1
$\begingroup$

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!

$\endgroup$
2
  • $\begingroup$ As a general statement without your constraints on $S$, this is clearly false. This is clearly false. For counterexample, take $$ K = \pmatrix{1&0\\0&2}, \quad S = \pmatrix{0&1\\1&0}. $$ $\endgroup$
    – Ben Grossmann
    Commented Apr 8, 2023 at 17:50
  • $\begingroup$ Certainly, it wouldn't hold for any matrix $S$. I was just curious for the specific case where $S$ is a diagonal. $\endgroup$
    – sudeep5221
    Commented Apr 9, 2023 at 0:06

1 Answer 1

Reset to default
2
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ Oh that was simple! Thanks! $\endgroup$
    – sudeep5221
    Commented Apr 9, 2023 at 0:05

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .