Given a real $n\times n$ matrix $A$ that is symmetric and negative definite. I am interested in the properties of the bilinear form associated with this matrix.
Specifically, could it ever be possible to choose two positive, real vectors $x, y\in \mathbb{R}^n_+$ such that $x^\top A y \geq 0$?
Clearly this would not be true for the quadratic form $x^\top A x$, since negative definiteness would imply that $x^\top A x < 0$ for all $x$.
But would it be possible, for example, to construct a negative definite matrix whose bilinear form is only negative when $x=y$ (or for $x$ in a small region around $y$?)
Consider the matrix $$A= \begin{pmatrix} -1 &0 \\ 0& -1 \end{pmatrix}$$ $$x=(-1,-1),y=(1,1), \langle Ax,y \rangle>0 $$ But if your matrix is written in diagonal form and $x,y \in \mathbb{R}_{+}^{n}$, then $$\langle Ax,y \rangle \leq 0$$ But if it is not in diagonal form, then there are counterxamples: $$B= \begin{pmatrix} -2 &\sqrt{1+\frac{2}{3}} \\ \sqrt{1+\frac{2}{3}} & -1-\frac{1}{3} \end{pmatrix}$$ $$\langle B(1,1),(1,1) \rangle \leq 0 $$ $$\langle B(0.001,1),(1,0.001)\rangle > 0$$ So in order to prevent such counterexamples , $\mathbb{R}_{+}^{n}$ shcould be replaced by the half space consisiting of positive linear combinations of eigenvectors of $A$. Concerning your second question:the map $(x,y)\to \langle Ax,y \rangle$ is continous , if it is negative on the diagonal ($x=y$),then it would be negative in a neighborhood of the diagonal.Remark that if $\langle Ax,y \rangle >0 \implies \langle Ax,-y \rangle <0 $,so the region where the bilinear form is negative is as large as the region where it is positive (the map $(x,y)\to (x,-y)$ is an isometry).