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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 $x \in \mathbb{R}^n?$

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    $\begingroup$ Are you sure you wrote what you meant? Just take $S = k I$ for sufficently large $k$. $\endgroup$
    – Robert Israel
    Commented Jul 24, 2016 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
    Commented Jul 24, 2016 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
    Commented Jul 24, 2016 at 18:02
  • $\begingroup$ @RobertIsrael Thanks a lot! I got it! $\endgroup$
    – Astor
    Commented Jul 24, 2016 at 18:17

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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

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

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