Does there exists a positive definite matrix $A^H = A\in \mathbb C^{n \times n}$ under some condition such that for a given $x,y \in \mathbb C^n$, where $x \neq 0$. $Ax = y$ holds good.
What I have tried : If $A^H = A$ then we can always found a rank one matrix $A = \frac{yy^H}{x^Hy} $,if and only if $x^Hy > 0$, here $A$ matrix is positive semidefinite and the proof is visible.
Obviously, $x^Hy>0$ is a necessary condition for $A$ to exist. Also, the cases where $n=1$ or $y$ is a scalar multiple of $x$ are trivial.
Suppose $n\ge2$, $x^Hy>0$ and $x,y$ are linearly independent. It suffices to consider the case where $n=2$, because we may simply leave invariant all vectors orthogonal to $x$ and $y$.
Let $z=x/\|x\|$. Since $x^Hy>0$, $y=az+bw$ for some real numbers $a>0, b\ge0$ and some unit vector $w\perp z$. Geometrically, $x$ lies on the positive $z$-axis and $y$ lies inside the first open quadrant of the $zw$-plane.
Since $\phi=\cos^{-1}\frac{x^Hy}{\|x\|\|y\|}<\frac\pi2$, for any angle $0<\theta<\frac\pi2-\phi$, if we denote by $R_\theta=\pmatrix{\cos\theta&\sin\theta\\ -\sin\theta&\cos\theta}$ the clockwise rotation by angle theta on the $zw$-plane and if we put $u=R_\theta z,\ v=R_\theta w$, then both $x$ and $y$ lie inside the first open quadrant of the $uv$-plane. Therefore, if we map the $u$- and $v$-components of $x$ to those of $y$, the matrix of the resulting linear map will be positive definite. In other words, if $Q$ is the $2\times2$ augmented matrix $[u|v]$, then $$ A=Q\operatorname{diag}\left(\frac{y^Hu}{x^Hu},\,\frac{y^Hv}{x^Hv}\right)Q^\ast $$ is a positive definite matrix such that $Ax=y$.