If we know that there are two vectors $x,y\in\mathbb{R}^d$ satisfying \begin{equation} \|x\|\ge c_1 \|y\|, \quad x^Ty\ge c_2\|x\|^2, \end{equation} where $c_1>0$ and $c_2>0$ are some given constants, can we always find a positive definite matrix $M\in\mathbb{R}^{d\times d}$ such that \begin{equation} x=My? \end{equation}
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Yes. We only need $x^T y > 0$, which is implied by your conditions if neither x nor y is zero. Let $s$ be the positive square root of $x^T y$, and $A$ a $d\times d$ real matrix who's first column equals $\frac1s x$. Let the remaining columns be a basis for the codimension-$1$ space perpendicular to y. Let $e_1$ be the first standard basis vector.
As x is not perpendicular to y, $A$ is non-singular, and $A \left(s e_1\right) = x$.
Also $A^T y = \left(\frac1s x^T y, 0, \dots, 0\right)^T = s e_1$.
Therefore $A A^T y = A \left(s e_1 \right) = x$. We let $M = A A^T$, which is positive definite.
