Imagine I have some values at $x=0,x=2,x=4,\cdots,x=10$. I want to interpolate to find values at $x=1,x=3,x=5,\cdots,x=9$. I do this $$\left[\begin{matrix}v_1 \\ v_3 \\ v_5 \\ v_7 \\ v_9\end{matrix}\right]=\left[\begin{matrix}\frac{1}{2} & \frac{1}{2} & 0 & 0 & 0 & 0 \\ 0 & \frac{1}{2} & \frac{1}{2} & 0 & 0 & 0 \\ 0 & 0 & \frac{1}{2} & \frac{1}{2} & 0 & 0 \\ 0 & 0 & 0 & \frac{1}{2} & \frac{1}{2} & 0 \\ 0 & 0 & 0 & 0 & \frac{1}{2} & \frac{1}{2} \end{matrix}\right]\left[\begin{matrix}v_0 \\ v_2 \\ v_4 \\ v_6 \\ v_8 \\ v_{10}\end{matrix}\right]$$ Now suppose I want to interpolate back from the odd-index points to the even-index points. Aside from some uncertainty about what to do for $v_0,v_{10}$, which are extrapolated now, a sensible way to do this is $$\left[\begin{matrix}v_0 \\ v_2 \\ v_4 \\ v_6 \\ v_8 \\ v_{10}\end{matrix}\right]=\left[\begin{matrix}\frac{1}{2} & \frac{1}{2} & 0 & 0 & 0 & 0 \\ 0 & \frac{1}{2} & \frac{1}{2} & 0 & 0 & 0 \\ 0 & 0 & \frac{1}{2} & \frac{1}{2} & 0 & 0 \\ 0 & 0 & 0 & \frac{1}{2} & \frac{1}{2} & 0 \\ 0 & 0 & 0 & 0 & \frac{1}{2} & \frac{1}{2} \end{matrix}\right]^T \left[\begin{matrix}v_1 \\ v_3 \\ v_5 \\ v_7 \\ v_9\end{matrix}\right]$$ This seems to be true in general, even in 2D or 3D (maybe only for regularly spaced grids): if we have an interpolation matrix from one grid to another, the interpolation back is given by the transpose of that matrix.
Why is this so? Is there a proof of this? Is there some sort of intuition behind the matrix transpose that makes this obvious? Does the transpose approximate some sort of pseudoinverse to make this work?
To me this looks more like an artifact of the bidiagonal structure of the interpolation matrix than anything suggestive of a deep relationship between the matrix and its transpose. The nonzero entries are all equal and lie on the two central diagonals. The transpose has the same bidiagonal structure, so aside from the first and last rows, you have the same averaging operations in each row that you had in the original matrix.