Karhunen-Loeve transform of a repeating process This question is inspired by measurements of an unsteady flow. I have some doubts about interpretation of principal component transform using Karhunen-Loeve theorem.
I have (centered $\equiv$ zero mean) data $X(\vec{x},t)$ and I decompose them using this method in such a way that:
$$
X(\vec{x},t) = \sum_k \Phi_k(\vec{x})\Psi_k(t)
$$
so there are coupled modes for a space only (topos) and a time only (chronos).


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*Are there any preconditions for a form and shape of the time modes (chronoi)?
E.g. when there is a process which would require a several peaks in frequency domain ("Fourier-like speaking") could it be decomposed in one non-pure-sine time mode only? (If the spatial action would be suitable for that.)


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*The modes are related to the covariance matrix eigenvalues. Is there any estimation of how much energy (in a signal processing meaning) is already decomposed based on eigenvalues of used modes?
 A: *

*Are there any preconditions for a form and shape of the time modes (chronoi)?
Yes, there are. The modes must be orthogonal, so for any $ij$ there must be $(\Psi_i,\Psi_j) = \delta_{ij}$. Hence there are in practical point of view limited options of shapes: If the first mode is "sine-like" than the second should be "cos-like", then the argument shoud be doubled etc.
For the spatial modes the same is valid but in higher dimensions and without the condition of such as "the time plot should end where it had started" way more options of shapes are naturally open.


*

*The modes are related to the covariance matrix eigenvalues. Is there any estimation of how much energy (in a signal processing meaning) is already decomposed based on eigenvalues of used modes?
Yes, there is. During the decompocition the eigenvalues $\lambda_k$ of the data covariance matrix $C_{ij} = \sum_{x} X(x,t_i)X(x,t_j)$ is calculated. The global energy stored in $N$ modes for presumed high $N$ is then defined:
$$
E = \sum_{k=1}^{N} \lambda_k^2
$$
The relative energy part for a number of modes $< N$ is then straightforward.
