I have a quadratic function$$W(x_1,x_2,\ldots,x_n)=A\sum_{i} x_i^2+ \sum_{i\neq j} x_ix_j.$$ Denote the input vector as $\textbf{x}$, in quadratic form, $W(\textbf{x})=\textbf{x}^TM\textbf{x}$, where $$M=\begin{bmatrix} A & 1 & 1 & \dots & 1 \\ 1 & A & 1 & \dots & 1 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 1 & 1 & 1 & \dots & A \end{bmatrix}.$$ Now I define a aggregator as $$W(\textbf{x},\textbf{y})=W(\textbf{x})+W(\textbf{y})$$ Is this aggregator "additive separable" in math language? Generally $$W(\textbf{x},\textbf{y})\neq W(\textbf{x}+\textbf{y})$$ as $$W(\textbf{x},\textbf{y})=A\sum_{i} (x_i^2+y_i^2)+ \sum_{i\neq j} (x_ix_j+y_iy_j).$$ While $$ W(\textbf{x}+\textbf{y})=A\sum_{i}(x_i+y_i)^2+\sum_{i\neq j} (x_i+y_i)(x_j+y_j).$$
Is the aggregator $W(\textbf{x},\textbf{y})$ also "quadratic"? I am not sure about the notion of "quadratic" aggregator here. Can it be written in a quadratic form using $M$ or another symmetric matrix?
I'm assuming we're working over the real numbers. Recall that a quadratic space is a vector space equipped with a quadratic form, typically written (q, V), and that the orthogonal sum of two quadratic spaces $(q_1, V_1)\perp (q_2, v_2)$ is the direct sum of the $V_1, V_2$ as vector spaces but equipped with the new quadratic form $q = q_1+q_2$.
By definition, it's clear that $W(\mathbf{x},\mathbf{y})$ is the orthogonal sum of two copies of the quadratic space $W(\mathbf{x})$. It's easily shown that this is again a quadratic space and that the matrix for the new quadratic form is the following:
$$ \begin{pmatrix} M & \mathbf{0} \\ \mathbf{0} & M \end{pmatrix}. $$
Because this is a quadratic form, we also get the property mentioned in the comments where $W(k\mathbf{x},k\mathbf{y})=k^2 W(\mathbf{x},\mathbf{y})$.