 Noninvertible spectral compression versus a coordinate change

Noninvertible spectral compression versus a coordinate change distinguishes discarding spectral coordinates from expressing the same measurements in different units. The covariance-consistent rescaling of whitened spectral coordinates article assumes an invertible transformation. A compression $\mathbf{z}=B\mathbf{y}$ with fewer independent rows than input coordinates instead has covariance $C_z=BC{B}^T$ and may lose information even when this covariance is propagated correctly.^[ref]

For positive-definite $C$ and a known signal shape $\mathbf{s}$, compare the squared noise-weighted signal sizesassuming $B$ has independent rows. Put $\mathbf{t}=C^{-1/2}\mathbf{s}$ and $M=B{C}^{1/2}$. ThenThe matrix $P$ is an orthogonal projector. This proves that compression cannot increase this signal measure under the stated covariance model. The condition for preserving a spectral signal under linear compression is that its whitened shape lie entirely in the retained subspace.

For an original example, let $C=I_2$, desired shape $\mathbf{s}=(1,1{)}^T$ and nuisance shape $\mathbf{x}=(1,-1{)}^T$. Keeping only the first channel, $B=(1,0)$, reduces $J$ from two to one and makes both shapes the same scalar. An unknown nuisance amplitude can then imitate the desired signal exactly.

Summing instead, $B=(1,1)$, gives desired response two, nuisance response zero and noise variance two. It retains $J_B=2$ for this particular signal while rejecting that nuisance. Thus a noninvertible map can preserve a chosen estimation task, but cannot in general be reversed to recover every original spectral component.

The primary Raman apparatus records array spectra before subsequent spectral processing.^[ref] Choose compression against the scientific signal and nuisance model, and propagate the full covariance. Correct uncertainty accounting does not restore a spectral distinction removed by the map.