# Obtain unknown matrix from eigendecomposition.

I have doubt with the mathematical tools used in a problem of Signal Analysis:

I have a complex observable series $$Y(t)$$ which is the result of summing two complex r.v $$X(t)$$ (unobservable) and a $$\epsilon(t)$$ (unobservable).

$$Y(t)=X(t)+\epsilon(t)$$

Assume that $$X$$ and $$\epsilon$$ are uncorrelated, and that $$\epsilon$$ is serially uncorrelated and has constant mean and variance.

I construct a Hankel Matrix of the observed series and then I create a sort of variance-covariance matrix by multiplying the later by its conjugate transpose.

$$R_Y=\mathcal{Y}\cdot \mathcal{Y^*}=\mathcal{X}\cdot \mathcal{X^*}+\mathcal{E}\cdot \mathcal{E^*}$$

After that I make the eigendecompositon of that matrix and I obtain:

$$U\cdot (\Lambda_X +\sigma^2\cdot I)\cdot U^T$$ Where $$U$$ is the orthogonal matrix of eigenvectors, $$\Lambda_X$$ is a diagonal matrix containing the eigenvalues (in decreasing order) of $$\mathcal{X}\cdot \mathcal{X^*}$$ and $$\sigma^2$$ represents the "variance" of $$\epsilon$$ (under my assumptions $$\mathcal{E}\cdot \mathcal{E^*}$$ is a diagonal matrix with $$\sigma^2$$ in its diagonal.

Assume that the matrix $$\mathcal{X}\cdot \mathcal{X^*}$$ has not full rank, and because of that it has some null eigenvalues.

By the eigendecomposition of $$R_Y$$ we could select which eigenvalues correspond to the $$\mathcal{X}\cdot \mathcal{X^*}$$ matrix, (by considering the first $$d$$ eigenvalues, HOW TO SELECT THE ORDER OF THE MODEL?) and then we can estimate $$\sigma^2$$.

Knowing $$\sigma^2$$ we could estimate $$R_X=\mathcal{X}\cdot \mathcal{X^*}$$.

Is there any way to factorize $$R_X$$ in order to obtain $$\mathcal{X}$$ and then by the one-to-one relation between Hankel matrices and Series, obtain $$X(t)$$? if this is possible, under which additional conditions? If the variables are real-valued, is it the same?

• Is this the kind of idea you are looking for? – AnonSubmitter85 Oct 29 '18 at 18:49
• Indeed it looks very promising, I'll give it a in-detail look. Thanks! – RScrlli Oct 29 '18 at 19:00