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
$$-1 \le a_{i,j} \le 0 \text{ for all } i \ne j$$
Is it possible to find a symmetric, positive definite matrix $S$ such that
$$\left< A x, x \right> \le \left< S x, x \right> \text{ for all }x \in \mathbb{R}^n$$
?
For any skew-symmetric matrix $K$, we have $x^TKx=(x^TKx)^T=x^TK^Tx=-x^TKx$ and therefore $x^TKx=0$.
It follows that if you split $A$ into the sum of its symmetric part $\frac{A+A^T}2$ and its skew-symmetric part $\frac{A-A^T}2$, then $\langle Ax,x\rangle=\langle \frac{A+A^T}2x,x\rangle$. In other words, it suffices to find a positive definite $S$ such that $S-\frac{A+A^T}2$ is positive semidefinite.
The easiest way is to take $S=tI$ for any positive number $t\ge\lambda_\max(\frac{A+A^T}2)$, because $\langle \frac{A+A^T}2x,x\rangle\le\lambda_\max(\frac{A+A^T}2)\|x\|^2$, but there are other ways to pick $S$ as well. For instance, since $\frac{A+A^T}2$ is symmetric, it can be orthogonally diagonalisable as $Q\operatorname{diag}(\lambda_1,\lambda_2,\ldots,\lambda_n)Q^T$. So, you may take $S=QDQ^T$ for any positive diagonal matrix $D$ that is entrywise $\ge\operatorname{diag}(\lambda_1,\lambda_2,\ldots,\lambda_n)$.
