Product of a positive semi-definite matrix and a diagonal matrix

Question

Let $A$ be a positive semi-definite matrix. $B$ is a diagonal matrix with positive diagonal entries. Let $x$ and $y$ be vectors. Is it true that $$ x'y > (<) 0 \Longrightarrow x'ABy \geq (\leq) 0 $$

Answer 1

No. Here is a counter-example. Consider $x = [1, 1]^T$, $y = [-0.1, 1]$, $A = I$, $B = diag(100, 1)$

Answer 2

$\def\ed{\stackrel{\text{def}}{=}}\ $ No. If $\ B\ $ is the identity matrix, for example, then the implication holds if and only if all the eigenvalues of $\ A\ $ are equal—that is, if and only if it's a (non-negative) multiple of the identity. Since $\ A\ $ is positive semi-definite, there must exist a complete orthonormal set $\ \big\{u_1,u_2,\dots,u_n\big\}\ $ of eigenvectors of $\ A\ .$ For each $\ i\ ,$ let $\ \lambda_i\ $ be the eigenvalue corresponding to $\ u_i\ ,$ and let $\ t\ $ be any real number with $\ t>1\ .$ Let \begin{align} x&\ed u_i+u_j\\ y&\ed tu_i-u_j\ . \end{align} Then $$ x^Ty=t-1>0\ , $$ so if the implication holds, then we must have $$ x^TAy=t\lambda_i-\lambda_j\ge0\ , $$ or, equivalently, $\ t\lambda_i\ge\lambda_j ,$ and since this holds for all $\ t>1\ ,$ it follows that $\ \lambda_i\ge\lambda_j\ .$ But this can only hold for every pair of eigenvalues if they are all equal—that is, $\ \lambda_i=\lambda_1\ $ for all $\ i\ ,$ and this implies that $\ A=\lambda_1I\ .$

Conversely, if $\ A=rI\ $ for some $\ r\ge0\ ,$ then $$ x^TAy=rx^Ty\ , $$ and the implication clearly holds.