Prove that if $A$ is positive, then $\exists\;\alpha>0$ such that $\langle Ax, x \rangle\geq \alpha|x|^2$ for $x\in\Bbb{R}^n$

Question

Let $M_{n\times n}(\Bbb{R}).$ We say that $A$ is positive if $\langle Ax, x \rangle>0$ for $x\neq 0.$

I want to prove that if $A$ is positive, then $\exists\;\alpha>0$ such that

$\langle Ax, x \rangle\geq \alpha|x|^2$ for $x\in\Bbb{R}^n.$

Please, I need help for this! Proofs and references will be highly regarded!

Answer 1

$A$ is positive implies that $\langle A(x),x\rangle=f(x)>0$. The restriction of $f$ on $S^{n-1}$ has a minimum $\alpha>0$. For every $x\in\mathbb{R}^n, f({x\over{\|x\|}})\geq\alpha$ implies that $\langle A({x\over{\|x\|}},{x\over{\|x\|}})\geq \alpha$ and $\langle A(x),x\rangle \geq \alpha\|x\|^2$.

Answer 2

Hint: $\alpha=\min_{|x|=1}\langle Ax,x\rangle$. Why does the minimum attained?

Answer 3

Let $A$ be positive, then $$\langle Ax, x \rangle>0$$ We can easily find $\alpha$ such that $$\langle (A - \alpha I) x, x \rangle>0$$ Let $\alpha = \lambda_{min}(A) - \epsilon$ for $0 < \epsilon < \lambda_{min}(A)$. For that choice, you can get from the equation $$\langle A x, x \rangle - \alpha \langle x, x \rangle>0$$ or $$\langle A x, x \rangle > \alpha \langle x, x \rangle$$

Answer 4

Let $\lambda>0$ be an eigenvalue of the matrix $A$ (recall that positive $A$ implies, there exists a $\lambda>0$ . Then $A x =\lambda x$

\begin{eqnarray*} x^{T} A x &=& \lambda x^{T} x &=& \lambda \left \lVert x \right\rVert^{2} \end{eqnarray*}

Hence,

\begin{eqnarray*} \frac{x^{T} A x } {\left \lVert x \right\rVert^{2}} &=& \lambda >0 \end{eqnarray*}