Let $A \in \mathbb R^{n\times n}$ be positive definite, i.e.
$$\langle A x{,}x \rangle > 0 \quad \text{for all } x\in\mathbb R^n\setminus\{0\}.$$
Why does this imply that the induced bilinear form of $A$ is strongly positive, i.e.
$$\text{there is an } \alpha > 0 \text{ such that }
\langle A x{,}x \rangle \geq \alpha \lVert x \rVert^2 \quad \text{for all } x\in\mathbb R^n?$$
Of course this makes positive definiteness and strong positivity equivalent since the inverse implication is trivial. But how to choose $\alpha$ independent of $x$?
Best Answer
There are several approaches that you can take here.
One approach is to note that, by the Rayleigh-Ritz theorem, we have $\langle Ax, x\rangle\geq \lambda_{\min}(A) \cdot \|x\|^2$, giving us an explicit $\alpha$ in terms of the eigenvalues of $A$. This assumes that $A$ is symmetric; if $A$ is possible non-symmetric, then note that the symmetric matrix $B = (A + A^T)/2$ satisfies $\langle x,Bx \rangle = \langle x,Ax \rangle$ for all $x \in \Bbb R^n$, which means that the same analysis can be applied to $B$.
Another approach is to note that the function $f(x) = \langle x,Ax \rangle$ is continuous over the (compact) spherical domain $S = \{x \in\Bbb R^n: \|x\| = 1\}$, which means that it must attain a minimum. Then, verify that this minimum is a suitable value for $\alpha$.