Does a symmetric diagonally dominant real matrix $A$ with nonnegative diagonal entries satisfy $(x^{2p-1})^T A x \geq 0$?

Question

In https://mathworld.wolfram.com/DiagonallyDominantMatrix.html, I find that

A symmetric diagonally dominant real matrix with nonnegative diagonal entries is positive semidefinite.

If $A \in \mathbb{R}^{N\times N}$ is symmetric diagonally dominant real matrix with nonnegative diagonal entries, is it still ture that \begin{align} (\mathbf x^{2p-1})^T A \mathbf x \geq 0, \quad \forall \mathbf x \in \mathbb{R}^N \end{align} where $p \geq 1$ is an integer, and the $(2p-1)$-th power of the vector $\mathbf{x}$ is element wise, i.e. $\mathbf x^{2p-1} = [x_1^{2p-1}, \cdots, x_N^{2p-1}]^T$.

EDIT 1 I wrote a short matlab code to verify the inequality

clear;
N = 10;
A0 = 2*rand(N, N) - 1;               % random value in [-1, 1]
A = A0 + A0';                        % construct symmetric matrix;
v = (sum(abs(A), 2) - abs(diag(A))); % diagonally dominant
for i = 1:N
    A(i,i) = v(i);                   % Assign v to the diagonal elements
end
xv = 2*rand(N, 1000000) - 1;
p = 3;
x = min(dot((xv.^p),  A * xv))

Thank you very much!

Answer 1

This is true. Denote the standard basis of $\mathbb R^N$ by $e_1,e_2,\ldots,e_N\}$. If $A$ has a nonzero off-diagonal entry $a_{ij}$, let $$ B=|a_{ij}|\left(e_i+\operatorname{sign}(a_{ij})e_j\right)\left(e_i+\operatorname{sign}(a_{ij})e_j\right)^T. $$ Then both $B$ and $A-B$ are diagonally dominant and their diagonals are nonnegative. Proceed recursively, we can extract positive multiples of matrices in the form of $(e_i+se_j)(e_i+se_j)^T$ (with $s=\pm1$) from $A$, until only a nonnegative diagonal matrix $D$ remains. Clearly $(x^{2p-1})^TDx =\sum_{i=1}^Nd_{ii}x_i^{2p}\ge0$. Also, when $s=\pm1$, we have \begin{align} (x^{2p-1})^T(e_i+se_j)(e_i+se_j)^Tx &=x_i^{2p}-sx_i^{2p-1}x_j-sx_ix_j^{2p-1}+x_j^{2p}\\ &\ge|x_i|^{2p}-|x_i|^{2p-1}|x_j|-|x_i||x_j|^{2p-1}+|x_j|^{2p}\\ &=(|x_i|-|x_j|)(|x_i|^{2p-1}-|x_j|^{2p-1})\\ &=(|x_i|-|x_j|)^2\sum_{k=0}^{2p-2}|x_i|^k|x_j|^{2p-2-k}\ge0. \end{align} It follows that $(x^{2p-1})^TAx\ge0$.