0
$\begingroup$

Let $A \in \mathbb{R}^{n \times n}$ be a normalized non-strict column diagonally dominant matrix, that is:

$$a_{j,j} = \sum_{i \ne j} \left|a_{i,j}\right|$$

where $0 \le a_{j,j} \le 1$ and $-1 \le a_{i,j} \le 0$ for all $i \ne j$. Is it possible to find a symmetric, positive definite matrix $S$ such that $\langle A x, x \rangle \le \langle S x, x \rangle$ for all $x \in \mathbb{R}^n?$

$\endgroup$
  • 1
    $\begingroup$ Are you sure you wrote what you meant? Just take $S = k I$ for sufficently large $k$. $\endgroup$ – Robert Israel Jul 24 '16 at 17:48
  • $\begingroup$ @RobertIsrael Thanks, I got the suggestion to take $k=\lambda_{max}(A+A^T)/2$ but do I not need $\left< A \cdot, \cdot \right>$ to be an inner product for that? $\endgroup$ – Astor Jul 24 '16 at 17:53
  • $\begingroup$ $\langle Ax, x \rangle$ is bounded on the unit sphere $\{x: \langle x, x \rangle = 1\}$. Its maximum there is $k$. $\endgroup$ – Robert Israel Jul 24 '16 at 18:02
  • $\begingroup$ @RobertIsrael Thanks a lot! I got it! $\endgroup$ – Astor Jul 24 '16 at 18:17
1
$\begingroup$

Take $S = k I $ with $k = \lambda_{max}(\frac{A + A^T}{2})$.

Because

$A = \frac{A + A^T}{2} + \frac{A - A^T}{2}$

But

$\left< \frac{A - A^T}{2}x,x \right> = 0$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.