Positive definite matrix acts on positive vector

Question

Suppose $A \succ 0$ is a positive definite matrix. I was asked whether it is true that $A x > 0$ if and only if $x >0$? where $x > 0$ denotes inequality component-wise.

I think the statement is false. For the reverse direction, I constructed an example with \begin{align*} A = \begin{pmatrix} 1 & -1.5 \\ -1.5 & 4 \end{pmatrix} \qquad x = (1, 1)^T. \end{align*} In this case, $x > 0$ but $Ax \not > 0$. Yet, I could not find a proper example for the forward direction.

Answer 1

You have found an $x>0$ such that $y=Ax \not >0 $

Hence we have $$A^{-1}y=x$$

Let $B=A^{-1}$, then we have found

$y \not > 0$, but $By >0$ and $B$ is a positive definite matrix.

Answer 2

Take the matrix $\left(\begin{array}{cc} 2 & 1\\ 1 & 2\end{array}\right)$ and the vector $(3,-1)$.