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Range restriction in singular-covariance spectral fitting requires a residual to lie in the subspace where the measurement model allows random error. Let $\mathbf{y}=\mathbf{m}(\theta )+\mathbf{ϵ}$ with zero-mean error covariance $C$. Every vector $\mathbf{v}$ in the kernel of $C$ satisfiesso ${\mathbf{v}}^T\mathbf{ϵ}=0$ almost surely. Thus a permissible residual obeyswhere $C^{+}$ is the Moore-Penrose inverse. The matrix $C{C}^{+}$ is the orthogonal projector onto the range of the covariance.

Only after enforcing this condition should a generalized least squares objective use $(\mathbf{y}-\mathbf{m}{)}^T{C}^{+}(\mathbf{y}-\mathbf{m})$. The quadratic form alone assigns zero penalty to the forbidden directions and therefore cannot impose the condition. Rank deficiency from deterministic spectral upsampling gives its physical spectral-grid origin through $C=A{C}_{\mathrm{in}}{A}^T$. Covariance propagation preserves those exact dependencies.^[ref]

For an original idealized two-channel example, take $C=\mathrm{diag}(1,0)$, mean $\mathbf{m}(\theta )=(\theta ,2\theta {)}^T$ and observation $(3,4{)}^T$. The second channel has no random error in this stipulated model, so the range condition forces $\theta =2$. The remaining squared residual is one. Minimizing only the pseudoinverse quadratic would instead choose $\theta =3$ and objective zero, while predicting the impossible second-channel value six.

Real measurements need not contain a perfect channel: deterministic rebinning produces exact noise-free combinations of otherwise noisy channels. Additional physical noise in such a combination must be modelled if present. If covariance ranges change with the parameters, pseudoinverse fitting with a parameter-dependent covariance range requires enforcing the appropriate support for each candidate; a single fixed projection need not apply.

Raman detector-array spectra are recorded before numerical redistribution onto another spectral grid.^[ref] Fitting the original independent measurement coordinates often makes these support restrictions explicit and avoids treating interpolation points as new observations.

Rank deficiency from deterministic spectral upsampling occurs because a deterministic map cannot manufacture independent measurements. For $\mathbf{y}=A\mathbf{n}$ and input covariance matrix $C$,When the output has more channels than the input, its covariance matrix is necessarily singular. Covariance of count-conserving rebinned Raman channels gives the conserved-count example and its full propagated uncertainty.^[ref]

Using that parent's two-to-three channel map, the output always satisfies $y_1-y_2+y_3=0$. If the input readings are $20$ and $30$, the rebinned vector is $(10,25,15{)}^T$. Recovering the original readings as $n_1=2y_1$ and $n_2=2y_3$ shows that the third displayed number is not a third independent observation.

For an original fitting example, suppose both input means equal an unknown $\theta$ and their independent measurement-error variances are known to be $25$. Fitting in the original channels gives $\hat{\theta }=(20+30)/2=25$, with variance $25/2=12.5$. A correct generalized least squares fit after rebinning gives the same result and uncertainty. It must use the singular covariance on its supported range, for example through the Moore-Penrose inverse, together with the exact linear constraints.

The range restriction in singular-covariance spectral fitting matters if a proposed output model fails those constraints. A residual along $(1,-1,1{)}^T$ cannot arise from this rebinning model at all. Merely inserting a pseudoinverse into a quadratic objective would assign zero penalty to that impossible direction. Either enforce the range condition or supply a physically justified additional noise model; adding a small diagonal term solely to make the matrix invertible changes the statistical assumptions.

The motivating Raman apparatus measures detector-array spectra before any subsequent numerical grid change.^[ref] Working on the original grid often avoids this singular fitting problem while retaining the actual measurement information.

If a design matrix $X$ has deficient rank, every ordinary least squares minimizer has the form $X^{+}Y+z$ with $z\in \ker X$, where $X^{+}$ is the Moore-Penrose inverse. Hence the fitted values are unique but the coefficients are not: the minimizers form an affine subspace of dimension $p-\mathrm{rank}X$.

For arbitrary real or complex matrices, the Moore-Penrose inverse need not reverse products:For example, $A=(11)$ and $B=(10{)}^T$ give $AB=1$, while $A^{†}=(1/21/2{)}^T$ and $B^{†}=(10)$. Thus $(AB{)}^{†}=1$ but $B^{†}{A}^{†}=1/2$.

A useful sufficient condition is that $A$ have full column rank and $B$ have full row rank, with their shared inner dimension $r$. Then $A^{†}A=I_r$ and $B{B}^{†}=I_r$. Put $C=AB$ and $X=B^{†}{A}^{†}$. Direct multiplication givesThe last two products are Hermitian orthogonal projection matrices. All four Penrose equations therefore hold, proving $X=C^{†}$. These rank conditions are sufficient, rather than necessary; checking the Penrose equations explains why the rule works in a particular case.

For zero cash interest, constant excess drift $\mu$, volatility $\sigma$, and both terminal and consumption utility $2\sqrt{x}$, set $A=\sigma {\sigma }^T$ and assume $\mu$ belongs to the image of a linear map $A$. Let $A^{+}$ be the Moore-Penrose inverse and $q={\mu }^T{A}^{+}\mu$. Then $h^{\prime }+qh+1=0$, $h(T)=1$, so $h(t)=e^{q(T-t)}+(e^{q(T-t)}-1)/q$ for $q>0$, and $h(t)=1+T-t$ for $q=0$. The Hamilton-Jacobi-Bellman equation has the displayed solution. Maximizing dollar investment and consumption are ${\theta }_t=2X_t{A}^{+}\mu$ and $C_t=X_t/h(t)$. The resulting positive portfolio wealth is a geometric Brownian motion with a deterministic time-dependent drift, and finite-horizon moment bounds justify equality in verification.

Now take one equation for two unknowns:The linear system $x_1+2x_2=3$ has an entire affine line of exact solutions. Since $A$ has full row rank, its Moore-Penrose inverse isThe kernel is $\mathrm{span}\{(-2,1{)}^T\}$, and all exact solutions areBecause $x_{\ast }$ is orthogonal to $(-2,1{)}^T$,The Euclidean norm is therefore smallest exactly at $t=0$. The zero residual alone cannot select a solution; the minimum-norm rule supplies the additional criterion.

This selection depends on the specified inner product. Rescaling one unknown changes what counts as a small Euclidean norm, so units and weighting matter in an application. A minimum-norm answer is mathematically canonical for the chosen geometry, rather than an automatic guarantee of physical plausibility.

In an eigenbasis of $G=X^{T}X$, ridge regression divides the corresponding component of $X^{T}Y$ by $\gamma +\lambda$. If $\gamma =0$, its eigenvector $v$ satisfies $Xv=0$, and hence $v^T{X}^{T}Y=0$ exactly. Only positive eigenvalues contribute to the limit, giving the Moore-Penrose inverse expression for the minimum-norm least-squares solution. No full-rank or sample-size assumption is needed.