Quadratic form with positive semi-definite matrix

Question

Suppose $A$ is a positive semi-definite matrix, and $x$ and $y$ are real vectors.

1. Under what conditions does the following result hold? $$ x'y > (<) 0 \Longrightarrow x'Ay \geq (\leq) 0 $$

2. Does there exist $\varepsilon>0$ such that the following result always holds? $$ x'y > \varepsilon (<-\varepsilon) \Longrightarrow x'Ay \geq (\leq) 0 $$

Answer 1

Your first problem is equivalent to wonder when $\langle x,y\rangle>0$ implies that $\lambda_1x_1y_1+\cdots+\lambda_nx_ny_n>0$ where $\lambda_i\geq 0$ for all $i.$ To see this, just diagonalize $A$ in an orthonormal basis. You are now facing with a problem in $R^{2n}$ with the quadratic forms $(x,y)\mapsto \langle x,y\rangle>0$ or $\lambda_1x_1y_1+\cdots+\lambda_nx_ny_n>0.$ Diagonalizing again, you wonder when $\|x\|^2-\|y\|^2>0$ implies that $$\lambda_1(x_1^2-y_1^2)+\cdots+\lambda_n(x_n^2-y_n^2)>0$$ which seems easier. In particular, without loss of generality you can assume $\|x\|^2=1.$ The problem for $n=2$ now seems quite tractable and gives the ideas for $n>2.$

Edit. Something simpler, using the formula

$$\langle x,y\rangle=\frac{1}{4}(\|x+y\|^2-\|x-y\|^2)$$ and $X=\frac{x+y}{2}$, $Y=\frac{x-y}{2}.$ Therefore $\langle x,y\rangle\geq 0\Leftrightarrow \|X\|^2\geq \|Y\|^2$ and $x^TAy\geq 0\Leftrightarrow X^TAX\geq Y^TAY.$ Without loss of generality we may assume that $$A=\mathrm{diag}(1/a_1^2,\ldots,1/a_n^2)$$ with $$0<a_1\leq a_2\leq \ldots\leq a_n$$ and denote for simplification $R^2(X)=X^TAX.$ The set

$$\mathcal{E}(R)=\{X\in E; R^2(X)\leq R^2\}$$ is an ellipsoid with axis $a_1 R,\ldots,a_n R.$

Proposition: Let $X\in E$ fixed. Then every $Y\in E$ which is such that $\|X\|^2\leq \|Y\|^2$ satisfies $R^2(Y)\leq R^2(X)$ if and only if $\|Y\|\leq a_1 R(X).$

The algebraic proof below may seem complicated, while the geometric intuition is simple: fix $X$, draw the ellipsoid $\mathcal{E}(R(X))$ which has $X$ on its boundary, the only acceptable $Y$ are in the ball of radius $a_1 R(X).$

Proof. $\Rightarrow$ If $\|Y\|> a_1R(X)$ then $Y_1=(\|Y\|,0,\ldots,0)$ satisfies $\|X\|^2\leq \|Y_1\|^2$ but also $$R^2(X)=\frac{x_1^2}{a_1^2}+\cdots+\frac{x_n^2}{a_n^2}< \frac{y_1^2+\cdots+y_n^2}{a_1^2}\leq \frac{x_1^2+\cdots+x_n^2}{a_1^2}$$ which implies $$x_2^2\left(\frac{1}{a_2^2}-\frac{1}{a_1^2}\right)+\cdots+x_n^2\left(\frac{1}{a_n^2}-\frac{1}{a_1^2}\right)>0$$ which is impossible since all left hand terms are non positive.

$\Leftarrow.$ Conversely $R^2(Y)\stackrel{(*)}{\leq} \frac{\|Y\|^2}{a_1^2}\stackrel{(**)}{\leq} R^2(X).$ (*) comes from $a_1^2\leq a_i^2$ and (**) comes from the hypothesis.