How to judge whether there is a positive definite matrix $A$,that $AX=B$,where $X,B$ are given column vectors.

Question

Problem

$X,B$ are n-dimension column vectors.

How to judge,if there is a positive definite matrix $A$,that

$$AX=B$$

What I have tried

I get a necessary condition:

$$X^TB \geq 0$$

Is it sufficient?

Thanks.

Answer 1

Hint. Consider the case where $x\ne0$. Let $x_1=x/\|x\|$. Then $b = px_1 + qx_2$ for some $p,q\in\mathbb{C}$ and some unit vector $x_2\perp x_1$. (Can you express $p$ in terms of $x$ and $b$?) Let $U$ be a unitary matrix whose first two columns are respectively $x_1$ and $x_2$, and let $e_i$ denotes the $i$-th vector in the standard basis of $\mathbb{C}^n$. Then \begin{align*} &Ax = b\\ \Leftrightarrow &Ax = px_1 + qx_2\\ \Leftrightarrow &AUe_1 = pUe_1 + qUe_2\\ \Leftrightarrow &(\underbrace{U^\ast AU}_M)e_1 = pe_1 + qe_2. \end{align*} Hence the question boils down to finding a positive definite matrix $M$ whose first column is equal to $(p,q,0,\ldots,0)^T$. Under what conditions can this happen? What about the case $x=0$?