystem with Fourier basis functions and diagonal matrix given by FFT of first row / column.
But what happens if we interpolate the filter? Can we say anything about the eigensystem then?
I made a small example with the identity filter (leaves vector unchanged). Then the 16x16 matrix interpolating linearly $[0,15] \to [0,14]$ assuming we have sampled at integers $\{0,1,2,\cdots,15\}$ First row : $$\left[\frac {15} {16},\frac 1 {16},0,0,0,\cdots\right]$$ second row: $$\left[0,\frac{14}{16},\frac{2}{16},0,0,\cdots\right]$$
This gives me the peculiar eigenvalue decomposition:
eigenvectors $[1,1,0,0,0,0]^{(*k)} \cdot [1,-1,1,-1,\cdots]$
Where $(\cdot)^{(*k)}$ denotes convolutional power ( iterated self convolution ). Corresponding eigenvalues seem to be exactly $\frac{16-k}{16}$. Also very curious.
Can we derive or explain this somehow?