# Non-strict column diagonally dominant matrix inner product

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

• Are you sure you wrote what you meant? Just take $S = k I$ for sufficently large $k$. – Robert Israel Jul 24 '16 at 17:48
• @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? – Astor Jul 24 '16 at 17:53
• $\langle Ax, x \rangle$ is bounded on the unit sphere $\{x: \langle x, x \rangle = 1\}$. Its maximum there is $k$. – Robert Israel Jul 24 '16 at 18:02
• @RobertIsrael Thanks a lot! I got it! – Astor Jul 24 '16 at 18:17

Take $S = k I$ with $k = \lambda_{max}(\frac{A + A^T}{2})$.
$A = \frac{A + A^T}{2} + \frac{A - A^T}{2}$
$\left< \frac{A - A^T}{2}x,x \right> = 0$