Does $v^Tw>0$ imply that $\exists A>0$ such that $v=Aw$? Let $v,w \in \mathbb{R}^n$ satisfy $v^Tw>0$.
Does there exist a symmetric positive-definite matrix $g$ such that $v=gw$?
The condition $v^Tw>0$ is necessary for the existence of such $g$:
$$
v^Tw=w^Tg^Tw=w^Tgw>0.
$$
I guess there should be an easy argument for the existence of such $g$, but I don't see it immediatley.

Motivation:
I am trying to understand whenever two vectors can be gradients w.r.t different metrics.
 A: Let $V$ be the subspace of ${\mathbb R}^n$ spanned by $v$ and $w$.  Supposing without loss of generality that $\|w\|=1$, let $B=\{e_1, e_2\}$ be an
orthonormal basis for $V$ such that $e_1=w$.
Writing the coordinates of $v$ relative to $B$ as $(a,b)$, we have that
$$
  a = e_1^Tv = w^Tv >0.
  $$
For every $c\in {\mathbb R}$ the matrix
$$
  g_0=\pmatrix {a & b \cr b & c}
  $$
is therefore symmetric, satisfies $g_0w=v$, and we claim that  $c$ can be chosen so that $g_0$ is also positive definite.
Recalling that a
$2\times 2$ matrix is positive definite iff its trace and determinant are positive, any positive  $c$ such that
$ac-b^2>0$, that is $c>b^2/a$, will make $g_0$ a positive definite matrix.
In order to obtain a positive definite matrix acting on the whole ${\mathbb R}^n$, we may split ${\mathbb R}^n=V\oplus V^\perp$ and let
$$
  g=g_0\oplus id.
  $$
A: Let $X\in\mathbb R^{n\times n}$ be any matrix whose first column is $v$ and whose remaining columns form a basis of $\{w\}^\perp$. Then $X$ is nonsingular and $g=\frac{1}{v^Tw}XX^T$ is a positive definite matrix such that $gw=v$.
A: Suppose such positive-definite $g$ exist, then $g = ODO^T$ for some rotation matrix $O$, and $D$ is diagonal with all entries positive. Or to rephrase, we can find a basis where $g$ is diagonal. If we write out the coordinates of $v$ and $w$ in that basis, then the $i$th coordinates of $v$ and $w$ are of the same sign.
Conversely, if we can find a rotation such that every coordinate of $v$ and $w$ have the same sign in the rotated basis, then we have found $g$. This rotation is always possible because the angle between the two vectors is acute. Just have to rotate them into the first quadrant.
