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Let's assume $N\times T$ series $Y_t$ is generated by the following equation. $$ Y_t = \begin{bmatrix}A_x & A_m\end{bmatrix}\begin{bmatrix}x_t \\ m_t \end{bmatrix}$$ Where $A_x$ and $A_m$ are $N\times 1$ matrix and $x_t$ and $m_t$ are $1\times T$ series and $$ Cov\left(\begin{bmatrix}x_t \\ m_t \end{bmatrix}\right)= \begin{bmatrix}\sigma_x^2 & 0 \\ 0& \sigma_m^2\end{bmatrix}$$

Can we say anything (expression in terms of $A_x$ , $A_m$ $\sigma_m^2$ and $\sigma_m^2$) about the eigenvalue and eigenvector of the $Cov(Y_t)$

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  • $\begingroup$ One thing you can say is that $\text{cov}(Y_t)$ has rank of at most 2. So it has at most 2 positive eigenvalues, and the rest are exactly 0. $\endgroup$ – jcz Jul 26 at 10:23
  • $\begingroup$ Yes, but can we not get any expression for these two eigenvalues? $\endgroup$ – Sudarshan Kumar Jul 26 at 11:52

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