PSD vector inner product with positive vectors Suppose $A \in \mathbb{R}^{n \times n}$ be a PSD matrix. Let $x,y \in \mathbb{R}^n$ such that $x = [x_1,...,x_n]^T$ and $y = [y_1,...,y_n]^T$. We require that $x,y$ are element-wise positive, that is $x_i >0$ and $y_i > 0$ for all $i \in \{1,2,...,n\}$. In that case can it be concluded that $x^T A y \geq 0$ in general? 
 A: No. Take any large positive $n$ and consider
$$
\pmatrix{1&n}\pmatrix{5&-2\\ -2&1}\pmatrix{1\\ 1}=3-n.
$$
A: No. Let 
$$M=\newcommand\bmat{\begin{pmatrix}}\newcommand\emat{\end{pmatrix}}\bmat 0 & 1 \\ -1 & 0 \emat,$$
then $$\bmat x & y \emat \bmat 0 & 1 \\ -1 & 0 \emat \bmat a \\ b\emat = bx-ay,$$
so in particular when $x=a$, $y=b$, we have that the quadratic form corresponding to $M$ is always zero on any vector, so $M$ is PSD.
However if $a=y=1$, $x=b=\frac{1}{2}$, then the product of the vectors with the matrix will be $\frac{1}{4}-1=\frac{-3}{4}$.
The Idea:
The idea is that rotation by 90 degrees is PSD, since the dot product of a vector and its 90 degree rotation will always be zero.
However, if we choose the vector to be rotated by 90 degrees to already be closer to the $y$-axis, and the vector to compare it to to be closer to the $x$-axis, then the final product will be a dot product of vectors with an obtuse angle between them, which will be negative.
The same idea allows us to replace $M$ with a matrix which is e.g. rotation by 45 degrees to get a positive definite matrix which also doesn't have $v^TMu>0$ for positive vectors $v$ and $u$.
A: I believe the matrix $A=\pmatrix{1&-1\\0&1}$ is PSD, however $\pmatrix{1&0}^TA\pmatrix{0&1}=-1$. You can replace $0$ by small enough $\epsilon>0$ and it will not change the fact that this results in a negative number.
