# Reducing eigenvalues of symmetric PSD matrix towards 0: effect on ratios of original matrix elements?

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.)

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.