Let $\boldsymbol{S}$ be $k \times k$ positive semi-definite real symmetric matrix with eigen decomposition $\boldsymbol{S} = \boldsymbol{X} \boldsymbol{\Lambda} \boldsymbol{X}'$ ($\boldsymbol{\Lambda}$ diagonal, $\boldsymbol{X}$ orthonormal matrix of eigenvectors). Assume that we reduce each eigenvalue $\lambda\_i$ by $\psi\_i \in [0, \lambda\_i]$, for $i = 1, \ldots, k$ with $\lambda\_i$ sorted so that $\lambda\_i \ge \lambda\_{i+1}$.


Define our ratio of interest $r\_{ij}^\psi$ in terms of elements of the new matrix $\boldsymbol{S}^{\psi}$:


$$ r\_{ij}^\psi = \frac{s\_{ij}^\psi}{\sqrt{s\_{ii}^\psi}\sqrt{s\_{jj}^\psi}} = \frac{\sum\_{l=1}^k x\_{il} x\_{jl} (\lambda\_l - \psi\_l)}{\sqrt{\sum\_{l=1}^k x\_{il}^2 (\lambda\_l - \psi\_l)} \sqrt{\sum\_{l=1}^k x\_{jl}^2 (\lambda\_l - \psi\_l)}}. $$


How does $r\_{ij}^\psi$ change as $\psi\_i$ grows? In particular, I am interested in the case when $\psi\_i \ge \psi\_{i+1}$ for $i = 1, \ldots, k$.


Initially my numerical experiments delivered an increase in absolute value, but now I have found the cases that yield a decrease instead. I can formulate a counterexample for some subset of parameter values using the fact that columns of $\boldsymbol{X}$ are length one and orthogonal to each other (e.g., when $\lambda\_k \approx 0$), so strictly speaking I'm done; but I wonder if it can be shown more generally and elegantly.


(If $r\_{ij}^\psi$ is thought as a correlation coefficient, then this problem has direct relation to statistics.)


[Updated after Armadillo Jim's response.]



Reducing eigenvalues of symmetric PSD matrix towards 0: effect on ratios of original matrix elements?
I don't think you can say whether the ratios $r\_{ij}$ increase or decrease. Consider the following two extremes when $S$ is positive definite with smallest eigenvalue $\lambda\_{min}>0$.


(1) Pick every $\psi$ the same positive number, and smaller than $\lambda\_{min}$. Then $S^\psi=S-\lambda\_{min}I$ since $\lambda\_{min}I$ commutes with $X$. Only the diagonal elements of $S$ will change, the diagonals will decrease, and the ratios $r\_{ij}$ will all increase.


(2) Pick the $\psi$s so that every adjusted eigenvalue is constant (and equal to $\lambda\_{min}$). That is, $\psi\_i=\lambda\_i-\lambda\_{min}$. Then $S^\psi$ is diagonal: it is $\lambda\_{min}I$. All the off-diagonals are zero, and the diagonals non-zero so $r\_{ij}=0$. That is all the ratios will decrease.


