The Rayleigh quotient for a covariance matrix $\mathbf{C}$ and a non-zero steering vector $\mathbf{a}$ is given by $$ R(\mathbf{C},\mathbf{a}) := \frac{\mathbf{a}^H\mathbf{C}\mathbf{a}}{\mathbf{a}^H\mathbf{a}}. $$

In this context, it can be shown that under the constraint $\mathbf{a}^H\mathbf{a}=1$ $$ \lambda_m \leq \mathbf{a}^H\mathbf{C}\mathbf{a} \leq \lambda_1, $$ where $\lambda_i, i = 1,...,m$ are the eigenvalues of $\mathbf{C}$ sorted in nonincreasing order. These equalities can be achieved when $\mathbf{a}$ is equal to the eigenvector of $\mathbf{C}$ associated with the smallest (i.e. $\lambda_m$) and the largest (i.e. $\lambda_1$) eigenvalue, respectively.

I was therefore thinking, that under the assumption that $$ \mathbf{a} = \exp(j \cdot \mathbf{k}x) $$ with $$ \mathbf{k} = [k_1,...,k_m]^T $$ I were able to solve for $\mathbf{x}$ by $$ x = (\tilde{\mathbf{a}}^H\tilde{\mathbf{a}})^{-1}\tilde{\mathbf{a}}^H\tilde{\mathbf{e}}_1, $$ where $$ \tilde{\mathbf{a}} = j \cdot \mathbf{k} $$ is the "model" contained in $\mathbf{a}$ and $$ \tilde{\mathbf{e}}_1 = \ln(\mathbf{e_1}) $$ is the logarithm of the eigenvector corresponding to the largest eigenvalue.

Unfortunately, I seem to have messed up something in my considerations: Actually, I expect $x$ to be a real-valued scalar result. However, as $\tilde{\mathbf{a}}$ is complex-valued (but containing an imaginary part only), $\tilde{\mathbf{a}}^H\tilde{\mathbf{a}}$ is a real-valued scalar. $\tilde{\mathbf{e}}_1$ is a complex-valued vector, i.e. $\tilde{\mathbf{a}}^H\tilde{\mathbf{e}}_1$ is a complex-valued scalar. Therefore, I will always receive a complex-valued scalar $x$ as a result.

Can anyone help me identifying my fallacy?

  • $\begingroup$ Your objective is not clear to me. Are you trying to maximize the Rayleigh Quotient subject to the constraint that $\textbf{a}$ is a complex sinusoid? $\endgroup$ – Mark Borgerding Jun 12 '14 at 17:39
  • $\begingroup$ Mark Borgerding: I would not really call it a constraint, but rather an assumption or model, but yes, this is basically what I'm intending to do. I just want to find the scalar x, which maximizes the Rayleigh quotient by equalling a and the eigenvector corresponding to the largest eigenvalue. $\endgroup$ – Michael Jun 13 '14 at 23:02
  • $\begingroup$ Could you rephrase your problem in a clearer way? What exactly do you want to solve? $\endgroup$ – Royi Jun 18 '18 at 5:54

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