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I am seeking clarification on a particular step in the derivation of the MUSIC algorithm as presented in a specific paper. Here, there is an intermediate step I cannot follow and I would appreciate any insights or explanations.

We start with the equation $$ \mathbf{X} = \mathbf{A}\mathbf{F} + \mathbf{W}, $$ and proceed to obtain the spectral matrix $\mathbf{S}$ as $$ \begin{aligned} \mathbf{S} &= \mathrm{E}\big[\mathbf{X} \mathbf{X}^{\mathsf{H}}\big] \\ &= \mathbf{A} \mathrm{E}\big[\mathbf{F} \mathbf{F}^{\mathsf{H}}\big] \mathbf{A}^{\mathsf{H}} + \mathrm{E}\big[\mathbf{W} \mathbf{W}^{\mathsf{H}}\big] \\ &= \mathbf{A} \mathbf{P} \mathbf{A}^{\mathsf{H}} + \lambda \mathbf{S}_0, \end{aligned} $$ where $\mathbf{P}$ and $\mathbf{S}_0$ are Hermitian matrices, with $\mathbf{P}$ being positive definite.

The matrices are defined as follows:

  • $\mathbf{S}, \mathbf{S}_0 \in \mathbb{C}^{M \times M}$,
  • $\mathbf{A} \in \mathbb{C}^{M \times D}$,
  • $\mathbf{P} = \mathrm{E}\big[\mathbf{F} \mathbf{F}^{\mathsf{H}}\big] \in \mathbb{C}^{D \times D}$, with $\mathbf{F} \in \mathbb{C}^{D}$,
  • $\mathbf{S}_0 = \frac{1}{\lambda}\mathrm{E}\big[\mathbf{W} \mathbf{W}^{\mathsf{H}}\big] \in \mathbb{C}^{M \times M}$, with $\mathbf{W} \in \mathbb{C}^{M}$.

Now, when $D < M$, the matrix $\mathbf{A} \mathbf{P} \mathbf{A}^{\mathsf{H}}$ becomes singular, leading to $$ \det\left(\mathbf{A} \mathbf{P} \mathbf{A}^{\mathsf{H}}\right) = \det\left(\mathbf{S} - \lambda \mathbf{S}_0\right) = 0. $$

This brings us to the point where $\lambda$ equals one of the eigenvalues of $\mathbf{S}$ in the metric of $\mathbf{S}_0$, implying that $\lambda$ is in the spectrum of the matrix pencil $(\mathbf{S}, \mathbf{S}_0)$.

The point of contention arises in the next part of the derivation. Given that $\mathbf{A}$ has full rank and $\mathbf{P}$ is positive definite, it follows that $\mathbf{A} \mathbf{P} \mathbf{A}^{\mathsf{H}}$ is nonnegative definite. However, the claim is made that $\lambda$ can only be the minimum eigenvalue $\lambda_{\min}$, leading to the equation $$ \mathbf{S} = \mathbf{A} \mathbf{P} \mathbf{A}^{\mathsf{H}} + \lambda_{\min} \mathbf{S}_0, \quad \lambda_{\min} \ge 0, $$ where $\lambda_{\min}$ is the smallest solution to $\det\left(\mathbf{S} - \lambda \mathbf{S}_0\right) = 0$.

I struggle to understand why $\lambda$ can only be $\lambda_{\min}$. Why must it only be the smallest eigenvalue, rather than the second smallest or another eigenvalue? Could someone provide a detailed explanation or clarification on this part of the derivation?

Is this step based on strict mathematical reasoning, or is it due to an underlying assumption that the noise is always smaller than the signal?

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  • $\begingroup$ By the way, this isn't just "a specific" paper on MUSIC. This is THE paper. Read the "Calculating a Solution" section and it will describe what Joseph said below. There are theoretically multiple minimum eigenvalues of the same value, but usually are instead very close to each other since the results come from measurements. $\endgroup$
    – Envidia
    Oct 5, 2023 at 17:46

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This is a simple consequence of previous insights. As was observed that $APA^H$ is positive semi-definite, this means that also $S-\lambda S_0$ has to be the same. Now if $\lambda$ is not the smallest eigenvalue, then for an eigenvector $v_\min$ of $\lambda_\min$ we get $$ 0\le \underbrace{v_\min^H(S-\lambda_\min S_0)v_\min}_{=0}+\underbrace{(\lambda_\min-\lambda)}_{<0}\underbrace{v_\min^HS_0v_\min}_{>0}<0 $$ which is impossible.

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  • $\begingroup$ Given that $\mathbf{S}_0 = \frac{1}{\lambda}\mathrm{E}\big[\mathbf{W} \mathbf{W}^{\mathsf{H}}\big]$ is positive definite, i.e., the noise correlation matrix is invertible $\endgroup$
    – Naetmul
    Oct 6, 2023 at 3:34
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This is something that I studied doing my PhD research on a competing algorithm. The number of eigenvalues corresponds to the number of sensors, the number of "large" eigenvalues corresponds to the number of sources, therefore the "noise" eigenvalues correspond to the number of sensors minus the number of sources. So, it may not only be the smallest one used, but for convenance it's often assumed that the noise eigenvalues are all equal in value, at least theoretically for the noise model.

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