# Spectral norm of modified matrix

Let $$K$$ be a symmetric, positive definite matrix and $$S$$ be a diagonal matrix such that all entries on its diagonal are greater than $$1$$. I am trying to prove the following relation.

$$\| S K S - K \| \leq \|K\| \left( \| S \|^2 - 1 \right),$$

where $$\| \cdot \|$$ denotes the spectral norm of a matrix.

I feel that intuitively the relation should hold but I haven't been able to prove it. Any hints or solutions (or even a counterexample) would be appreciated. Thanks!

• As a general statement without your constraints on $S$, this is clearly false. This is clearly false. For counterexample, take $$K = \pmatrix{1&0\\0&2}, \quad S = \pmatrix{0&1\\1&0}.$$ Commented Apr 8, 2023 at 17:50
• Certainly, it wouldn't hold for any matrix $S$. I was just curious for the specific case where $S$ is a diagonal. Commented Apr 9, 2023 at 0:06

It is true. Since $$S\succeq I$$, we have $$\|S-I\|=\|S\|-1$$. Therefore \begin{aligned} \|SKS-K\| &\le\|SKS-KS\|+\|KS-K\|\\ &\le\|S-I\|\|K\|\|S\|+\|K\|\|S-I\|\\ &=(\|S\|-1)\|K\|\|S\|+\|K\|(\|S\|-1)\\ &=\|K\|(\|S\|^2-1). \end{aligned} Note that $$K$$ doesn't need to be positive definite. It can be a general square matrix. $$S$$ also doesn't need to be a diagonal matrix. As long as $$S\succeq I$$, the proof above remains valid.