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I want to find the eigendecomposition of a 3-dimensional discretely sampled signal $X$, where each sample $X_{i,j,k}$ is treated as a vector $\langle i, j, k\rangle$ (with the origin at the middle of the signal) and with an associated weight $X_{i,j,k}$. (I need to figure out the rotation that best aligns one such sampled 3D grid with another, hence the eigendecomposition.)

Since I'm working with the Fourier transform of $X$, i.e. $\mathcal{F}(X)$, I wanted to see if it was possible to calculate the eigendecomposition directly in the Fourier domain.

My initial thought was that $N$-dimensional rotation in the signal domain (for $N \gt 2$) results in the same rotation in the frequency domain -- so the eigenvectors should be directly correlated. However, I assume there must be some sort of rotational offset between the eigenvector basis in each of the two domains, since calculating eigenvectors in the frequency domain will produce vectors dependent upon the magnitude of each frequency, whereas calculating eigenvectors in the signal domain will produce vectors dependent upon the position (phase) of values.

Questions:

  1. Is there any way to compute the eigendecomposition directly in the frequency domain?

  2. Is there any shortcut for computing the eigendecomposition as described, i.e. treating each signal-domain sample as a vector or point $\langle i, j, k\rangle$ with weight $X_{i,j,k}$, without having to create a large matrix listing all the indices $i$, $j$, and $k$ of each point in the signal, with their associated weight?

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