If $a_1 I_n \leq A \leq a_2 I_n$, where $A \in \mathbb{R}^{n\times n}$ and $a_1,a_2$ are positive scalars. Then what can I say about $A$?

Question

I have, \begin{equation} a_1 I_n \leq A \leq a_2 I_n \end{equation} where, $A \in \mathbb{R}^{n\times n}$, and $a_1,a_2$ are positive scalars. Then can be say that $A$ is a Positive definite matrix? I have developed a rudimentary proof: \begin{equation} \begin{split} a_1x^T I_n x& \leq x^T A x \leq a_2 x^T I_n x \\ a_1\|x\|^2 & \leq x^T A x \leq a_2\|x\|^2 \end{split} \end{equation} where $x \in \mathbb{R}^n$ is any vector. Now looking at the last equation, and using the fact that $a_1,a_2 \geq 0$, can we claim that $A$ is PD?

Answer 1

Yes, $A\le B$ implies $x^HAx\le x^HBx$ for all $x$, by means of the definitory inequality $x^H(B-A)x\ge 0$.