I'm reading a paper and it defines $\lambda(\omega)$ as the largest eigenvalue of the determinantal equation: $$|A'f^{re}(\omega)A-\lambda A'VA|=0$$ Where $f^{re}(\omega)$ is the real part of $f(\omega)$ is the spectral density of stationary time series $X_{t}$ at frequency $\omega$, $V$ is the the variance-covariance of $X_{t}$ and $A=[I_{k-1}|0]'$, where $I_{k-1}$ is the $(k-1)\times(k-1)$ identity matrix and $0$ is the $(k-1) \times 1$ vector of zeros. What matrix are we calculating its eigenvalues?

From what I know, to extract the eigenvalues $\lambda$s of a matrix $X$ we calculate the roots of $(X-\lambda I)=0$. Where $I$ is the identity matrix.

  • $\begingroup$ Presumably $A^\prime V A$ is invertible. Multiply everything by its inverse: what do you get? $\endgroup$ – whuber Jan 22 '17 at 21:46
  • $\begingroup$ @whuber the identity matrix? $\endgroup$ – Toney Shields Jan 23 '17 at 19:43

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