Does $v^Tw>0$ imply that $\exists A>0$ such that $v=Aw$?

Question

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.

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Motivation:

I am trying to understand whenever two vectors can be gradients w.r.t different metrics.

Answer 1

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 $$ g=g_0\oplus id. $$

Answer 2

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 $gw=v$.