Operator norm of a soft thresholded symmetric matrix

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!

• I don't think so: imagine a large matrix of all 1's with the (1,1) entry replaced by -1.1. The vector corresponding to the largest singular vector will be all positive. Changing -1.1 to -1 increases the norm of the image. – Anthony Quas Jan 1 at 19:45
• Thanks for your comment. The soft thresholding operator shrinks all the entries at the same time, right? So, I don't understand how does you example fits into the framework. – Student Jan 1 at 19:59
• @AnthonyQuas Mathematica indicates that for $n=50$ the eigenvalues decrease. Here is the relevant code (N[Eigenvalue[ConstantArray[1,{50,50}]-2.1*SparseArray[{{1,1}->1},{50,50}]]], N[Eigenvalue[ConstantArray[.9,{50,50}]-1.9*SparseArray[{{1,1}->1},{50,50}]]]). It is certainly true if the entries are all positive (and greater than $\lambda$) in $A$. – Josiah Park Jan 1 at 19:59
• @JosiahPark Yes, the claim is true if all the entries are non-negative. I don't think we need the part regarding greater than $\lambda$. When all the entries are non-negative then all the entries of the leading eigenvector has to be non-negative as well. So, shrinking operation reduces the norm. – Student Jan 1 at 20:12
• It is true for $n=2$, but it is likely you have already checked the trivial cases. – Josiah Park Jan 2 at 20:37