Since \( \lambda \geq 0 \), it means that compared to OLS, we have $$ \frac{\sigma_j^2}{\sigma_j^2+\lambda} \leq 1. $$
Ridge regression finds the coordinates of \( \boldsymbol{y} \) with respect to the orthonormal basis \( \boldsymbol{U} \), it then shrinks the coordinates by \( \frac{\sigma_j^2}{\sigma_j^2+\lambda} \). Recall that the SVD has eigenvalues ordered in a descending way, that is \( \sigma_i \geq \sigma_{i+1} \).
For small eigenvalues \( \sigma_i \) it means that their contributions become less important, a fact which can be used to reduce the number of degrees of freedom. Actually, calculating the variance of \( \boldsymbol{X}\boldsymbol{v}_j \) shows that this quantity is equal to \( \sigma_j^2/n \). With a parameter \( \lambda \) we can thus shrink the role of specific parameters.