Subspace frequency estimation methods like MUSIC or ESPRIT decompose the signal correlation matrix into a signal and a noise subspace. Assume the signal model is given by

$$\boldsymbol{s} = \boldsymbol{H}(\boldsymbol{f})\boldsymbol{a} + \boldsymbol{n}$$ where $\boldsymbol{s} \in \mathbb{C}^{M \times 1}$ contains the observations, $\boldsymbol{f} \in \mathbb{R}^{P \times 1}$ is the vector of frequencies to be estimated, $\boldsymbol{a} \in \mathbb{C}^{P \times 1}$ are the (complex) amplitudes and $\boldsymbol{n}$ is zero-mean white Gaussian noise with variance $\sigma^2$ that is uncorrelated with $\boldsymbol{a}$. The signal correlation matrix is then given by

$$ \boldsymbol{R}_{\boldsymbol{ss}} = \mathbb{E}[\boldsymbol{s} \boldsymbol{s}^H] = \boldsymbol{H}(\boldsymbol{f}) \boldsymbol{R}_{\boldsymbol{aa}} \boldsymbol{H}^H(\boldsymbol{f}) + \sigma^2 \boldsymbol{I}.$$ whereby $\boldsymbol{R}_{\boldsymbol{aa}} = \mathbb{E}[\boldsymbol{a} \boldsymbol{a}^H]$. Since $ \boldsymbol{R}_{\boldsymbol{ss}}$ is Hermitian, a separation into signal subspace $\boldsymbol{Q}_s$ and noise subspace $\boldsymbol{Q}_n$ according to

$$\tag{1} \boldsymbol{R}_{\boldsymbol{ss}} = [\boldsymbol{Q}_s, \boldsymbol{Q}_n] \textrm{diag}(\lambda_1, \cdots, \lambda_P, \underbrace{\sigma^2, \cdots, \sigma^2}_{M-P \textrm{ times}}) [\boldsymbol{Q}_s, \boldsymbol{Q}_n]^H$$

can be performed, whereby $\boldsymbol{Q} = [\boldsymbol{Q}_s, \boldsymbol{Q}_n]$ is an unitary matrix. It is relatively straightforward to show (and it is often shown in the literature) that $\sigma^2$ occurs ${M-P}$ times as eigenvalue, provided that $\boldsymbol{R}_{\boldsymbol{aa}}$ has full rank $P$. The remaining $P$ eigenvalues $\lambda_1, \cdots, \lambda_P$ are then associated with the signal subspace, and they are often assumed to be bigger than $\sigma^2$ such that a separation into $\boldsymbol{Q}_s$ and $\boldsymbol{Q}_n$ can be performed based on sorting and comparing the eigenvalues of $\boldsymbol{Q}$ against each other. However, I was wondering if it is in general (or in some special cases) also possible to compute the values of $\lambda_1, \cdots, \lambda_P$ - the eigenvalues of the signal subspace.

To this end, I try to use the knowledge that the signal subspace $\boldsymbol{Q}_s$ is spanned by the column vectors of $\boldsymbol{H}(\boldsymbol{f})$, i.e. there is some regular transformation matrix $\boldsymbol{T} \in \mathbb{C}^{P \times P}$ such that $$\boldsymbol{Q}_s = \boldsymbol{H}(\boldsymbol{f}) \boldsymbol{T}$$ holds. Also, since $\boldsymbol{R}_{\boldsymbol{aa}}$ is a Hermitian matrix, it may be written as $$\boldsymbol{R}_{\boldsymbol{aa}} = \boldsymbol{M} \textrm{diag}(\lambda_{a,1}, \cdots, \lambda_{a,P}) \boldsymbol{M}^H$$ with some unitary matrix $\boldsymbol{M} \in \mathbb{C}^{P \times P}$.

Inserting these relations in the equation above yields

$$ \boldsymbol{R}_{\boldsymbol{ss}} = \boldsymbol{Q}_s \boldsymbol{T}^{-1} \boldsymbol{M} \textrm{diag}(\lambda_{a,1}, \cdots, \lambda_{a,P}) \boldsymbol{M}^H \boldsymbol{T}^{-H} \boldsymbol{Q}_s^{H} + \sigma^2 \boldsymbol{I}$$

Unfortunately, $\boldsymbol{T}^{-1} \boldsymbol{M} \textrm{diag}(\lambda_{a,1}, \cdots, \lambda_{a,P}) \boldsymbol{M}^H \boldsymbol{T}^{-H}$ is in general no diagonal matrix such that a direct comparison with (1) seems not valid. In the case of uncorrelated amplitudes, $\boldsymbol{M}$ would be the identity matrix, but then there is still $\boldsymbol{T}$ "in the way" and I could not come up with a condition when this is also the identity matrix. Anyhow, this would then constitute another special case, where the initial idea is to be as general as possible. Maybe there is another way to show it, but I have not found it yet.

Any suggestions on how (or in which special cases) this can be shown? Thank you for your contributions.

  • $\begingroup$ Interesting question. I suspect the answer is "no" to the question: Is it possible to compute the values of the eigenvalues of the signal subspace? --- at least not without just calculating them numerically. The reason is because they're going to depend on the signal components, the relative phases between the signal components, and the effect of the noise on the signal components. However, I'll think about it a bit more. Just thought i'd share my thoughts first here as a comment. $\endgroup$
    – Peter K.
    Commented Feb 8, 2022 at 21:58


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