# eigen value decomposition of co-variance a series generated by factor model

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 $$Cov(Y_t)$$

• 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. – jcz Jul 26 at 10:23
• Yes, but can we not get any expression for these two eigenvalues? – Sudarshan Kumar Jul 26 at 11:52