Let $A$ be a symmetric real-valued $n\times n$ matrix and let ${\left\|A\right\|_{2\rightarrow 2}} := \max_{\left\|u\right\|_{2}\leq 1} \left\|Au\right\|_{2}$ denotes its operator norm (largest singular value). Furthermore, for a strictly positive scalar $\lambda$, define the soft thresholding function by

$$f_{\lambda}(x) := (\left\lvert x \right\rvert-\lambda)_{+} \mathop{\mathrm{sign}}(x) = \max(\left\lvert x \right\rvert-\lambda, 0) \mathop{\mathrm{sign}}(x).$$

Finally, I define the soft thresholded version of $A$ by $A_{\lambda} := \Big[f_{\lambda}(A_{ij})\Big]^n_{i,j=1}$.

Is it true to say that the soft thresholding operation reduces the operator norm, i.e. ${\left\|A_{\lambda}\right\|_{2\rightarrow 2}} \leq {\left\|A\right\|_{2\rightarrow 2}}$. Note that this claim is obviously true for all entry-wise norms such as Frobenius or $\sup$ norms.

I could not find any counter-example for $10^5$ randomly generated Gaussian matrices!