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.]
