For any positive-definite matrix $A$, there exists a $x\in\Bbb R^n$, such that each component of $Ax$ is greater than $0$.

Question

For any positive-definite matrix $A$, show that there exists a $x\in\Bbb R^n$, such that each component of $Ax$ is greater than $0$.

How to prove it? If $A$ is just diagonal, say $diag(a_1,\cdots,a_n)$, then it is OK by $x=(1,\cdots,1)'$. For the general, any ideas?

Answer 1

That is quite trivial. Any (strictly) positive definite matrix has to be injective and thus also bijective and invertible.

Else there would be an $x\neq0$ so that $Ax = 0$, so $x^\mathrm T A x = 0$, which would contradict positive definiteness. But if $A$ is invertible then in fact for each $y\in\mathbb R^n$ there exists an $x$ so that $Ax = y$. So you are free to choose any $y>0$ you like.