Let $u,v\in\mathbb{R}^n$ be such that $\langle u,v\rangle>0.$ My question is whether or not there exists a symmetric positive definite matrix $Q$ such that $v=Qu$. If such a matrix $Q$ exists, how to construct $Q$ from $u,v$?
Thank you for all solutions.
Best Answer
Extend $u_1=\frac{u}{\|u\|}$ to an orthonormal basis $\{u_1,u_2,\ldots,u_n\}$ of $\mathbb R^n$. Let $a=\langle v,u_1\rangle$ and $$ V=\pmatrix{\frac{1}{\sqrt{a}}v&u_2&\cdots&u_n}. $$ Since $\langle v,u_1\rangle>0$, $v$ is independent of $\{u_2,\ldots,u_n\}$. Thus $V$ is nonsingular and $$ Q=\frac{1}{\|u\|}VV^T $$ is positive definite. Also, $Qu=VV^Tu_1=V(\sqrt{a}e_1)=v$ where $e_1=(1,0,\ldots,0)^T$.