If $A$ is positive definite and $Ax\le b$ can I say that $x\le A^{-1}b$?

Question

If I have a vector enequiality $Ax\le b$ where $x,b\in\mathbb{R}^n$ and $A$ is a $n\times n$ positive definite matrix, can I say that the following vector inequality is true?

$$x\le A^{-1}b$$

Answer 1

Counterexample:

Let $A= \begin{bmatrix} 2 & 1 \\ 1 & 2 \end{bmatrix}$, $x = \begin{bmatrix} 1 \\ 0 \end{bmatrix}$, then $Ax = \begin{bmatrix} 2 \\ 1 \end{bmatrix}$. Let $b = \begin{bmatrix}2 \\ 2 \end{bmatrix} $.

We have $A^{-1}b = \begin{bmatrix}\frac23 \\ \frac23 \end{bmatrix} $.

Answer 2

Given $A$, for $A x \le b$ to imply $x \le A^{-1} b$, it's necessary and sufficient that $b \ge 0$ implies $A^{-1} b \ge 0$, and that in turn is equivalent to $A^{-1}$ having all nonnegative entries. That's not always true for positive definite matrices $A$ (e.g. Syong Thye Goh's counterexample where $A^{-1} = \pmatrix{2/3 & -1/3\cr -1/3 & 2/3}$).