# Maximum Eigenvalue of Autocorrelation Matrix Bound

I am reading Detection and Estimation by Kay. In problem 2.7, one of the parts is to show that the maximum eigenvalue of the autocorrelation matrix is less than the maximum value of the PSD.

The autocorrelation matrix for a wide-sense stationary signal $X$ is defined as $$\mathbf{R} = \begin{bmatrix} r_{xx}(0) & r_{xx}(1) & \cdots & r_{xx}(N-1) \\ r_{xx}(1) & r_{xx}(0) & \cdots & r_{xx}(N-2) \\ \vdots & \vdots & \ddots & \vdots \\ r_{xx}(N-1) & r_{xx}(N-2) & \cdots & r_{xx}(0) \end{bmatrix}.$$

The PSD, $P_{xx}(f)$, is defined as: \begin{align*} P_{xx}(f) & = \sum_{k = -\infty}^{+\infty} r_{xx}(k) \exp(-j2\pi fk) \\ & = \sum_{k = -(N-1)}^{N-1} r_{xx}(k) \exp(-j2\pi fk) \end{align*} where $j = \sqrt{-1}$.

The following statement uses the Rayleigh quotient:

$$\lambda_{\text{MAX}} = \max_{\mathbf{u}} \frac{\mathbf{u}^T \mathbf{R} \mathbf{u}}{\mathbf{u}^T \mathbf{u}}$$ where $\lambda_{\text{MAX}}$ represents the largest eigenvalue of the matrix $\mathbf{R}$. It turns out the equation for $\lambda_{\text{MAX}}$ can also be expressed in another form, after using $\mathbf{u}^T \mathbf{R} \mathbf{u} = \int_{-1/2}^{1/2} |U(f)|^2 P_{xx}(f) \, \mathrm{d}f$ in the numerator and $\mathbf{u}^T \mathbf{u} = \int_{-1/2}^{1/2} |U(f)|^2 \, \mathrm{d}f$ in the denominator. Finally,

$$\lambda_{\text{MAX}} = \max_{\mathbf{u}} \frac{\mathbf{u}^T \mathbf{R} \mathbf{u}}{\mathbf{u^T} \mathbf{u}} = \frac{\int_{-\frac12}^{\frac12} |U(f)|^2 P_{xx}(f)\, \mathrm{d}f}{\int_{-\frac12}^{\frac12} |U(f)|^2 \, \mathrm{d}f}$$

From the last equation, I am supposed to be able to show $\lambda_{\text{MAX}} < P_{xx}(f)_{\text{MAX}}$, where $P_{xx}(f)_{\text{MAX}}$ is the maximum value of $P_{xx}(f)$. How am I supposed to show that? It appears the numerator (and denominator) is a function of the vector $\mathbf{u}$. The variable $f$ disappears after integration is performed.

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I think there may an argument where in the maximization of the fraction, the vector $\mathbf{u}$ will "fill up" where-ever $P_xx(f)$ is larger. I'm somewhat curious how this would work without constraints on $\mathbf{u}$. –  jrand Sep 19 '12 at 4:05

The answer is easy: Since $\lvert U(f)\rvert^2 P_{xx}(f) \leq \lvert U(f) \rvert^2 P_{xx}(f)_{max}$, monotonicity of the integral implies \begin{align} \int_{-\frac12}^{+\frac12} \lvert U(f) \rvert^2 P_{xx}(f)\,\mathrm{d}f &\leq \int_{-\frac12}^{+\frac12} \lvert U(f) \rvert^2 P_{xx}(f)_{max}\,\mathrm{d}f\\ &= P_{xx}(f)_{max} \int_{-\frac12}^{+\frac12} \lvert U(f) \rvert^2\,\mathrm{d}f \end{align}
Note that $P_{xx}(f)_{max}$ is simply a constant that can be taken out of the integral. Thus, $$\underbrace{\frac{\int_{-\frac12}^{+\frac12} |U(f)|^2 P_{xx}(f)\,\mathrm{d}f}{\int_{-\frac12}^{+\frac12} |U(f)|^2\,\mathrm{d}f}}_{\substack{\text{True for any u.} \\ \text{Suppose u is set so that this quantity is maximized.}}} \leq P_{xx}(f)_{max}$$
Then, $$\lambda_{MAX} = \max_u \frac{\int_{-\frac12}^{+\frac12} |U(f)|^2 P_{xx}(f)\,\mathrm{d}f}{\int_{-\frac12}^{+\frac12} |U(f)|^2\,\mathrm{d}f} \leq P_{xx}(f)_{max}$$ That's all.