No direction on a cycle

$\hat{\mat A}$ is a circulant matrix, and its eigenvectors are the discrete Fourier modes. For the second part, compare transposes.

Every node has degree $3$ with its self-loop, so $\hat{\mat A}$ has $\tfrac13$ on the diagonal and on the two neighbouring positions of each row, cyclically. On the Fourier mode $x_i=e^{\mathrm i\theta i}$ with $\theta=2\pi j/6$ it acts as multiplication by

so the eigenvalues are $1$, $\tfrac23$, $\tfrac23$, $0$, $0$ and $-\tfrac13$. High frequencies are damped and the lowest kept: a low-pass filter.

The difference operator $\mat D$, with $(\mat D\vec x)_i=x_{i+1}-x_{i-1}$, satisfies $\mat D^{\mathsf T}=-\mat D$: it is antisymmetric. Every power of the symmetric $\hat{\mat A}$ is symmetric, so every polynomial $\sum_kc_k\hat{\mat A}^k$ is symmetric. A matrix that is both symmetric and antisymmetric is zero, and $\mat D$ is not zero. So no polynomial in $\hat{\mat A}$ equals $\mat D$. $\blacksquare$

A convolution of width three, as in I.11, computes $\mat D$ with weights $(-1,0,1)$. The graph convolution cannot, because it has no way to tell the neighbour at $i+1$ from the one at $i-1$. Features that encode position around the cycle, which the bare graph does not supply, would be needed.